s

Transkrypt

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• solid-state implementations of quantum computing
• quantum tomography
• quantum error correction
• quantum algorithms
• quantum teleportation
• quantum cryptography
• quantum entanglement
• quantum logic gates
Keywords
summer semester 2008
http://zon8.physd.amu.edu.pl/∼miran
e-mail: [email protected]
Adam Miranowicz
(with an emphasis on optical implementations)
Methods of Quantum Information Processing
2
1
8. Quantum Computing and Communications
by Sandor Imre, Ferenc Balazs
7. Quantum Computing by Josef Gruska
6. Quantum Information Processing
edited by Gerd Leuchs, Thomas Beth
5. Explorations in Quantum Computing
by Colin P. Williams, Scott H. Clearwater
4. Quantum Computing by Mika Hirvensalo
3. Introduction to Quantum Computation and Information
edited by Hoi-Kwong Lo, Tim Spiller, Sandu Popescu
2. Approaching Quantum Computing
by Dan C. Marinescu, Gabriela M. Marinescu
1. Quantum Computing by Joachim Stolze, Dieter Suter
Other Textbooks on Quantum Computing
4. Quantum mechanical description of linear optics
J. Skaar, J.C.G. Escartin, H. Landro,
Am. J. Phys. 72, 1385-1391 (2005).
4
3. Quantum optical systems for the implementation of quantum information
processing, T.C. Ralph, free downloads at http://arxiv.org/quant-ph/0609038.
2. Linear optics quantum computation: an overview, C.R. Myers, R. Laflamme,
free downloads at http://arxiv.org/quant-ph/0512104.
1. Linear optical quantum computing,
P. Kok, W.J. Munro, K. Nemoto, T.C. Ralph, J. P. Dowling, G.J. Milburn,
free downloads at http://arxiv.org/quant-ph/0512071.
Review articles on quantum-optical computing:
by Michael A. Nielsen and Isaac L. Chuang (Cambridge, 2000)
1. Quantum Computation and Quantum Information
Basic textbook:
3
9. An Introduction to Quantum Computing Algorithms
by Arthur O. Pittenger
10. Quantum Computing by M. Nakahara, Tetsuo Ohmi
11. Principles of Quantum Computation and Information - Vol.1
by Giuliano Benenti, Giulio Casati, Giuliano Strini
5
12. Principles of Quantum Computation And Information: Basic Tools And
Special Topics by Giuliano Benenti
13. Quantum Optics for Quantum Information Processing
edited by Paolo Mataloni
14. Lectures on Quantum Information
edited by Dagmar Bruß, Gerd Leuchs
15. A Short Introduction to Quantum Information and Quantum Computation
by Michel Le Bellac
6
16. Quantum Computation and Quantum Communication by Mladen Pavicic
17. Scalable Quantum Computers: Paving the Way to Realization
edited by Samuel L. Braunstein, Hoi-Kwong Lo
18. Fundamentals of Quantum Information edited by Dieter Heiss
19. Experimental Aspects of Quantum Computing
edited by Henry O. Everitt
20. Quantum Computing: Where Do We Want to Go Tomorrow
edited by Samuel L. Braunstein
21. Quantum Information by Gernot Alber et al.
22. The Physics of Quantum Information
edited by Dirk Bouwmeester, Artur K. Ekert, Anton Zeilinger
23. Temple of Quantum Computing by Riley T. Perry
(free downloads at www.toqc.com)
24. Quantum Computation edited by Samuel J. Lomonaco, Jr.
(free downloads at www.cs.umbc.edu/∼lomonaco)
25. Lecture Notes for Physics: Quantum Information and Computation
by John Preskill
(free downloads at www.theory.caltech.edu/people/preskill/ph229).
|gi - ground state
|ei - excited state
– physical notation
0
,
1
|ei =
1
0
|gi =
0
1
σ̂z ≡
7
8
σ̂z = σ̂ †σ̂−σ̂σ̂ † = |eihe|−|gihg|
1 0
.
0 −1
a two-level atom/system
– information notation
|ei =
0 −i
,
i 0
1
,
0
σ̂y ≡
|gi =
0 1
,
1 0
• Pauli matrices
σ̂x ≡
• rasing (σ̂ †) and lowering (σ̂) energy operators,
σ̂x
= atomic-transition operators
σ̂ − iσ̂y
+ iσ̂
x
y
= |eihg|, σ̂ =
= |gihe|,
2
2
σ̂ † =
qubit = quantum bit
• the smallest unit of quantum information
• physically realized by a 2-level quantum system
whose two basic states are conventionally labelled |0i and |1i
• By contrast to classical bits,
a qubit can be in an arbitrary superposition of ’0’ and ’1’:
|ψi = α|0i + β|1i
with normalization condition |α|2 + |β|2 = 1.
• matrix representation of qubit states:
1
0
|0i =
, |1i =
0
1
1
0
α
+β
=
0
1
β
|ψi = α|0i + β|1i = α
cn|ni
Ŷ (α|0i + β|1i) =
0 −i
i 0
k = 0, 1
01
10
examples
0
1
X̂|1i = X̂
=
= |0i
1
0
01
α
β
X̂(α|0i + β|1i) =
=
= β|0i + α|1i
10
β
α
X̂|ki = |1 ⊕ ki,
X̂ ≡ σ̂x =
Pauli X̂ gate = quantum NOT gate = bit flip
they are for quantum computers what
classical logic gates are in conventional computers
k = 0, 1
example
0 −i
α
−β
=i
= −iβ|0i + iα|1i
i 0
β
α
1 1
1 −1
Ĥ|−i = |1i
examples
1
1
1
1
1
1
Ĥ|0i = √2
= √12 (|0i + |1i) ≡ |+i
= √2
1 −1
1
0
1 1
1
0
1
1
√
√
Ĥ|1i = 2
= √12 (|0i − |1i) ≡ |−i
= 2
1 −1
−1
1
1
1
1
2
1
1
1
1
√
=2
Ĥ|+i = √2
=
= |0i
1 −1 2 1
0
0
Ĥ =
√1
2
Hadamard gate
Ŝ =
1 0
0 i
basic quantum circuits
operating on a small number of qubits
Ŷ |ki = i(−1)k |1 ⊕ ki,
Ŷ ≡ σ̂X =
phase gate
10
example
1 0
α
α
=
= α|0i − β|1i
0 −1
β
−β
Pauli Ŷ gate = phase flip + bit flip
Ẑ(α|0i + β|1i) =
k = 0, 1
1 0
0 −1
Ẑ|ki = (−1)k |ki,
Ẑ ≡ σ̂z =
Pauli Ẑ gate = phase flip
quantum (logic) gates
optical qudits are spanned in d-dimensional Fock space
...
5D qudit = (qu)quintit
4D qudit = (qu)quartit = ququart
3D qudit = qutrit
2D qudit = qubit
special cases
n=0
n=0
with normalization condition
d−1
X
|cn|2 = 1
|ψid =
d−1
X
d-dimensional quantum states, generalized qubits
qudits = qunits
9
12
11
x
|ψ
y
2
|1i
|0
|1
|0i + eiφ sin
2
|0 + |1
θ
θ
|0
Bloch representation/sphere
z
θ
φ
|ψi = cos
2
qubit rotations
• rotations on Bloch sphere
R̂n(2θ) ≡ exp (−iθn · σ̂ )
= σ̂I cos θ − in · σ̂ sin θ
= σ̂I cos θ − i(nxσ̂x + ny σ̂y + nz σ̂z ) sin θ,
• Pauli matrices (and identity matrix)
0 1
σ̂x ≡
= |0ih1| + |1ih0|,
1 0
0 −i
σ̂y ≡
= −i|0ih1| + i|1ih0|,
i 0
1 0
σ̂z ≡
= |0ih0| − |1ih1|,
0 −1
1 0
= |0ih0| + |1ih1|
0 1
σ̂I ≡ Iˆ =
σ̂ = (σ̂x, σ̂y , σ̂z )
13
+ i |1
2
14
• rotations about k = x, y, z axes – special cases of R̂n(θ)
R̂k (θ) = exp −i θ2 σ̂k = σ̂I cos 2θ − iσ̂k sin 2θ
or explicitly
cos 2θ −i sin 2θ
X̂(θ) ≡ R̂x(θ) =
−i sin 2θ cos 2θ
cos 2θ − sin θ2
Ŷ (θ) ≡ R̂y (θ) =
sin θ2 cos 2θ
e−iθ/2 0
0 eiθ/2
Ẑ(θ) ≡ R̂z (θ) =
• Do we need all XY Z-rotations? No!
e.g. we can omit Z-rotation as
Ẑ(θ) = X̂( π2 )Ŷ (θ)X̂(− π2 ) = Ŷ ( π2 )X̂(−θ)Ŷ (− π2 ).
• any single-qubit gate
can be written as a decomposition ZY (or, equivalently, XY or XZ)
Û = eiαR̂n(θ) = eiαẐ(θ1)Ŷ (θ2)Ẑ(θ3)
θ
θ
|0i + eiφ sin
|1i
2
2
Bloch vector (Pauli vector) for a qubit
ρ11 ρ12
ρ21 ρ22
= 12 (σ̂I + rBloch · σ̂)
hσ̂i
Tr[ρ̂σ̂]
(Tr[ρ̂σ̂x], Tr[ρ̂σ̂y ], Tr[ρ̂σ̂z ])
(2Reρ21, 2Imρ21, ρ11 − ρ22)
= 21 (σ̂I + rxσ̂x + ry σ̂y + rz σ̂z )
(so-called Pauli basis)
rBloch ≡ (rx, ry , rz ) = (sin θ cos φ, sin θ sin φ, cos θ)
|ψi = cos
• in any pure state
we have
ρ̂ =
• in any mixed state
we get
rBloch
=
=
=
=
15
16
then
let
⊗
a11 a12
,
a21 a22
a11 a12
a21 a22
b11 b12
b21 b22
B=
b11 b12
b21 b22


b11 b12
b11 b12
a
,
a
12
 11 b21 b22

b21 b22 
=


b b
b b
a21 11 12 , a22 11 12
b21 b22
b21 b22


a11b11 a11b12 a12b11 a12b12
 a11b21 a11b22 a12b21 a12b22 

=
 a21b11 a11b12 a22b11 a12b12 
a21b21 a11b22 a22b21 a12b22
A⊗B =
A=
Kronecker tensor product
|rBloch| = 1 – for pure state,
|rBloch| < 1 – for mixed state
• Properties:
= ρ12 + ρ21 = ρ∗21 + ρ21 = 2Re ρ21 = 2Re ρ12
0 1
·
1 0
+ 2rx + 0 + 0)
ρ11 ρ12
ρ21 ρ22
ρ12 ρ11
= Tr
ρ22 ρ21
Tr[ρ̂σ̂x] = Tr
moreover
= rx
=
1
2 (0
= 12 (Tr[σ̂x] + rxTr[σ̂I ] + ry Tr[−iσ̂z ] + rz Tr[iσ̂y ])
= 21 (Tr[σ̂I σ̂x] + rxTr[σ̂xσ̂x] + ry Tr[σ̂y σ̂x] + rz Tr[σ̂z σ̂x])
= 21 Tr[σ̂I σ̂x + rxσ̂xσ̂x + ry σ̂y σ̂x + rz σ̂z σ̂x]
Tr[ρ̂σ̂x] = Tr[ 21 (σ̂I + rxσ̂x + ry σ̂y + rz σ̂z )σ̂x]
• Exemplary proof:
18
17
=α

α
β

=
γ
δ

 
 
 
 
1
0
0
0
0
1
0
0
 

 
 
= α
 0  + β  0  + γ  1  + δ 0 
0
0
0
1
1
1
1
0
0
1
0
0
⊗
+β
⊗
+γ
⊗
+δ
⊗
0
0
0
1
1
0
1
1
|ψi = α|00i + β|01i + γ|10i + δ|11i
matrix representation
 
1
0
1
1

|00i =
⊗
=
 0 ,
0
0
0
 
 
 
0
0
0
0
0
1
 
 

|01i = 
 0 , |10i =  1 , |11i =  0 ,
1
0
0
matrix representation
|ψi = |ψ ′iA ⊗ |ψ ′′iB ≡ |ψ ′iA|ψ ′′iB ≡ |ψ ′ψ ′′iAB ≡ |ψ ′ψ ′′i
notation of two-mode states
|ψi = α|00i + β|01i + γ|10i + δ|11i
two-qubit pure states
20
19
Universal set of gates for quantum computing
• rotations of single qubits
• any nontrivial two-qubit gate (e.g., CNOT, NS, CZ)
CZ
|0, 0i
|0, 1i
|1, 0i
−|1, 1i
CNOT
|0, 0i
|0, 1i
|1, 1i
|1, 0i
CZ and CNOT gates
CZ = controlled phase gate (CPhase)
control bit target bit
|0i
|0i
|1i
|0i
|1i
|0i
|1i
|1i
= controlled sign gate (CSign)
or equivalently
CZ :|q1, q2i → (−1)q1q2 |q1, q2i
CNOT : |q1, q2i → |q1, q1 ⊕ q2i
where qk = 0, 1 and q1 ⊕ q2 = mod (q1 + q2, 2).
Introduction to quantum-optical cryptography
basic cryptographic terms
Vernam protocol
quantum key distribution
BB84 protocol
security and no-cloning theorem
a note
quantum cryptography is, probably,
the most important application of quantum optics nowadays
22
21
24
23
basic cryptographic terms (I)
Plaintext
a string of numbers (letters) of our digital alphabet
Encryption
a computation which is usually quick and easy to perform
Decryption
quick and easy computation is only when some
additional information is available
otherwise it is very long and time consuming computation
Key
a set of instructions to encrypt and decrypt a message,
e.g., randomly chosen series of numbers known
to Alice and Bob only
25
01
06
11
16
21
26
31
36
A
D
H
Ł
Ó
Ś
X
_
Ą
E
I
M
P
T
Y
-
03
08
13
18
23
28
33
38
B
Ę
J
N
Q
U
Z
?
04
09
14
19
24
29
34
39
C
F
K
Ń
R
V
ś
,
cleartext: A D A M
plaintext: 01060117
02
07
12
17
22
27
32
37
05
10
15
20
25
30
35
40
exemplary digital alphabet
Ć
G
L
O
S
W
Ź
.
cleartext
plaintext
ciphertext
plaintext
cleartext
27
Decryption using key k2 (private key)
Encryption using key k1 (public key)
28
Decryption using the same key k
key distribution
Encryption using key k
asymmetric algorithms
= public-key algorithms
cleartext
cryptology = cryptography + cryptoanalysis
26
plaintext
ciphertext
plaintext
cleartext
symmetric algorithms
cryptosystem = cryptographic algorithm
+ all keys + all cryptograms + all cleartexts
cipher
1. encryption scheme
= cryptographic algorithm
= math. function for encryption and decryption
using a key
2. encrypted message = ciphertext = cryptogram
basic cryptographic terms (II)
eavesdropping on quantum system
31
first paper on quantum cryptography
29
1983 (1970 !)
monitoring disturbs
quantum information
no-cloning theorem
quantum teleportation seems to be also impossible ???
superluminal communication by using entangled states is impossible
quantum cryptography is secure
This is one of the most fundamental theorems of
quantum mechanics and quantum information
[Wootters, śurek and Dieks (1982)]
It is impossible to make a copy of an unknown quantum state.
32
Eve cannot clone the information
as she does not know the state
of the „carrier” of information
by Stephen Wiesner
first description of quantum coding
How to print banknotes,
which cannot be counterfeited
How to combine two messages such that
by reading one of them,
the other is automatically destroyed
30
eavesdropping on classical system
two steps:
1. Eve makes a copy (clone) of
the information carrier
2. and reads information from the copy
passive monitoring of
classical information is possible
photons with slanted
polarization
photons polarized
vertically
photons polarized
horizontally
crystal
enables discrimination of polarizations perpendicular to each other
for example Iceland spar (calcite) or BBO
a birefringent crystal
34
1. It is possible to disinguish two
polarizations at a = 0o and 90o
2. It is possible to disinguish two
polarizations at a = 45o and 135o
3. It is possible to change quickly
polarization (e.g. by Pockels cell)
4. BUT it is impossible to measure
simultaneously four polarizations at
a=0o, 90o, 45o, 135o
Heisenberg uncertainty principle
passive eavesdropping is impossible
33
BB84
cipher
Bob
PROBLEM:
How can Alice and Bob establish
their common key?
2 stages:
1. quantum key distribution
2. classical encryption
e.g.Vernam protocol
Alice
1. quantum private for key distribution
2. classical public (e.g. internet)
for cipher transmission
key
36
35
scheme of Bennett & Brassard (1984)
= BB84 protocol
two channels:
Vernam cipher/protocol (1918)
= Che Guevara cipher
= one-time pad
algorithm
addition modulo N (e.g. 40)
key
1. a series of randomly chosen number
2. physically safe
3. not shorter than the length of message
Vernam cipher is unbreakable
if the above conditions are satisfied.
example of Vernam protocol
A
M
_
M
I
R
A
N
O
W
I
C
Z
_
Z
37
38
O
N
16 19 24 03 13 24 07 25 10 23 20 19 22 38 14 16 12 16 11
Key is a series of random numbers:
D
cleartext:
A
plaintext:
01 06 01 17 36 17 12 24 01 18 20 30 12 04 33 36 33 20 18
--------------------------------------------------------------------------------------------
Sum:
17 25 25 20 49 41 19 49 11 41 40 49 34 42 47 52 45 36 29
Sum mod (40):
17 25 25 20 09 01 19 09 11 01 40 09 34 02 07 12 05 36 29
cipher
BB84 (1)
convention
and / (a = 45o )
| (a = 90o ) and \ (a = 135o )
- (a = 0o )
07 08 09 10 11 12 13 14 15
1.Alice sends photons
01 02 03 04 05 06
39
=> bit 1
=> bit 0
+ + x + x x x + x + + x x + x base
40
| - \ | / / \ | / - | \ \ - \ polarization
1 0 1 1 0 0 1 1 0 0 1 1 1 0 1 bit
BB84 (2)
2. Bob randomly chooses the measurement base
polarization of
Alice’s photons
07 08 09 10 11 12 13 14 15
| - \ | / / \ | / - | \ \ - \
Bob’s base
01 02 03 04 05 06
+ x + + x x + + x + x x + + x
| \ - | / / | | / - / \ - - \
polarization of
photons after
measurement
42
07 08 09 10 11 12 13 14 15
1 . . 1 0 0 . 1 0 0 . 1 . 0 1 Bob’s series
1 . . 1 0 0 . 1 0 0 . 1 . 0 1 Alice’s series
01 02 03 04 05 06
4. Alice and Bob keep only those results
obtained for the same bases
BB84 (4)
test
OK
OK
BB84 (6)
OK
Bob’s series
* . . 1 * 0 . 1 0 * . 1 . * 1
1 0 1 0 1 1
thus the secret key known only to Alice and Bob is
Alice’s series
07 08 09 10 11 12 13 14 15
44
* . . 1 * 0 . 1 0 * . 1 . * 1
01 02 03 04 05 06
6. They reject the bits for the tested photons
OK
1 . . 1 0 0 . 1 0 0 . 1 . 0 1 Bob’s series
+ x + + x x + + x + x x + + x Bob’s bases
y n n y y y n y y y n y n y y
1 . . 1 0 0 . 1 0 0 . 1 . 0 1 Alice’s series
+ + x + x x x + x + + x x + x Alice’s bases
07 08 09 10 11 12 13 14 15
01 02 03 04 05 06
01 02 03 04 05 06
07 08 09 10 11 12 13 14 15
3. Alice and Bob publicly compare bases
BB84 (5)
43
5. Alice and Bob publicly check results
for some photons say 1, 5, 10, 14th.
BB84 (3)
41
Test of agreement
1. Alice and Bob compare arbitrary subset of their data.
(Obviously, the tested subset is then not used for the key.)
2. Methods:
(a) testing bit after bit
(b) testing parity
e.g 20 times ⇒ (1/2)20 ∼ 0.000001
3. If the subset reveals eavesdropping,
all the data of Alice and Bob are rejected and
the BB84 protocol is repeated.
4. privacy amplification
45
46
via e.g. Bennett-Brassard-Robert, Ekert et al. or Horodecki et al. schemes
Eve’s strategy of eavesdropping (I)
Eve measures a photon sent by Alice
(e.g. by using another calcite crystal)
QUESTION
What is the probability that
a single photon was measured by Eve,
but Alice and Bob have not realized it?
ANSWER
P = 43
the probability is surprisingly high!
Base
Polarization
Probability
47
= 1/8
strategy of eavesdropping
Alice
+
Eve
+
1/2
Bob
+
|
1/2
= 1/8
Bob
+
Alice
+
|
Eve
x
/
1/2*1/2
Bob
+
|
1/2
48
= 1/2
Alice
+
|
Eve
x
\
1/2*1/2
security of BB84
for 1 photon P1=3/4
so for n photons Pn=(3/4)n
e.g.
P2=(3/4)2 ~ 0.56
P10=(3/4)10 ~ 0.06
P20=(3/4)20 ~ 0.003
P100=(3/4)100 ~ 10-13
and for 1000 photons
P1000=(3/4)1000 ~10-125
49
1997 Grover algorithm for searching databases:
How to search the keys more effectively?
1994 Shor algorithm for finding discrete logarithms:
How to break ElGamal cryptosystem?
1994 Shor algorithm for number factorization:
How to break cryptosystems of
RSA, Rabin, Williams, Blum-Goldwasser,…?
1985 Deutsch (Deutsch-Jozsa / DJ) algorithm:
How to see both sides of a coin simultaneously?
quantum algorithms and cryptography
50
2004 Englert et al. protocol claimed to be the
most efficient nowadays (Singapore protocol)
1992 Bennett protocol using any two
nonorthogonal states (B92)
1992 Bennett-Brassard-Mermin protocol
a la Ekert protocol but without Bell’s inequality
1991 Ekert protocol based Bell’s inequality (E91)
1984 Bennett-Brassard protocol (BB84)
famous protocols of
quantum key distribution
A bipartite mixed state is
P
separable if ρAB = i piρiA ⊗ ρiB
P
entangled if ρAB 6= i piρiA ⊗ ρiB
P
where i pi = 1 and pi ≥ 0 for all i.
separable if |ψAB i = |ψAi ⊗ |ψB i
entangled if |ψAB i =
6 |ψAi ⊗ |ψB i
A bipartite pure state is
definition
have to be described with reference to each other.
the quantum states of two or more objects (possibly spatially separated)
it is a quantum phenomenon in which
(Schrödinger 1935)
quantum entanglement
= inseparability
= Verschränkung = entwinement
quantum cryptography is
a branch of physics
- mainly of quantum optics
classical cryptography is
a branch of mathematics
„Information is inevitably tied to
a physical representation and
therefore to restrictions and possibilities
related to the laws of physics”
[R. Landauer]
information is physical
52
51
quantum entanglement
& Einstein-Podolsky-Rosen paradox
→
→
S 1+ S
2
= 0
2
S 1y + S 2 y = 0
S
→
53
How to ‘violate’ uncertainty relation ?
Can we measure exactly both compontents of spin?
2
!2
S 1z
4
1
var S 1 x var S 1 y ≥
→
S
S 1 x + S 2 x = 0,
54
one of the
triplet states
singlet state
So let’s measure S 1 x i S 2 y to determine S 2 x i S 1 y
Bell states (EPR states)
maximally entangled two-qubit states
ΦA = Ψ ( − )
ΦB = Ψ ( + )
ΦC = Φ ( − )
1
=
( 0 x 1 y − 1 x 0 y)
2
1
=
( 0 x 1 y + 1 x 0 y)
2
1
( 0 x 0 y − 1 x 1 y)
=
2
1
( 0 x 0 y + 1 x 1 y)
2
ΦD = Φ ( + ) =
55
GHZ (Greenberger-Horne-Zeilinger) states
maximally? entangled states of 3 qubits
'
ΨGHZ
' ''
ΨGHZ
√1 
2
56
1
1
''
(
( 001 ± 110 )
000 ± 111 )
=
ΨGHZ
=
2
2
1
iv
1
( 100 ± 011 )
Ψ
(
010 ± 101 )
=
GHZ =
2
2
W states
1
( 100 + 010 + 001 )
ΦW =
3
1
( 011 + 101 + 110 )
3
ΨW =
√1 (|00i
2
 
0
 
 1 ,
|Ψ(+)i = √12 (|01i + |10i) = √12 
1
0
 
1
0

− |11i) =
|Φ(+)i =
0
1
Bell states in matrix representation
√1 
2


0


 1 ,
|Ψ(−)i = √12 (|01i − |10i) = √12 
−1 
0


1
 0 
,
− |11i) =
|Φ(−)i =
0 
−1
√1 (|00i
2
a GHZ state in matrix representation
√1
2
√1 ([10000000]T
2
+ [00000001]T ) =
√1 [10000001]T
2
|Ψ′i = √12 (|000i + |111i)
1
1
1
0
0
0
⊗
⊗
+
⊗
⊗
0
0
0
1
1
1
=
=
quantum limit: λ/(2N ) for N -photon absorption
classical limit: λ/2
• interferometric lithography beyond diffraction limit [Boto et al.’00]
• better clock synchronization [Jozsa et al.’00, Chuang’00]
• to improve frequency standards [Huelga et al. ’97]
• quantum noise reduction in spectroscopy [Wineland et al.’92]
to increase precision of quantum measurements:
• remote state preparation [Bennett et al.’01]
• telecloning [Murao et al. 99]
in quantum state engineering:
• to simplify communication complexity [Cleve & Buhrman’97]
in quantum communication:
Applications of quantum entanglement (II)
• quantum watermarking [Imoto et al.’05]
• quantum authentication [Ljunggren et al.’00]
• secret sharing [Żukowski et al.’98]
• privacy amplification [Bennett et al.’96, Deutsch et al.’96]
• quantum key distribution [Ekert’91, Bennett et al.’92]
in quantum cryptography:
• quantum error correction [Shor’95, Steane’96]
• superfast [Shor’94] and fast [Grover’97] algorithms
• entanglement swapping [Żukowski et al.’93]
• quantum teleportation [Bennett et al.’93]
• dense coding [Bennett and Wiesner’92]
in quantum information theory:
Applications of quantum entanglement (I)
58
57
|01i−|10i
√
2
60
where
= aa′|00i + ab′|01i + ba′|10i + bb′|11i
= (a|0i + b|1i) ⊗ (a′|0i + b′|1i)
which admits no solution.
1
1
aa′ = 0, ab′ = √ , ba′ = − √ , bb′ = 0
2
2
So we get set of equations
a, a′, b, b′ ∈ C
|a|2 + |b|2 = |a′|2 + |b′|2 = 1
|01i−|10i
√
2
proof that it cannot be written as a product state
|ψi =
example of inseparable state
- singlet state
1
1
|ψi = [|0iA(|0iB + |1iB ) + |1iA(|0iB + |1iB )] = (|0iA + |1iA)(|0iB + |1iB )
2
2
|0iA + |1iA |0iB + |1iB
√
√
·
≡ |+iA|+iB
=
2
2
1
|ψi = (|00i + |01i + |10i + |11i)
2
proof:
example 2
|+iB
1
|0iB + |1iB
1
√
= |1iA|+iB
|ψi = √ (|1iA|0iB +|1iA|1iB ) = √ |1iA(|0iB +|1iB ) = |1iA
2
2
| {z2 }
proof:
1
|ψi = √ (|10i + |11i)
2
example 1
examples of separable states
59
example of inseparable (Bell-like) state
proof:
=|−iB
1
|ψi = (|00i + |01i + |10i − |11i)
2
=|+iB
|0i − |1i |0+i + |1−i
|0i + |1i
1 B
B
B
B
√
√
√
+|1iA
=
|ψi = √ |0iA
2
2
{z2 }
| {z2 }
|
=|−i
this state can be obtained from a Bell state
by applying Hadamard gate to 2nd qubit
|00i + |11i
√
|ψi = HB
2
=|+i
1
= √ (|0iA(H|0i)B + |1iA(H|1i)B )
| {z }
| {z }
2
1
= √ (|0iA|+iB + |1iA|−iB )
2
generation of all Bell states from one of them
|00i + |11i
√
2
e.g. from
|Φ(+)i =
flip qubit B = apply NOT gate to qubit B:
|00i + |11i |01i + |10i
√
√
= |Ψ(+)i
=
XB
2
2
phase flip qubit B:
|00i + |11i |00i − |11i
√
√
=
= |Φ(−)i
2
2
flip + phase flip qubit B:
|00i + |11i
|01i − |10i
√
√
=i
= i|Ψ(−)i
2
2
ZB
YB
61
62
How to generate Bell states from |00i
Bell state |Φ(+)i :
=
#1
|0i + |1i
|00i + |10i
√
= UCNOT √
|0i = U CNOT
2
2
|0, 0 ⊕ 0i + |1, 0 ⊕ 1i
√
=
2
|00i + |11i
√
= |Φ(+)i
2
UCNOTHA|00i = UCNOT| + 0i
Bell state |Φ(−)i :
#2
|0, 0 ⊕ 0i − |1, 0 ⊕ 1i |00i − |11i
√
√
=
= |Φ(−)i
2
2
UCNOTHAXA|00i = UCNOTHA|10i = UCNOT| − 0i
=
|01i + |10i
√
= |Ψ(+)i
2
|0, 1 ⊕ 0i + |1, 1 ⊕ 1i
√
2
How to generate Bell states from |00i
Bell state |Ψ(+)i :
=
= UCNOT| + 0i =
UCNOTHAXB |00i = UCNOTHA|01i
Bell state |Ψ(−)i :
UCNOTHAXAXB |00i = UCNOTHA|11i
|0, 1 ⊕ 0i − |1, 1 ⊕ 1i
√
2
|01i − |10i
√
= |Ψ(−)i
2
= UCNOT| − 1i =
=
63
64
l
qubit B
H
CNOT
Bellkl
|0li + (−1)k |1, 1 ⊕ li
√
= |Bellkl i
2
k
qubit B
H
CNOT
Bellkl'
CNOT’
in | out
-------00 | 00
01 | 11
10 | 10
11 | 01
SWAP
in | out
-------00 | 00
01 | 10
10 | 01
11 | 11
66
65
CNOT
CNOT’
SWAP
ÛSWAP|ψini = c0|00i + c2|01i + c1|10i + c3|11i ≡ c0|0i + c2|1i + c1|2i + c3|3i
′
|ψini = c0|00i + c3|01i + c2|10i + c1|11i ≡ c0|0i + c3|1i + c2|2i + c1|3i
ÛCNOT
ÛCNOT|ψini = c0|00i + c1|01i + c3|10i + c2|11i ≡ c0|0i + c1|1i + c3|2i + c2|3i
|ψini = c0|00i + c1|01i + c2|10i + c3|11i ≡ c0|0i + c1|1i + c2|2i + c3|3i
CNOT
in | out
-------00 | 00
01 | 01
10 | 11
11 | 10
CNOT and SWAP gates
l
qubit A
′
UCNOT
HB |lki = |Bell’kl i
obviously, we can obtain Bell states by:
k
qubit A
UCNOTHA|kli =
general formula (k, l = 0, 1)
Generation of Bell states
0
0
1
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
1
0
0
0
0
1
   
0
c0
c0
 c1   c2 
0
  =  
0  c 2   c 1 
1
c3
c3
   
0
c0
c0
 c1   c3 
1
  =  
0  c 2   c 2 
0
c3
c1
   
0
c0
c0
 c1   c1 
0
  =  
1  c 2   c 3 
0
c3
c2
xy
=
(
1
↑ x ↓ y− ↓ x ↑
2
y
)
xy
=
(
y
y
)
)
68
xy
=
1
2
(0
x
1 y + e iχ 1 x 0
y
)
projection of separable state onto entangled one
Φ
output light of beam splitter 50:50 with a single input photon
Ψ
1
H x V y + e iχ V x H
2
1
Φ xy =
H x H y + e iχ V x V
2
(
light generated by parametric down conventer (PDC II)
Ψ
decay of a particle with spin 0 into 2 particles with spin ½
Generation of Bell states
 
c0
1
 c1   0
 
ÛSWAP
 c2  =  0
0
c3

 
c0
1



c
′
 1=0
ÛCNOT
 c2   0
c3
0

 
c0
1
 c1   0
 
ÛCNOT
 c2  =  0
c3
0

CNOT and SWAP gates
in matrix representation
67
Entanglement
from a mystery to a physical resource
69
• An entangled wave-function does not describe the physical reality in a
complete way. [Einstein,Podolsky,Rosen]
• For an entangled state is the best possible knowledge of the whole does
not include the best possible knowledge of its parts. [E. Schrödinger]
70
• Entanglement is a “fundamental resource of Nature, of comparable importance to energy, information, entropy, or any other fundamental resource.” [M. Nielsen, I. Chuang]
Entanglement is...
• a correlation that is stronger than any classical correlation. [J. Bell]
• a correlation that contradicts the theory of elements of reality.
[D. Mermin]
• a trick that quantum magicians use to produce phenomena that cannot be
imitated by classical magicians. [A. Peres]
• a resource that enables quantum teleportation. [C. Bennett]
• a global structure of the wavefunction that allows for faster algorithms.
[P. Shor]
• a tool for secure communication. [A. Ekert]
• the need for first applications of positive maps in physics.
[Horodecki family]
[collected by Dagmar Bruß, quant-ph/0110078]
→
S
1
→
S 1+ S
→
S
= 0
2
→
2
− | ↓i | ↑i
| ↑↓i − | ↓↑i
1
2
√
√
≡
2
2
1
2
71
b
72
filter 2
Measurement of entangled spins
and Bell inequalities
Stern-Gerlach filter
a
filter 1
N++(α, β) – number of outcomes when spin +1 was measured
in filter 1 at angle α and in filter 2 at angle β
N – total number of outcomes
|
↑i1|
↓i2
probability P++(α, β) = N++(α, β)/N
• QM predictions
|ψ (−)i =
for electrons in singlet state
then
α−β
1
P++(α, β) = P−−(α, β) = sin2
2
2
1 2 α−β
cos
2
2
P+−(α, β) = P−+(α, β) =
spins of opposite values
1
4
P+−(α, α) = P−+(α, α) =
⇒
measurement of 2 independent spins
+α
P++(α, α) = P−−(α, α) = 0,
⇒
special cases:
π
2
P++(α, β) = P+−(α, β) = P−+(α, β) = P−−(α, β) =
⇒
1. α = β
2. β =
⇒
74
73
• “the deepest discovery in history of science” [Stapp]
as enables to decide experimentally a dispute between Einstein and Bohr.
• violation of a Bell inequality implies that
the world is nonrealistic or/and nonlocal
– measurement by filter in position 1 does not influence
the result of measurement by filter in a distant place 2
3. locality
– physical objects have properties independent of whether we measure
them or not
2. realism
– this is very natural assumption
– from large number of measurements probability can be determined
1. reasoning by induction
Bell assumptions
P[a_, b_] := 1/2*Sin[(a - b)/2]^2
LHS = P[0, Pi/4] + P[Pi/4, Pi/2]
RHS = P[0, Pi/2]
Mathematica:
LHS = sin2(π/8) = 0.146 · · · ≥
\ RHS = 0.25
occurs, e.g., for α = 0, β = π/4, γ = π/2
• violation of Bell inequality
P++(α, β) + P++(β, γ) ≥ P++(α, γ)
N++(α, β) + N++(β, γ) ≥ N++(α, γ)
• Bell inequality for 3 series of measurements
PM
C
C
PM
IˆB −
C
75
hIÂ+IˆB+i + hIÂ−IˆB−i + hIÂ+IˆB−i + hIÂ−IˆB+i
h(IÂ+ − IÂ−)(IˆB+ − IˆB−)i
h(IÂ+ + IÂ−)(IˆB+ + IˆB−)i
hIÂ+IˆB+i + hIÂ−IˆB−i − hIÂ+IˆB−i − hIÂ−IˆB+i
76
S = E(φ′A, φ′B ) − E(φ′A, φ′′B ) + E(φ′′A, φ′B ) + E(φ′′A, φ′′B )
should be satisfied for realistic local theories.
where
|S| ≤ 2
• Bell inequality according to Clauser-Horne-Shimony-Holt
at any angles φA, φB .
=
E(φA, φB ) =
• What is measured in the experiment?
PM – photon multipliers
PA, PB – polarizing beam splitters = optical analogues of Stern-Gerlach filters
Iˆk ∼ n̂k – intensity of kth beam
S – source of photons
C – correlation systems
Key:
C
IÂ−
Aspect et al. tests of Bell-CHSH inequalities (1984)
PA ˆ
PB ˆ
I B+
IA
IˆB
IÂ+
PM
PM
S
|HiA|V iB − |V iA|HiB
√
|Ψ(−)i =
2
• polarization singlet state
• QM predicts that
77
E(φA, φB ) = P++(φA, φB ) + P−−(φA, φB ) − P+−(φA, φB ) − P−+(φA, φB )
where Pjk (φA, φB ) in the former slides
• maximal violation of Bell inequality
√
max |S| = 2 2
II. loopholes in BI assumptions
1. independence assumption:
79
There are physical processes independent of the Bell experiment that can
be used as an effective source of randomness.
III. interpretational loopholes
Accepting that Bell’s theorem is true then either locality or realism might be
false.
1. universe is non-local but real
e.g. Bohmian hidden variable theory is explicitly non-local and
contextual, but still fairly natural looking.
QM is just a set of recipes for calculating probabilities of measurement
results.
the standard Copenhagen interpretation:
2. operationalism of quantum mechanism (QM)
which can be obtained for the following angles of polarizers
π
π
π
′
′
′′
′′
φA
= 0, φB
= , φA
= , φB
=3
4
2
4
as
π √
π
3π
3π
S = −3 cos2
+ 3 sin2
= −2 2 ≈ −2.8
+ cos2
− sin2
8
8
8
8
loopholes in Aspect’s et al. experiment (1982)
1. a fair-sample assumption
• BI test more ideal than ever by Zeilinger et al. (PRL 1998) on
“Violation of Bell’s inequality under strict Einstein locality conditions”.
The necessary space-like separation of the observations was not achieved by
sufficient physical distance between the measurement stations.
(ii) no true space-like separation
Aspect et al. switched the directions of polarization analysers after the photons left the source, but they used periodic sinusoidal switching, which is predictable.
Analyzers were not randomly rotated during the flight of the particles.
2. causality
(i) no true randomization
All experiments so far detect only a small subset of all pairs created thus it
is necessary to assume that the pairs registered are a fair sample of all pairs
emitted.
80
It says nothing about an underlying physical reality, which may not even
exist, and therefore nothing about its locality.
78
• Note: the same setup of Aspect was used for testing
violation of Schwarz inequality
possible loopholes of Bell’s inequalities (BI)
• in the present experimental BI tests
• in the proof or assumptions of BI
• in the BI interpretation
I. experimental loopholes
there has been no loophole-free experimental test of BI.
1. detection efficiency and fair-sample assumption:
None experimental test does not detect 100% the particle pairs emitted.
Thus it is not clear that the pairs registered are a fair sample of all pairs
emitted.
2. causality:
The choice of measurement settings of Alice and Bob should be
truly random and placed at sufficient physical distance.
i pi
⊗
ρiB
= 1 and pi ≥ 0 for all i.
i
piρiA
pi|ψAi ihψAi | ⊗ |ψBi ihψBi |
i pi
P
i
X
i pi
i
p
||Â|| = Tr|Â| = Tr( Â†Â)
trace norm
If ρ̂AB acts in finite-dimensional Hilbert spaces
then the approximation part is redundant.
systems of finite dimensions
where
P
A state ρ̂AB is separable iff
it can be written or approximated (in, e.g., trace norm) by
X
ρ̂AB =
piρiA ⊗ ρiB
General definition of bipartite entanglement
These definitions are equivalent, as ρiA (and ρiB )
can be expanded in terms eigenvectors
X
ρjA =
qj |ψAj ihψAj |
P
where j qj = 1 and qj ≥ 0 for all i.
where
ρAB =
Def. 2. A state ρAB is separable iff
where
P
ρAB =
X
Two definitions of entanglement
Def. 1. A state ρAB is separable iff
82
81
ihΨ
(+)
|=
i=1
5
X
pi|aiihai| ⊗ |biihbi|
1−p
1̂ ⊗ 1̂ + p|Ψ(±)ihΨ(±)|
4
|01i ± |10i
√
2

B(ρ̂W ) > 0 for all
√1
2
<p≤1
and violates Bell inequality
E(ρ̂W ) > 0 for all 13 < p ≤ 1
the Werner state is entangled
1
0
1̂ ⊗ 1̂ ≡ 1̂2 ⊗ 1̂2 = 1̂4 = |00ih00| + |01ih01| + |10ih10| + |11ih11| = 
0
0
and maximally mixed state
|Ψ(±)i =
is mixture of maximally entangled state
(±)
ρ̂W =
Werner state
thus it is not a convex combination of product states
and the state is entangled.
note that p4 = p5 < 0
|0i + |1i
√
2
|0i + ei2π/3|1i
√
|a2i = |b2i =
2
|0i + e−i2π/3|1i
√
|a3i = |b3i =
2
|a4i = |0i, |b4i = |1i, |a5i = |1i, |b5i = |0i
|a1i = |b1i =
2
1
p1 = p2 = p3 = , p4 = p5 = −
3
2
|Ψ
(+)
e.g. the ‘triplet’ state
Can we decompose Bell states
into separable states?
0
1
0
0
0
0
1
0

0
0

0
1
84
83
=
i=1
7
X
pi|aiihai| ⊗ |biihbi|
decomposition of Werner state
(+)
ρ̂W
2
1 − 3p
1−p
p1 = p2 = p3 = , p4 = p5 =
, p6 = p7 =
3
4
4
=
=
=
=
|0i,
|1i,
|0i,
|1i,
|0i + |1i
|a1i = |b1i = √
2
|0i + ei2π/3|1i
√
|a2i = |b2i =
2
|0i + e−i2π/3|1i
√
|a3i = |b3i =
2
|b4i = |1i
|b5i = |0i
|b6i = |0i
|b7i = |1i
|a4i
|a5i
|a6i
|a7i
clearly, the state is separable only for p ≤ 13 when all pi ≥ 0.
1−p
1̂ ⊗ 1̂ + p|ψihψ|
=
4
Werner and Werner-like states
ρ̂W (p)
|01i−|10i
√
2
original Werner state
|ψi → |Ψ(−)i =
|00i+|11i
√
2
isotropic state (Werner-like state)
|ψi → |Φ(+)i =
|ψi → |Ψ(+)i =
|00i−|11i
√
2
|01i+|10i
√
2
other Werner-like states
|ψi → |Φ(−)i =
1−p
d2
(+)
1̂⊗d + p|Φd ihΦd |
(+)
two qudit (d-dimensional) isotropic state
ρ̂W (p) =
√1
d
where − d21−1 ≤ p ≤ 1
Pd−1
i=0 |iii
(+)
|Φd i =
85
86
properties of Werner states
(−)
87
88
1. They are so-called maximally entangled mixed states (MEMS)
• entanglement E(ρW ) cannot be increased by
any unitary operations,
• entanglement E(ρW ) is maximal
for a given linear entropy (and vice versa)
[Ishizaka, Hiroshima’00, Munro et al.’01]
(−)
2. Original Werner state exhibits U ⊗ U invariance:
Û ⊗ Û ρ̂W = ρ̂W
√
3. All entangled Werner states (even for p ∈ (1/3, 1/ 2i)
can be used for quantum-information processing
including teleportation [Popescu’94, Lee,Kim’00]
nonlocality
quantum states are called nonlocal if
there is no local unhidden variable model of their behavior.
Thus a measurement of the whole can reveal
more information about the system’s state
than any sequence of classically coordinated measurements of the parts.
1. nonlocality without entanglement
[Bennett et al. 1999]
There are orthogonal sets of product states of two qutrits
that cannot be reliably distinguished by a pair of separated observers
ignorant of which of the states has been presented to them,
even if the observers are allowed to communicate by LOCC.
Note: LOCC = LQCC
local quantum operations and classical communication
1 √ |uiatom|aliveicat + |diatom|deadicat
2
where |di – decayed state, |ui – undecayed state.
• state of the isolated atom after a time equal to its half-time is
1 √ |ui + |di
2
– automatic device to release the poison when the atom decays
– radioactive atom
– a bottle with poison gas
– a cat
image a chamber containing:
• gedanken experiment of Schrödinger cat
– can we talk about of superposition, entanglement or smearing of classical
objects?
92
• Note: the above states are not mixtures of coherent states
1
ρ̂mix = |αihα| + | − αih−α| 6= |ψihψ|
2
1
|ψYSi = √ |αi ± i| − αi
2
– Yurke-Stoler states for ϕ = ±π/2:
∞
X
1
α2n+1
p
|ψ−i = N− |αi − | − αi = p
|2n + 1i ≡ |1iL
sinh(α2) n=0 (2n + 1)!
– odd coherent state for ϕ = π:
∞
X
α2n
1
p
|2ni ≡ |0iL
|ψ+i = N+ |αi + | − αi = p
cosh(α2) n=0 (2n)!
– even coherent state for ϕ = 0:
• questions
Schrödinger cats:
N = [2 + 2 cos ϕ exp(−2α2)]−1/2
where N is a normalization constant assuming α ∈ R:
|ψi = N [|αi + eiϕ| − αi]
• e.g. superposition of two coherent states
special cases of single-mode
– how to interpret superposition of states and entanglement?
91
Note: the definition can be applied to a single mode, thus not necessarily has
to be related to entanglement
it is a superposition of two macroscopically distinct states
(modern) definition of Schrödinger cat (state)
the „quantum chamber” has to be opened to check the state of the „cat”.
to understand it, one has to include the observer or measurement process:
• Copenhagen interpretation
so the state is
• but the atom is entangled with the cat via the apparatus
Schrödinger cat paradox
90
Note: two-qubit pure states violate Bell inequality iff they are entangled.
thus they can be described in terms realistic local theories
are entangled and satisfy Bell inequality
e.g. Werner states and isotropic states for
√
p ∈ (1/3, 1/ 2i
two-qubit mixed states can be entangled without violating Bell inequality
[Werner 1989]
2’. entanglement without violating Bell inequality
(although there are some serious doubts about it)
are often identified
nonlocality and violation Bell-inequality
2. entanglement without nonlocality
89
Problems
93
two-mode Schrödinger cats
Hamiltonians
(for simplicity assume g = |g|)
1. hamiltonian for a degenerate process
Ĥdeg = h̄g[â2 + (â2)†]
2. hamiltonian for a nondegenerate process of type I
Ĥ1 = h̄gâs† âi† + h.c.
Note: signal (s) and idler (i) have the same polarization
ωp
pump
r
ks
r
kp
χ ( 2)
nonlinear crystal
r
ki
momentum conservation
idler
95
ω s signal
ωi
ωs
ωi
energy conservation
ωp
spontaneous parametric down-conversion (SPDC)
96
a nonlinear optical phenomenon in which a nonlinear crystal splits incoming
photons into pairs of photons of lower energy.
spontaneous parametric down-conversion (SPDC)
– How to generate Bell polarization states?
– How to generate single photons?
94
continuous variable (CV) version of two-qubit Bell states
1
|Φ(+)i = √ (|00iL + |11iL) = N (|α, βi + | − α, −βi)
2
1
|Φ(−)i = √ (|00iL − |11iL) = N (|α, −βi + | − α, βi)
2
1
|Ψ(+)i = √ (|01iL + |10iL) = N (|α, βi − | − α, −βi)
2
1
|Ψ(−)i = √ (|01iL + |10iL) = N (|α, −βi − | − α, βi)
2
assumption α = β
exemplary proof of a CV Bell state
definition of logical qubits (α = β):
|0iL = N+ |βi + | − βi
|1iL = N− |βi − | − βi
so
2N+N− |α, βi + | − α, −βi
+|α, βi − |α, −βi − | − α, βi + | − α, −βi
N N− h
+
√
|α, βi + |α, −βi + | − α, βi + | − α, −βi
2
√
i
i
1
N N− h
+
√ (|00iL+|11iL) = √
(|αi+|−αi)(|βi+|−βi)+(|αi−|−αi)(|βi−|−βi)
2
2
=
=
97
signal
pump
idler
o-ray
e-ray
98
pump
"red"
"green"
"blue"
type II parametric down converter
orange
red
deep red
type I parametric down converter
pump
type II parametric down converter
pump
type I parametric down converter
|ψ(0)i = |0iV s|0iHs|0iV i|0iHi
iτ
|ψ(t)i = (1 − 12 τ 2)|0is|0ii − √ |1is|1ii
2
|ψ(0)i = |0is|0ii
|1iV s|0iHs ≡ |V is,
|0iV s|1iHs ≡ |His,
1
|Ψ(+)i = √ (|V is|Hii + |His|V ii)
2
the second term is a polarization Bell state
iτ
|ψ(t)i = (1 − 12 τ 2)|0is|0ii − √ (|V is|Hii + |His|V ii)
2
thus we have
and analogously for mode i
|0iV s|0iHs ≡ |0is,
• notation:
|ψ(t)i = (1 − 12 τ 2)|0iV s|0iHs|0iV i|0iHi
iτ
− √ (|1iV s|0iHs|0iV i|1iHi + |0iV s|1iHs|1iV i|0iHi)
2
– type II process:
where τ = gt
– type I process:
"
2#
t
1 t
1
|ψ(0)i
Ĥt |ψ(0)i ≈ 1 + Ĥ +
Ĥ
|ψ(t)i = exp
ih̄
ih̄
2 ih̄
short-time evolution
Note: signal and idler modes have perpendicular polarizations
â†V s – creation operator for signal mode of vertical polarization etc.
V (H) – linear vertical (horizontal) polarization
where
Ĥ2 = h̄g(â†V sâ†Hi + â†Hsâ†V i) + h.c.
3. hamiltonian for a nondegenerate process of type II
100
99
101
How to generate other polarization Bell states?
– place half-wave plate (HWP) on the path of one beams:
1
1
|Ψ(+)i = √ (|V is|Hii + |His|V ii) → √ (|His|Hii + |V is|V ii) = |Φ(+)i
2
2
– use phase shifter or simply rotate crystal to change α:
1
|Ψ(+)i → |ψ (α)i = √ (|V is|Hii + eiα |His|V ii)
2
• this is how all four Bell states can be obtained including:
1
|Ψ(−)i = √ (|V is|Hii − |His|V ii)
2
1
|Φ(−)i = √ (|His|Hii − |V is|V ii)
2
the experimental method was developed by Kwiat, Zeilinger et al. (1995)
using crystals of BBO type II.
102
standard matrix representations of qubit states
1 1
1 −1
so far we have been applying the orthonormal basis
1
0
|0i =
, |1i =
0
1
√1
2
satisfying the identity resolution
1
0
1 0
0 0
1 0
[10] +
[01] =
+
=
0
1
0 0
0 1
0 1
Iˆ = |0ih0| + |1ih1| =
0 1
1 0
then the NOT gate is
X̂ = |0ih1| + |1ih0| =
and the Hadamard gate is
Ĥ = |0ih+| + |1ih−| = |+ih0| + |−ih1| =
can we define other representations?
1
2
√1
2
1
1
1 1
1 −1
1 0
[11] + 21
[1, −1] = 12
+ 21
=
1
−1
1 1
−1 1
0 1
identity resolution
then
1
=0
h0|1i = 21 [1, 1]
−1
|0i =
let’s use the orthonormal basis
1
1
|1i = √12
1
−1
another matrix representation of qubit states
Iˆ =
NOT gate
1
2
0
1
1
1
X̂ = |0ih1| + |1ih0| = 12
[1, −1] + 21
[1, 1]
1
−1
1 −1
1 1
1 0
+ 21
=
1 −1
−1 −1
0 −1
=
Hadamard gate
|+i =
√1 (|0i
2
√1 (|0i
2
− |1i) =
note that
1
+ |1i) =
0
|−i =
so
phase-flip gate
Ĥ = |0ih+|
+
|1ih−| 1
1 1
1
[10] + √
[01]
= √
2 1 2 −1 1 0 1
1 1 1
1 1 0
+√
=√
= √
2 1 0
2 0 −1
2 1 −1
1
2
1
1
Ẑ = |0ih0| − |1ih1| = 1
[1, 1] − 1
[1, −1]
2
2
1
−1
1 1
1 −1
0 1
− 21
=
1 1
−1 1
1 0
=
103
104
or equivalently
′
|0i
cos θ sin θ
|0i
=
− sin θ cos θ
|1i
|1i′
orthonormal basis
cos θ
− sin θ
, |1i =
sin θ
cos θ
CNOT
a 00 + b 11
|1i = c|xi + d|yi
=
ac|xxi + ad|xyi + bc|yxi + bd|yyi − ac|xxi − ad|yxi − bc|xyi − bd|yyi
√
2
ad(|xyi − |yxi) − bc(|xyi − |yxi)
√
=
2
|xyi − |yxi
|xyi − |yxi
a b |xyi − |yxi
√
√
√
= det
= eiφ
= (ad − bc)
c d
2
2
2
|01i − |10i (a|xi + b|yi) ⊗ (c|xi + d|yi) − (c|xi + d|yi) ⊗ (a|xi + b|yi)
√
√
=
2
2
proof
|0i = a|xi + b|yi,
unitary transformation
proof
|00, 900i − |900, 00i |θ, θ + 900i − |θ + 900, θi
√
√
=
2
2
it does not exist a general quantum cloning device
as the device can clone only orthogonal states.
thus
either |ψ1i = |ψ2i or |ψ1i is orthogonal to |ψ2i
equation x = x2 has only trivial solutions: x = 0 and x = 1
so
Û |ψ2i ⊗ |si = |ψ2i ⊗ |ψ2i
so
†
Û |ψ2i ⊗ |si = hψ1|ψ2i
LHS = hs| ⊗ hψ1|Û
RHS = hψ1| ⊗ hψ1| |ψ2i ⊗ |ψ2i = hψ1|ψ2i2
Û |ψ1i ⊗ |si = |ψ1i ⊗ |ψ1i
let us clone two states
It is impossible to copy (clone, reproduce) an unknown quantum pure state.
108
Measurements of the polarization singlet state at different angles of the crystal
now measure one of the qubits...
= a2|00i + ab(|01i + |10i) + b2|11i
Û a|0i + b|1i ⊗ |si → a|0i + b|1i ⊗ a|0i + b|1i
requirement for a cloning device
It is a cloning device iff ab = 0.
now measure one of the qubits...
0
a 0 +b 1
Quantum no-cloning theorem:
106
CNOT as a cloning device?
ÛCNOT a|0i + b|1i ⊗ |0i = a|0, 0 ⊕ 0i + b|1, 0 ⊕ 1i = a|00i + b|11i
107
singlet state and rotations of analyzer
Iˆ =
cos θ
− sin θ
[cos θ, sin θ] +
[− sin θ, cos θ]
sin θ
cos θ
cos2 θ sin θ cos θ
sin2 θ
− sin θ cos θ
1 0
=
+
=
sin θ cos θ sin2 θ
− sin θ cos θ
cos2 θ
0 1
− sin(2θ) cos(2θ)
NOT gate: X̂ = cos(2θ) sin(2θ)
−
sin(2θ)
+
cos(2θ),
sin(2θ)
+
cos(2θ)
1
Hadamard gate: Ĥ = √2 sin(2θ) + cos(2θ), sin(2θ) − cos(2θ)
identity resolution
in terms of standard qubit representations |0i = [1; 0] and |1i = [0; 1]
|0i =
rotated matrix representation of qubit states
105
Further questions
1. Can we clone unknown mixed quantum states?
Impossible!
2. What about cloning by non-unitary operations?
Impossible!
3. What about optimal approximate cloning machines?
For example, a cloning machine of Bužek and Hillery (1996)
4. Can we clone known quantum states?
Yes.
5. Can we clone known quantum states by local operations?
In general impossible!
109
Quantum no-deleting theorem:
Ûdel|ψi|ψi|Ai = |ψi|Si|Aψ i
we try to find a unitary operation Ûdel such that
proof
It is not (time) reverse of quantum no-cloning theorem.
note
It is impossible to delete an unknown quantum state against a copy.
where
|ψi = a|0i + b|1i - arbitrary qubit state
|Ai – initial state of ancilla
|Aψ i – final state of ancilla (dependent on |ψi)
|Si – some standard final state of the qubit
Let us analyze polarization states
|ψi = a|Hi + b|V i
We expect
Ûdel|Hi|Hi|Ai = |Hi|Si|AH i
Ûdel|V i|V i|Ai = |V i|Si|AV i
110
|1>
Quantum cloning and stimulated emission
ω 01
Ûdel
Thus, we have
LHS =
=
=
=
RHS = a|Hi|Si|Aψ i + b|V i|Si|Aψ i
Ûdel|ψi|ψi|Ai
Ûdel(a|Hi + b|V i) ⊗ (a|Hi + b|V i)|Ai
Ûdel[a2|Hi|Hi + ab(|Hi|V i + |V i|Hi) + b2|V i|V i]|Ai
√
a2|Hi|Si|AH i + 2ab|Bi + b2|V i|Si|AV i
where |Bi is some combined input-ancilla state.
|Hi|V i + |V i|Hi
√
|Ai = |Bi
2
|0>
For each perfect clone there is also
one randomly polarized, spontaneously emitted, photon.
Another question
Suppose that we have several copies of a photon in an unknown state.
Is it possible to delete the information content of
one or more of these photons by a physical process?
having only solution for
|Aψ i = a|AH i + b|AV i
∀ψ hAψ |Aψ i = 1 ⇒ hAH |AV i = 0
111
112
|Bi =
|Hi|Si|AV i + |V i|Si|AH i
√
2
114
[it might be you...]
4. quantum no-deleting for mixed states ???
Given a general mixed state for a quantum system, there are no physical
means for broadcasting that state onto two separate quantum systems, even
when the states need only be reproduced marginally on the separate systems.
specifically
It is impossible to copy an unknown quantum mixed state.
[Barnum, Caves, Fuchs, Jozsa, Schumacher (1996)]
3. quantum no-broadcasting
It is impossible to delete an unknown quantum pure state.
[Pati, Braunstein (2000)]
2. quantum no-deleting
It is impossible to copy an unknown quantum pure state.
[Wootters, Żurek, Dieks (1982)]
1. quantum no-cloning
no-go theorems
|Aψ i = a|AH i + b|AV i.
“deleting” is just swapping onto 2D subspace of ancilla
as
113
√ |Hi|Si|AV i + |V i|Si|AH i
√
a2|Hi|Si|AH i + 2ab
+ b2|V i|Si|AV i
2
a2|Hi|Si|AH i + ab|Hi|Si|AV i + ab|V i|Si|AH i + b2|V i|Si|AV i
a|Hi|Si a|AH i + b|AV i + b|V i|Si a|AH i + b|AV i
a|Hi + b|V i |Si|Aψ i = RHS
Thus it is seen that
=
=
=
LHS =
so
and
EPR
EPR
B
PM
PM
115
(
1
H
2
V
A
−V
B
A
H
B
)=
(
1
+
2
A
−
B
− +
EPR
EPR
EPR
EPR
PM
PM
PM
PM
116
A
−
B
)
BIT 0
BIT 0
BIT 1
BIT 1
a patent for superluminal communication
Ψ =
PM = photon multiplier
EPR = source of the polarization singlet state (e.g. BBO)
A
superluminal communication?
Alice
Bob
(super)dense coding
[Bennett and Wiesner 1992]
+ |1iA|1iB
√
2
+ |1iA|1iB
√
2
117
Ẑ|Φ(+)i = |Φ(−)i
|Φ(+)i
⇒
118
X̂|Φ(+)i = |Ψ(+)i
iŶ |Φ(+)i = |Ψ(−)i
⇒
⇒
⇒
How to send 2 bits of information by transmitting 1 qubit?
|0iA|0iB
1. Alice and Bob share 2 qubits in an EPR state e.g.
|Φ(+)i =
2. if Alice wants to send a message:
00 - then she does nothing
01 - then flips phase of her qubit (Pauli Z gate)
10 - then flips her qubit (Pauli X gate)
11 - then flips & phase-flips her qubit (Pauli iY gate)
3. Alice sends her qubit to Bob
4. Bob measures both qubits using Bell state analyzer
superdense coding - a detailed analysis
|0iA|0iB
1. Alice takes qubit A and Bob takes qubit B from an EPR pair:
|Φ(+)i =
2. Alice encodes her message by applying proper gate to her qubit A:
|00i → IÂ|Φ(+)i = |Φ(+)i
|10i + |01i
√
= |Ψ(+)i
2
|10i + |01i −|10i + |01i
√
√
=
= |Ψ(−)i
2
2
|00i − |11i
√
=
= |Φ(−)i
2
|01i → X̂ A|Φ(+)i =
|10i →
Ẑ A|Φ(+)i
|11i → ẐA X̂A|Φ(+)i = ẐA
2. Alice sends qubit A to Bob. So, he has now two entangled qubits A and B.
•
H
•
89:;
?>=<
fe
fe
|βmni
|mi
|ni
119
120
|00i + |11i |0, 0 ⊕ 0i + |1, 1 ⊕ 1i |00i + |10i |0i + |1i
√
√
√
=
=
= √
|0i = |+, 0i
2
2
2
2
3. Bob applies CNOT to remove entanglement between qubits A and B:
ÛCNOT
|10i + |01i |11i + |01i |1i + |0i
√
√
=
= √
|1i = |+, 1i
ÛCNOT
2
2
2
|00i − |11i |00i − |10i |0i − |1i
√
√
=
= √
|0i = |−, 0i
ÛCNOT
2
2
2
|01i − |10i |01i − |11i |0i − |1i
√
√
=
= √
|1i = |−, 1i
2
2
2
ÛCNOT
4. Bob applies Hadamard gate to qubit A:
ĤA|+, 0i = |00i
ĤA|+, 1i = |01i
ĤA|−, 0i = |10i
ĤA|−, 1i = |11i
4. Bob measures qubits A and B in the standard basis.
H
q-circuit for Bell-state generation
|ni
|mi
|βmni
89:;
?>=<
for m, n = 0, 1
|β10i = |Φ(−)i,
where
|β01i = |Ψ(+)i,
or compactly
|0, ni + (−1)m|1, 1 − ni
√
2
|β11i = |Ψ(−)i
q-circuit for Bell-state analysis
|β00i = |Φ(+)i,
|βmn i =
121
in
1
BSM
ABCD
Alice
2
EPR source
3
ABCD
Bob
U
out
Bennett, Brassard, Crepeau, Jozsa, Peres, Wootters (1993)
principles of teleportation
122
any quantum circuit can be realized using only teleportation and singlequbit operations.
• Universal quantum computation via teleportation:
• Teleportation can provide the quantum channel for communication between quantum computers.
• Super-luminal communication via teleportation is impossible. (why?)
• Teleportation does not violate the no-cloning theorem. (why?)
• Teleportation is not a transfer of an object itself, neither its energy etc.
• Teleportation is a transfer of information about a quantum system without
measuring it!
Remarks:
and transmission of some classical information.
over large distances via entangled particles
a method to transfer (information about) unknown quantum states
What is quantum teleportation?
= (a 0 1 + b 1 1)⊗
2
0 21 3− 1 2 0
3
12
12
ΦC
ΦD
where
12
12
3
x → (−1) x +1 x ⊕ 1
phase flip + bit flip
bit flip x → − x ⊕ 1
phase flip x → (−1) x x
OK
124
Φ A = Ψ ( − ) , ΦB = Ψ ( + ) , ΦC = Φ ( − ) , Φ D = Φ ( + )
⇒ − iσy (a 1 3 − b 0
)
)
3
σz (a 0 3 − b 1 3 )
a 0 3+b1 3
⇒ − σx (− a 1 3 − b 0
⇒
⇒
measurement in Bell basis
ΦB
ΦA
23
1
1
ΦA 12 ⊗ (a 0 3 + b 1 3 ) − ΦB 12 ⊗ (a 0 3 − b 1 3 )
2
2
1
1
+ ΦC 12 ⊗ (a 1 3 + b 0 3) + ΦD 12 ⊗ (a 1 3 − b 0 3 )
2
2
=−
in 1 ⊗ ΦA
Problem: How to teleport state of qubit 1 to qubit 3 ?
Assumption: qubits 2 & 3 are in the singlet state
explanation of teleportation
123
12hΨ
Quantum teleportation - a detailed analysis
Let us analyze teleportation of a polarization state
as in Zeilinger’s experiment
|φi1 = a|0i1 + b|1i1 = a|Hi1 + b|V i1
via the singlet state
1
|Ψ(−)i23 = √ (|HV i23 − |V Hi23)
2
main idea
Let’s expand the total initial state |φi1|Ψ(−)i23 in the Bell basis
1
|Ψ(±)i = √ (|HV i ± |V Hi)
2
1
|Φ(±)i = √ (|HHi ± |V V i)
2
note that
125
126
|Ψ(+)ihΨ(+)| + |Ψ(−)ihΨ(−)| + |Φ(+)ihΦ(+)| + |Φ(−)ihΦ(−)| = I4 ≡ id(4)
which leads to
|φi1|Ψ(−)i23
= (|Ψ(−)ihΨ(−)| + |Ψ(+)ihΨ(+)| + |Φ(−)ihΦ(−)| + |Φ(+)ihΦ(+)|)12|φi1|Ψ(−)i23
= |Ψ(−)i12(12hΨ(−)|φi1|Ψ(−)i23) + |Ψ(+)i12(12hΨ(+)|φi1|Ψ(−)i23)
+ |Φ(−)i12(12hΦ(−)|φi1|Ψ(−)i23) + |Φ(+)i12(12hΦ(+)|φi1|Ψ(−)i23)
|φi1|Ψ(−)i23
Term 1:
(−)
1
1
= √ ( hHV | − hV H|)(a|Hi + b|V i ) √ (|HV i − |V Hi )
12
12
1
1
23
23
2
2
1
=
a|HHV i − a|HV Hi + b|V HV i − b|V V Hi
12hHV | − 12hV H|
2
1
− a 12hHV |HV Hi − b 12hV H|V HV i
=
2
1
= − a|Hi3 + b|V i3
2
1
= − |φi3
2
12hΨ
|φi1|Ψ(−)i23
Term 2:
(+)
since
Term 3:
(−)
|φi1|Ψ(−)i23
12hΦ
since
Term 4:
(+)
|φi1|Ψ(−)i23
12hΦ
1 0
0 −1
a
b
=
a
−b
= a|Hi − b|V i
127
1
=
12hHV | +12 hV H|
2
1
=
− a 12hHV |HV Hi + b 12hV H|V HV i
2
1
1
= − (a|Hi3 − b|V i3) = − Ẑ|φi3
2
2
Ẑ|φi3 =
0 1
1 0
a
b
=
b
a
= b|Hi + a|V i
128
1
=
12hHH| −12 hV V |
2
1
a hHH|HHV i + b 12hV V |V V Hi
=
12
2
1
1
(a|V i3 + b|Hi3) = X̂|φi3
2
2
=
X̂|φi3 =
a
b
=
0 −1
1 0
a
b
=
−b
a
= −b|Hi + a|V i
1
=
12hHH| +12 hV V |
2
1
a hHH|HHV i − b 12hV V |V V Hi
=
12
2
1
1
(a|V i3 − b|Hi3) = (−iŶ )|φi3
2
2
=
0 −i
i 0
since (−iŶ = X̂ Ẑ)
−iŶ |φi3 = −i
1 (−)
|Ψ i12|φi3 + |Ψ(+)i12Ẑ|φi3 − X̂|Φ(−)i12|φi3 + |Φ(+)i12(iŶ )|φi3
2
1
−|Ψ(−)i12|φi3 − |Ψ(+)i12Ẑ|φi3 + X̂|Φ(−)i12|φi3 + |Φ(+)i12(−iŶ )|φi3
2
thus we have
|φi1|Ψ(−)i23 =
= −
Now, Alice performs a Bell-state measurement (BSM)
- a projective measurement in the Bell-state basis (on particles 1 and 2).
n
2
3
|Ψ(−)i23
Xn
•
Zm
•
|φi3
PL
P
D1
H
fe
fe
n
m
4
BBO
1
BS
3
2
Alice
D2
D4
130
D3
EOM
PBS
Bob
D
2
in
(no photon)
Alice
=
a 0
+
b 1
BS1
0
out
D1 (1 photon)
=
BS2
a 0
Bob
teleportation of optical qubits
using quantum scissors
PBS – polarizing beam splitter
b 1
1
+
132
PC – polarization controller to correct extra rotation of polarization in fibres
Xn
BS – beam splitter
EOM – electro-optic modulator to perform Bob’s unitary operation
|φi3
131
BBO – β-barium borate used to generate entangled photon pair
(wavelength 788 nm) by spontaneous parametric down-conversion (PDC)
PL – pulsed laser (wavelength 394 nm, rate 76 MHz)
RF - unit – classical channel
F – optical fibre (length 800m) – quantum channel
qubit – polarization states |Hi and |V i of photon
Key:
experiment of Zeilinger at al. (2004)
teleportation across the River Danube
•
Zm
•
Zeilinger's experiment
i23
89:;
?>=<
•
D0 trigger
|Ψ
(−)
|φi1
or more explicitly
where m, n ∈ {0, 1} and BSM stands for the Bell state measurement
BSM
m
1
|φi1
q-circuit for quantum teleportation
129
FD
VA
0
PDC
D0
out
BS1
in
realization of quantum scissors
BS
delay - ‘trombone’ system
D0 - acts as a trigger
D0, D1, D2 – photon-counting detectors
BS, BS1, BS2 – beam splitters
CCL – coincidence counter and logic
VA – variable attenuator with narrow-band filter
PDC – parametric down conversion crystal
FD – frequency doubler
PL – pulsed laser
experimental quantum scissors
for teleportation and generation of qubit states
Key:
PL
ay
D
el
BS2
D1
133
D2
134
CCL
1
1
1
Bell state generator
Z
π
2
hide
detector
hide
detector
-1
hide
-1
2
π
experiment of Blatt et al. (2004)
UX
π
2
,
i0 + 1
135
detector
2
= 100 ms , T delay = 10 ms
136
Z X UX-1
deterministic teleportation of
atomic states
hide
Blatt et al. experiment (2004)
1 ≡ S1/ 2 ( m J = − 1 / 2 )
3 ions of 40Ca+
states
0 ≡ D5 / 2 (m J = −1 / 2)
2
0 + 1
H ≡ D5 / 2 (m J = −5 / 2)
Ux 1 → 1 , 0 ,
.lifetime
(66.7%<) 75% (<87%)
T teleport = 2 ms , T Bell.state
fidelity:
|e>n
laser mode
La
D2
(L)
wn
Ca
BS
Alicja
Dn
wn
(C)
Cb
138
|g>n
cavity mode
Lb
Ab
|r>n
3-level atom
Aa
D1
Bob
teleportation via cavity decay
137
4
139
4
1
ABCD
2
EPR sources
BSM
3
ABCD
entanglement swapping
U
140
5. Thus, although Alice and Claire never interacted with each other, their
particles A and C are now entangled.
4. Claire, as in the standard teleportation protocol, applies proper unitary
operation to C. So, the state of B1 is teleported to C.
3. Bob performs BSM on his particles B1 and B2 and informs Claire about
his measurement results.
2. Initially particles A and B1 are entangled, and so B2 and C.
1. Alice has particle A, Bob two particles B1, B2, and Claire particle C.
Simple description
• quantum repeaters can be based on entanglement swapping
• it is quantum teleportation of unknown entangled quantum state
Remarks:
a method to establish perfect entanglement between remote parties
What is the entanglement swapping?
3
141
What can be interesting about linear optics?
Can we construct a quantum computer composed only of:
teleportation - theory and experiments
T 1935 quantum entanglement - Schrödinger, and Einstein, Podolsky, Rosen
• beam splitters
• phase shifters
• single-photon sources
• photodetectors
[3] Briegel and Raussendorf [PRL 2001]:
“A one-way quantum computer”
[4] Zeilinger et al. [Nature 2005]:
“Experimental one-way quantum computing”
How to encode optical qubits
• single-rail qubits
logical values of qubits are encoded by number of photons
|0iL = |0i – vacuum
|1iL = |1i – single-photon state
• two-rail qubits
encoding in spatial modes
|0iL = |1i ⊗ |0i = |10i – photon in the first mode
|1iL = |0i ⊗ |1i = |01i – photon in the second mode
or vice versa
• polarization qubits
encoding in polarization modes of qubits
|0iL = |Hi = |1h, 0v i – photon with horizontal polarization
143
144
“[Experimental] demonstration of an all-optical quantum controlled-NOT gate”
[2] O’Brien et al. [Nature 2003]:
“A scheme for efficient quantum computation with linear optics”
[1] Knill, Laflamme and Milburn [Nature 2001]:
Yes!
T 1982 quantum no-cloning - Wootters, Żurek and Dieks
T 1993 quantum teleportation - Bennett,Brassard,Crepeau,Jozsa,Peres,Wootters
T 1993 entanglement swapping - Żukowski, Zeilinger, Horne, Ekert
E 1997 limited (conditional) optical teleportation - Zeilinger et al.
E 1998 unconditional optical teleportation - Furusawa, Kimble, Polzik et al.
E 1998 optical entanglement swapping - Zeilinger et al.
T 1999 universal quantum computation via teleportation - Gottesman and Chuang
E 2004 unconditional teleportation of atomic states
- (i) Barrett, Wineland at al. and (ii) Riebe, Blatt at al.
E 2006 unconditional teleportation between light and matter - Polzik et al.
E 2006 two-qubit teleportation - Zhang, Pan et al.
Key: T - theory, E - experiment
142
Introduction to linear-optical quantum computing
encoding optical qubits
linear-optical elements
scattering matrices
conditional measurements
optical single-qubit gates
optical two-qubit gates
a motto
efficient quantum computation with linear optics is possible!
|1iL = |V i = |0h, 1v i – photon with vertical polarization
or vice versa
B′′ =
B = eiθ0
t ir
ir t
t exp(iθt)
r exp(iθr )
−r exp(−iθr ) t exp(−iθt)
b̂1
â2
1
2
n1
n2
ψ out
light transmitted from any side of BS does not change its phase
light reflected from the ‘black’ surface does not change its phase
the phase of light reflected from the ‘white’ surface is π-shifted
• convention
|ψini - input state, e.g. |ψini = |n1i1|n2i2 ≡ |n1, n2i
|ψouti - output state,
b̂1,2 – annihilation operators for output modes
â1,2 – annihilation operators for input modes
b̂2
â1
scheme of (asymmetrical) BS
θt,r,0 – phase shifts
1 = T + R = t2 + r 2
R ≡ ρ = r – reflectance = reflectivity = reflection coefficient
2
T ≡ τ = t2 – transmittance = transmittivity= transmission coefficient
r – [probability] amplitude of reflection
t – [probability] amplitude of transmission
where
t r
,
−r t
the most general form of BS matrix
B′ =
typical BS matrix
Beam splitter (BS)
= partially-transparent mirror
146
145
b̂1
b̂2
ψ in
S
b̂N
b̂2
b̂1
â1
â2
BS
j=1
N
X
Sij âj
|ψini = |n1, . . . , nN i ≡ |ni
for N -input Fock states
b̂ ≡ [b̂1, b̂2, . . . , b̂N ]T
â ≡ [â1, â2, . . . , âN ]T
• in vector notation
b̂j – annihilation operator for the jth output mode
âj – annihilation operator for the jth input mode
Û – unitary operator describing evolution of N input modes
Sij – elements of unitary scattering matrix S
where
âi → b̂i = Û †âiÛ =
|ψouti = Û |ψini
transformations of states and operators in MP
S – scattering matrix
â N
â1
â2
linear optical multiport (MP)
â2
â1
equivalent schemes of BS
148
b̂1
ψ
b̂2 out
147
the MP
so
âi† =
j
X
j
Sjib̂j†  Û †
Sjib̂j†
â† = Û b̂†Û † = ST b̂†
b̂ = Û †âÛ = S â
transformations can be written compactly:
or explicitly
N
i=1
N
X
SjiÛ b̂j† Û †
Sjiâj†
– is the total number of photons in system.
n2
nN
– multiple sum j1, j2, ..., jM
n1
149
150
• How to express output operators in terms of input operators without knowing
explicitly Û ?


j
X
j
X
Û âi†Û † = Û 
=
=
general transformations in MP
i ni
Y 1
√ (Û âi†Û †)ni |0i
=
ni !
i=1
 ni

N
N
N M
Y
X
X
Y
1
1
√ 
Sjiâj†  |0i = √
Sj x â† |0i
ni! j=1
n1! · · · nN ! j=1 l=1 l l jl
=
(â† )n2
(â† )nN
(â† )n1
= Û √1 Û †Û √2 Û †Û · · · Û †Û √N Û †Û |0i
n1 !
n2 !
nN !
i=1
Y (â†)ni
(â† )n1 (â† )n2
(â† )nN
√i |0i = Û √1 √2 · · · √N |0i
ni !
n1 ! n2 !
nN !
input state |ψini = |n1, . . . , nN i ≡ |ni
output state
P
|ψouti = Û |ψini = Û
where M =
j
{xl } ≡ (1,
..., 1}, 2,
..., 2}, ..., |N, {z
..., N}) for l = 1, ..., M
| {z
| {z
P
→
t r
−r t
=|00i
Sj1âj† |00i
|ψouti = t|10i − r|01i
S = B′ =
|ψini = |10i
j
√
2rt(|20i − |02i) + (t2 − r2)|11i
151
152
Exemplary transformations of states by beam-splitter
assuming that
• example 1:
proof:
|ψouti = Û |ψini = Û |10i = Û â1† |00i
=1
= Û â1† |{z}
Û †Û |00i
P
= Û â1† Û † Û |00i
| {z } | {z }
=
= S11â1† |00i + S21â2† |00i = S11|10i + S21|01i
= t|10i − r|01i
so
|10i → t|10i − r|01i
√
in special case for 50:50 BS (i.e., t = r = 1/ 2)
|01i → r|10i + t|01i
1
|10i → √ (|10i − |01i) ≡ |Ψ(−)i (the singlet state)
2
• example 2:
|11i →
for 50:50 BS
1
|01i → √ (|10i + |01i) ≡ |Ψ(+)i (the ′triplet′ state)
2
• example 3:
for 50:50 BS
1
|11i → √ (|20i − |02i)
2
note: state |11i is not generated – it is so-called
photon coalescence or photon ‘bunching’ (but not in Mandel’s sense)
√
2rt|11i + r2|02i
√
1
|20i → |20i − 2|11i + |02i
2
|20i → t2|20i −
√
1 √
3|30i − |21i − |12i + 3|03i
|21i → √
2 2
153
â2
â1
â2
â1
for 50:50 BS
r
3
1
1
|0, 4i − |2, 2i +
2
2
2
r
3
|4, 0i
2
t − r 
r t 


 t r
S=

− r t 
convention 1.
â2
â1
b̂2
b̂1
â2
â1
b̂1
b̂2
r t 
t − r 


− r t 
 t r


convention 2.
scattering matrices for BS
1
|22i →
2
b̂2
b̂1
b̂2
b̂1
154
i
√ h
|22i → (r4 −4r2t2 +t4)|2, 2i+ 6rt rt|0, 4i+(r2 −t2)(|1, 3i−|3, 1i)+rt|4, 0i
• example 6:
for 50:50 BS
• example 5:
√
√
|21i → 3rt2|30i + t(t2 − 2r2)|21i + (r3 − 2rt2)|12i + 3r2t|03i
for 50:50 BS
• example 4:
e iθ

Sθ =  0
0

r
0
S=
t

0
â4
â3
0 t
1 0
0 −r
0 0
BS
θ
BS1
BS2
S = S 2 Sθ S1 = S 2 S1Sθ
t2
r2
0
b̂2
b̂1
b̂3
0 0
− r2
1 0 0 



1 0, S1 = 0 r1 t1 , S 2 =  t 2
 0
0 t1 − r1 
0 1
â3
â2
â1
b̂4
b̂3
0
0
0

1
exemplary scattering matrix (II)
1 0 0 
S = 0 r t 
0 t − r 
b̂3
b̂2
â2
b̂2
â2
â3
b̂1
â1
b̂1
â1
BS
exemplary scattering matrices (I)
0
0
1
156
155
159
Example of von Neumann projective measurement
157
von Neumann projective measurement
a projection synthesis via conditional measurements
general two-qubit pure state
prob1(0) ≡ prob(|0i1) = |c0|2 + |c1|2
c |0i + c1|1i
p0
≡ |φ0i
|c0|2 + |c1|2
√
... is the renormalization.
and post measurement state is
where
2) probability of measuring 1st qubit in |1i:
prob(|1i1) = |c2|2 + |c3|2
if |ψi = c0|00i + c1|01i + c2|10i + c3|11i
the post measurement state is
c |0i + c3|1i
p2
≡ |φ1i
|c2|2 + |c3|2
160
let us assume that the kth subsystem (e.g. mode, qubit) of a bipartite quantum
system is represented by complete orthonormal set of states |mi then:
|ψi = c0|00i + c1|01i + c2|10i + c3|11i
normalization
projector
= projection operator = projection valued (PV) measure
= orthogonal measurement operator is
(k ′ )
[P̂m(k), P̂m′ ] = 0
1) probability of measuring 1st qubit in |0i:
1 = |c0|2 + |c1|2 + |c2|2 + |c3|2
158
P̂m(k) = |mikk hm|
for the kth subsystem corresponding to the measurement outcome m
probability of the measurement outcome m
ˆ
prob1(m) = hψ|P̂m(1) ⊗ I|ψi
prob2(m) = hψ|Iˆ ⊗ P̂m(2)|ψi
(2)
state after projection/measurement
(1)
ˆ
P̂m ⊗ I|ψi
Iˆ ⊗ P̂m |ψi
|φ(1)i = q
, |φ(2)i = q
(1)
(2)
ˆ
hψ|P̂m ⊗ I)|ψi
hψ|Iˆ ⊗ P̂m |ψi
√
... is the renormalization.
where
requirements for the projectors
then
= δmm′ ,
non-negative, Hermitian, orthogonal and summing up to identity:
X
ˆ P̂m(k) = (P̂m(k))†,
= I,
P̂m(k)
m
′
k hm|m ik
X
3) probability of measuring 2nd qubit in |0i:
prob(|0i2) = |c0|2 + |c2|2
prob(m) = 1
if |ψi = c0|00i + c1|01i + c2|10i + c3|11i
m
measurement outcomes corresponding to non-orthogonal states
do not commute and thus are not simultaneously observable
- analogously
4) probability of measuring 2nd qubit in |1i:
c |0i + c3|1i
p1
|c1|2 + |c3|2
the post measurement state is
Note: number of projectors = dimension of Hilbert space
• Can we apply a more general type of measurement?
Yes!
positive operator valued measures (POVM, POM)
describe measurement outcomes associated with non-orthogonal states
... and we will discuss it later!
161
#
⇒
b̂†1
b̂†2
"
ST =
=S
T
#
S11 S21
S12 S22
â†1
â†2
ψ
φa
no photons
no photons
ψ
Answer: Proper conditional measurements should be performed
|φai ∼ c1|0i + c2|1i and |φbi ∼ c5|0i + c7|1i?
Problem: How to get single-mode states
φb
no photons
1 photon
|ψi = c1|000i+c2|001i+c3|010i+c4|011i+c5|100i+c6|101i+c7|110i+c8|111i
0
1
ψ
S = S3 S 2 S1
0
1 0
0, S 3 = 0 r3
0 t3
1
0 
t3 
− r3 
t2
r2
0
− r2
1 0 0 
S1 = 0 r1 t1 , S 2 =  t 2
 0
0 t1 − r1 
BS3
b̂2
„0”
„1”
ψ'
b̂3
BS1
BS2
b̂1
â3
â2
â1
|ψi = c0|0i + c1|1i + c2|2i −→ |ψ ′i = c0|0i + c1|1i − c2|2i
example: Let’s analyze a three-mode state
163
164
|100i → S11â†1|000i + S21â†2|000i + S31â†3|000i = S11|100i + S21|010i + S31|001i
e.g.

 
  
 † 
â†1
b̂†1
â†1
â
S11 S21 S31
 †
 †
 †
 †1 
 â2  →  b̂2  = ST  â2  =  S12 S22 S32  â2 
S13 S23 S33
â†3
b̂†3
â†3
â†3

• Analogously for a three-mode system (six-port system) holds:
probabilistic quantum state engineering via conditional measurements
162
S11 S12
S21 S22
→
"
|01i → S12â†1|00i + S22â†2|00i = S12|10i + S22|01i
#
|10i → S11â†1|00i + S21â†2|00i = S11|10i + S21|01i
S=
â†1
â†2
nonlinear sign gate (NS)
p
p
prob(|0i1)|0i ⊗ |φ0i + prob(|1i1)|1i ⊗ |φ1i
e.g.
where
"
• As we know, for a two-mode system (four-port system) holds:
projection synthesis
=
p
c0|0i + c1|1i p 2
c2|0i + c3|1i
= |c0|2 + |c1|2 |0i ⊗ p
+ |c2| + |c3|2 |1i ⊗ p
|
{z
}
{z
}
|c0|2 + |c1|2 |
|c |2 + |c |2
|
| 2 {z 3 }
{z
}
= |0i ⊗ (c0|0i + c1|1i) + |1i ⊗ (c2|0i + c3|1i)
|ψi = c0|00i + c1|01i + c2|10i + c3|11i
another description of projection synthesis
• scattering matrix
Mathematica:
S1 := {{1, 0, 0}, {0, r1, t1}, {0, t1, -r1}}
S2 := {{-r2, t2, 0}, {t2, r2, 0}, {0, 0, 1}}
S3 := {{1, 0, 0}, {0, r3, t3}, {0, t3, -r3}}
S = S3.S2.S1
thus, after multiplication, one gets

 

S11 S12 S13
−r2,
t2 r 1 ,
t1 t2
S =  S21 S22 S23  =  t2r3, t1t3 + r1r2r3, −t3r1 + t1r2r3  ↔ Û
S31 S32 S33
t2 t3 , t3 r 1 r 2 − t1 r 3 , t1 t3 r 2 + r 1 r 3
• probability amplitudes hn10|Û |n10i
Problem: find such reflection amplitudes r1, r2 and r3 for BS that
p
1 − ri2
h010|Û |010i = h110|Û |110i = −h210|Û |210i ≡ c
Note: transmission amplitudes for BS are calculated from ti =
165
166
Û |010i = (S12â1† + S22â2† + S32â3† )|000i = S12|100i + S22|010i + S32|001i
• |n10i → c|n10i for n = 0
so
h010|Û |010i = S12h010|100i+S22h010|010i+S32h010|001i = S22 = t1t3+r1r2r3 = c
k,l=1
= (S11S22 + S21S12)|110i + · · ·
= (S11S22â1† â2† + S21S12â2† â1† + · · · )|000i
3
X
S S
√k1 l2 âk† âl†|000i
Û |110i =
1!1!
• |n10i → c|n10i for n = 1
so
h110|Û |110i = (S11S22 + S21S12)h110|110i + · · ·
= S11S22 + S21S12
= −r2c + (t2r3)(t2r1)
= −r2c + r1r3t22
3
X
S
k1Sl1Sm2 † † †
√
âk âl âm|000i
2!1!
k,l,m=1
2|210i
+ 2S11S21S12) â12†â2† |000i + · · ·
| √ {z }
167
2† †
† 2†
† † †
√1 (S 2 S22â â +S21S11S12â â +S11S21S12â â â +· · · )|000i
1 2
2 1
1 2 1
2 11
√1 (S 2 S22
2 11
2
= (S11
S22 + 2S11S21S12)|210i + · · ·
=
=
Û |210i =
• |n10i → −c|n10i for n = 2
so
2
h210|Û |210i = S11
S22 + 2S11S21S12
= r22c − 2r2(t2r3)(t2r1)
= r22c − 2r1r2r3t22
finally, we get set of equations
168
c = t1t3 + r1r2r3 = −r2c + (t2r3)(t2r1) = −(r22c − 2r1r2r3t22)
√
for which the solutions are r1 = r3 = √ 1 √ , r2 = 2 − 1
4−2 2
simplified implementation of NS gate
1
ψ
BS2
ψ'
„1”
„0”
BS2
ψ'
0
BS1
„0”
ψ
1
„1”
0
BS1
(less number of optical elements but also lower probability of success!)
(a)
(b)
NS
BS 50:50
π
π
BS2
BS1
BS3
„1”
„1”
t1 = t2 = t3 = cos θ where θ = 54.74 and t4 = cos φ where φ = 17.63
transmission amplitudes for BSs:
target
qubit
1
1
control
qubit
another implementation of CZ gate
BS 50:50
NS
implementation of CZ gate
BS4
170
169

n2
a^ 2
a^ 4
B2
P2
P6
B3
P3
P4
^
b1
B4
P5
φ
b2
^

1 0
 0 t3
B3 = 
 0 −r3
0 0

0 0 0
1 0 0 

0 t5 r 5 
0 −r5 t5

Mk = diag[exp(iζδ1k ), exp(iζδ2k ), exp(iζδ3k ), exp(iζδ4k )]
mirror
Pk = diag[exp(iξk ), 1, 1, 1] dla k = 1, 2, 6,
Pk = diag[1, exp(iξk ), 1, 1] dla k = 3, 4,
P5 = diag[1, 1, exp(iξ5), 1].
phase shifters
r2
0
t2
0

0
0
,
0
1

1
0
B5 = 
0
0
0
1
0
0
beam splitters
S = P6B5P5B4P4B3P3B2P2B1P1
0
r3
t3
0
D3
b^ 4
D2
b3
B5
^
unitary transformations
ψ
a^3
B1
P1

0
t2
 0
0
 , B2 = 
 −r2
0
1
0


1 0 0 0
 0 t4 0 r 4 

B4 = 
 0 0 1 0 ,
0 −r4 0 t4
0
0
1
0
n3
r1
t1
0
0
1
t1
 −r1
B1 = 
 0
0
n1
a^
multiport Mach-Zehnder interferometer

0
0
,
0
1
172
D4
171
output state
cn(d)(T, ξ)
=
hnN2N3N4|Û |n1n2n3ni
|φouti1 = N 2hN2| 3hN3| 4hN4|Û |n1i1|n2i2|n3i3|ψini4 = N
where amplitudes
depend on:
transmittances T ≡ [t12, t22, t32, t42, t52]
phase shifts ξ ≡ [ξ1, ξ2, ξ3, ξ4, ξ5]
input Fock states |n1i, |n2i and |n3i3
measurements results N2, N3, N4
a^1
B1
a^ 2
B2
a^3
>
a^ 4
B3
P6
^
b1
n=0
d−1
X
B5
b^ 3
>
173
174
cn(d)γn|ni
b^ 4
1 photon
1 photon
1 photon
|φ
^
b
2
B4
for exemplary input states and measurement outcomes
|1>
|1>
|1>
|ψ
a
{
α|10iab + β|01iab
{
175
176
α|Hi + β|V i
interchanging polarization qubits and two-rail qubits
α|Hi + β|V i
b
PBS
– polarizing beam splitter (e.g. calcite crystal)
transmits |Hi and reflects |V i (or vice versa)
HWP(π/4)
– half-wave plate (polarization rotator)
changes photon polarization |V i ↔ |Hi.
How to rotate polarization qubits?
wave plate = waveplate = retarder
a birefringent crystal (with a properly chosen thickness)
changes polarization of a light
by shifting its phase between two perpendicular polarization components.
cos(2β) sin(2β)
sin(2β) − cos(2β)
half-wave plate (HWP)
Ŝλ/2(β) =
exp (i 3 π) cos(2β) − i
sin(2β)
√4
sin(2β) − cos(2β) − i
2
quarter-wave plate (QWP)
Ŝλ/4(β) =
cos(2β) sin(2β)
0 1
1 0
≡ σ̂X
1 0
0 −1
≡ σ̂Z
rotation of polarization in a waveplate
1 1 1
Ŝλ/2(π/8) = √
≡ Ĥ
2 1 −1
Hadamard gate
Ŝλ/2(0) =
Pauli Z gate = phase flip
Ŝλ/2(π/4) =
NOT gate = Pauli X gate = bit flip
Ŝλ/2(β) =
special values β for HWP
source: Wikipedia
178
177
V
HWP 2
45-PBS
c
ψ in
b
d
ψ out
ψ out
PBS
ψ in
45-PBS
HH + VV
control
a
PBS
control
implementation of polarization CNOT gate
β ≈ 61.50 for HWP2
HWP 1
β ≈ 150.50 for HWP1
ψ in
PBS
„0”
implementation of polarization NS gate
target
target
„V”
PBS
180
179
ψ
out
â1
â2
â1
â2
b̂2
â1
â2
â1
â2
b̂1
b̂2
b̂1
ir t 


 t ir 
− r t 


 t r
50%
?
181
182
convention 2.
symmetric vs asymmetric BSs
convention 1.
 t ir 
S=

ir t 
 t r
S=

− r t 
BS
?
beam-splitter and interference
BS
classical coin tossing
50%
0
1
BS
but here classical intuition fails
0
1
(all BSs are 50-50)
b̂1
b̂2
b̂1
b̂2
1
0
1
0
1
100%
0%
100%
0%
100%
0%
183
0
184
symmetric Mach-Zehnder interferometers
so
setup 3:
so
setup 2:
as
this is the
|01i ≡ |1iL
|0iL → |1iL.
= iX̂
1 1 1
1 −1 1
B1 = √
, B2 = √
2 1 −1
2 1 1
0 −1
S = B2B1 =
= −iŶ
1 0
|0iL → |0iL.
i 1
0 1
=i
1 i
1 0
|0iL → i|1iL ∼
= |1iL
1 i 1
2 1 i
1 −1 1
√
B1 = B2 =
2 1 1
1 0
S = B2B1 =
= Iˆ
0 1
S = B2B1 =
1 i 1
B1 = B2 = √
2 1 i
|0iL → S11|0iL + S21|1iL.
√
N OT gate!
setup 1:
one gets
|10i ≡ |0iL,
|10i → S11|10i + S21|01i.
Equivalently, in terms of two-rail qubit notation
transforms
We have already shown that 2x2 scattering matrix
S11 S12
S=
S21 S22
a single photon in Mach-Zehnder interferometer
186
185
0
1
0
1
0
1
θ
θ
θ
θ
θ
θ
θ 
sin 2  
2
θ 
cos 2  
2
θ 
cos 2  
2
θ 
sin 2  
2
 θ 188
sin 2  
2
θ 
cos 2  
2
Mach-Zehnder interferometers with phase shifter
187
Pθ =
1 0
0 eiθ
S = B2Pθ B1
+
S21|1iL
=
f−|0iL
+
if+|1iL
189
1
where f± = (eiθ ± 1)
2
Mach-Zehnder interferometer with phase shifter
total scattering matrix
→ |ψi =
S11|0iL
θ
2
– prob. of click in upper detector
θ prob(0L) = |Lh0|ψi|2 = |f+|2 = cos2
2
θ 2
prob(1L) = |Lh1|ψi|2 = |f−|2 = sin2
1 −1 1
1 0
1 1
f−, −f+
=
0 eiθ
1 −1
f+, −f−
2 1 1
|0iL → f−|0iL + f+|1iL
190
– prob. of click in lower detector
f+
1 −1 1
1 0
−1 1
f+ f−
=
0 eiθ
1 1
2 1 1
f−
θ
2
i|0i + |1i
i|0i + eiθ |1iL
L
L
L
√
√
→
→ Pθ →
→ B2 → f−|0iL+if+|1iL
2
2
|0iL
1 i 1
1 0
i 1
f− if+
S=
=
,
1 i
if+ −f−
0 eiθ
2 1 i
setup 1:
so
→
B1
or explicitly
|0iL
thus
prob(0L) = |Lh0|ψi|2 = |f−|2 = sin2
S=
|0iL → |ψi = S11|0iL + S21|1iL = f+|0iL + f−|1iL
S=
prob(1L) = |Lh1|ψi|2 = |f+|2 = cos2
setup 2:
so
and
setup 3:
so
and
θ
prob(0 ) = sin2
L
2
θ
2
prob(1L) = cos2
1
1
QIP with simple linear-optical systems
1. quantum key distribution (QKD)
based on Mach-Zehnder interferometer
using Bennett’s protocol B92
2. quantum-state engineering and teleportation
of qubit and qutrit states
using quantum scissors device
Bob
θB
θ B = 00 ,900
Dlow
Dup
B92 protocol (Bennett 1992)
Alice
θA
θ A = 00 ,900 ,1800 ,2700
191
192
Dup
Dlow
0o
90o
0o
90o
0o
90o
0o
90o
1
2
1
2
1
−
−
194
(n = 0, 1),
6. In case of full agreement of the tested photons, the secret key is formed
by the remaining bits. Otherwise the protocol is repeated.
Those photons are rejected from the key.
5. Alice and Bob publicly check their results for some photons.
Alice and Bob keep only those deterministic results.
which implies the deterministic detection of photons.
θA − θB = n 180o
4. Alice publicly informs Bob, when their phase shifts fulfill the condition
3. Bob through a public channel informs Alice about his phase shifts
without, of course, telling his measurement outcomes.
2. Bob, just before measuring each Alice’s photon,
randomly applies phase shift 90o or 0o.
1. Alice sends photons each time randomly choosing one of 4 phase shifts
(rotations): 0o, 90o 180o, 270o.
1
1
2
1
2
1
2
1
2
0
0
0
1
1
0
0
1
2
1
2
0
−
−
bit
1 1
θB Pup Plow
QKD using B92 protocol
0o
0o
90o
90o
180o
180o
270o
270o
θA
probabilities of click in upper and lower detectors
θ + θ θ + θ A
B
A
B
Pup = sin2
, Plow = cos2
2
2
quantum key distribution (QKD)
using B92 protocol
193
D
D
2
2
=
a 0
+
b 1
BS1
0
out
D1 (1 photon)
=
BS2
a 0
Bob
in
(1 photon)
Alice
=
a 0
+
b 1
BS1
D1 (0 photon)
0
out
=
BS2
a 0
Bob
phase flip in teleportation
in
(no photon)
Alice
teleportation of optical qubits
using quantum scissors
b 1
1
−
b 1
196
1
+
195
+
out
D1(1 photon)
+
b2 2
BS1 (79:21)
b1 1
≅
b1 1
Bob
+
197
1
b2 2
198
BS2 (79:21)
b0 0
1
+
teleportation of optical qutrits
b0 0
Alice
≅
D (1 photon)
2
in
quantum scissors device
Fundamental Concepts
• Entanglement
• Non-locality
• Conditional measurement
Possible Applications
• Qubit generation
• Teleportation of superposition states
• Quantum state engineering
• Conversion of a classical state into a non-classical one
• Cryptography
quantum scissors
Input: coherent state
2
0 +α1
1+ α
B2
B2
ψ
D2
1
b̂3
2
b̂
b̂
â1
φ
199
D2
D1
D1
200
n
2
∞
α
− α /2
n
α =e
∑
n!
n=0
2
2

α

− α /2 
2 + .... 
=e
 0 +α 1 +
2


φ
Φ =
â
1
â
2
â3
B1
B1 â
2
n2
3
â
Output: qubit state
n3
ψ
n2
n3
B1
â3
â2
â1
n2
B1
â2
θ
θ
B2
B2
r2

S = B2 Pθ B1 = t 2
0

1 0
1 0 0 


B1 = 0 − r1 t1 , Pθ = 0 eiθ
0 0
0 t1 r1 
θ
ψ
b̂3
b̂2
b̂1
â1
0
r2

0, B2 = t 2
 0
1
B2
e r1r2
t1
iθ
− eiθ r1t 2
0
0
1
202
eiθ t1t 2 

− e iθ r2t1 
r1 
B1 :
1, 0, 0 , 0, r1, t1 , 0, t1, r1
B2 :
r2, t2, 0 , t2, r2, 0 , 0, 0, 1
P:
1, 0, 0 , 0, Exp i theta , 0 , 0, 0, 1
S B2.P.B1
MatrixForm
0
− r2
t2
φ
N 2 clicks
N1 clicks
N1 clicks
201
N 2 clicks
scattering matrix for quantum scissors
n3
n2
ψ
B1
Mathematica:
n3
â3
φ
203
h010|Û |010i = S22 = eiθ r1r2
= (S21S32 + S31S22)|011i + · · ·
iθ
|φ10
01i ∼ e t1t2 γ0 |0i + r1 r2 γ1 |1i,
01
|φ01i ∼ −eiθ t1r2γ0|0i + r1t2γ1|1i,
iθ
|φ10
10i ∼ −e r1t2 γ0 |0i + t1 r4 γ1 |1i
Analogously, we can show
Other examples
where x ∼ y means that x and y are equal up a renormalization constant.
iθ
|φ01
10i ∼ e r1 r2 γ0 |0i + t1 t2γ1 |1i
h011|Û |110i = S21S32 + S31S22 = t2t1 + 0 · S22 = t1t2
thus we have
so
204
3
X
Sk1Sl2 † †
√
âk âl |000i = (S21S32â†2â†3 + S31S22â†3â†2 + · · · )|000i
Û |110i =
1!1!
k,l=1
• |n10i → c|01ni for n = 1
so
U |010i = (S12â†1 + S22â†2 + S32â†3)|000i
= S12|100i + S22|010i + S32|001i
after projective measurements of N1 and N2 photons in the respective modes.
Example for |φ01
10i:
• |n10i → c|01ni for n = 0
1 N2
|φi ≡ |φN
n2n3 i =1 hN1 | 2 hN2 |Φi
single-mode output state:
|Φi
three-mode output state (before projection):
|ψ, n2, n3i where |ψi ∼ γ0|0i + γ1|1i
input state:
output states of the quantum scissors
D
2
PPB quantum scissors
a 0
+
b 1
+
c 2 + ...
BS1
D1(1 photon)
0
BS2
205
206
1
out ∝ a 0 + b 1
Bob
- truncation and teleportation of qubit states
Alice
=
(no photon)
in
quantum state truncation to qubit states
Note that the same solutions are for
|ψi ∼ γ0|0i + γ1|1i + γ2|2i + · · ·
Why?
Since, in this case, the single-mode output state |φi is
always a superposition of |0i and |1i for any |ψi.
conditions for perfect qubit-state truncation and teleportation
01
10
t1 = r2 and θ = 0 for |φ10
i and |φ01
i
01
10
t1 = t2 and θ = π for |φ01
i and |φ10
i
optimized solutions
correspond to the highest probability of successful truncation and teleportation.
=
t2
1
=√
2
In the all 4 cases, the optimized solutions are for the 50:50 BSs:
t1
=
+
b 1
+
+
1
!
b 1
Bob
+
d 3 + ...
207
1
c 2
208
BS2 (79:21)
∝ a 0
c 2
BS1 (79:21)
out
D1 (1 photon)
a 0
r1 t1
1 − 3(r1t1)2
+
truncation and teleportation of qutrit states
Alice
D (1 photon)
2
in
|φi ∼ γ0|0i + γ1|1i + γ2|2i
quantum scissors for qutrit states
general output state
assuming n1 = n2 = N1 = N2 = 1 is
11
|φ11
i ∼ 2r1t1r2t2 e2iθ2 γ0|0i + γ2|2i + eiξ2 (r12 − t12)(r22 − t22)γ1|1i
01
as can be shown analogously to |φ10
i.
1±p
truncation and teleportation to qutrit states
1
2
11
corresponds to |φ11
i assuming
t22 =
and θ2 = 0 or = π (show this!).
√
√
3) ≈ 0.21 or t12 = 61 (3 + 3)/6 ≈ 0.79
optimized 4 solutions
for t12 = 16 (3 −
and t22 = t12 if ξ4 = 0 or t22 = 1 − t12 if ξ4 = π.
D2
BS1
0
in
BS2
1
BS1 ’
D1’
0
We cannot measure simultaneously q and p,
thus we cannot measure directly the Wigner function W (q, p).
7. tomography of a nuclear spin I=3/2
212
problem
this is the quantum tomography
based on homodyne detection
for reconstruction of Wigner function of optical fields
optical homodyne tomography
this is the tomography applied to quantum objects
quantum tomography
a method to reconstruct the shape of a hidden object from
shadows (projections) at different angles
tomography
Can we reconstruct a hidden object (“real thing”)
from its shadows?
question
[Πλατ ωνoς Πoλιτ ǫια - Plato’s Republic]
We are like slaves in a dark cave watching only shadows on a
wall. The shadows are projections of the “real things”.
We think the shadows are real because we do not know better.
Plato’s allegory
211
6. tomography of nuclear spins I=1/2
5. tomography of a single qudit
4. maximum-likelihood method
3. tomography of two qubits
2. tomography of a single qubit
1. optical-homodyne tomography of a single-mode field
Outline
How to reconstruct density matrix of a quantum state?
1
210
BS2 ’
Bob
out
Introduction to quantum tomography
α
D1
D2’
Alice
generation and teleportation of qubit states
209
phase-space quasiprobability distributions
in quantum mechanics
uncertainty principle
makes the concept of phase space in quantum mechanics problematic:
213
a particle cannot simultaneously have a well defined position and momentum,
thus one cannot define a probability that
a particle has a position q and a momentum p
quasiprobability distribution functions
= quasiprobabilities = quasidistributions
functions which bear some resemblance to phase-space distribution functions
useful not only as calculational tools
examples:
• Wigner (Wigner-Ville) W function
Z ∞
i
1
W (q, p) =
dxhq + x2 |ρ̂|q − x2 i exp
px
2πh̄ −∞
h̄
– eigenstates of position operator q̂
What is the Wigner function?
214
• Cahill-Glauber s-parametrized W (s) function
• Glauber-Sudarshan P function
• Husimi (Husimi-Kano) Q function
but can also reveal the connections between classical and quantum mechanics.
where |q ±
x
i
2
• Wigner function can be negative
• density matrix ρ can be calculated from Wigner function.
Why is the Wigner function so important?
Z
∞
∞
−∞
Z
−∞
W (q, p)dq
W (q, p)dp
marginal distributions of Wigner function correspond to classical distributions
• position distribution
pr(q) ≡ hq|ρ̂|qi =
• momentum distribution
pr(p) ≡ hp|ρ̂|pi =
10 formulations of quantum mechanics:
1. matrix formalism of Heisenberg (1925)
2. wave-function formalism of Schrödinger (1926)
3. density-matrix formalism of von Neumann (1927)
4. second-quantization formalism of Dirac, Jordan and Klein (1927)
5. variational formalism of Jordan and Klein (1927)
6. phase-space formalism of Wigner (1932)
7. path-integral formalism of Feynman (1948)
8. pilot-wave formalism of de Broglie and Bohm (1952)
9. many-worlds interpretation (MWI) of Everett (1957)
W (α) =
2
exp[−2|α − α0|2]
π
simple examples of Wigner function
coherent state |α0i:
10. action-angle quantum formalism of Hamilton and Jacobi (1983)
♣
Fock state |ni:
where α = q + ip
♣
2
W (α) = (−1)nLn(4|α|2) exp[−2|α|2]
π
W±(α = q + ip) =
215
216
f0 = 2 exp[−2q 2 − 2p2]
f1 ± f0 cos(4pα0)
π[1 + exp(−2α02)]
Schrödinger cat states |ψ±i ∼ |α0i ± | − α0i:
where Ln(x) is Laguerre polynomial
♣
where α0 ∈ R and
f1 = exp[−2(q − α0)2 − 2p2] + exp[−2(q + α)2 − 2p2],
2
0
Imα
1
−1
−2
−3
−3
0
0.2
0.4
0.6
0.8
1
−2
−0.4
3
−0.2
0
0.5
−0.5
3
W
1
2
0
Reα
1
−1
−2
−3
−3
0
0.5
−2
−0.5
3
W
2
−1
0
Imα
1
Reα
0
−1
1
−2
2
−3
3
−3
−2
−0.8
3
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
−1
2
0
Reα
0
Imα
1
1
−1
2
−2
−3
3
−3
−2
−1
Reα
0
1
2
3
218
217
2
−1
0
Reα
1
Imα
0
−1
1
−2
2
−3
3
−3
−2
−1
3
−0.5
0
0.5
−1
2
0
Imα
0
Reα
1
1
−1
2
−2
−3
3
−3
−2
−1
Imα
0
1
2
3
Wigner function for Schrödinger cat states (a) |ψ+i, (b) |ψ−i
and (c) mixture of coherent states ρ̂mix = |α0ihα0|+| − α0ih−α0|
0
3
0.1
0.2
0.3
W
0.4
0.5
0.6
0.7
W
Wigner function for Fock states |0i, |1i, |2i
W
W
3
2
1
0
0.2
0.4
0.6
0.8
1
−1
−0.8
−0.6
−0.4
−0.2
0
3
−1
2
−2
1
−3
0
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
3
−1
2
−2
1
−3
pr(x) ≡ hx|ρ̂|xi =
∞
−∞
Z
W (x, p)dp,
0
−1
pr(p) ≡ hp|ρ̂|pi =
marginal distributions of Wigner function
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.6
0.4
1
0.8
1
0.8
∞
−∞
Z
−2
Wigner function for (a) |ψ+i, (b) |ψ−i and (c) ρ̂mix
220
W (x, p)dx
−3
219
♣
signal
aˆ ' 2
I2
aˆ '1
detector
I1
balanced homodyne detection
detector
aˆ s
BS
aˆ L ≅ α L
laser
(local oscillator)
What is measured in homodyne detection?
• intensity Ik (k = 1, 2) is proportional to the number of photons n̂k :
I1 ∼ n̂1 = â1′†â1′
I2 ∼ n̂2 = â2′†â2′
⇒
âL ≈ αL
• parametric approximation for laser field:
hn̂Li ≫ 1
• beam splitter (BS) is 50:50 (T = R = 1/2),
thus it is called ‘balanced’ detection.
• relations between input, âk , and output, âk′ , annihilation operators:
1
1
â1′ = √ (âS − âL) ≈ √ (âS − αL)
2
2
1
1
â2′ = √ (âS + âL) ≈ √ (âS + αL )
2
2
221
222
• we measure difference of intensities, which is proportional to:
1
1
n̂2 − n̂1 = (âS† + αL∗ )(âS + αL) − (âS† − αL∗ )(âS − αL)
2
2
= αL∗ âS + αLâS†
= |αL|(e−iθ âS + eiθ âS† ) = 2|αL|X̂(θ)
where θ = arg(αL) and
1
X̂(θ) = (âS e−iθ + âS† eiθ )
2
is the generalized quadrature operator
X̂(0) = q̂
– momentum operator
– position operator
• special cases:
X̂( π2 ) = p̂
Z
−∞
∞
W (q cos θ − p sin θ, q sin θ + p cos θ)dp
223
224
marginal distribution of Wigner function at angle θ
pr(X, θ) =
describes probability of measurement result of the quadrature
X̂(θ) = q̂ cos θ − p̂ sin θ
Remarks:
• this is so-called Radon transformation
• pr(X, θ) is directly measured in homodyne detection
1
(2π)2
Z
∞
−∞
Z
0
π
Z
−∞
∞
pr(X, θ) exp[iξ(q cos θ +p sin θ −X)]|ξ|dXdθdξ
Can we reconstruct Wigner function from its
marginal distributions?
W (q, p) =
this is the inverse Radon transformation
[K. Vogel and H. Risken, 1989]
−∞
∞
W (q cos θ − p sin θ, q sin θ + p cos θ)dp
226
225
3. first experimental reconstruction of single-photon state and
its superposition with vacuum (qubit state) – A. Lvovsky and J. Mlynek (2002)
2. first experimental reconstruction of nonclassical state
[i.e., squeezed vacuum] – D. Smithey, M. Raymer et al. (1993)
1. theoretical proposal – K. Vogel and H. Risken (1989)
history
e.g., in the first experiment, Smithey et al. performed 4000 measurements at
27 angles θ, so in total 108 000 measurements; in the second experiment
they performed 160 000 measurements.
number of measurements
one repeats measurements on many copies
i.e. identically prepared quantum objects
Note: x ≡ q, y ≡ p, pr(X, 0) = hq|ρ̂|qi, pr(X, π2 ) = hp|ρ̂|pi
pr(X, θ) =
Z
marginal distribution of Wigner function at angle θ
(|Hi − i|V i) – right-circular polarization
(|Hi + i|V i) – left-circular polarization
√1
2
√1
2
|Ri =
|Di+|Di
√
,
2
√
− i|Di) = eiπ/4 |Di−i|Di
,
2
=
1+i
2 (|Di
|Ri+|Li
√
2
√
,
|Di = eiπ/4 |Ri−i|Li
2
|Ri =
|Hi =
thus we readily have the inverse relations:
228
|Di−|Di
√
2
√
|Di = eiπ/4 |Li−i|Ri
2
√
|Li = eiπ/4 |Di−i|Di
2
√
=
|V i = i |Ri−|Li
2
(|Hi + |V i) – linear-diagonal polarization at 1350
√1
2
|Di =
|Li =
(|Hi − |V i) – linear-diagonal polarization at 450
√1
2
|Di =
|V i – vertical polarization
|Hi – horizontal polarization
single-qubit tomography by measuring Stokes parameters
• notation:
Optical-qubit tomography (part II)
In principle, the same method can be applied,
but there more effective methods, which will be described in the following.
4. How to reconstruct finite-dimensional states?
hidden object ←→ Wigner function
quantum “shadows” ←→ marginal distributions
of Wigner function
3. By applying the inverse Radon transformation, we can reconstruct Wigner
function and thus the density matrix.
2. But we can measure its marginal distributions pr(X, θ) at various angles θ
1. We cannot measure simultaneously q and p, thus we cannot measure directly
Wigner function.
Summary
227
|Li−i|ri hL|+ihR|
√
√
2
2
=
1
2
moreover
hR|+hL|
1
√
√
|HihH| = |Ri+|Li
=
2 |RihR| + |RihL| + |LihR| + |LihL|
2
2
hR|−hL|
1
√
√
=
|V ihV | = |Ri−|Li
2 |RihR| − |RihL| − |LihR| + |LihL|
2
2
hR|+ihL|
1
√
√
|DihD| = |Ri−i|Li
|RihR|
+
i|RihL|
−
i|LihR|
+
|LihL|
=
2
2
2
|LihL| − i|RihL| + i|LihR| + |RihR|
|DihD| =
Pauli operators
σ̂0 = |RihR| + |LihL|
σ̂1 = |RihL| + |LihR|
σ̂2 = i(|LihR| − |RihL|)
σ̂3 = |RihR| − |LihL|
four measurements of intensity, e.g.:
(1) with a filter that transmits 50% of the incident radiation,
regardless of its polarization:
N
N
n0 =
(hH|ρ̂|Hi + hV |ρ̂|V i) =
(hR|ρ̂|Ri + hL|ρ̂|Li)
2
2
(2) with a polarizer that transmits only photons in state |Hi:
n1 = N hH|ρ̂|Hi
(3) with a polarizer that transmits only photons in state |Di:
n2 = N hD|ρ̂|Di
(4) with a polarizer that transmits only photons in state |Ri:
n3 = N hR|ρ̂|Ri
≡
≡
≡
≡
2n0
2 (n1 − n0)
2 (n2 − n0)
2 (n3 − n0)
Stokes parameters via photon counts
S0
S1
S2
S3
229
230
3
1 X Si
σ̂i,
2 i=0 S0
where
1
1 + hσ̂3i hσ̂1i − ihσ̂2i
2 hσ̂1i + ihσ̂2i 1 − hσ̂3i
Si = N hσ̂ii = N Tr{σ̂iρ̂}
ρ̂ =
231
Single-qubit density matrix via Stokes parameters
ρ̂ =
Why do we need S0?
for renormalization of the count statistics
to compensate experimental inefficiencies (e.g. of detectors)
232
single polarization-qubit measurement operators
there are infinitely many sets of such projectors
µ̂0′′ = µ̂0′ ,
µ̂0′ =
=
µ̂1′ =
∼
µ̂2′ =
∼
=
µ̂3′ =
∼
µ̂1′′ = µ̂1′ ,
µ̂2′′ = |DihD|,
µ̂3′′ = µ̂3′
|HihH| + |V ihV |
|LihL| + |RihR|
|HihH|
|RihR| + |RihL| + |LihR| + |LihL|
|DihD|
|LihL| − i|RihL| + i|LihR| + |RihR|
|HihH| − |HihV | − |V ihH| + |V ihV |
|RihR|
|HihH| − i|HihV | + i|V ihH| + |V ihV |
the common ones are e.g.
or
µ̂0
µ̂1
µ̂2
µ̂3
=
=
=
=
N σ̂0
N σ̂1
N σ̂2
N σ̂3
= µ̂′0
= µ̂′1 − µ̂′0
= µ̂′2 − µ̂′0
= µ̂′3 − µ̂′0
(k = 0, ..., 3)
µ̂00 = N σ̂0 ⊗ σ̂0,
µ̂01 = N σ̂0 ⊗ σ̂1,
···
µ̂33 = N σ̂3 ⊗ σ̂3
µ^ 1
µ^ 2
µ^ 3
µ^ 0
µ^ 1
µ^ 2
µ^ 3
µ^ 0
µ^ 1
µ^ 0
µ^ 1
µ^ 2
µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 3
µ^ 3
234
233
................................................................................................................................................................................................................................................
3 qubits – 64 measurement operators: µ̂i ⊗ µ̂j ⊗ µ̂k
2 qubits – 16 measurement operators: µ̂i ⊗ µ̂j
n qubits – 4n measurement operators: µ̂i1 ⊗ µ̂i2 ⊗ · · · ⊗ µ̂in
···
µ^ 2
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 2
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 1
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0 µ^ 1µ^ 2µ^ 3
µ^ 0
µ^ 0
1 qubit – 4 measurement operators: µ̂i
••
•
3 qubits
2 qubits
1 qubit
measurement operators required for tomography
standard 16 projectors:
two-qubit measurement operators
N = 1 ⇒ µ̂k = σ̂k
assuming loss-free detection (perfect detectors and wave plates)
or
Q2
2
H2 P
D2
CCL
H1, H2 – HWPs
output
235
cos(2β) sin(2β)
cos(2β) − i
sin(2β)
sin(2β) − cos(2β) − i
′
Ŝλ/2
(β) = Ŝλ/2(−β),
′
Ŝλ/4
(β) = −Ŝλ/4(−β)
slightly modified defs.
Ŝλ/4(β) ∼
√1
2
quarter-wave plate (QWP)
Ŝλ/2(0) = σ̂Z
Ŝλ/2(22.50) = Ĥ
Ŝλ/2(450) = σ̂X
special cases of HWPs
Ŝλ/2(β) =
half-wave plate (HWP)
wave plates - a reminder
CCL – coincidence counter and logic
236
P1, P2 – polarizers transmits only photons of one polarization (say, vertical)
Q1, Q2 – QWPs,
black box – a source of photon pairs in an unknown state
black box
D1
P
1
Q1 H1
tomography of a polarization state of two photons
|Hi − i|Vi
√
2
237
238
• rotation matrix 4 × 4 of 2 beams by 2 HWPs and 2 QWPs
rotation[45, 0, 45, 0];
rotation[45, 0, 0, 0];
rotation[0, 0, 0, 0];
rotation[0, 0, 45, 0];
rotation[45/2, 0, 45, 0];
rotation[45/2, 0, 0, 0];
rotation[45/2, 45, 0, 0];
rotation[45/2, 45, 45, 0];
rotation[45/2, 45, 45/2, 0];
rotation[45/2, 45, 45/2, 45];
rotation[45/2, 0, 45/2, 45];
rotation[45, 0, 45/2, 45];
rotation[0, 0, 45/2, 45];
rotation[0, 0, 45/2, 90];
rotation[45, 0, 45/2, 90];
rotation[45/2, 0, 45/2, 90]
239
240
rotation[h1_, q1_, h2_, q2_] :=
Kron[HWP[h1 Degree].QWP[q1 Degree], HWP[h2 Degree].QWP[q2 Degree]]
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
• our sequence of rotations
R[1]
R[2]
R[3]
R[4]
R[5]
R[6]
R[7]
R[8]
R[9]
R[10]
R[11]
R[12]
R[13]
R[14]
R[15]
R[16]
• rotated ρ̂(n) = R̂(n)ρ̂(R̂(n))† for the nth measurement
RhoRotated[n_] := R[n].rho .hc[R[n]]
• examples
MF
2,0
3,0
1,0
0,0
2,1
3,1
0,1
1,1
#
%
%
%
%
3,1
2,1
"
|Ri ≡
3,2
2,2
0,1
1,1
MF
2,0
3,0
0,0
1,0
!
!
|Hi + i|Vi
√
,
2
RhoRotated 1
2,3
0,2
1,2
"
"
.
3,3
1,3
0,3
0,3
3,3
2,3
RhoRotated 2
2,2
3,2
0,2
!
!
|Li ≡
1,3
tomography of polarization state of two photons
No. Mode 1 Mode 2 HWP 1 QWP 1 HWP 2 QWP 2
1
|Hi
|Hi
45o
0
45o
0
2
|Hi
|Vi
45o
0
0
0
3
|Vi
|Vi
0
0
0
0
4
|Vi
|Hi
0
0
45o
0
5
|Ri
|Hi
22.5o
0
45o
0
6
|Ri
|Vi
22.5o
0
0
0
7
|Di
|Vi
22.5o
45o
0
0
8
|Di
|Hi
22.5o
45o
45o
0
9
|Di
|Ri
22.5o
45o
22.5o
0
10
|Di
|Di
22.5o
45o
22.5o
45o
11
|Ri
|Di
22.5o
0
22.5o
45o
12
|Hi
|Di
45o
0
22.5o
45o
13
|Vi
|Di
0
0
22.5o
45o
14
|Vi
|Li
0
0
22.5o
90o
15
|Hi
|Li
45o
0
22.5o
90o
16
|Ri
|Li
22.5o
0
22.5o
90o
|Hi − |Vi
√
,
|Di =
2
Sin 2 t
Cos 2 t
"
Mathematica program for the tomography scheme
%
%
%
%
%
%
%
%
%
$
Sin 2 t
Cos 2 t
0,3
1,3
;
1,2
"
"
• def. of Kronecker tensor product (Kron) and Hermitian conjugate (hc)
Cos 2 t
Sin 2 t
1,2
2,3
0,2
2,2
<< LinearAlgebra‘MatrixManipulation‘
f[x_, y_] := x*y;
Kron[matrix1_, matrix2_]:= BlockMatrix[Outer[f, matrix1, matrix2]]
hc[x_] := Transpose[Conjugate[x]]
Cos 2 t
Sin 2 t
1
2
0,1
1,1
2,1
• def. of rotation by HWP and QWP and def. of [ρnm]4×4
!
!
"
"
"
"
3,3
!
!
3,2
3,1
HWP t_ :
0,0
1,0
#
%
%
%
%
%
%
%
%
%
%
%
%
$
2,0
3,0
!
!
!
!
"
"
"
QWP t_ :
rho
"
"
"
9
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P 15
Q
psi[1]
psi[2]
psi[3]
psi[4]
psi[5]
psi[6]
psi[7]
psi[8]
psi[9]
psi[10]
psi[11]
psi[12]
psi[13]
psi[14]
psi[15]
psi[16]
|ψ1i = |HHi,
...,
|ψ16i = |RLi
Kron[ketH, ketH];
Kron[ketH, ketV];
Kron[ketV, ketV];
Kron[ketV, ketH];
Kron[ketR, ketH];
Kron[ketR, ketV];
Kron[ketDbar, ketV];
Kron[ketDbar, ketH];
Kron[ketDbar, ketR];
Kron[ketDbar, ketDbar];
Kron[ketR, ketDbar];
Kron[ketH, ketDbar];
Kron[ketV, ketDbar];
Kron[ketV, ketL];
Kron[ketH, ketL];
Kron[ketR, ketL];
|ψ2i = |HVi,
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
= (projection) measurement states = projectors
our choice of tomographic states
where |Hi=ketH, hH|=braH, etc.
Z
P 14
I
Z
3,3
Z
3,2
Z
c
2,3
c
c
2,2
^
1
2
^
P 13
^
1,1
^
1,0
^
0,1
^
0,0
^
1
2
^
P 12
d
3,3
^
d
3,2
^
d
3,1
^
`
1
2
_`
3,3
^
`
3,2
^
3,0
_`
3,1
^
2,3
`
3,0
^
2,2
^
2,3
^
2,1
^
2,2
^
2,0
^
2,1
^
1,3
^
2,0
^
1,2
]
1,3
`
1,1
^
1,2
^
1,1
`
1,0
`
1,0
^
0,3
^
0,3
A
3,3
^
0,2
3,2
Z
^
0,2
3,1
^
^^
0,1
3,0
[
b
0,1
2,3
^
^
0,0
2,2
^
^
0,0
2,1
^
^
1
4
2,0
^
^
P 11
1,3
^
^^
1,2
^
^
1,1
^
^
1,0
^
^
0,3
^
Z
0,2
^
Z
0,1
^
c
0,0
^
Z
1
4
^
c
P 10
Z
d
P 9
^
c
2,2
^
d
2,0
^
^
0,2
^
^
0,0
^^
^
1
4
^
Z
P 8
^
^
3,3
^
^
3,1
^
^
1,3
^
^
1,1
[
^
1
2
\
^
P 7
Z
^
3,3
^
3,1
^
1,3
^
1,1
^
1
2
^
P 6
^
2,2
^
2,0
`
0,2
^
_`
0,0
^
`
1
2
`
2,2
^
1
2
^
P 5
^
P 4
^
3,3
^
P 3
^
1,1
^
^
P 2
^
^
, n, 1, 16
a
", P n
^
ketV
ketV
^
0,0
^
d
2
2
hc ketH ; braV hc ketV ;
hc ketD ; braDbar hc ketDbar ;
hc ketR ; braL hc ketL ;
^
Do Print "P ", n, "
^
^
P 1
\
_`
all1
E
242
^
ketH
^
b
braH
braD
braR
]
`
; ketR :
`
2
^
ketH
ketH
^
ketV
1
0
; ketV :
;
0
1
ketH ketV
; ketDbar :
2
^
ketL :
ketD :
ketH :
^
2,2
3,2
2,2
^
2,0
C
^
0,2
B
^
0,0
A
^
1
2
C
^
P 5
B
^^
1,1
C
^
P 2
B
^
0,0
3,0
3,2
2,2
a
FS
A
0
1
B
^
ketVV : Kron ketV, ketV ;
braVV : hc ketVV
P n_ : braVV.RhoRotated n .ketVV
P 1
A
2,0
D
`
0
;
1
1,2
A
^
ketV :
A
0,2
3,0
2,0
H
^
so
1,2
0,2
H
(n)
D
3,1
1,0
0,0
E
0,3
1,3
1
2
1
2
1
2
1
2
E
H
0,1
1,1
D
3,2
3,3
2,3
D
H
2,2
E
3,0
3,1
2,1
@
H
2,0
1,3
0,3
A
1,2
D
0,1
1,1
A
H
0,2
1
2
1
2
1
2
1
2
@
1,0
3,2
C
H
0,0
A
2,2
A
H
2,3
3,0
A
H
3,3
A
2,0
E
2,1
1,2
H
3,1
A
0,2
H
0,3
@
H
1,3
1,0
@
0,0
@
H
0,1
1
2
1
2
1
2
1
2
@
2,3
3,3
@
2,1
@
0,3
MF
3,1
H
1,1
FS
1,3
F
0,1
1,1
equivalent approach using tomographic states
Z
measured probabilities Pn = hV V |ρ̂(n)|V V i = ρ33 where |V i =
1
2
1
2
1
2
1
2
;
RhoRotated 5
241
^
^
^
^
^
^
^
^
^
^
^
^
^^
E
8
8
8
7
7
8
8
7
/
6
/
6
7
/
6
/
4
8
/
4
/
4
/
/
6
5
/
6
5
3
/
/
6
/
/
5
4
5
/
/
6
/
/
4
6
/
6
5
/
/
6
5
/
4
/
4
/
6
/
4
5
/
/
5
/
6
5
/
/
4
/
/
4
4
/
6
4
4
/
6
5
/
/
/
6
4
X
6
/
6
4
8
7
8
7
5
/
/
6
/
6
4
7
5
/
4
5
/
/
5
/
4
/
5
/
4
/
6
/
6
0
2
2
2
2
2
2
SS
-
2
244
;
;
243
p1 = hHH|ρ̂|HHi, p2 = hHV |ρ̂|HV i , etc.:
Kron[braH, braH]
.rho.Kron[ketH, ketH];
Kron[braH, braV]
.rho.Kron[ketH, ketV];
Kron[braV, braV]
.rho.Kron[ketV, ketV];
Kron[braV, braH]
.rho.Kron[ketV, ketH];
Kron[braR, braH]
.rho.Kron[ketR, ketH];
Kron[braR, braV]
.rho.Kron[ketR, ketV];
Kron[braDbar, braV]
.rho.Kron[ketDbar, ketV];
Kron[braDbar, braH]
.rho.Kron[ketDbar, ketH];
Kron[braDbar, braR]
.rho.Kron[ketDbar, ketR];
Kron[braDbar, braDbar].rho.Kron[ketDbar, ketDbar];
Kron[braR, braDbar]
.rho.Kron[ketR, ketDbar];
Kron[braH, braDbar]
.rho.Kron[ketH, ketDbar];
Kron[braV, braDbar]
.rho.Kron[ketV, ketDbar];
Kron[braV, braL]
.rho.Kron[ketV, ketL];
Kron[braH, braL]
.rho.Kron[ketH, ketL];
Kron[braR, braL]
.rho.Kron[ketR, ketL];
245
p
p
p
q
q
q
0
0
0
0
0
q
q
p
p
0
0
0
0
0
0
0
0
0
q
q
f
f
g
g
e
h
FS , n, 1, 16
j
p n
g
i
i
P (n) = p(n)
g
g
k
g
possible methods:
• method 1:

ρ1

 ρ5
ρ̂ = 
ρ9
ρ13
ρ3
ρ7
ρ11
ρ15


 

ρ4
ρ1
ρ00


 

ρ8 
 ρ.2  ≡  ρ.01 
7→ 
  . 
ρ12 
.
ρ16
ρ16
ρ33
• another labelling:




x
x + ix x + ix x + ix
x1
1
2
11
3
12
4
13




x
x + ix x + ix 
5
6
14
7
15
 x2 − ix11
 x2 
ρ̂ = 
 7→  .. 
x
x
x
3 − ix12 x6 − ix14
8
9 + ix16
x16
x4 − ix13 x7 − ix15 x9 − ix16
x10
ρ2
ρ6
ρ10
ρ14
247
248
represent operators of Hilbert space as superoperators in Liouville space
convert 4x4 matrix into 16-dim column vector
f
", P n
f
p ", n, "
f
h
both methods give the same probabilities
p
p
Do Print "P ", n, "
p
p
test
p
p
p 1
p 2
p 3
p 4
p 5
p 6
p 7
p 8
p 9
p
p
P 1
P 2
P 3
P 4
P 5
P 6
P 7
P 8
P 9
q
q
0
q
q
0
q
q
p 10
p 11
p 12
p 13
p 14
p 15
q
q
f
p 16
q
q
P 10
P 11
P 12
P 13
P 14
P 15
P 16
q
q
p
p
reconstruction analysis
q
q
e
projection measurements
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
246
q
q
p
p
o
enable determination of the probabilities
p[1]
p[2]
p[3]
p[4]
p[5]
p[6]
p[7]
p[8]
p[9]
p[10]
p[11]
p[12]
p[13]
p[14]
p[15]
p[16]
all2 := Do[Print[n, " -> ", p[n] // FS], {n, 1, 16}]
projection measurements
hc[psi[1]] .rho.psi[1] ;
hc[psi[10]].rho.psi[10];
enable determination of the probabilities
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
:=
p1 = hψ1|ρ̂|ψ1i = hHH|ρ̂|HHi, p2 = hψ2|ρ̂|ψ2i = hHV |ρ̂|HV i, etc.:
p[1]
p[2]
p[3]
p[4]
p[5]
p[6]
p[7]
p[8]
p[9]
p[10]
p[11]
p[12]
p[13]
p[14]
p[15]
p[16]
all2 := Do[Print[n, " -> ", p[n] // FS], {n, 1, 16}]
q
p
o
q
p
q
p
p
q
p
p
q
p
p
q
p
p
q
p
p
q
p
p
l
n
m
m
m
o
o
m
m
o
o
m
m
o
o
m
m
o
o
m
n
l
l
l
l
m
m
n
n
l
l
l
l
m
m
l
n
n
m
l
l
l
l
n
m
m
l
n
n
l
l
m
l
q
p
m
m
o
o
m
m
o
o
m
m
o
o
m
l
l
l
l
l
l
l
n
m
l
l
n
m
m
l
n
n
m
l
l
n
m
m
n
n
m
l
l

¡

¡
¡
¡
¡
¡
¡

0
0
0
0
0
1
¡
0

¡
¡
¡
¡
¡
¡
¡
¡

¡
¡

r
s
r
r
s
r
s
s
w
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
{
z
z
z
z
z
z
z
z
z
z
z
z
z
t
z
z
z
z
z
z
z
z
z
z
z
z
z
z
|
{{
z
{{
|
{
{
{
{{
|
{
{
{
{
{
{
{
{
{{
{
z
z
z
z
z
z
z
z
z
z
z
1
4
{
{
0
{{
{{
z
z
z
z
z
1
4
0
{{
0
v
0
{{
|
{
z
z
z
z
0
0
z

z

{
{{

|
z

z

{
{
{
{{
{
{{
{
{
{
{
z

z

|

z

{
{{
z

z

z
1
2

1
2

1
0.487213
0.00454278
0.502019
0.00622529
0.228877
0.245661
0.188455
0.236968
0.251423
0.449062
0.212165
0.241693
0.18467
0.240739
0.234458
0.470907

0

0
0
0

1
2
0
0
{{
0
{
0
{{
0
{
1
{{
0
0
0

0
{
0
|
0
s
0
0

{
1
2
{
{

1
2
|
{{

0
|
0
{
{
0
{
{

0
{
{

1
2
1
4
{{

{{
{{
{{
{{
{{
{{
{{
{{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
0
{
{

0
{{

0
{
{
{

0
{
{

0
{{
{
1
2
1
2
{
{

1
{

1
2
1
2
|

0
v
0
|
0
{{
{
{
0
{{
{

0
{
{

0
{

0
|

0
{

0
{
{

0
{

0
u
{

0
{{

0
0

1
2
0

1
2

0
{

0
{
0

1
2

0
0
{
1
2

0

0
{{

0

0
{{

0

1
{{

1
2
1
2

0
1
2
0

1
2
0

0
1
2

1
2
1
2

0
{{

1
2
1
4
{
0

0
{

0
{{

1
2
1
2
|

1
2

1
4
1
4
1
4

0
0

1
2
1
2
0

0
0

1
4
1
4
1
4
1
2
0

0
{
{
0

1
2
1
2
{{

0
{
{
0

0

1
2
1
4
1
4
1
4

0

Nexp : Transpose
34749, 324, 35805, 444, 16324, 17521, 13441,
16901, 17932, 32028, 15132, 17238, 13171, 17170, 16722, 33586
Norm Sum Nexp i , i, 1, 4
1
Pexp 1. Nexp Norm;
MF
Pexp
ν=1

0

r
0

1
u

0
0

1
2
1
4
1
4
1
4
1
2
{

0
1

0
0

0

1
2
1
2

0

1

0
0
0
0
0
0

1
2
1
2
0
0
0
0
0

0
0
0
0
0
1
where nν = N hψν |ρ̂|ψν i so
4
X
nν = N hHH|ρ̂|HHi+N hHV |ρ̂|HV i+N hV H|ρ̂|V Hi+N hV V |ρ̂|V V i = N

0
0
0
0
0
0

0
0
0
1
0
0

0
0
0
0
0
0

1
2
0
0
0
1

0
0
0
0
0
0

0
0
1
0
0
0

0
t
0
0
0
0
0
}

1
2
r
0
0
0
0
0
r
0
0
0
0
0
252
251
experimental estimations of probabilities Pexp = {Pexp1, ..., Pexp16} = { nN1 , ..., nN16 }
£
1
0
0
0
0
0
¤
a n_, m_ : Coefficient P n , x m
A : Table a n, m
1, 1 , n, 1, 16 , m, 1, 16
A
MF
0
0
¤
experimental coincidence counts Nexp = {n1, ..., n16}
¤
where P = [P1, P2, · · · , P16]T , x = [x1, x2, · · · , x16]T , A = [am,n]16×16
P
P1 = x1 = 1 · x1 + 0 · x2 + · · · + 0 · x16 = [1, 0, · · · , 0] · x = n a1,nxn, etc.
¤
experimental data [James et al. 2002]
0
¤
0
0
¤
0
0
¤
0
0
¤
0
0
1
2
1
2
0
0
¤
0
¤
0
0
0
0
0
0
1
¤
1
¡
¤
0
1
2
1
2
0
0
0
0
¤
0
¤
1
2
¤
0
¤
0
¡
1
1
2
1
2
0
0
1
2
1
2
1
0
1
0
0
0
1
2
1
2
1
2
1
2
1 0
0
0
0
0
¡
¤
0
1
2
1
2
0
0
0
0
0
0
0
0
0
0
¤
0
1
2
1
2
1
0
0
1
0
¤
1
2
0
0
0
0
0
0
0
0
0
1
¤
1
2
1
2
1
2
0
1
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
2
0
¤
1
2
1
2
0
0
0
0
0
0
1
2
0
0
1
0
0
0
¤
1
2
0
0
1
2
0

¤
set of equations P = Ax
250
1
2
1
2
¤
2 x 16

1
0
0
1
2
0
1
2
0
¡
¤
2 x 15

0
1
2
1
¤
2 x 12

0
¤
2 x 11

x 15
¤
x 10

x 14
¤
x 8

¤
2x 6

¤
x 5

¤
2x 4

¡
¤
x 1

1
2
1
2
1
2
0
0
0
0
¤
2 x 11

0
0
1
2
0
0
0
0
¤
x 5

0
0
¡
0
1
1
2
1
2
0
0
0
0
¤
x 1

x 13

0
¤
2 x 16

0
¤
x 10

0
¤
x 8

0
¤
x 10

¤
2x 9

¤
x 5

¤
2x 2

¤
2 x 12

¤
x 10

¤
2x 9

¤
x 8

¤
x 5

¤
2x 2

x 10

2x 9

x 8

2x 7

2x 6
x 5
2x 4
¡
2x 3

0
¡
0
0
1
2
¡
1
¤
1
4

1
2
0
¤
1
2

1
2
¡
0
0
¤
1
2

¤
x 8

¤
2x 2

0
1
2
0
0
¡
¤
x 1

0
1
0
0
0
1
2
1
2
0
0
¤
x 1

0
1
2
0
0
1
0
0
1
2
0
0
0
0
0
¡
¤
x 1

0
1
2
0
1
0
0
0
0
¡
x 16
0
0
0
0
x 14
1
2
0
0
0
x 13

0
0
0
0
2 x 11

¤
P 16

¡
P 15

¤
1
2

¤
x 10
¤
x 8
¤
2x 7
¤
P 14

¤
P 13

1
2
1
2
1
2
0
0
¡
¤
x 5
¤
1
2

1
0
0
0
0
0
1
2
0
¤
2x 3
1
2
1
2
1
2
1
¤
1
4

¤
x 1

¤
x 8

invA : Inverse A
invA
MF
0
0
x = A−1P where A−1 is equal to
exists if A is non-singular and it is simply given by
solution of the set of equations

¤
P 12

¤
P 11

¤
P 10

¤
1
4

¤
2x 3
, n, 1, 16
¤
x 10
1, 1
¡
¤
1
4

¤
x 1

¤
1
2

¤
2x 7

¤
x 5

x 15

x 10
x 12

x 5
x 8
", P n

¤
1
2

x 1

1
2

¤
P 9

1
2

¤
P 8

x 8

¤
P 7

x 10

¤
P 6

P 5

P 4

x 5

P 3

x 1

P 2

P 1
Do Print "P ", n, "

In[81]:= all1
249
¢

z

x
z

~
{{
{{
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{
{
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5
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§
3
6
9
x
¦
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¹
¹
¹
¦
¦
x
x
§
¦
¦
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¦
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12
14
16
§
§
x
x
x
4
7
9
x
x
x
10
§
©
¦
¦
¦
¦
¦
13
15
16
¸
±
³
³
³
Tr {γ̂ν · γ̂µ} = δν,µ
16
X
¦
16
X
¹
0.519209 0.0380247
0.0648257 0.00762037
0.0694526 0.013383
0.502019
¦
®
§
x
¦
©
γ̂ν rν ,
253
254
1 0
;
0 1
0
0
1
0
1
0
0
0
0
1
0
0
0
0
0
1
0
−i
0
0
0
0
0
−1
1
¾
0
0
0
0
1
2
0
0
0
0
2
¾
0
0
À
4m
0 1
;
1 0
Do
MF
MF
Å
Æ
½
2
n
¾
γ̂6
1
2

0

0
= 
0
i
1
2
−i
0
0
0
0
0
−i
0
0
0
1
0
−i
0
0
0


−i
0


0 
 0
 , γ̂ = 1 
10
2
0 
 0
0
−1

1

0
,
0
0

0

i
γ̂2 = 
0
0
2
0
¿


0
0


0 
i
 , γ̂ = 1 
14
2
−1 
0
0
0
¿
¾
0
0
0
i
0
i
0
0
0
1
0
0
Á
1
0
Â
2
0
À
;
Kron
½
3
¾
m,
½
½
n
¿
Ç
1
0
Á
0
1
À

−i

0 
,
0 
0
γ̂7
1
2
0
0
0
−1
0
−1
0
0
0
−1
0
0
0
0
0
−i

0

0
= 
1
0
1
2


0
1


0
0
 , γ̂ = 1 
15
2
i
0
0
0
0
0
1
0
1
0
0
0
;
Ê

0

0
,
0
1
γ̂8
1
2
Ç
255
1
0
0
0
−i
0
0
0
0
0
1
0
0
0
0
i
0
1
0
0
0
0
−1
0

1

0
= 
0
0
1
2
0
0
0
1
256

0

0
= 
i
0
γ̂16
0
1
0
0

0

0
γ̂4 = 
1
0
1
2
Ë


0
1


i
0
 , γ̂ = 1 
12
2
0
0
0
0

0

0 
,
0 
−1
0
0
−1
0
−i
0
0
0

0

−1 
,
0 
0
, m, 0, 3 , n, 0, 3

1

0
γ̂3 = 
0
0
Å
Â

0

0 
,
−i 
0
È


−1
0


0 
0
 , γ̂ = 1 
11
2
0 
i
0
0
0
0
0
−i
É
ÈÈ
½
exemplary generators of the Lie algebra SU (2) ⊗ SU (2)
1
2
1
2
1
0
0
0
0
0
i
0

0

0
= 
0
1
1
2

0

1
= 
0
0
1
2
Ê
È
0
0
Ç
Simplify
Å

0

0
,
1
0
É
È
½
GenerateGamma
¾
0
1
2
16 :
1
0
0
Ñ
Ä
generators of the Lie algebra SU (2) ⊗ SU (2)
Ð
Ä
Ã
¿
Ç
Ì
Ì
Ä
Simplify
1
2
Ó
Ó
Ó
Ó
Ó
Ó
ÐÐ
0
Õ
0
ÐÐ
Ð
0
ÐÐ
2
Ð
0
Ð
0
Ð
Ó
Ó
Ó
Ó
Ó
Ó
1
2
0
0
2
0
2
0
0
Ð
Å

0

1
γ̂1 = 
0
0
ÐÐ
Ó
Ó
Ó
Ó
Ó
Ò
Ð
Ì
Ì
Ì
Ì
Ð
Ì
Ç
Ì
Ç
Í
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Pexp
8
x
¸
¦
Ä
Ä
x = A−1
11
x
x
x
¹
ν=1
Ô
©
14
15
§
®
§
®
γ̂ν Tr {γ̂ν · ρ̂} ≡
Ñ
x
¦
®
x
x
x
§
ν=1
ÐÐ
2
6
7
¸
À
Ð
§
§
§
¯
¯
¯
§
§
§
¦
¦
¦
¦
¦
¦
§
§
§
§
§
§
§
³
³
³
x
x
x
©
ÐÐ
ÐÐ
§
¦
¦
¦
¦
¦
¦
®
§
®
§
§
Å
Å
reconstructed ρ̂ by linear tomography
§
¾
Ð
Ð
final solution for x assuming the analyzed experimental data:
¦
Ï
Î




x1
x1
x2 + ix11 x3 + ix12 x4 + ix13




x5
x6 + ix14 x7 + ix15 
 x2 
 x2 − ix11
 .  7→ ρ̂ ≡ ρ̂exp = 

.
x
x
x
3 − ix12 x6 − ix14
8
9 + ix16
x16
x4 − ix13 x7 − ix15 x9 − ix16
x10
11
12
13
¸
∀ρ̂ ρ̂ =
Õ
Flatten invA.Pexp
¦
x
1
¦
§
x
x
x
¦
x
¦
0.487213, 0.00418524, 0.00975155, 0.519209, 0.00454278, 0.0271305, 0.0648257,
0.00622529, 0.0694526, 0.502019, 0.01142, 0.0178416, 0.0380247, 0.0145958, 0.00762037, 0.013383
rhoexp :
2
3
4
¦
Ð
Ð
Ó
Ó
Ó
Ó
Ó
Ó
Ó
Ó
¦
°
are generalized Pauli operators or their products chosen as,
Õ
x
x
x
MF
¹
¦
§
§
§
0.00975155 0.0178416
0.0271305 0.0145958
0.00622529
0.0694526 0.013383
¹
¯
¯
¯
³
³
³
³
²
Í

0

0
= 
0
i
γ̂5
Ð
¦
«
¥
ª
e.g., generators of the Lie algebra SU (4) or just SU (2)⊗SU (2).
Ð
¦
¦
¯
¯
¯
(m, n = 0, ..., 3)
Ô
¦
¦
§
°
§
°
§
§
§
§
¦
¦
¦
¦
¦
¦
¯
¯
¯
§
°
§
°
§
§
§
§
§
°
§
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ï
Ó
Ó
Ó
Ó
§
§
§
0.00418524 0.01142
0.00454278
0.0271305 0.0145958
0.0648257 0.00762037
¹
§
1
γ̂4m+n = σ̂m ⊗ σ̂n,
2
γ̂9
Õ
rhoexp
©
¦

¬
where σ̂n are the standard Pauli operators.
ÐÐ
´
0.487213
0.00418524 0.01142
0.00975155 0.0178416
0.519209 0.0380247
©
©
´
• method 2: ρ̂ 7→ [rν ]16×1
©
©
Ï
¦
¸
¥
find a set of 16 linearly independent 4 × 4 matrices {γ̂ν } satisfying:
¸
¨
for convention we denote γ̂16 = γ̂0
γ̂13
Ð
Ó
Ó
Ó
Ó
Ó
Ò
¹
¹
Ï
Ï
Ï
Ï
Ï
º
¼
¼
¼
¼
¼
¼
¼
¼
¼
»
Ï
Î
§
¹
¹
µ
·
·
·
·
·
·
·
·
·
¶

0

1
,
0
0

0

−i 
,
0 
0

0

0 
,
0 
−1

0

0
.
0
1
×
Ö
×
Ö
Ö
×
×
Ö
Ö
×
Ù
Ø
Ö
Û
Ú
×
Ü
ß
ß
ß
ß
ß
ß
ß
ß
ß
ß
ß
á
ß
àà
àà
àà
à
à
à
à
ß
ß
ß
ß
ß
à
ß
à
à
à
à
à
ß
ß
ß
ß
ß
ß
ß
ß
ß
ß
ß
ß
á
ß
àà
á
àà
ß
ß
ß
ä
×
ß
ä
à
ß
ä
á
ß
ä
àà
àà
àà
à
à
à
à
ß
ä
ß
ä
ß
ä
ß
ä
ß
ä
ß
ä
àà
ß
ä
ß
ä
ß
ä
ß
ä
à
Ø
à
ß
ä
ß
ä
àà
ß
ä
ß
ä
ß
ä
ß
ä
à
ß
ä
ß
ä
ß
ä
àà
à
à
à
à
à
à
à
à
á
á
á
á
ß
ä
ß
ä
ß
ä
ß
ä
à
ß
ä
á
ß
ä
àà
á
àà
ß
ä
ß
ä
ß
ä
ß
ä
à
ß
ä
á
ß
ä
àà
ß
ä
ß
ä
ß
ä
ß
ä
ß
ä
ß
ä
á
ß
ä
àà
á
à
à
à
à
à
à
à
àà
à
ß
ä
ß
ä
ß
ä
à
ß
ä
ß
ä
ß
ä
àà
àà
à
à
à
á
à
ß
ß
1
2
0
1
2
1
2
0
0
0
1
ò
2
2
2
0
0
1
0
0
M 16
î
ä
ß
î
ä
ß
î
ä
ß
î
ä
ß
î
ä
ä
à
à
ß
ß
î
ä
ß
0
ä
0
ä
î
0
1
0
0
MF
1
2
1
2
MF
0
0
2
1
1
0
0
0
2
2
2
õ
î
ä
à
ß
0
ä
î
0
ä
î
0
ä
í
õ
à
0
à
à
ä
î
0
á
ä
î
0
àà
à
ä
î
õ
1
2
àà
à
ä
î
0
×
ä
î
0
àà
ä
î
õ
0
á
ä
î
õ
1
2
à
ä
î
0
àà
ä
î
õ
0
à
ä
î
0
ä
î
õ
0
ä
î
0
ä
î
õ
0
à
à
ä
î
õ
0
à
à
ä
î
àà
à
0
àà
ä
î
0
ä
î
0
ä
ì
0
à
ä
M 2
í
0
à
à
ä
î
õ
0
àà
ä
î
à
1
2
1
2
1
2
1
2
ó
0
à
ä
î
0
ä
î
0
ä
î
0
ä
î
æë
0
ä
î
ï
0
ä
î
ë
0
ä
î
ïï
0
ä
î
ð
0
â
î
0
î
0
î
ïï
1
2
1
2
1
2
î
ï
0
î
0
î
0
î
0
ì
0
ü
à
1
2
1
2
ü
1
2
1
2
1
2
1
2
ü
0
ü
0
ü
0
ü
0
ü
å
0
ü
0
ü
ï
0
ü
0
þ
ï
0
ü
0
ü
ïï
0
ü
0
ü
ë
0
ü
0
ü
0
ü
ò
0
ü
ï
0
ü
ïï
0
ü
å
ï
0
ü
ï
0
ü
ïï
ï
0
ü
æë
0
ü
ï
ï
0
ü
ï
ï
0
ü
ïï
ï
0
ü
ïï
0
ü
ïï
ï
0
ü
ð
0
ü
ï
ï
ñ
0
ü
ïï
0
ü
ò
1
1
1
1
0
0
0
0
0
1
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1 0 0 0
1
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 2
0 0 0
1 0 0 0
1 2 0 0
1 0 2 0
0 0 0
0
0
1
0 0 0
0 0 0
0
0
0
0
0
0
2
0
0
2
0
1
2
2
ï
1
2
1
2
0
0
0
0
0
1
2
m, n
2
õ
0
ü
å
Ú
à
àà
á
0
æë
õ
1
2
ë
0
ú
å
ï
õ
1
2
æ
ï
ñ
0
ý
ïï
õ
0
þ
ýý
ï
0
ý
ý
ñ
õ
0
ü
ýý
ï
õ
0
ý
ïï
õ
1
2
ý
ï
ï
õ
0
ýý
ñ
ñ
õ
0
ý
ý
ò
õ
0
ü
ýý
ï
ð
õ
0
ý
ý
ï
ò
0
ý
ý
ï
0
ý
ý
ï
õ
1
2
1
2
1
2
1
2
ý
ý
ïï
0
ý
ý
ïï
õ
0
ý
ý
ï
0
ý
ý
ñ
0
ý
ý
ï
0
ý
ý
ñ
õ
0
ù
ïï
ï
0
ü
þ
ïï
ï
õ
0
ý
ï
0
ü
þ
ç
M n_ : Sum invB
M 1
MF
ù
ñ
ï
0
ý
ï
õ
0
ý
ýý
ïï
õ
0
ý
ý
ý
ó
0
ü
ýý
õ
0
ý
ý
õ
0
ý
ý
ï
õ
0
ý
ý
ñ
õ
0
ý
ý
à
àà
1
2
ý
ý
ï
0
ý
ý
ð
0
ý
ý
ñ
0
ý
ïï
0
ü
þ
ï
à
1
2
ü
å
õ
0
ý
ý
è
0
ü
ýý
ý
ýý
õ
0
ý
ý
ï
õ
0
ýý
ýý
ñ
õ
0
ý
ý
ò
õ
0
ý
ý
å
õ
0
ý
ý
õ
0
ý
ý
0
ý
ý
0
ý
ý
õ
0
ü
þ
å
õ
à
1
2
1
2
ü
þ
ýý
0
ý
ý
0
ü
ý
õ
1
2
1
2
1
2
1
2
ü
ýý
ï
õ
0
ü
ýý
ý
ñ
õ
0
ý
ý
õ
0
ý
ý
ï
0
ý
ý
ó
0
ý
ý
æ
0
ý
ý
æ
0
ý
ý
m , m, 1, 16
æ
0
ý
ý
0
1
0
0
0
1
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0
1 1
0
1
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
1
0
1
1
0
0
0
0
0
0
−1
ν=1 Bν,µ γ̂µ for our set of tomographic states
ý
ý
P16
ý
ý
é
0
ü
ýý
ý
0
ü
ýý
0
ü
ýý
þ
0
ü
þ
0
ü
ý
þ
0
ü
ýý
ý
ý
0
ü
ýý
ý
ý
0
ü
ýý
ý
ý
0
ü
ýý
ýý
0
ü
ýý
ý
0
þ
þ
0
ü
ýý
þ
0
ý
ý
0
ü
þ
ýý
ý
ý
0
ý
ý
ý
ýý
ýý
0
ü
ýý
ý
0
ü
ýý
0
ýý
þ
0
þ
0
ý
2
ýý
à
1
2
1
2
ü
ý
0
ü
þ
þ
0
ü
ýý
ýý
ý
0
ü
ýý
ý
1
2
1
2
ü
ýý
0
ü
þ
0
ü
ýý
0
ü
þ
0
þ
0
ý
1
2
ýý
0
ý
0
ü
þ
0
þ
þ
0
ü
ýý
þ
0
ýý
þ
0
ý
þ
0
ý
0
þ
ýý
0
ýý
ý
ý
þ
0
ýý
þ
0
ü
ýý
ý
ý
0
ýý
0
þ
0
þ
ý
0
ýý
ýý
ý
ý
0
ü
ýý
0
ü
ýý
ýý
0
þ
þ
0
ýý
ý
0
þ
0
ü
ýý
0
ü
þ
0
ý
0
ü
þ
1
ýý
0
ý
0
ü
þ
0
ü
þ
0
ýý
0
ü
þ
0
ü
þ
0
ýý
0
ü
ýý
0
ü
þ
0
ýý
þ
0
þ
0
ý
ý
0
ýý
0
ýý
ý
ý
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
ýý
M̂ν =
ýý
1
2
1
2
1
2
1
2
þ
1
2
1
2
1
2
1
2
ý
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
ýý
ý
ý
MF
ýý
ýý
B
ýý
ý
Ö
Ö
Ö
b m_, n_ : hc psi m . n .psi m
B : Table b m, n
1, 1 , m, 1, 16 , n, 1, 16
258
ýý
where Bm,n = hψm|γ̂n|ψmi
ü
B = [Bm,n]16×16
ü
non-singularity of B ↔ sensitivity of method
ü
with Bν,µ = hψν |γ̂µ|ψν i
ü
B = [Bν,µ]16×16
ü
ν=1
ü
γ̂ .
ν,µ µ
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
þ
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
þ
B −1
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
M̂ν =
ü
16
X
ü
where
ü
reconstructed ρ
P
ρ̂ = (N )−1 16
ν=1 M̂ν nν
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
nν = N hψν |ρ̂|ψν i
ö
invB : Inverse B
invB
MF
ø
average number of coincidence counts
inverse matrix B−1
÷
µ̂ν = |ψν ihψν |
projection measurement
reconstruction analysis
257
ÿ
æ
ê
å
ô
ï
ïï
ï
ïï
ô
ï
ïï
ï
ïï
ï
ïï
ò
ï
ïï
õ
ô
ïï
ï
ò
ï
ü
û
ì
î
í
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
àà
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
à
ä
ä
Ý
ß
ã
àà
à
àà
à
à
á
à
àà
à
àà
à
á
àà
à
á
àà
Û
àà
àà
á
à
à
à
àà
á
à
à
àà
à
à
à
à
à
à
à
àà
á
à
àà
àà
à
àà
à
á
×
á
à
àà
àà
á
àà
à
á
àà
àà
àà
à
à
à
à
à
àà
à
Ü
á
Þ
260
259
another common set of tomographic states
M 2
1
2
1
2
2
2
MF
−1
ν=1 Bν,µ
2
0
γ̂µ for the new set of tomographic states
1
2
2
0
0
2
1
2
1
2
0
0
0
2
1
2
0
P16
1
2
0
2
0
2
1
2
2
1
1
2
0
1
0
0
0
2
2
2
1
2
1
2
0
1
2
M̂ν =
No. Mode 1 Mode 2
1
|Hi
|Hi
2
|Hi
|Vi
3
|Vi
|Vi
4
|Vi
|Hi
5
|Ri
|Hi
6
|Ri
|Vi
7
|Di
|Vi
8
|Di
|Hi Note: we have just replaced D by D.
9
|Di
|Ri
10
|Di
|Di
11
|Ri
|Di
12
|Hi
|Di
13
|Vi
|Di
14
|Vi
|Li
15
|Hi
|Li
16
|Ri
|Li
b m_, n_ : hc psi m . n .psi m
B : Table b n, m
1, 1 , n, 1, 16 , m, 1, 16
invB : Inverse B
M n_ : Sum invB m, n
m , m, 1, 16
M 1
MF
261
262

2 −β −α

i
 −α 0
= 
 −β −i 0
1
0
0
1
2
0
0
α
0

1

0
,
0
0

2i −α

β 0 
,
0 0 
0 0


0 0
0
1


i −α 
0 0
= 
,
 0 −i 0 −β 
1 −β −α 2
1
2

0 −α 1

0 i
0 
,
−i 2 −β 
0 −α 0


0 −β 0 1


 −α 2 i −α 
= 
,
 0 −i 0 0 
1 −α 0 0
1
2

0

 0
= 
 −β
1
1
2


0
0 0 −α


0 β 2i 
 0
M̂6 = 12 
,
0
α 0 0 

−β −2i 0 0
M̂4
M̂2
16 matrices M̂n for the latter tomographic states
M̂1
M̂3

0

 0
M̂5 = 21 
−2i

−β

i

0
,
0
0

i

0
,
0
0
0
0
1
0
0
1
0
0

1

0
,
0
0


0
2
0 −α


0 −α 0 
 2
M̂12 = 12 
,
0
−β
0
0


−β 0
0
0

0

0
M̂10 = 
0
1




0
0
0 −α
0
0
2 −α




0 −β 2 
0 −β 0 
 0
 0
M̂7 = 21 
 , M̂8 = 21 
,
0
−α
0
0
2
−α
0
0




−β 2
0
0
−β 0
0
0
0
i
0
0
0
−i
0
0
0
0
−i
0

0 0

 0 0
M̂9 = 
 0 i
−i 0

0

 0
M̂11 = 
 0
−i
0 −β
β 0
0 0
0 0



,


0

0
M̂16 = 
0
1
0
0
−1
0
0
−1
0
0

1

0
,
0
0




0
0
0 −α
0
0
0 −β




0 −α 0 
0 −β 0 
 0
 0
M̂13 = 
 , M̂14 = 1 
,
2
2
 0 −β 0
 0 −α 0 −2i 

−β 0
2
0
−α 0 2i 0
1
2

0 −2i

 2i 0
M̂15 = 
α
 0
−α 0
1
2
where α ≡ 1 + i; β ≡ 1 − i.
Note: other M̂n’s are obtained for other sets of rotations.
263
264
with e.g.
By the Schur decomposition, any normal matrix Â (i.e., Â†Â = ÂÂ†)
can be represented as Â = Û T̂ Û † in terms of a triangular T̂ and a unitary Û .
For simplicity, we neglect Û .
of lower-triangular (or upper-triangular) form, which is easily invertible

0
0
,
0
t4
Tr{T̂ †T̂ }
T̂ †T̂
t1
0
0
 t5 + it6
t
0
2
T̂ = 
 t11 + it12 t7 + it8
t3
t15 + it16 t13 + it14 t9 + it10

ρ̂phys ≡ ρ̂phys(t1, t2, . . . t16) =
e.g.
• positive
ν=1
=
2
N
min L(t)
2 t
hψν |ρ̂p(t)|ψν i − nNν
hψν |ρ̂p(t)|ψν i
where
2
is a useful ‘likelihood’ function.
L(t) =
16
X
N hψν |ρ̂p(t)|ψν i − nν
−
max
t
2N hψν |ρ̂p(t)|ψν i
ν=1
16
X
or, equivalently, analyze logarithm of P (n, t):
for a given measured data n:
"
2 #
16
Y
N hψν |ρ̂p(t)|ψν i − nν
exp −
max
.
t
2N hψν |ρ̂p(t)|ψν i
ν=1
find numerically a maximum of P (n, t)
• normalized
• Hermitian
3. optimize ρ̂phys(t):
1. construct explicitly a “physical” density matrix ρ̂phys
Nnorm(t) = const – normalization assumed to be independent of t
maximum-likelihood method
266
267
268
n̄ν (t) = N hψν |ρ̂phys (t) |ψν i – number of counts expected for νth measurement
p
σν (t) ≈ n̄ν (t) – standard deviation
P
N = 4ν=1 nν ≫ 1 – normalization, so n̄Nν corresponds to a probability
where
(nν − n̄ν (t))2
P (n, t) =
exp −
Nnorm(t) ν=1
2σν2(t)
1
16
Y
is
for given ρ̂phys(t) with t = (t1, t2, . . . t16)
n = (n1, n2, . . . n16)
the probability of obtaining a set of counts
2. construct a likelihood function
assuming e.g. Gaussian noise of the measurements,
maximum-likelihood method
maximum-likelihood (MaxLik) method
this is not a physical density matrix!
Tr ρ̂2 = 1.053
eig ρ̂ = {1.02155, 0.0681238, −0.065274, −0.024396}
problems with semi-definiteness and normalization

0.4872
0.0042 + i0.0114 0.0098 − i0.0178 0.5192 − i0.0380
 0.0042 − i0.0114
0.0045
0.0271 + i0.0146 0.0648 − i0.0076 

ρ̂ = 
 0.0098 + i0.0178 0.0271 − i0.0146
0.0062
0.0695 + i0.0134 
0.5192 + i0.0380 0.0648 + i0.0076 0.0695 − i0.0134
0.5020

reconstructed ρ̂ by linear tomography
n13 = 13171, n14 = 17170, n15 = 16722, n16 = 33586
n9 = 17932, n10 = 32028, n11 = 15132, n12 = 17238,
n5 = 16324, n6 = 17521, n7 = 13441, n8 = 16901,
n1 = 34749, n2 = 324, n3 = 35805, n4 = 444,
counts for 16 projection measurements
experimental example of James et al.
265
‘likelihood’ function
t5
t11
t15
$
%
"
!
&
#
!
!
;
thus, we need initial values tini.
r
M(0)
(1)
M00
(1)
M01
q
(1) (2)
M00 M00,11
(2)
M
q01,12
√
(2)
M00,11
ρ33
ρ
√ 30
ρ33
s
0
(1)
(2)
M00
M00,11
(2)
M
q00,12
√
(2)
M00,11
ρ33
ρ
√ 31
ρ33
r
0
0
(2)
M00,11
ρ33
ρ
√ 32
ρ33
0
0
ρ33
0
√



















How to estimate initial ‘physical’ matrix T̂ (tini)?










T̂ (tini) = 








By calculating T̂ from our “unphysical” ρ̂:
where
M(0) = Det(ρ̂),
(1)
Mij = Det(ρ̂without row i & col j),
(2)
Mij,kl = Det(ρ̂without rows i,k & cols j,l)
i, j, k, l = 0, ..., 3
269
270
FindMinimum[
L[t1,t2,t3,t4,t5,t6,t7,t8,t9,t10,t11,t12,t13,t14,t15,t16],
{t1,tini1},{t2,tini2},{t3,tini3},{t4,tini4},{t5,tini5},{t6,tini6},
{t7,tini7},{t8,tini8},{t9,tini9},{t10,tini10},{t11,tini11},
{t12,tini12},{t13,tini13},{t14,tini14},{t15,tini15},{t16,tini16}]
t1
0
0
0
t6
t2
0
0
;
t12
t7
t8
t3
0
t16 t13
t14 t9
t10 t4
L t1_, t2_, t3_, t4_, t5_, t6_, t7_, t8_, t9_, t10_, t11_,
t12_, t13_, t14_, t15_, t16_ : Module T, RhoPhys, sum, Nmean ,
T:
RhoPhys : hc T .T Tr hc T .T ;
sum 0;
Do
Nmean Re Norm hc psi m . phys .psi m ;
sum sum
Nmean Nexp m
^ 2 2 Nmean , m, 1, 16
Return Re sum
;
numerical optimalization
'
for our experimental data
271


0.5948i
0
0
0
 0.8706 + 0.7282i

0.4060
0
0 
T̂ (tini) = 

−0.0215
+ 0.9933i −0.2827 − 0.2982i
0.0615i
0 
0.7328 + 0.0536i 0.0915 + 0.0107i 0.0981 − 0.0189i 0.7085


0.7870
0.0325 − 0.0014i 0.0328 − 0.0051i 0.1289 − 0.0094i
 0.0325 + 0.0014i

0.0850
−0.0024 − 0.0050i 0.0161 − 0.0019i 
ρ̂phys(tini) = 

0.0328
+
0.0051i
−0.0024
+
0.0050i
0.0034
0.0173
+ 0.0033i 
0.1289 + 0.0094i 0.0161 + 0.0019i 0.0173 − 0.0033i
0.1247
L(tini) = 3.0695
272
by applying the optimalization procedure we can diminish this value to
L(topt) = 0.0104
for the following matrix ...
our matrix reconstructed by MaxLik method


0.5028
0.0230 + 0.0117i 0.0279 − 0.0130i 0.4686 − 0.0346i


0.0051
0.0050
+
0.0001i
0.0333
−
0.0082i 
 0.0230 − 0.0117i
ρ̂phys(topt) = 
0.0279
+
0.0130i
0.0050
−
0.0001i
0.0066
0.0416
+ 0.0102i 
0.4686 + 0.0346i 0.0333 + 0.0082i 0.0416 − 0.0102i
0.4854
is a physical density matrix as it is
• positive (semidefinite)
eig ρ̂phys = {0.9687, 0.0313, 0, 0}
• Hermitian
†
ρ̂phys = ρ̂phys
• normalized
Tr{ρ̂phys} = 1
2
} = 0.9394 ≤ 1
Tr{ρ̂phys
273
mn
vh
vh
m
hv
m
hv
vh
vh
hh
hh
n
hv
n
hv
vh
vh
vh
vh
−0.05
hh
0
0.05
0
hh
0.5
vh
vh
m
hv
m
hv
vh
vh
hh
hh
1

3
1+
√
√
3
3hσ̂3i)
2 (hσ̂8i +
3
(hσ̂
i
+
ihσ̂
1
2i)
2
3
(hσ̂4i + ihσ̂5i)
2
1+
3
i)
2√(hσ̂1i − ihσ̂
√2
3
(hσ̂
i
−
3hσ̂
3 i)
8
2
3
(hσ̂6i + ihσ̂7i)
2
3
2 (hσ̂4 i −
3
−
2 (hσ̂6 i√
1−
vh
vh
vh
vh
274

ihσ̂5i)

ihσ̂7i) 
3hσ̂8i
n
hv
n
hv

1 0 0
√1  0 1 0  .
3
0 0 −2

Analogously, a qudit density matrix can be reconstructed
via measurement of generators of the Lie algebra SU (d)





0 0 −i
0 0 0
0 0 0
σ̂5 =  0 0 0  , σ̂6 =  0 0 1  , σ̂7 =  0 0 −i  , σ̂8 =
i 0 0
0 1 0
0 i 0

which can be chosen as generators of the Lie algebra SU (3)








0 1 0
0 −i 0
1 0 0
0 0 1
σ̂1 =  1 0 0  , σ̂2 =  i 0 0  , σ̂3 =  0 −1 0  , σ̂4 =  0 0 0  ,
0 0 0
0 0 0
0 0 0
1 0 0
via measurement of the generalized Pauli operators
ρ̂ =

reconstruction of a qutrit density matrix
−0.05
hh
0
0.05
0
hh
0.5
ρ̂ by linear tomography (left) and ρ̂phys by MaxLik tomography (right figures)
comparison of reconstructed matrices
phys
mn
mn
Re ρ
mn
Im ρ
Re ρ
Im ρphys
simulation of gates by various circuits
Toffoli and CCU gates
reversible gates
(part II)
Quantum gates
276
Copies of ρ̂ are generated from the same (known) initial state(s) by applying
the same transformations in the same experimental setup.
No! The term “copies” refers to identically prepared quantum objects.
Answer
Thus, do we violate the no-cloning theorem?
But unknown ρ̂ cannot be copied.
For complete reconstruction of ρ̂ we need to repeat measurements on many
copies of ρ̂.
Quantum tomography and no-cloning theorem:
Question
275
277
Ĥ(c0|+i + c1|−i) = c0|0i + c1|1i
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
0 0
0 0
0 1
0 0
1 0
0 0
0 0 0
0 0 1
1 0 0
0 1 0
0 0 0
0 0 0
0 0 0
0 0 0
1
0
0
0
0
0
0
0
|a〉
|a〉
|a〉
|b〉
|a〉
|b〉
|ab ⊕ c〉
|a ⊕ b〉
|c〉
|b〉
CNOT and Toffoli gates
1 0
0 1
0 0
0 0
0 0
0 0
0 0
0 0
=
V
V†
V
Note: Toffoli gate is a special case of CCU
U
Any CCU gate can be built from
CNOT, CV, and CV† gates, where V2 = U
theorem of Barenco,Bennett et al. (1995)
replacement for CCU gate
0 0
0 0
1
2
3
3
3
0
0
0
reversible computing
→
+
c3|11i
278
.
/
- a computational process that is (or almost is) time-invertible (time-reversible).
Landauer’s principle
- a computational process to be physically reversible,
it must also be logically reversible.
logically-reversible process
- a discrete, deterministic computational process for which the transition function that maps input states to output states is a one-to-one function.
examples of reversible quantum gates
+
c2|10i
all unitary quantum gates are reversible, e.g.:
+
c1|01i
Ĥ(c0|0i + c1|1i) = c0|+i + c1|−i
=
c0|00i
XOR=CNOT
ab | c
-------00 | 0
01 | 1
10 | 1
11 | 0
R-XOR
ab | abc
ac
bc
-----------------00 | 000
00
00
01 | 011
01
11
10 | 101
11
01
11 | 110
10
10
-
-
-
-
-
-
-
*
*
*
*
*
*
*
+
,
|ψini
ÛCNOT|ψini = c0|00i + c1|01i + c2|11i + c3|10i ≡ |ψouti
ÛCNOT|ψouti = c0|00i + c1|01i + c2|1, 1 ⊕ 1i + c3|1, 0 ⊕ 1i = |ψini
but measurement is irreversible
examples of irreversible classical gates
OR
ab | c
-------00 | 0
01 | 1
10 | 1
11 | 1
logic gates in a classical computer, other than NOT gate, are irreversible, e.g.:
AND
ab | c
-------00 | 0
01 | 0
10 | 0
11 | 1
how to make these gates reversible?
R-OR
ab | abc
---------00 | 000
01 | 011
10 | 101
11 | 111
keep input(s) together with the standard output, e.g.:
R-AND
ab | abc
---------00 | 000
01 | 010
10 | 100
11 | 111
(
)
279
280
281
5
V
V†
|x
V
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
7
4
7
>
@
=
@
=
@
=
@
=
@
=
@
@
=
4
=
@
@
=
4
4
4
4
4
=
@
=
@
;
?
4
MF
ket 0, 1 .bra 0, 1
v0,1
v1,1
ket 1, 0 .bra 1, 1
v0,0
v1,0
ket 1, 1 .bra 1, 0
CV gate – control unitary V =
0
0
1
0
9
:
:
=
<
284
4
4
4
B
C
D
B
C
D
C
B
B
C
C
C
B
B
A
B
4
4
4
4
4
E
4
4
4
0
0
0
1
0
0
0 v0,0 v0,1
0 v1,0 v1,1
B
1
0
0
0
cv :
ket 0, 0 .bra 0, 0
ket 0, 1 .bra 0, 1
v0,0 ket 1, 0 .bra 1, 0
v0,1 ket 1, 0 .bra 1, 1
v1,0 ket 1, 1 .bra 1, 0
v1,1 ket 1, 1 .bra 1, 1
cv
MF
cv = |00ih00| + |01ih01| + v0,0|10ih10| + v0,1|10ih11| + v1,0|11ih10| + v1,1|11ih11|
0
0
0
1
8
0
1
0
0
6
1
0
0
0
cnot :
ket 0, 0 .bra 0, 0
cnot
MF
8
U
T
CNOT gate – cnot = |00ih00| + |01ih01| + |11ih10| + |10ih11|
W
C
4
U|x
5
U
4
V
5
X
V|x
5
U
4
V†
X
B
C
V|x
U
V
X
C
B
|x
U
D
B
4
4
4
U|x
X
C
|x
U
|1
X
|1
U
|0
X
|0
U
|1
X
|1
U
?
X
=
U
|1
X
|1
U
|1
X
|1
U
|1
X
|1
S
|1
5
V
|1
L
9
|1
T
8
|1
W
V
U
V†
X
V
U
U
X
|x
U
V† |x
X
|x
U
|x
X
|x
U
9
|x
X
|0
U
|0
X
|1
U
7
|1
X
8
|0
U
7
|0
N
?
M
=
M
|0
Q
|0
N
|1
Q
7
|1
M
8
|1
M
7
|1
L
8
|1
L
7
|1
N
8
|1
M
7
|1
X
282
L
|x
M
V|x
R
M
|x
R
|x
L
R
0
0
0
1
N
R
ket 1, 1
L
n2_ : Module state ,
ZeroMatrix 4, 1 ;
2 n1 n2 1, 1
1;
state
n2_ : Transpose ket n1, n2
MF
M
replacement for CCU gate (proof)
U
U
|1
X
|1
U
|1
1
0
0
0
X
|1
L
ket n1_,
state
state
Return
bra n1_,
ket 0, 0
U
|1
|0
|x
bra[n1, n2] = hn1, n2| → row vector
X
|1
|0
V
|0
283
ket[n1, n2] = |n1, n2i → 4-element column vector with ‘1’ at pos. 2n1 + n2 + 1
M
|x
?
=
|0
|x
|0
|0
S
|1
|0
V†
|0
|0
V
|1
|0
|x
|0
|0
• two-qubit states
implementing gates in M athematica
L
|0
5
L
|0
V
|0
|0
5
L
|x
|0
|0
5
M
|x
?
=
|0
5
LO
|0
U
|0
|0
5
P
|x
5
|0
|0
replacement for CCU gate (proof)
5
M
8
D
C
C
B
D
I
K
H
K
K
H
H
K
H
K
K
H
4
4
4
4
H
K
K
H
H
K
K
H
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
H
K
H
K
F
H
J
E
G
0
0
0
0
1
0
0
0
Y
[
0, 1, 1
0, 1, 1
1, 1, 1
1, 1, 1
i
f
e
e
d
e
f
dg
d
h
e
u
u
~
287

288

e

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0 v0,0 v0,1
0 v1,0 v1,1
v
285
0
0
0
0
1
0
0
0
t
d
3. CV23=kron(id,cv)
t
• three-qubit states
v
i
CV23new : Kron id, cv
CV23new
MF
s
‘1’ at position 4n1 + 2n2 + n3 + 1
u
KET n1_, n2_, n3_ : Module state ,
state ZeroMatrix 8, 1 ;
1;
state 4 n1 2 n2 n3 1, 1
Return state
BRA n1_, n2_, n3_ : Transpose KET n1, n2, n3
KET 0, 0, 0
MF

or explicitly
t

1
0
0
0
0
0
0
0
MF
v
MF
t
f

0
0
0
1
0
0
0 v0,0 v0,1
0 v1,0 v1,1
0
0
0
0
0
0
0
0
0
0
0
0
u
e

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0 v0,0 v0,1
0 v1,0 v1,1
t
v
v
u
u
1
0
0
0
0
0
0
0
u
i
0 1 0 0 0 0 0 0
286
u
t
t
t
0
0
0
0
1
0
0
0
u
t
u
u
u
u
1 0
t
t
t
t
t
v
v
v
v
u
u
u
u
t
CV23 :
KET 0, 0, 0 .BRA 0, 0, 0
KET 0, 0, 1 .BRA 0, 0, 1
v0,0 KET 0, 1, 0 .BRA 0, 1, 0
v0,1 KET 0, 1, 0 .BRA
v1,0 KET 0, 1, 1 .BRA 0, 1, 0
v1,1 KET 0, 1, 1 .BRA
KET 1, 0, 0 .BRA 1, 0, 0
KET 1, 0, 1 .BRA 1, 0, 1
v
KET 1, 1, 0 .BRA 1, 1, 0
v0,1 KET 1, 1, 0 .BRA
0,0
v1,0 KET 1, 1, 1 .BRA 1, 1, 0
v1,1 KET 1, 1, 1 .BRA
t
0
0
0
1
0
0
0 v0,0 v0,1
0 v1,0 v1,1
0
0
0
0
0
0
0
0
0
0
0
0
CV23
1
0
0
0
0
0
0
0
u
e
KET 1, 0, 0 .BRA 1, 0, 0
KET 1, 1, 0 .BRA 1, 1, 1
t
t
t
t
v
u
u
u
u
v
t
t
t
t
BRA 0, 0, 1
\

e
MF
Z

f
0 0 0 0 0 0 0 1
[

d
BRA 1, 1, 1
Z

e
KET 0, 0, 1 .BRA 0, 0, 1
KET 0, 1, 0 .BRA 0, 1, 1
KET 1, 1, 1 .BRA 1, 1, 0
\
[
u
u
\
[

d
d
control gates for 3 qubits
0
0
0
0
0
0
1
0
[
\
[
[
\
[
\
+|100ih100| + |101ih101| + |111ih110| + |110ih111|
Z
0
0
0
0
0
0
0
1
Z
Z
[
[
Z
Z
Z
Z
Z
Z

e
j
BRA[n1, n2, n3] = hn1, n2, n3| → row vector
0
0
0
0
0
1
0
0
w
j
j
j
1. CNOT23=kron(id,cnot) where id = 0 1
MF
0
0
1
0
0
0
0
0
w
KET[n1, n2, n3] = |n1, n2, n3i → 23-element column vector with
0
0
0
1
0
0
0
0
a
\
[
[
r
r
n
d
d
j
e
k
m
d
d
d
j
e
p
p
p
m
m
m
m
m
p
p
p
p
p
p
m
m
m
m
m
p
p
p
p
p
m
m
m
m
m
m
p
p
p
p
p
m
m
m
m
m
p
p
p
p
p
p
m
m
m
m
m
m
p
p
p
p
p
o
m
m
l
q
q
or CNOT23 = |000ih000| + |001ih001| + |011ih010| + |010ih011|
Z
CNOT23 :
KET 0, 0, 0 .BRA 0, 0, 0
KET 0, 1, 1 .BRA 0, 1, 0
KET 1, 0, 1 .BRA 1, 0, 1
Z
0
1
0
0
0
0
0
0
]
[
[
Z
CNOT23
1
0
0
0
0
0
0
0
{
]
Z
Z
[
[
2. CNOT12=kron(cnot,id) – analogously
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
|
x
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
y
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
b
^
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
_

1
1
1
1

CV13= SWAP12 CV23 SWAP12 = kron(swap,id) CV23 kron(swap,id)

tricky way to calculate CV13
0,
0,
1,
1,

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0 v0,0 v0,1
0
0
0 v1,0 v1,1
0
0
0
0
0
v0,0 v0,1
0
0
0
v1,0 v1,1

0
0
1
0
0
0
0
0

0
1
0
0
0
0
0
0

MF

CV13
CV13 :
KET 0, 0, 0 .BRA 0, 0, 0
KET 0, 0, 1 .BRA 0, 0, 1
KET 0, 1, 0 .BRA 0, 1, 0
KET 0, 1, 1 .BRA 0, 1, 1
v0,0 KET 1, 0, 0 .BRA 1, 0, 0
v0,1 KET 1, 0, 0 .BRA 1,
v1,0 KET 1, 0, 1 .BRA 1, 0, 0
v1,1 KET 1, 0, 1 .BRA 1,
v0,0 KET 1, 1, 0 .BRA 1, 1, 0
v0,1 KET 1, 1, 0 .BRA 1,
v1,0 KET 1, 1, 1 .BRA 1, 1, 0
v1,1 KET 1, 1, 1 .BRA 1,

¡
1
0
0
0
0
0
0
0
¡
¡
¡
¡
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0 u0,0 u0,1
0 u1,0 u1,1
¤
¡
¤
¡
¤
¡
¤
¡

¤
¡

¡
¤
£

U
•
¢

¡

¡
¤
¤
¡
¤
¡

¤

©

¡
¥
£
¢
¥

¤
¡
¡

¤

£
¨

¨
v0,1 v1,0
v0,0 v0,1
£
v1,0 v1,1
¢
¡
v0,1 v1,0

MF
v21,1
v0,1 v1,1
v0,0 v0,1
v0,0 v0,1
.
;
v1,0 v1,1
v1,0 v1,1
0
0
0
1
0
0
0
0
¤
u0,0 u0,1
u1,0 u1,1
0
1
0
0
0
0
0
0
¤
¬
©
v0,0 v1,0
PQRS
WVUT
×
×
¤
v20,0
•
PQRS
WVUT
•
=
5. CCU gate (U = V 2)
=
PQRS
WVUT
•
×
¤

CV13=SWAP23 CV12 SWAP23 = · · ·
¡
¤
ª
or
¡
¤
£
0
¡
¤
CV13new
¡
¤
¤
Max CV13
¡
¤
u0,0 u0,1
u1,0 u1,1
CCU :
¡
¤
¤
test
U
•
¤
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0 v0,0 v0,1
0
0
0 v1,0 v1,1
0
0
0
0
0
v0,0 v0,1
0
0
0
v1,0 v1,1
¡
¤
0
0
1
0
0
0
0
0
×
×
¡
¤
0
1
0
0
0
0
0
0
×
×
¤
1
0
0
0
0
0
0
0
=
×
substitution circuit for SWAP gate
U
•
substitution circuits for CU13 gate
¢
CV13new : Kron swap, id .CV23.Kron swap, id
CV13new
MF
290
+v0,0|110ih110| + v0,1|110ih111| + v1,0|111ih110| + v1,1|111ih111|

1
0
0
0
0
0
0
0
289
+v0,0|100ih100| + v0,1|100ih101| + v1,0|101ih100| + v1,1|101ih101|
CV13 = |000ih000| + |001ih001| + |010ih010| + |011ih011|
4. CV13 gate
¢
¬
¬
¨
¬
«
©
©
¡
¦
¨
§

×
×
292
291
V
293
properties of unitary matrices
V =
thus for
v0,0 v0,1
v1,0 v1,1
we have
∗
∗
v0,0
v0,1 + v1,0
v1,1 = 0
v0,0
v0,0
v0,1
v1,0
v0,0
v0,0
v0,1
v1,0
etc.
v0,0
v0,0
v0,1
v1,0
v0,1
v1,0
v0,0
v0,0
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
∗
∗
v0,0 + v0,1
v0,1 = 1
v0,0
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
Conjugate
v1,0
v0,1
v1,1
v1,1
v1,0
v0,1
v1,1
v1,1
v1,0
v0,1
v1,1
v1,1
v1,1
v1,1
v1,0
v0,1
·
1
1
1
1
0
0
0
0
a matrix is unitary iff its row and column vectors are orthonormal
r1 :
r2 :
r3 :
r4 :
r5 :
r6 :
r7 :
r8 :
·
a substitution circuit for CCU gate
V†
...finally
·
È
È
È
È
È
È
È
È
Ç
Ç
Ç
Ç
Ç
Ç
Ç
Ç
Å
Å
Å
Å
Å
Å
Å
Å
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Ä
V
gate1 gate2 gate3 gate4 gate5
294
·
Ä
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
v0,1
0
0
0
0
0
0
v0,1 v1,0
v0,0
v1,1
2
v1,1
v0,1
295
296
CCU1 : CCU1temp . r1 . r2 . r3 . r4 . r5 . r6 . r7 . r8
CCU1
MF
0
1
0
0
0
0
0
¶
substitution circuit for CCU gate
0
0
;
1
0
1
0
0
0
0
0
0
v1,0
v1,0 v1,1
¼
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
¾
»
0
0
0
1
·
Â
Â
Â
Æ
Æ
Æ
Æ
Æ
Æ
Æ
Æ
Å
Å
Å
Å
Å
Ã
Ã
Ã
Ã
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Å
v0,1
»
0
1
0
0
·
Å
Å
MF
»
1
0
0
0
·
2
v0,0
·
0
0
; cnot
0
1
·
CCU1
·
±
Â
Â
Â
Â
Â
Ã
Ã
Ã
Ã
Max CCU
Á
¾
·
0 0 0 0 0 0 v0,0 v1,0
test
0
À
½
»
·
we have shown that our gate CCU1 simulates the original CCU gate
¶
¸
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
º
0
1
0
0
µ
0
0
1
0
´
1
0
0
0
µ
º
º
º
º
º
º
º
º
º
º
º
º
³
³
³
³
³
³
IdentityMatrix 2 ;
´
±
0
0
0
1
0
0
; id
0 v0,0 v0,1
0 v1,0 v1,1
´
´
´
¹
¯
³
°
µ
³
°
²
³
°
°
µ
³
°
°
µ
°
°

°
´
²
³
®
³
°
³
³
³
³
³
³
³
³
³
²
swap :
cv

³
³
³
³
±
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³
µ
®
°
°
°
°
°
°
°
°
°
°
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³
µ
´

1
0
0
0

¿

gate1 : Kron id, cv
gate2 : Kron cnot, id
gate3 : Kron id, hc cv
gate4 : gate2
gate5 : Kron swap, id .gate1.Kron swap, id
CCU1temp : gate5.gate4.gate3.gate2.gate1

®
°
°
°
°
°
°
°
°
°
°
°
¯
Ê
Ë
Ê
Ë
Ê
CCU2temp : CV13.Kron cnot, id .hc CV23 .Kron cnot, id .CV23
or more explicitly
Ë
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
v20,0
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Ð
Ð
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Ë
Ô
É
qubits cannot be measured without disturbing quantum information they carry,
i.e., by measuring
Steane’s ECC
fault-tolerant gates
without quantum ECCs construction of
practical quantum computers would be impossible
quantum information cannot be copied with perfect fidelity
3. no cloning
we get either |0i with probability |a|2 or |1i with probability |b|2
a|0i + b|1i
2. measurement destroys quantum information
Shor’s ECC
a note
there are infinitely many quantum errors
300
correction of bit-flip and phase-flip errors
can we really correct quantum errors?
thus we need a quantum ECC – an analog of classical ECC
quantum states are very fragile to decoherence and dissipation,
error-correcting code (ECC)
there is a continuous set of quantum errors
thus
which can also correspond to
removing the qubit and replacing it with garbage!
3. small errors
√
√
a|0i + b|1i → a2 + ǫ|0i + b2 − ǫ|1i
e.g. |+i → |−i & |−i → |+i (orthogonal)
|0i → |0i & |1i → −|1i
2. phase-flip errors = phase errors
|0i → |1i & |1i → |0i
1. bit-flip errors = amplitude errors
errors in quantum computers
0→1&1→0
bit-flip errors
errors in classical computers
299
1. errors are continuous
298
297
quantum vs classical errors
Introduction to
quantum error-correction codes
MF
Ó
CCU2
Ó
Max CCU
Ó
test
Ó
v21,1
v0,1 v1,1
Ó
v0,1 v1,0
v0,0 v0,1
0
0
0
0
0
0
Ó
v1,0 v1,1
0
0
0
0
0
0
v0,1 v1,0
Ó
0
0
0
1
0
0
0
0
Ñ
0 0 0 0 0 0 v0,0 v1,0
0
1
0
0
0
0
0
É
1
0
0
0
0
0
0
CCU2 : CCU2temp . r1 . r2 . r3 . r4 . r5 . r6 . r7 . r8
CCU2
MF
É
Ì
Ì
Ì
Ì
Ì
Ì
Ì
Ì
301
bit-flip error
|ψ2i = |ψ1i|00i = a|00100i + b|11000i
24
14
|ψ i = ÛCNOT
ÛCNOT
|ψ2i
3
= a|0, 0, 1, 0 ⊕ 0 ⊕ 0, 0i + b|1, 1, 0, 0 ⊕ 1 ⊕ 1, 0i
= a|0, 0, 1, 0, 0i + b|1, 1, 0, 0, 0i = |ψ2i (sic!)
45hM
′
, M ′′|ψ4i = a|0, 0, 1i + b|1, 1, 0i
so let us measure modes 4 & 5:
35
25
|ψ i = ÛCNOT
ÛCNOT
|ψ3i
4
= a|0, 0, 1, 0, 0 ⊕ 0 ⊕ 1i + b|1, 1, 0, 0, 0 ⊕ 0 ⊕ 1i
= a|0, 0, 1, 0, 1i + b|1, 1, 0, 0, 1i
a|0, 0, 1i + b|1, 1, 0i ⊗ |0, 1i
=
|ψ5i =
in our case the measurement result (i.e. error syndrome) is:
M ′ = 0, M ′′ = 1
error correction = recovery
so we know that we have to flip the 3rd qubit
|ψ6i = X̂3|ψ5i = a|0, 0, 0i + b|1, 1, 1i = |ψ0i
M′
M′
M′
M′
=
=
=
=
so in general
0, M ′′ = 1 =⇒ |ψ0i = X̂3|ψ5i
1, M ′′ = 0 =⇒ |ψ0i = X̂2|ψ5i
1, M ′′ = 1 =⇒ |ψ0i = X̂1|ψ5i
0, M ′′ = 0 =⇒ |ψ0i = |ψ5i
error syndrome =⇒ error correction
303
Analogously, we can find a proper action for other measurement results:
304
error detection = error syndrome diagnosis/measurement
quantum repetition code
in general
+b 1
+ b|1i)|0i⊗(N −1) = a|0i⊗N + b|1i⊗N
a 00 + b 11
⊗N
a 000 + b 111
a0
302
|ψ0i = a|000i + b|111i → 3rd bit flip → |ψ1i = a|001i + b|110i
⊗N
|ψi = a|0i + b|1i
12
12
ÛCNOT
|ψi|0i
=
ÛCNOT
(a|00i + b|10i)
= a|0, 0 ⊕ 0i + b|1, 0 ⊕ 1i
= a|00i + b|11i
1n
ÛCNOT
(a|0i
13
12
12
13
ÛCNOT
ÛCNOT
= ÛCNOT
ÛCNOT
note that
13
12
13
ÛCNOT
Û
CNOT|ψi|0i|0i = ÛCNOT(a|00i + b|11i)|0i
= a|0, 0, 0 ⊕ 0i + b|1, 1, 0 ⊕ 1i
= a|000i + b|111i
i=2
N
Y
a 0 +b 1
0
a 0 +b 1
0
0
⊗( N −1)
a 0 +b 1
0
ˆ
|ψ0i = X̂|3M ′′−2M ′||ψ5i, where X̂0 = I,
|ψi
|ψi
|0i
|0i
|0i
45hM
′
•
M
M
_
•
•
R
Recovery
|ψ0i
35
25
24
14
, M ′′|ÛCNOT
ÛCNOT
ÛCNOT
ÛCNOT
|ψ1i|00i
•
//
syndrome detect
error correct
decode
89:;
?>=<
•
//
89:;
?>=<
//
//
•
|0i
|0i
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
•
GF
@A
M ′′BCED
•
MBCED′
GF
@A
R
example: a quantum bit-flip ECC
306
89:;
?>=<
•
|ψi
89:;
?>=<
•
where −//− denotes quantum channel where an error can occur
encode
|0i
•
_
a typical error correction code (ECC)
|ψ0i = X̂|3M ′′−2M ′|
|ψ1i
•
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
GF
ED
?>=<
89:;
?>=<
89:;
′
@A BC
GF
ED
89:;
?>=<
89:;
?>=<
@A ′′
BC
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Syndrome Measurement
bit-flip error correction |ψ1i → |ψ0i
|ψ1i = a|001i + b|110i, a|010i + b|101i, a|100i + b|011i or a|000i + b|111i
|ψ0i = a|000i + b|111i
bit-flip error |ψ0i → |ψ1i
305
|ψi
initial encoding
|−i → phase flip → |+i
|0i + |1i
|0i − |1i
√
→ phase flip → |−i = √
2
2
13
12
|ψ0i = Ĥ ⊗3ÛCNOT
ÛCNOT
(a|0i + b|1i)
= Ĥ1Ĥ2Ĥ3(a|000i + b|111i)
= a| + ++i + b| − −−i
or explicitly
a|0i + b|1i → |ψ0i = a|000i + b|111i
→ |ψ0′ i = a| + ++i + b| − −−i
|+i =
so
a|0i + b|1i → phase flip → a|0i − b|1i
phase-flip error
appears as an amplitude error (bit flip) and vice versa.
a phase error (phase flip) in a rotated basis
thus
1 1 1
0 1 1 1 1
1 0
√
Ĥ σ̂xĤ = √
=
= σ̂z
1 0
0 −1
2 1 −1
2 1 −1
1 1 1
1 1 1
1 0
√
Ĥ σ̂z Ĥ = √
1 −1
0 −1
2 1 −1
2
1 1 −1
1 1
=
1 −1
2 1 1
0 1
=
= σ̂x
1 0
phase vs amplitude errors
308
307
encoded initial state
|ψ0′ i = a| + ++i + b| − −−i
phase-flip error |ψ0′ i → |ψ1′ i
|ψ1′ i = a| + +−i + b| − −+i
or a| + −+i + b| − +−i
or a| − ++i + b| + −−i
or a| + ++i + b| − −−i
and analogously
a| + −+i + b| − +−i → a|010i + b|101i
a| − ++i + b| + −−i → a|100i + b|011i
a| + ++i + b| − −−i → a|000i + b|111i
so now our standard bit-flip ECC can be applied
•
Syndrome Measurement
•
89:;
?>=<
•
•
89:;
?>=<
•
•
R
Recovery
phase-flip error correction |ψ1′ i → |ψ0i
H
H
H
89:;
?>=<
89:;
?>=<
GF ′′
@A
M BCED
GF BC
@A
MED′
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ |0i
|0i
|ψ0i
′
35
25
24
14
, M ′′|ÛCNOT
ÛCNOT
ÛCNOT
ÛCNOT
Ĥ ⊗3|ψ1i|00i
309
310
apply Hadamard gates to |ψ1′ i → |ψ1i
|ψ1i = Ĥ ⊗3|ψ1′ i = Ĥ1Ĥ2Ĥ3(a| + +−i + b| − −+i) = a|001i + b|110i
|ψ1′ i
45 hM
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
|ψ0i = X̂|3M ′′ −2M ′|
•
89:;
?>=<
•
89:;
?>=<
a|000i + b|111i
encoding qubit against bit-flip error
a|0i + b|1i
|0i
|0i
•
89:;
?>=<
•
89:;
?>=<
H
H
H
a| + ++i + b| − −−i
encoding qubit against phase-flip error
a|0i + b|1i
|0i
|0i
How to encode qubit against both errors?
•
89:;
?>=<
•
89:;
?>=<
|0i
|0i
|0i
|0i
H
H
H
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
89:;
?>=<
•
encoding code for Shor’s nine-qubit ECC
a|0i + b|1i
|0i
|0i
|0i
|0i
311
312
first real error correction code against both bit-flip and phase-flip single error
(|0i + |1i)(|0i + |1i)(|0i + |1i)
1
(|0i − |1i)(|0i − |1i)(|0i − |1i)
a|0i + b|1i −→ a|0iL + b|1iL
(|000i − |111i)(|000i − |111i)(|000i − |111i) ≡ |1iL
23/2
H
H
|0i
|0i
g
g
s
|0i
|0i
a|0i + b|1i
|0i
H
|0i
g
g
g
s
g
g
g
s
g
g
g
s
314
a|0iL + b|1iL
encoding code for Steane’s seven-qubit ECC
• there are other more concise ECCs
• thus Shor’s code can correct combined bit and phase flips on a single qubit
• the corrections of bit and phase flip errors are independent
Note that:
23/2
1
→ | − −−i =
→
23/2
(|000i + |111i)(|000i + |111i)(|000i + |111i) ≡ |0iL
23/2
1
|1i → |111i
→
→ | + ++i =
76
75
|ψ2i = ÛCN
OT ÛCN OT |ψ1i = |ψ1i
|ψ1i = Ĥ1Ĥ2Ĥ3|0i⊗6|ψ0i = |+, +, +, 0, 0, 0, 0i
34
35
36
|ψ5i = ÛCN
OT ÛCN OT ÛCN OT |ψ4i
1
(|0, 0, +, 0, 0, 0, 0i + |0, 1, +, 1, 1, 0, 1i
2
+|1, 0, +, 1, 0, 1, 1i + |1, 1, +, 0, 1, 1, 0i)
1
= √ (|0, 0, 0, 0, 0, 0, 0i + |0, 0, 1, 0 ⊕ 1, 0 ⊕ 1, 0 ⊕ 1, 0i
8
+|0, 1, 0, 1, 1, 0, 1i + |0, 1, 1, 1 ⊕ 1, 1 ⊕ 1, 0 ⊕ 1, 1i
+|1, 0, 0, 1, 0, 1, 1i + |1, 0, 1, 1 ⊕ 1, 0 ⊕ 1, 1 ⊕ 1, 1i
+|1, 1, 0, 0, 1, 1, 0i + |1, 1, 1, 0 ⊕ 1, 1 ⊕ 1, 1 ⊕ 1, 0i)
=
24
25
27
|ψ4i = ÛCN
1
= (|0, 0, +, 0, 0, 0, 0i + |0, 1, +, 0 ⊕ 1, 0 ⊕ 1, 0, 0 ⊕ 1i
2
+|1, 0, +, 1, 0, 1, 1i + |1, 1, +, 1 ⊕ 1, 0 ⊕ 1, 1, 1 ⊕ 1i)
final step 5:
step 4:
14
16
17
|ψ3i = ÛCN
1
= √ (|0, +, +, 0, 0, 0, 0i + |1, +, +, 0 ⊕ 1, 0, 0 ⊕ 1, 0 ⊕ 1i)
2
1
= √ (|0, +, +, 0, 0, 0, 0i + |1, +, +, 1, 0, 1, 1i)
2
step 3:
step 2:
step 1:
|ψ0i = |0i⊗6|ψi
|ψi = |0i
let
|0i → |000i
1
encoding code for Steane’s seven-qubit ECC
encoding code for Shor’s nine-qubit ECC
313
316
315
so finally
|ψi = |1i → |ψ5i ≡ |1iL
1
|0iL ≡ |ψ5i = √ (|0000000i + |0011110i + |0101101i + |0110011i
8
+|1001011i + |1010101i + |1100110i + |1111000i)
X
1
= √
|νi
8 even ν∈Hamming
Analogously
where
a|0i + b|1i→a|0iL + b|1iL
1
|1iL = √ (|1111111i + |1110000i + |1001100i + |1000011i
8
+|0101010i + |0100101i + |0011001i + |0010110i)
X
1
= √
|νi
8 odd ν∈Hamming
Thus, in general
s
s
s
s
g g g g NM
|0i
s
s
s
s
|0i
s
s
s
317
318
g g g g NM
s
bit-flip syndrome detection in Steane’s 7-qubit ECC
|0i
g g g g NM
phase-flip syndrome detection
- the same circuit but with extra Hadamard gates, as for Shor’s code
fault-tolerant syndrome detection
to make the circuit fault tolerant, each ancilla qubit must be replaced
by four qubits in a suitable state.
fault-tolerant device
- a device that works effectively even
when its elementary components are imperfect.
s
g
s
g
g
s
g
s
ancilla
data
s
g
s
g
correct!
g
s
fault-intolerant and fault-tolerant circuits
data
ancilla
wrong!
most general single-qubit error
the most general single-qubit unitary error transformation
(apart from irrelevant global phase factor) can be expanded to order ǫ as
Ûerror = ǫiIˆ + Ô(ǫ) = ǫiIˆ + ǫxσ̂x + ǫy σ̂y + ǫz σ̂z
discrete set of quantum errors
fundamental result of quantum error correction theory:
correcting just a discrete set of errors
(bit flip, phase flip and combined bit-phase flip)
a quantum error-correcting codes can correct a continuous set of errors.
Shor’s and Steane’s ECCs
• they protect not only against bit and phase flip errors
but against arbitrary errors affecting only a single qubit
• by measuring the error syndrome,
the state collapses into one of the four states
σ̂x|ψi, σ̂y |ψi, σ̂z |ψi, |ψi
which are correctable with the codes.
319
s
g
320
x
↔
Rθ
g
s
g
s
↔
i
u
|yi
|xi
|x ⊕ yi
CNOT
g
s
|xi
|yi
|xi
89:;
?>=<
•
•
89:;
?>=<
•
•
89:;
?>=<
•
•
naive solution but not fault-tolerant
|x ⊕ 1i
|xi
g
s
s
|z ⊕ xyi
|yi
|xi
322
Toffoli gate
Toffoli gate = controlled-controlled-NOT (CCNOT)
NOT
g
g
s
3. fault-tolerant Toffoli gate
data
data
can also be applied bitwise with majority vote
2. fault-tolerant CNOT (XOR) = transversal CNOT
Rθ
Rθ
Rθ
can be applied bitwise with majority vote
1. fault-tolerant single qubit-gates
fault-tolerant gates
so far we have analyzed coding for
fault-tolerant storing of quantum information, but we also need:
fault-tolerant memory
321
e |z ⊕ xyi
e |xi
e r |yi
NM
NM
NM
6 6 6
Z
Z e
r r
Note: for convenience we neglect normalizations
error-threshold theorem
requirements for scalable QIP
perfect ECC with the smallest number of ancillas
(part II)
Error correction codes
324
• each line represents a block of 7 qubits!
• |ψ cati = √12 (|0i⊗7 + |1i⊗7), |0i = |0i⊗7, |1i = |1i⊗7, |xi = |xi⊗7, etc.
• gates are implemented transversally
• if a given measurement outcome is 1, the arrow points to the set of gates to
be applied; no action is taken if the outcome is 0.
r
r
r
r
|0i H
r
r
r
r
|0i H
r
r
e
e
|0i H r
|ψcat i
Z H e 6
NM
|ψcat i
Z H e NM
|ψcat i
Z H e NM
e
|xi
e
|yi
r H
|zi
Shor’s construction of fault-tolerant Toffoli gate
323
Hilbert space for ECC
subspaces corresponding to different errors should be orthogonal
thus the total Hilbert space for ECC should be
large enough to contain all the orthogonal subspaces.
number of subspaces is 2(3n + 1)
325
Orthogonality requires a subspace for each of the three errors every qubit can
suffer and another one for the unperturbed logical state.
encoded qubits
|0iL = |B1i|00i − |B3i|11i + |B5i|01i + |B7i|10i
|1iL = −|B2i|11i − |B4i|00i − |B6i|10i + |B8i|01i
in terms of the (unnormalized) 3-qubit Bell states:
2
|B 1 i = |000i ± |111i
4
|B 3 i = |100i ± |011i
|B 5 i = |010i ± |101i
89:;
?>=<
326
•
π
@ABC
GFED
8
@ABC
GFED
7
•
6
|0iL =
=
=
=
|1iL =
=
=
=
|B1i|00i − |B3i|11i + |B5i|01i + |B7i|10i
(|000i + |111i)|00i − (|100i + |011i)|11i
+(|010i + |101i)|01i + (|110i + |001i)|10i
|00000i + |11100i − |10011i − |01111i
+|01001i + |10101i + |11010i + |00110i
(lexicographic order)
|00000i + |00110i + |01001i − |01111i
−|10011i + |10101i + |11010i + |11100i
encoded qubits explicitly
Thus, all the allowed encodings have the same sign pattern:
with two minus signs in one of the logical states and four in the other.
Other allowed encodings can be found from those by
permutations of bits and coordinated signs.
8
6
We must double this to have enough space to accommodate both logical states
and their erroneous descendants.
26 ≤
/ 16 for n = 4 & 32 ≤ 32 for n = 5
|B 7 i = |110i ± |001i
⇒
minimum number of encoded qubits is n = 5 for ECC
2(3n + 1) ≤ 2n
1. 5-bit encoder
•
•
@ABC
GFED
•
•
89:;
?>=<
@ABC
GFED
5
•
it requires the minimum number of ancillas
Los Alamos ECC
⇒
number of encoded qubits for ECC
n = 9 – Shor ECC
n = 7 – Steane ECC
n = 5 – Los Alamos ECC
H
•
π
4
@ABC
GFED
π
3
•
2
•
[Laflamme, Miquel, Paz, Zurek (1996)]
|ai
H
1
H
|bi
|Qi
|ci
|di
gates
−|B2i|11i − |B4i|00i − |B6i|10i + |B8i|01i
−(|000i − |111i)|00i − (|100i − |011i)|11i
−(|010i − |101i)|01i + (|110i − |001i)|10i
−|00011i + |11111i − |10000i + |01100i
−|01010i + |10110i + |11001i − |00101i
(lexicographic order)
−|00011i − |00101i − |01010i + |01100i
−|10000i + |10110i + |11001i + |11111i
327
328
8
7
5
4
@ABC
GFED
•
syndrome
|a′b′c′ d′ i
none 0000
X3Z3
1101
X5Z5
1111
X2
0001
Z3
1010
Z5
1100
X2Z2
0101
X5
0011
Z1
1000
Z2
0100
Z4
0010
X1
0110
X3
0111
X4
1011
X1Z1
1110
X4Z4
1001
error
−a|1i − b|0i
−a|0i − b|1i
a|0i − b|1i
resulting state
′
|Q i
a|0i + b|1i
−a|1i + b|0i
−a|0i + b|1i
2
π
π
3
•
•
•
89:;
?>=<
•
errors and their syndromes
6
@ABC
GFED
@ABC
GFED
89:;
?>=<
1
H
H
H
input state |Qi = a|0i + b|1i, Xi – bit flip error, Zi – phase flip error
gates
@ABC
GFED
•
•
•
@ABC
GFED
π
•
•
(reverse of the encoder)
Los Alamos ECC
2. decoder
|b′i
|a′i
330
|d′i
|c′i
|Q′i
329
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
R̂ =
R11 R12
R21 R22
R
•
•
double-controlled rotation: CCR
0
0
1
0
0
0
0
0
0
0
0
1
332
CCR= [ 1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
R(1,1)
R(2,1)
0
0
0
0
0
0
R(1,2)
R(2,2)];
ˆ
CCR = I+|110ih110|(R
11−1)+|110ih111|R12 +|111ih110|R21 +|111ih111|(R22 −1)
CCNOT=[
1
0
0
0
0
0
0
0
];
CNOT=[
1
0
0
0
];
I=eye(2);
% identity operator
H=[1 1;1 -1]; % Hadamard gate - for simplicity we neglect 1/sqrt(2)
R=-I;
% pi-rotation
Matlab program for Los Alamos ECC
331
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
R̂ =
•
•
•
R
R11 R12
R21 R22
triple-controlled rotation: CCCR
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
333
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
R(1,1) R(1,2)
R(2,1) R(2,2)];
0])*(R(1,1)-1)+...
1])*R(1,2)
+...
0])*R(2,1)
+...
1])*(R(2,2)-1);
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
CCCR gate explicitly
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
334
ˆ
CCCR = I+|1110ih1110|(R
11 −1)+|1110ih1111|R12+|1111ih1110|R21+|1111ih1111|(R22−1)
[
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
CCCR=eye(16)+...
ket([1 1 1 0])*bra([1
ket([1 1 1 0])*bra([1
ket([1 1 1 1])*bra([1
ket([1 1 1 1])*bra([1
CCCR=
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
R̂ =
89:;
?>=<
•
89:;
?>=<
R
R11 R12
R21 R22
another triple-controlled rotation: cCcR
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
R(1,2)
R(2,2)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0])*(R(1,1)-1)+...
1])*R(1,2)
+...
0])*R(2,1)
+...
1])*(R(2,2)-1);
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
cCcR gate explicitly
0
0
0
0
R(1,1)
R(2,1)
0
0
0
0
0
0
0
0
0
0
335
336
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0];
ˆ
cCcR = I+|0100ih0100|(R
11 −1)+|0100ih0101|R12 +|0101ih0100|R21+|0101ih0101|(R22−1)
[
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
cCcR=eye(16)+...
ket([0 1 0 0])*bra([0
ket([0 1 0 0])*bra([0
ket([0 1 0 1])*bra([0
ket([0 1 0 1])*bra([0
cCcR=
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
3
];
SWAP=[
89:;
?>=<
•
1
0
0
0
=
0
0
1
0
0
1
0
0
0
0
0
1
89:;
?>=<
×
×
× • ×
SWAP34=tensor_product(I,I,SWAP,I);
CNOT45=tensor_product(I,I,I,CNOT);
gate4=SWAP34*CNOT45*SWAP34;
5
qubit 1
2
3
4
• gate 4
gate1=tensor_product(H,H,I,H,I);
gate2=tensor_product(I,CCCR);
gate3=tensor_product(I,cCcR);
1
π
π
|di
89:;
?>=<
•
H
•
89:;
?>=<
|ci
•
•
H
|bi
|Qi
H
|ai
Los Alamos encoding circuit in Matlab
• gates 1, 2, 3
338
337
5
4
3
2
1
89:;
?>=<
89:;
?>=<
•
=
89:;
?>=<
•
89:;
?>=<
•
=
×
×
89:;
?>=<
•
×
×
××
89:;
?>=<
•
×
×
339
6
7
8
•
•
•
89:;
?>=<
π
89:;
?>=<
•
340
†
Ûdecoder = Ûencoder
U_encoder=gate8*gate7*gate6*gate5*gate4*gate3*gate2*gate1;
U_decoder=gate1*gate2*gate3*gate4*gate5*gate6*gate7*gate8;
thus the encoding and decoding circuits are described by the operators
gate=tensor_product(I,I,CCR);
gate8=swap_qubits(gate,5,3);
gate6=swap_qubits(CNOT45,5,3);
gate7=swap_qubits(CNOT45,4,2);
5
4
3
1
2
• gates 6,7,8
CNOT45=tensor_product(I,I,I,CNOT);
gate5 =swap_qubits(swap_qubits(CNOT45,5,3),4,1)*...
swap_qubits(CNOT45,4,1);
Alternatively, by defining a function swap_qubits(Û , n1, n2), which
temporarily swaps n1th and n2th qubits in a matrix Û , we can simply write
gate5=SWAP23*CNOT12*SWAP23*SWAP25*CNOT12*SWAP25;
CNOT12=tensor_product(CNOT,I,I,I);
SWAP23=tensor_product(I,SWAP,I,I);
SWAP34=tensor_product(I,I,SWAP,I);
SWAP45=tensor_product(I,I,I,SWAP);
SWAP25=SWAP23*SWAP34*SWAP45*SWAP34*SWAP23;
• gate 5
(z = −1)
(z = −1)
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
U_encoder= z
0
0
0
z
0
1
0
0
0
0
1
0
1
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
U_decoder= 0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
0
1
0
0
0
0
0
z
1
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
0
0
0
z
0
1
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
1
0
0
0
0
0
z
0
0
1
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
0
z
0
1
0
0
0
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
1
0
0
0
0
0
1
1
0
0
0
0
0
z
0
0
0
z
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
z
0
z
0
0
0
0
z
0
1
0
0
0
z
0
0
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
1
0
z
0
0
0
0
0
0
1
0
1
0
0
0
z
0
0
0
0
0
z
1
0
0
0
0
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
1
0
z
0
0
0
0
1
0
1
0
0
0
z
0
0
0
0
0
z
0
0
1
0
0
0
0
0
z
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
0
0
z
0
z
0
0
0
0
0
0
z
0
1
0
0
0
z
0
0
0
0
0
1
1
0
0
0
0
0
1
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
1
0
0
0
0
0
1
0
0
z
0
0
0
0
0
1
0
0
0
z
0
1
0
0
0
0
z
0
z
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
0
0
0
0
z
z
0
0
0
0
0
z
0
0
0
1
0
1
0
0
0
0
0
0
1
0
z
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
0
0
0
0
z
0
0
z
0
0
0
0
0
z
0
0
0
1
0
1
0
0
0
0
1
0
z
0
0
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
1
0
0
0
0
0
1
z
0
0
0
0
0
1
0
0
0
z
0
1
0
0
0
0
0
0
z
0
z
0
0
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
z
0
z
0
0
0
0
1
0
z
0
0
0
z
0
0
0
0
0
1
0
0
z
0
0
0
0
0
z
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
1
0
z
0
0
0
0
0
0
z
0
z
0
0
0
z
0
0
0
0
0
z
z
0
0
0
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
1
0
z
0
0
0
0
z
0
z
0
0
0
z
0
0
0
0
0
z
0
0
z
0
0
0
0
0
1
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
0
z
0
z
0
0
0
0
0
0
1
0
z
0
0
0
z
0
0
0
0
0
1
z
0
0
0
0
0
z
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
z
0
0
0
1
0
z
0
0
0
0
z
0
z
0
0
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
z
1
0
0
0
0
0
1
0
0
0
z
0
z
0
0
0
0
0
0
1
0
z
0
0
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
1
0
0
0
0
0
z
0
0
1
0
0
0
0
0
1
0
0
0
z
0
z
0
0
0
0
1
0
z
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
0
z
0
0
0
1
0
z
0
0
0
0
0
0
z
0
z
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
0
0
0
z
0
z
0
0
0
0
z
0
1
0
0
0
1
0
0
0
0
0
z
0
0
z
0
0
0
0
0
z
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
0
1
0
z
0
0
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
0
1
z
0
0
0
0
0
1
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
0
0
0
1
0
z
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0
0
z
0
0
0
0
0
1
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
z
0
z
0
0
0
0
0
0
z
0
1
0
0
0
1
0
0
0
0
0
z
z
0
0
0
0
0
z
0
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
1
0
0
0
0
0
z
z
0
0
0
0
0
z
0
0
0
z
0
z
0
0
0
0
0
0
z
0
1
0
0
341
1
0
0
0
0
0
1
0
0
z
0
0
0
0
0
1
0
0
0
1
0
z
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
342
0
0
0
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
1
0
0
0
0
0
z
0
0
z
0
0
0
0
0
z
0
0
0
z
0
z
0
0
0
0
z
0
1
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
0
1
0
0
0
0
0
1
z
0
0
0
0
0
1
0
0
0
1
0
z
0
0
0
0
0
0
1
0
1
0
0
0
z
0
1
0
0
0
0
0
1
0
z
0
0
0
0
0
1
0
z
0
0
0
0
0
z
0
1
0
0
0
0
0
0
0
z
0
z
0
0
0
0
1
0
z
0
0
0
1
0
0
0
0
0
z
0
0
1
0
0
0
0
0
1
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
z
0
0
0
0
0
0
z
0
z
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
0
z
0
0
1
0
1
0
0
0
0
0
z
0
z
0
0
0
0
0
z
0
z
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
z
0
0
0
0
z
0
z
0
0
0
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
z
0
0
0
0
0
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0
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0
0
1
0
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0
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0
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0
0
0
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0
0
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z
1
0
0
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0
0
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0
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0
0
0
1
0
z
0
0
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z
0
1
0
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0
z
0
1
0
0
0
0
0
1
0
z
0
input state to encoder
|ψini = a|00000i + b|00100i
psi_in=a*ket([0 0 0 0 0])+b*ket([0 0 1 0 0])
or explicitly
psi_in’
ans = [a000b000000000000000000000000000];
output state
|ψouti = Ûencoder|ψini
psi_out=U_encoder*psi_in
1. encoding qubit |0iL
special case: a = 1, b = 0
psi_out’
ans =[100000100100000z000z010000101000]
343
344
show_state(psi_out,’|0>_L = ’)
ans =|0>_L = +|00000>+|00110>+|01001>-|01111>...
-|10011>+|10101>+|11010>+|11100>
2. encoding qubit |1iL
input state to encoder: |ψini = |1i
special case for a = 0, b = 1
psi_in’
ans = [00001000000000000000000000000000];
output state |ψouti = |1iL
psi_out’
ans = [000z0z0000z01000z000001001000001]
or explicitly
show_state(psi_out,’|1>_L = ’)
ans = |1>_L = -|00011>-|00101>-|01010>+|01100>...
-|10000>+|10110>+|11001>+|11111>
thus we have got |0iL and |1iL in agreement with our
definition
Q
345
decoder
′
|1i → −|0i
|0i → −|1i
A1→1 = −1
A0→1 = 0
⇒
⇒
|1i → −|1i
|0i → |0i
psi_in
= ket([0 0 Q
0 0]);
psi_syndrome = ket([0 1 Q_prime 1 0]);
amplitude=psi_syndrome’*U_decoder*U_error*U_encoder*psi_in
|Q′i → σ̂z → |Qi
How to correct it?
|Qi = a|0i + b|1i → |Q′i = a|0i − b|1i
A1→0 = 0,
A0→0 = 1,
psi_syndrome=ket([1 0 Q_prime 1 0]);
transition amplitudes |Qi → |Q′i
syndrome
sigma_z=[1,0;0,-1]
U_error=tensor_product(I,I,sigma_z,I,I);
error
Ûerror = Iˆ ⊗ Iˆ ⊗ σ̂z ⊗ Iˆ ⊗ Iˆ
example 2: phase flip of the 3rd qubit
|Q′i → R̂(π)σ̂x → |Qi
How to correct it?
AQ→Q′ = hψsyndrome|ÛdecoderÛerrorÛencoder|ψini
|ψsyndromei = |a b Q c d i = |01Q 10i
′ ′ ′
346
⇒
⇒
so
⇒
′ ′
Q
so
A1→1 = 0
A0→1 = −1
|Qi = a|0i + b|1i → |Q′i = −b|0i − a|1i
A1→0 = −1,
A0→0 = 0,
AQ→Q′
transition amplitudes |Qi → |Q′i
transition amplitude
|a b c d i = |0110i
′ ′ ′ ′
syndrome
|ψini = |abQcdi = |00Q00i
input state to encoder
U'
U ' ≡ U a 'b 'c ' d '
c'
d'
sigma_x=[0,1;1,0];
U_error=tensor_product(sigma_x,I,I,I,I);
error
Ûerror = σ̂x ⊗ Iˆ ⊗ Iˆ ⊗ Iˆ ⊗ Iˆ
Errors and their syndromes
example 1: bit flip of the 1st qubit
encoder
Q'
a'
b'
syndromes
basic elements of the perfect ECC
1 error
348
347
Requirements for scalable QIP
[Knill, Laflamme, Zurek et al., 2002]
1. Scalable physical systems:
the ability to support any number of independent qubits.
2. State preparation:
the ability to prepare any qubit (or at least large fraction of them)
in the standard initial state |0i.
3. Measurement:
the ability to measure any qubit (or at least large fraction of them)
in the logical basis.
349
350
Note: sometimes the standard projective measurement can be
replaced by weak measurements that return a noisy number whose
expectation is the probability that a qubit is in the state |1i.
4. Errors:
The error probability per gate must be below a threshold and satisfy
independence and locality properties.
• For the most pessimistic independent, local error models, the error
threshold is above ∼ 10−6.
For some special error models, the threshold is substantially higher.
For example:
• For the independent depolarizing error model, it is believed
to be better than ∼ 10−4.
• For the independent ‘erasure’ error model, where error events
are always detected, the threshold is above .01.
• The threshold is also above .01 when the goal is only to transmit
quantum information through noisy quantum channels.
5. Quantum control:
351
the ability to implement a universal set of unitary quantum gates acting on a small number (usually at most two at a time) of qubits.
• For most accuracy thresholds, it is necessary to be able to apply the
quantum control in parallel to any number of disjoint pairs of qubits.
This parallelism requirement can be weakened if a nearly noiseless
quantum memory is available.
352
• The universality assumption can be substantially weakened by replacing some or all unitary quantum gates with operations to prepare special states or by having additional measurement capabilities.
accuracy-threshold theorem
Assuming the above requirements for scalable QIP:
If the error per gate is less than a threshold,
then it is possible to efficiently quantum compute arbitrarily accurately.
This is one of the most important results in
quantum ECC and fault-tolerant computation.
Introduction to quantum algorithms
(part I)
quantum algorithms and classical cryptography
public key cryptography (PKC)
354
f−1(out) = ?
example
but
x mod 3 = 2 => x = 2, 5, 8, 11 ,14 , 17, 20, ... (not unique result)
y = 17 mod 3 => y = 2 (unique result)
f(in) = out,
• without additional information, it is not always possible to decrypt by
repeating encryption operations in reverse order
2. private, confidential key
1. public, open key
two keys in PKC
and independently by Diffie & Hellman (1976)
invented by Ellis (1970, British CESG)
public-key cryptography (PKC)
= asymmetric cryptography
= non-secret encryption
1997 Grover algorithm for searching databases:
How to search the keys more effectively?
1994 Shor algorithm for finding discrete logarithms:
How to break ElGamal cryptosystem?
plaintext
1994 Shor algorithm for number factorization:
How to break cryptosystems of
RSA, Rabin, Williams, Blum-Goldwasser,…?
cleartext
plaintext
ciphertext
plaintext
cleartext
355
Decryption using key k2 (private key)
Encryption using key k1 (public key)
356
Decryption using the same key k
key distribution
Encryption using key k
asymmetric algorithms
= public-key algorithms
cleartext
plaintext
ciphertext
cleartext
symmetric algorithms
1985 Deutsch (Deutsch-Jozsa / DJ) algorithm:
How to see both sides of a coin simultaneously?
quantum algorithms and cryptography
353
e
c
secure
channel
unsecured
channel
Eve
357
358
plaintext
decryption
d=e-1
Ed(c)=m
m
Bob
Encryption with a secure channel for key exchange
Alice
key source
e
encryption
Ee(m)=c
m
plaintext
c
unsecured
channel
unsecured
channel
Eve
Why do we need PKC?
Number of keys
a problem of huge number of keys for symmetric algorithms.
N (N −1)
2
359
360
• How many keys should be generated for N correspondents if everyone
wants to communicate with all others using symmetric algorithms?
nsym(N ) = (N2 ) =
• How many keys are required for 6 or 1 milion correspondents?
1012
2
nsym(6) = 15
nsym(106) ≈
nasym(N ) = 2N
• How many keys are required asymmetric algorithms?
ElGamal
computational problem
integer factorization problem
popular public-key encryption schemes
based on integer factorization problem
decryption
d=e-1
Ed(c)=m
m
generalized ElGamal
public-key encryption scheme
RSA
(Rivest-Shamir-Adleman & Cocks)
Rabin
Williams
Blum-Goldwasser probabilistic
Goldwasser-Micali probabilistic
plaintext
Bob
key source
Encryption with a unsecured channel for key exchange
Alice
e
encryption
Ee(m)=c
m
plaintext
quadratic residuosity problem
or integer factorization problem
discrete logarithm problem
generalized discrete logarithm problem
361
p
log n(log log n))
our estimations
Nclas(n) = c3 exp[c(log n)1/3(log log n)2/3]
by having only one key it is practically impossible to calculate the other key
using practically available computers over practically long period of time.
364
363
for simplicity, we choose: c3 = 1
with c = 2 in the fastest version of number field sieve method due to Lenstra
NShor(n) = c1(log n)2 log(log n) + c2 log n
hard problems in PKC
Thus, we estimate numbers of operations required to factorize n as:
and log n steps of post processing on a classical computer.
• practically, exceeding the specified bound on the number of operations or
memory corresponding to a specified security parameter
• require super-polynomial time or space
∼ (log n)2 log log n
• optimized Shor algorithm
∼ (log n)3
• original Shor algorithm
polynomial in time, poly(log n)
hard = computationally infeasible
– perhaps seconds or milliseconds.
• practically, within a certain number of machine operations or time units
• in polynomial time and space
effectiveness of quantum factorization methods
exponential in time, exp(log n)
should be interpreted relative to a specified frame of reference.
easy = computationally feasible
effectiveness of classical factorization methods
This is the fastest publicly available ;-) classical method for r ≥ 150.
5. number field sieve method
p
∼ exp(c 3 log n(log log n)2)
∼ exp(c
thus
3,4. Fermat method and quadratic sieve method
√
∼ exp(c r log r), where r ∼ log n is the number of bits
∼ n1/4 log3 n
2. Monte Carlo method (Pollard method)
1. direct method
√
∼ n = exp( 21 log n)
What is the time or number of bit operations required to factorize n?
Motivation for Shor’s Algorithm
– effectiveness of classical integer factorization methods
“easy” and “hard” computational problems
362
computational problem
linear code decoding problem
or error-correction-code problem
Merkle-Hellman
knapsack subset sum problem
cracked by Shamir & Zippel!
Graham-Shamir
cracked!
Lu-Lee
cracked by Adleman & Rivest!
Goodman-McAuley
cracked!
Powerline System
(a simple version of Chor-Rivest)
cracked by Lenstra!
Chor-Rivest
publicly not cracked!
public-key encryption scheme
McEliece
popular public-key encryption schemes
not based on integer factorization problem
examples of factorization times
MIPS = millions of instructions per second
• 1-10 MIPS in a modest PC
• hundreds-thousands of MIPS in a supercomputer
/60
min
/60
hours
/24
days
/365
years
• say, N ′ = 106 MIPS are available in contemporary computers
/N’
sec
⇒
Nclas ∼ 7.0 · 1020 MIPS → 22.5 mln years
Nclas ∼ 3.5 · 1012 MIPS → 40.5 days ≈ 1 month
Nclas ∼ 5.9 · 1013 MIPS → 687.5 days ≈ 2 years
⇒
⇒
time required to factorize n
t(n)= N(n)
examples
1. n = 10130
2. n = 10150
3. n = 10300
How old is the universe?
Tuniverse ∼ 1.25 · 1010 = 12.5 billion years
What is the largest factorizable integer within Tuniverse?
⇒
T ≈ 3.17 · 1010 > Tuniverse
1. direct method
N = 1060
⇒
T ≈ 1.6 · 1010 > Tuniverse
2. Monte Carlo method (Pollard method)
N = 1091
⇒
T ≈ 1.36 · 1010 > Tuniverse
3,4. Fermat method and quadratic sieve method
N = 1094
⇒
⇒
T ≈ 1.18 · 1010 < Tuniverse
T ≈ 8.1 · 1010 > Tuniverse
5. number field sieve method
N = 10400
N = 10375
365
366
(part II)
classical RSA algorithm
quantum Deutsch algorithm
quantum Deutsch-Jozsa (DJ) algorithm
quantum Hadamard transform
quantum Fourier transform
Rivest-Shamir-Adleman (RSA) algorithm (1978)
I. generation of RSA keys (by Bob)
and
φ = (p − 1)(q − 1)
1. choose two large primes p 6= q
2. calculate
n = pq
3. choose randomly an integer e (1 < e < φ) coprime to φ,
i.e. their greatest common divisor is one, GCD(e, φ) = 1.
n - modulus
367
368
4. using the extended Euclidean algorithm, calculate d (1 < d < φ) such that
ed ≡ 1 mod φ
5. thus
public key - (n, e)
private key - d
• cryptographic terms:
e – encryption exponent
d – decryption exponent
2=
32
10
2
0
116 mod 42 =
42 mod 32 =
32 mod 10 =
10 mod 2 =
Answer: 2
a
b
x2
x1
y2
y1
372
a,b,x,y,d - integers
GCD(116,42) = d = ax + by ?
initial x1=y2=0; x2=y1=1
while b>0
q = Int(a/b)
r =a-qb
x1 = x2 -q x1 ; x2 = x1
y1 = y2 -q y1 ; y2 = y1
a=b
b=r
return (d,x,y)=(a, x2, y2)
extended Euclidean algorithm
GCD(a,b) = d =ax+by x,y,d=?
371
42
1
0
0
42 32 = 0
1=
1
[1
1-2
11
6
16
*0
mo
/4
d
2]
42
116
initial values
-2 =
0-2
*1
1
-------------------------------------
q
GCD(116,42) = ?
how to calculate
the greatest common divisor (GCD)?
Euclidean algorithm
370
It has been revealed only very recently that the RSA algorithm has been devised
already in 1973 by Cocks for the British security agency CESG.
A note
m = cd mod n
use your private key d to calculate
III. RSA decryption (by Bob)
3. send a cipher c to Bob
c = me mod n
2. calculate
m ∈ h0, n − 1i
1. change cleartext into plaintext represented by an integer
encrypt your message with the public key (n, e)
II. RSA encryption (by Alice)
369
a
b
x2
x1
y2
y1
373
QCD(116,42) = d = ax + by ?
q
42
0
1
1
0
-2
1
0
3
-2
1
374
------------------------------------116
32
-1
42
1
2
10
-11
32
3
1
4
2
-1
10
58
3
-21
0
5
-11
||
y
4
||
x
2
||
d
answer:
multiplicative or modular
inverse of an integer
problem: 4-1 mod 9 = ?
4y=1 mod 9 => 9x + 4y =1 mod 9
q
a
b
y2
y1
-------------------9
4
0
1
2
4
1
1
-2
4
1
0
-2
.
so 9*1 + 4*(-2)=1 & y=-2=7 mod 9
thus 7 is the inverse of 4 mod 9
4*7=1 mod 9
inverse of an integer modulo 9
problem: 8-1 mod 9 = ?
8y=1 mod 9 => 9x + 8y =1 mod 9
q
a
b
y2
y1
-------------------9
8
0
1
1
8
1
1
-1
8
1
0
-1
.
so 9*1 + 8*(-1)=1 & y=-1=8 mod 9
thus 8 is the inverse of itself mod 9
8=8-1 mod 9
8*8 = 1 mod 9
inverse of integers mod 9
2-1 mod 9 = 5
5-1 mod 9 = 2
1-1 mod 9 = 1
3-1 mod 9 = ?
3y=1 mod 9
9x + 3y =1 mod 9
3(3x + y) =1 mod 9 => no solution
6-1 mod 9 = ? => no solution
all invertible elements modulo 9
Z*9={1,2,4,5,7,8}
n is invertible GCD(n,9)=1
375
376
0
1
-12
37
y2
1
-12
37
.
y1
2
2
2
2
2
5
GCD(φ, e) =GCD(160, 13) =GCD(25 · 5, 13) = 1
|
|
|
|
|
|
|
02
12
22
32
A˛
I
P
Y
03
13
23
33
B
J
q
Z
04
14
24
34
C
K
R
Ż
05
15
25
35
Ć
L
S
Ź
06
16
26
36
D
Ł
Ś
_
07
17
27
37
E
M
T
-
E˛
N
U
?
09
19
29
39
F
Ń
v
,
c = me mod n = 11713 mod 187 =?
2. Alice calculates cipher using the public key (n, e) = (187, 13):
m = (01, 17) = 117
1. Alice converts her cleartext into the plaintext:
A
H
Ó
X
08
18
28
38
10
20
30
40
Assume that Alice wants to encrypt the following cleartext: “AM” :-)
II. RSA encryption
public key (n, e) = (187, 13)
private key d = 37
or vice versa
public key (n, d) = (187, 37)
private key e = 13
5. so Bob has generated the keys:
de = 13 · 37 = 481 ≡ 1 (mod 160)
01
11
21
31
OK
13
4
1
0
e
160
80
40
20
10
5
• inefficient direct method
1
⇒
d = 37
160
13
4
1
φ
4. he double checks that d is the multiplicative inverse of e:
⇒
12
3
4
q′
using, for example, the following public alphabet:
378
GCD(160, 13) = 1
3. Bob calculates d using the extended Euclidean algorithm
⇒
160 mod 13 = 4
13 mod 4 = 1
4 mod #1 = 0
• efficient Euclidean algorithm
2. Bob checks whether φ are e are coprime:
φ = (p − 1)(q − 1) = 10 · 16 = 160
n = pq = 11 · 17 = 187
1. Bob calculates
(artificially small numbers so completely insecure)
Bob chooses p=11, q=17, e=13
I. RSA key generation
Example of RSA application
• Mathematica: b=PowerMod[a,k,n]
return(b)
end
end
b = b · Aki mod n
A = A2 mod n
for i = 1 to j do begin
b = Ak0
i
i=0 ki 2
Pj
if n ∈ I; a, k ∈ Zn
convert k into binary form k =
A=a
else begin
if k = 0 then b = 1
Algorithm
b = ak mod n =?
How to calculate powers modulo n?
377
G
O
W
.
380
379
13 = (1101)2
i
0
1
2
3
ei
1
0
1
1
382 ≡135 1352 ≡86 (mod 187)
A2 117 1172 ≡38
c 117
117 117*135≡87 87*86≡2 (mod 187)
so the cipher is c = 2
• Mathematica: PowerMod[117, 13, 187]֒→ 2
III. RSA decryption
1. Bob calculates m = cd mod n = 237 mod 187
0
1
2
2
1 2
3
4
5
0 1
0
0
1
4 16 162 ≡69 692 ≡ 86
862 ≡ 103 (mod 187)
2 32
32
32 32*103 ≡ 117 (mod 187)
d = 37 = (100101)2
i
di
A2
m
2. thus he finds Alice’s message to be
m = 117 = (01, 17) = „AM”
lengths of public keys
381
382
1. The above PKC examples were given for artificially small numbers
and thus such cryptosystems are not secure at all.
2. To insure security of a PKC system, the lengths of public keys should be
of hundreds of decimal digits.
3. To determine the required key length you should consider:
• intended security
• lifetime of the key
• current state-of-the-art of factoring.
4. The wise cryptographer is ultra-conservative when choosing publickey key lengths as history teaches us a lot:
• “I shall be surprised if anyone regularly factors numbers of size 1080
without special form during the present century” (R. Guy, 1976).
• “Factoring a 125-digit number would take 40 quadrillion years”,
tj. 4 · 1019 lat (R. Rivest, 1977).
• ... but already in 1994 a 129-digit number was factorized.
Recommended lengths of public keys
length in bits
(a)
(b)
(c)
768 1280 1536
1024 1280 1536
1280 1536 2048
1280 1536 2048
1536 2048 2048
=>
383
length in decimal digits
(a)
(b)
(c)
231
386
463
308
386
463
386
463
617
386
463
617
463
617
617
to be secure against attacks of
(a) a single person, (b) private agencies, (c) national security agencies:
year
1995
2000
2005
2010
2015
[Bruce Schneier: “Applied Cryptography”]
384
• American National Security Agency (NSA) recommends key lengths from
512 up to 1024 bits (so from 154 do 308 decimal digits) in their Digital Signature Standard (DSS).
integer length (number of digits)
length Lb of integer n with base b is given by
Lb = [ logb n]] + 1 = [ ln n/ ln b]] + 1
How Eve can crack the RSA cryptosystem?
By factorizing n:
If Eve can factorize n = pq then she can calculate φ = (p − 1)(q − 1)
and thus can calculate Bob’s private key d
Why is the RSA believed to be secure for large integers?
There is seemingly no efficient
(i.e., polynomial-time and polynomial-space)
classical algorithm for integer factorization.
RSA hypothesis (1978)
Any general method of cracking the RSA cryptosystem
which enables finding private key d from public key (n, e)
requires an efficient algorithm for integer factorization.
Note
There is still no proof of this hypothesis.
Shor’s algorithm
a Deutsch problem
on a DNA computer.
Adleman’s algorithm
constant f1(0) = 0
f2(0) = 1
balanced f3(0) = 0
f4(0) = 1
f1(1) = 0
f2(1) = 1
f3(1) = 1
f4(1) = 0
balanced and constant functions
Can you check whether a coin is real or fake by a single observation?
or
How many questions one should ask the oracle to check whether f is
constant or balanced?
Given an oracle (black box) that computes a Boolean function
f (x) : {0, 1} → {0, 1}
•
or
on a large quantum computer
•
386
would be insecure if we could implement efficiently:
RSA and majority of public-key cryptosystems
385
f
•
X
H
H
f
•
H
|−i
|0i
H
H
H
H
H
f
•
(−1)f (x)
Zk
f
•
f
•
=
Uf
•
=
388
|−i
Uf
|0i for constant f & |1i for balanced f
|δk1i (k = 0, 1)
|−i
|0i − |1i
different symbols for the same gate
or
|0i
main idea
|−i
|0i
or 1 query + 2 auxiliary operations
|1i
|0i
H
f (0) ⊕ f (1)
f
387
1
•
1 query + 3 auxiliary operations
quantum solution
1
0
2 queries + 1 auxiliary operation
classical solution
0
1
ψ1
H
H
x
y
Uf
x
y ⊕ f ( x)
ψ3
Deutsch algorithm
ψ2
Deutsch algorithm
1. prepare input state
|ψ1i = |0i|1i
H
2. apply Hadamard gates (create equal superposition)
|0i + |1i |0i − |1i
√
|ψ2i = Ĥ ⊗2|ψ1i = |+i|−i = √
2
2
1
= (|00i − |01i + |10i − |11i)
2
|xi|yi → Ûf → |xi|y ⊕ f (x)i
3. apply Ûf gate
|ψ3i = Ûf |ψ2i
ψ4
389
390
∼ Ûf |00i − Ûf |01i + Ûf |10i − Ûf |11i
= |0, 0 ⊕ f (0)i − |0, 1 ⊕ f (0)i + |1, 0 ⊕ f (1)i − |1, 1 ⊕ f (1)i
We neglect normalization thus sign ‘∼’ is used.
|0, 0 ⊕ 0i − |0, 1 ⊕ 0i + |1, 0 ⊕ 0i − |1, 1 ⊕ 0i
|0, 0i − |0, 1i + |1, 0i − |1, 1i
|0i(|0i − |1i) + |1i(|0i − |1i)
|+i|−i
case I: f (0) = f (1) = 0
|ψ i ∼
3
=
=
∼
|ψ3i ∼
=
=
∼
|0, 0 ⊕ 1i − |0, 1 ⊕ 1i + |1, 0 ⊕ 1i − |1, 1 ⊕ 1i
|0, 1i − |0, 0i + |1, 1i − |1, 0i
−|0i(|0i − |1i) − |1i(|0i − |1i)
−|+i|−i
case II: f (0) = f (1) = 1
so
f (0) = f (1) ⇒ |ψ3i = ±|+i|−i
391
392
|0, 0 ⊕ f (0)i − |0, 1 ⊕ f (0)i + |1, 0 ⊕ f (1)i − |1, 1 ⊕ f (1)i
|0, 0 ⊕ 0i − |0, 1 ⊕ 0i + |1, 0 ⊕ 1i − |1, 1 ⊕ 1i
|0, 0i − |0, 1i + |1, 1i − |1, 0i
|0i(|0i − |1i) + |1i(|1i − |0i)
|0i|−i − |1i|−i
|−i|−i
case III: f (0) = 0, f (1) = 1
|ψ i ∼
3
=
=
=
∼
∼
|0, 0 ⊕ f (0)i − |0, 1 ⊕ f (0)i + |1, 0 ⊕ f (1)i − |1, 1 ⊕ f (1)i
|0, 0 ⊕ 1i − |0, 1 ⊕ 1i + |1, 0 ⊕ 0i − |1, 1 ⊕ 0i
|0, 1i − |0, 0i + |1, 0i − |1, 1i
−|0i(|0i − |1i) − |1i(|1i − |0i)
−|−i|−i
case IV: f (0) = 1, f (1) = 0
|ψ i ∼
3
=
=
=
∼
so
f (0) 6= f (1) ⇒ |ψ3i = ±|−i|−i
393
H
•
H
|0i or |1i
Uf
H
is a common pattern in quantum algorithms
H
• lower qubit is an ancilla – it is not measured
and can be discarded after function evaluation
• a sequence of gates
Hadamard - function evaluation - Hadamard
f
|−i
|−i
• relative phases are introduced by the function evaluation
|0i
• top qubit undergoes a single-qubit interference
(−1)f (x)
Some remarks on Deutsch algorithm
QED
x = 0 ⇒ f (0) = f (1)
x = 1 ⇒ f (0) 6= f (1)
5. measure only the 1st register
|ψ4i = ±|f (0) ⊕ f (1)i|−i ≡ |xi|yi
thus we have for any case:
394
ˆ
f (0) 6= f (1) ⇒ |ψ4i = ±Ĥ ⊗ I|−i|−i
= ±|1i|−i = ±|f (0) ⊕ f (1)i|−i
cases III & IV:
ˆ
f (0) = f (1) ⇒ |ψ4i = ±Ĥ ⊗ I|+i|−i
= ±|0i|−i = ±|f (0) ⊕ f (1)i|−i
cases I & II:
ˆ 3i
|ψ4i = Ĥ ⊗ I|ψ
4. apply Hadamard gates
0
ψ1
H
H
⊗n
ψ2
y
x
Uf
y ⊕ f ( x)
x
ψ3
H
⊗n
|ψ2i = Ĥ ⊗(2+1)|ψ1i
= |+i⊗2|−i
|0i + |1i |0i + |1i |0i − |1i
√
√
√
=
2
2
2
1
|0i − |1i
= (|00i + |01i + |10i + |11i) √
2
2
∼ |000i + |010i + |100i + |110i
−(|001i + |011i + |101i + |111i)
|ψ1i = |0i⊗2|1i = |001i
1. prepare input state
Deutsch-Jozsa (DJ) algorithm for 3 qubits
1
⊗n
ψ4
Deutsch-Jozsa (DJ) algorithm
396
395
3. apply Ûf gate
|ψ3i ∼ |00, 0 ⊕ f (00)i + |01, 0 ⊕ f (01)i
+|10,
0 ⊕ f (10)i + |11, 0 ⊕ f (11)i
− |00, 1 ⊕ f (00)i + |01, 1 ⊕ f (01)i
+|10, 1 ⊕ f (10)i + |11, 1 ⊕ f (11)i
ˆ 3i
|ψ4i = Ĥ ⊗2 ⊗ I|ψ
*. let’s analyze all cases for different functions f
number of cases
Ncases = C04 + C44 + C24 = 1 + 1 + 6 = 8
where |{00, 01, 10, 11}| = 4
Cnm – binomial coefficient
case 1: (trivial)
f (00) = f (01) = f (10) = f (11) = 0
|ψ4i = (Iˆ ⊗ Ĥ)|ψ1i = |00−i
|ψ3i = Ûf |ψ2i
= Ûf |00i + |10i + |01i + |11i |0i − |1i
= |00i + |10i + |01i + |11i |1i − |0i
= −|ψ2i
f (00) = f (01) = f (10) = f (11) = 1
case 2: (trivial)
so
|ψ4i = −(Iˆ ⊗ Ĥ)|ψ1i = −|00−i
397
398
case 3:
f (00) = 0, f (01) = 0, f (10) = 1, f (11) = 1
= |000i + |010i + |101i + |111i
−(|001i + |011i + |100i + |110i)
= (|00i − |10i + |01i − |11i)(|0i − |1i)
= (|0i − |1i)(|0i + |1i)(|0i − |1i)
∼ |−, +, −i
399
−(|00, 1 ⊕ 0i + |01, 1 ⊕ 0i + |10, 1 ⊕ 1i + |11, 1 ⊕ 1i)
|ψ3i ∼ |00, 0 ⊕ 0i + |01, 0 ⊕ 0i + |10, 0 ⊕ 1i + |11, 0 ⊕ 1i
so
400
balanced
answer
constant
ˆ 3i ∼ (Ĥ ⊗2 ⊗ I)|−,
ˆ
|ψ4i = (Ĥ ⊗2 ⊗ I)|ψ
+, −i = |1, 0, −i
|x, yi
|0−i
|0−i
|2−i
|1−i
|3−i
|3−i
|1−i
|2−i
all cases in DJ algorithm for 3 qubits
x ≡ (x1, x2) ∼
= 2x1 + x2
f (x) is balanced
f (x) is constant
case f(00) f(01) f(10) f(11) |x1, x2, yi
1.
0
0
0
0
|00−i
2.
1
1
1
1
|00−i
3.
0
0
1
1
|10−i
4.
0
1
0
1
|01−i
5.
0
1
1
0
|11−i
6.
1
0
0
1
|11−i
7.
1
0
1
0
|01−i
8.
1
1
0
0
|10−i
⇒
⇒
5. measure x
if x = 0
if x > 0
x=0
|xi
x
X
=
x∈{0,1}n
X
=
x1=0 x2=0
1 X
1
X
...
xn =0
1
X
=
x=0
N
−1
X
1 X
Ĥ ⊗n|0i⊗n = Ĥ ⊗n|0i = √
|xi
N x
1
2
where N = 2n and x = 2nx1 + 2n−1x2 + . . . xn
in general
≡
x∈{0,1}
3
X
|0i + |1i |0i + |1i
√
√
Ĥ ⊗2|00i =
2
2
1
= (|00i + |01i + |10i + |11i)
2
1 X
≡
|xi
2
2
1. How to generate equal (equally-weighted) superposition
quantum Hadamard transform
end;
402
psi4=Norm*tensor_product(H,H,H)*psi3; % for clarity we add extra H gate
vector2ket(psi4)
psi3=ket([0,0,f00])+ket([0,1,f01])+ket([1,0,f10])+ket([1,1,f11])+...
-(ket([0,0,1-f00])+ket([0,1,1-f01])+ket([1,0,1-f10])+ket([1,1,1-f11]));
f00=f(n,1);f01=f(n,2);f10=f(n,3);f11=f(n,4);
H=[1 1;1 -1];Norm=1/8;
f=[
0 0 0 0
1 1 1 1
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
];
for n=1:8,
Matlab program
401
1
403

1
1 X
Ĥ|x1i = √
(−1)x1·z1 |z1i
2 z1=0
1
|x1x2...xni = √
N
z1 ,z2,...,zn=0
(−1)x1·z1+x2·z2+...+xn·zn |z1z2...zni
or compactly
|xi = |x1, x2, ..., xni
|zi = |z1, z2, ..., zn i
hx|zi ≡ x · z = x1z1 + x2z2 + ... + xnzn
is the scalar product.
and
x = (x1, x2, ..., xn),
z = (z1, z2, ..., zn),
1 X
Ĥ ⊗n|xi = √
(−1)hx|zi|zi
N z
where N = 2n is the Hilbert-space dimension
Ĥ
⊗n
1
X
1
1
1 XX
√
(−1)x1·z1+x2·z2 |z1z2i
=
2
2 z1 =0 z2=0
• thus in general for n qubits
where
1
|0i − |1i
1 X
√
(−1)1·z |zi
=√
2
2 z=0
|0i + |1i
1 X
√
(−1)0·z |zi
=√
2
2 z=0


1
1
X
X
1
1
Ĥ ⊗2|x1x2i =  √
(−1)x1·z1 |z1i  √
(−1)x2·z2 |z2i
2 z1 =0
2 z2=0
• for two qubits
combining together
Ĥ|1i =
Ĥ|0i =
404
2. How to compactly write the quantum Hadamard transform
• for a single qubit
DJ algorithm for n qubits
1. initialize (n + 1)-qubit state
|ψ1i = |0i⊗n|1i
|xi|−i
2. generate equal superposition by applying Hadamard gates
X
N x
1
|ψ2i = Ĥ ⊗(n+1)|ψ1i = (Ĥ ⊗n|0i⊗n)|−i = √
3. calculate f by applying Ûf gate
1 X
(−1)f (x)|xi|−i
|ψ3i = Ûf |ψ2i = √
N x
DJ algorithm for n qubits
4. apply Hadamard gate
ˆ 3i
|ψ4i = (Ĥ ⊗n ⊗ I)|ψ
is constant
is balanced
1 X
Ĥ ⊗n|xi = √
(−1)f (x) Ĥ ⊗n|xi |−i
N x
1 XX
=
(−1)hx|zi+f (x)|zi|−i
N x z
5. measure |zi :
z = 0 ⇒ f (x)
z > 0 ⇒ f (x)
405
406
quantum Fourier transform (QFT)
(N = 2n)
• QFT on group (Z2)n
= quantum Hadamard transform
1 X
QF T ′|xi = Ĥ ⊗n|xi = √
(−1)hx|yi|yi,
N y
group Z2 = the set {0, 1} with addition modulo 2 (⊕)
group (Z2)n = the set {0, 1}n with addition modulo 2 (⊕) bit by bit
• QFT on group ZN
2πi
1 X
exp
xy |yi
QF T |xi = √
N
N y
or just without the bold face:
N −1
1 X
2πi
xy |yi
QF T |xi = √
exp
N
N y=0
group ZN = the set {0, 1, ..., N − 1}n with addition modulo N (⊕)
Quantum Fourier Transform
N −1
1 X
2πi
|xi → QF T → √
exp
xy |yi
N
N z=0
or compactly
N −1
1 X xy
β |yi
|xi → QF T → √
N z=0
where
2πi
β = exp
N
QFT of arbitrary pure state of n qubits
N −1
N −1 N −1
X
1 X X
|ψi =
cx|xi → QF T → √
cxβ xy |yi
N x=0 y=0
x=0
407
408
= exp
2πi
N



 
2
1

1 
1 0
1−i
= √
QF T |ψi = QF T √ 




0
0
2
2 2
1+i
1
eiπ/4
1
e−iπ/4
|01i +
|11i
= √ |00i +
2
2
2
|ψi =
|00i + |11i
1
√
= √ [1001]T
2
2
π
=i
N = 22 = 4 ⇒ s = 3, β = exp i
2




1 1 1 1
1 1 1 1


1  1 β β2 β3 
 1  1 i −1 −i 
QF T = √ 
2
4
6  = 2


1
β
β
β
1
−1
1
−1
4
1 β3 β6 β9
1 −i −1 i
example: QFT of a Bell state
s = N − 1 = 2n − 1
β
where
QF T |ψi =
x=0
cx|xi

1 1 1 ... 1
c0

 
2
s
 1 β β ... β   c1 
 
1 
 1 β 2 β 4 ... β 2s   c2 
√
 
s+1
 .. .. .. . . . ..   .. 

 
s
2s
ss
1 β β ... β
cs

|ψi =
N
−1
X
Quantum Fourier Transform
in matrix representation
410
409
|x0i
|x1i
|x2i
|x3i
H
H
H
H
•
R2
QFT for n=4
|x0i
|x1i
|x2i
QFT for n=3
|x0i
|x1i
"
•
R2
#
H
•
R3
•
R2
•
R2
•
R4
•
R3
H
1 1 1
,
=√
2 1 −1
H
including SWAP
where
|x0i
|x1i
H
QFT for n=2
|x0i
QFT for n=1
H
•
R2
•
R2
× |y0i
H


•
R3
H
H
|y2i
|y1i
|y0i
•
R2
0 1 1

=√ 
i
2 0 exp 2π
k
2
× |y1i
Rk
|y1i
|y0i
Circuits for QFT
H
412
411
|y3i
|y2i
|y1i
|y0i
H
S
•
T
•
H
S
•
H
|y2i
|y1i
|y0i
Seemingly another implementation of QFT for n=3
|x2i
|x1i
|x0i
#
"
R̂3 = T̂
1
0
T̂ =
0 exp(iπ/4)
#
given in terms of the phase gate (S gate) and π/8 (sic!) gate (T gate)
"
10
Ŝ =
,
0 i
but it is exactly our scheme as
R̂2 = Ŝ,
413
414
problem for Grover’s algorithm
(the needle in a haystack problem)
assumptions
• given an oracle which calculates f (x):
f (x) = 1 if x = x0
f (x) = 0 if x 6= x0
• f (x) can be calculated using ordinary (reversible) computer code.
problem
find the element x0 in the least number of oracle queries.
average number of evaluations of f
classically – N/2
quantumly – ???
examples
• searching a phone book for a name and phone number
• searching a database for a key to crack a cryptosystem
oracle queries
1. oracle query in DJ algorithm
f
Ûf = |xi|yi 7→ |xi|x ⊕ yi
a note
and f (x) is encoded in the sign of the control register |xi.
(part III)
Grover’s algorithm for searching database
2. oracle query in Grover’s algorithm
415
416
in fact, the content of the target register |yi is unchanged in the DJ algorithm
Shor’s algorithm for integer factorization
Ûf |xi = (−1)f (x)|xi
or
Ûf = 1 − 2|x0ihx0|
it is just equivalent to the query in DJ algorithm
⇒
Ûf → Ûf′
1
|ψII i = √
|xi.
N x=0
N
−1
X
apply Hadamard gate Ĥ ⊗n to all qubits to get
II. Generation of equally-weighted superposition
|ψI i = |0i⊗n
prepare n qubits in state
I. Initialization
ALGORITHM:
find x0 among N = 2n elements encoded by n-qubits
PROBLEM:
n-Qubit Grover’s algorithm
|yi = |−i
Note:
|xi → Uf′ → (−1)f (x)|xi
2.
quantum phase-oracle Ûf′
|xi|yi → Uf → |xi|y ⊕ f (x)i
f : {0, 1} → {0, 1}
1. quantum oracle Ûf of a Boolean function
oracle = a black-box unitary operation
two types of quantum oracles
418
III. Grover iteration √
repeat the following subroutine Int(π N /4) times:
419
|2>
|3>
|2>
|3>
−1
|1>
−1
|0>
−0.5
0
0.5
1
−0.5
0
0.5
1
1. oracle phase flip of |x0i
−1
|1>
−1
|0>
−0.5
0
0.5
1
−0.5
0
0.5
1
initialization
|1>
|2>
|0>
|1>
|2>
2. Hadamard gate
|0>
Hadamard gate
|3>
|3>
two-qubit Grover’s search of |x0i = |2i
420
operations 2,3,4 are called the inversion about the average
Note:
4. Apply Hadamard gate Ĥ ⊗n.
f0 : |xi 7→ −(1 − 2δ0,x)|xi.
3. Flip the phase of all eigenstates |xi except |0i:
2. Apply Hadamard gate Ĥ ⊗n.
where δy,x is Kronecker delta.
fx0 : |xi 7→ (1 − 2δx0,x)|xi,
1. Call the oracle to flip the phase of eigenstate |x0i:
<x|ψ>
<x|ψ>
417
<x|ψ>
<x|ψ>
two-qubit Grover’s search of |x0i = |2i
4. Hadamard gate
3. phase flip of all |xi except |0i
III.1
f Xo
III.2
H
H
III.3
f0
421
III.4
H
H
0
−0.5
−1
1
0.5
0
−0.5
−1
1
0.5
0
−0.5
−1
|0>
|3>
|4>
|5>
initialization
|2>
|6>
|2>
|3>
|4>
|5>
|6>
|7>
|7>
0
−0.5
−1
1
0.5
0
−0.5
−1
|0>
|0>
|1>
|2>
|2>
|3>
|3>
|4>
|4>
|5>
|5>
|6>
|6>
I.2 Hadamard gate
|1>
|7>
|7>
three-qubit Grover’s search of |x0i = |2i
Hadamard gate
|1>
|1>
|2>
|3>
|4>
|5>
|6>
|7>
|1>
|2>
|3>
|4>
|5>
|6>
|7>
1
0.5
0
−0.5
−1
1
0.5
0
−0.5
−1
|0>
|0>
|1>
|2>
|2>
|3>
|3>
|4>
|4>
|5>
|5>
|6>
|6>
II.2 Hadamard gate
|1>
|7>
|7>
423
424
I.4 Hadamard gate - end of 1st cycle
I.3 phase flip of all |xi except |0i
|0>
I.1 oracle phase flip of
|1>
1
|0>
1
|3>
1
|2>
1
|1>
0.5
0
−0.5
−1
|0>
0.5
|3>
0.5
|2>
step II
H
H
|x0i
0.5
0
−0.5
|1>
- end of 1st Grover iteration
√
Nqueries = Int(π N /4)
where N = 2n = 4 so
Nqueries = 1
422
recommended number of repetitions (or queries)
−1
<x|ψ>
quantum circuit for two-qubit Grover’s search
qubit A
qubit B
|0>
II.1 oracle phase flip of |x0i
1
0.5
0
−0.5
−1
<x|ψ>
<x|ψ>
<x|ψ>
<x|ψ>
|0>
<x|ψ>
<x|ψ>
<x|ψ>
<x|ψ>
<x|ψ>
|0>
|1>
|2>
|3>
|4>
|5>
|6>
|7>
|1>
|2>
|3>
|4>
|5>
|6>
|7>
|0>
|1>
|2>
|3>
|4>
|5>
|0>
|1>
|2>
|3>
|4>
|5>
end of 5th cycle
|6>
|7>
|6>
|7>
−1
−0.5
0
0.5
1
|0>
|0>
|2>
|3>
|4>
|5>
|1>
|2>
|3>
|5>
426
|4>
end of 6th cycle
|1>
c_0
+
(k)
c5 |101i
+
(k)
c6 |110i
(k)
c_1
c_2
c_3
c_4
+
c_5
(k)
c7 |111i
(k)
+ c2 |010i + c3 |011i
after the kth Grover we get:
(k)
+c4 |100i
+
state
(k)
c1 |001i
c_6
c_7 ]
|7>
|7>
|ψ (6)i =[ -.01
|ψ (5)i =[ -.25
|ψ (4)i =[ -.38
|ψ i =[ -.31
-.01
-.25
-.38
-.31
-.09
.18
-.01
-.25
-.38
-.31
-.09
.18
(6)
c5
-.74
-.11
.57
.97
.88
-.01 -1.00
-.25
-.38
-.31
-.09
.18
= h110|ψ (6)i = .99989 · · · ≈ 1
after the 6th iteration we get
-.01
-.25
-.38
-.31
-.09
|ψ (2)i =[ -.09
(3)
.18
.18
|ψ i =[
(1)
-.01
-.25
-.38
-.31
-.09
.18
-.01 ]
-.25 ]
-.38 ]
-.31 ]
-.09 ]
.18 ]
__________________________________________________________
|ψ i =[
(k)
|ψ (k)i =
(k)
c0 |000i
|6>
|6>
end of 3rd cycle
425
numerical data for three-qubit Grover’s search of |x0i = |5i
−1
|0>
−1
0
0.5
1
−0.5
<x|ψ>
−0.5
0
0.5
1
end of 4th cycle
−1
−1
0
0.5
1
−1
0
0.5
1
II.4 gate Ĥ ends 2nd cycle
−0.5
<x|ψ>
−0.5
0
0.5
1
II.3 phase flip all |xi except |0i
<x|ψ>
−0.5
<x|ψ>
<x|ψ>
<x|ψ>
O(N
√ ) classical search algorithms
O( N )
Grover’s algorithm
How many oracle queries are required
to search an N -element database?
Mathematica: Sqrt[2ˆ56]/10.ˆ6/60 ֒→ 4.47 min
• quantum computer needs about 4 minutes
• classical computer needs 1,000 years
assuming that 1 mln keys per second can be checked than
What is the time required to find the correct DES key?
428
requires basically to search among N = 256 = 7.2 × 1016 possible keys.
Attack on DES
Grover’s algorithm and cryptography
- it is the Digital Signature Standard (DSS) of US Federal Government
- applied for digital signatures
- asymmetric
3. DSA (Digital Signature Algorithm) ↔ Shor’s algorithm
- applied for encryption/decryption and digital signatures
- asymmetric
2. RSA (Rivest-Shamir-Adleman) ↔ Shor’s algorithm
- used by US army
- American and international cryptographic standard
- applied for encryption and decryption
- symmetric
427
1. DES (Data Encryption Standard) ↔ Grover’s algorithm
The most popular cryptosystems and their cryptoanalysis
Quantum speedup
429
Grover’s search is quadratically faster than classical search
Note: It speeds up any kind of database search.
However the maximum advantage is gained in unsorted databases.
Can we find faster quantum-search algorithms?
Shor’s algorithm is exponentially faster than classical ones,
√ N iterations.
so it possible to find also a search algorithm that fast?
Optimality theorem:
Grover’s algorithm is optimal!
The search problem cannot be solved in less than O
⇒
430
But can we find an algorithm that would run, say, twice faster?
Possibly yes, but it is not the issue of the optimality theorem.
quantum entanglement and quantum speed-up
Q: Quantum entanglement is a key resource for QIP. But
do we need it for quantum speed-up in e.g. Grover’s algorithm?
A: "Entanglement is neither necessary for Grover’s algorithm
itself, nor for its efficiency." [Bhattacharya et al., 2002]
Q: Really?
A: Inversion about the average amplitude is a classical process.
Q: Can Grover’s algorithm be implemented classically?
A: Grover’s algorithm has already been experimentally
implemented using classical Fourier optics.
quantum entanglement and database size
Q: Is entanglement useful for Grover’s algorithm at all?
A: Yes. Lack of entanglement limits the database size,
which scales linearly with the beam diameter D
(or D2 for a 2D version)
⇒ number of qubits scales only as ∝ log2 D
assume D equal to the size of the universe, ∼ 1026m
⇒ it is equivalent to ∝ 86 qubits.
431
This limitation exists for any database containing classical information.
Q: Anyway, it seems that entanglement is not necessary
for quadratic speed-up.
But do we need it for exponential speed-up?
A: Most probably, yes.
432
How to implement Grover’s algorithm classically?
via classical optical interference
A classical implementation of Grover’s search
[Amsterdam’s experiment of Bhattacharya et al. (2002)]
• quantum probability amplitudes
֒→ a transverse laser beam profile
= a complex electric field amplitude E(x)
• quantum states, which label items of the database
֒→ continuous coordinate x
• sought item x0
֒→ narrow area around the “item position” x0
434
433
transmitted through one of the mirrors with transmission of 2%
֒→ after each roundtrip, a moving photodiode records light
• readout
Φ0(x′) = 0 elsewhere.
Φ0(x′) = φ if x′ is in a narrow area around 0
E(x) → E(x) exp[iΦ0(x′)]
(IAA = inversion about the average amplitude)
which imprints phase profile in the Fourier plane
֒→ phase plate (called the IAA plate) like the oracle plate
• gate Ûf (0) which marks |0i by phase flipping all |xi except |0i
specifically: T=13.5ns
֒→ a roundtrip of a pulse between the cavity mirrors
• each Grover’s iteration
Φx0 (x) = 0 elsewhere.
Φx0 (x) = φ if x is in a narrow area around x0
E(x) → E(x) exp[iΦx0 (x)]
which imprints a phase profile Φx0 (x) on the beam
֒→ phase plate (called the oracle plate)
• oracle Ûf (x0), which marks the item x0
by phase flipping all |xi except |x0i
with focal lengths f1 = 40 cm, f2 = 60 cm
specifically: spherical, achromatic doublet lenses
֒→ classical Fourier transform implemented by lenses
• quantum Hadamard transform
x
f1
Ff
F
f1
IAA
plate
L1
f2
F
L2
f2
M2
Iteration
order
2 2 2
1
_5 _3 _
pulses
out
0
1
⊗n
H
H
⊗n
y
x
Uf
y ⊕ f ( x)
x
H
H
⊗n
used in: • Simon algorithm
• Grover algorithm
• Bernstein-Vazirani algorithm
• and others
quantum DJ algorithm
is a common quantum subroutine
F – Fourier transform (instead of Hadamard transform) via lenses L1,2
436
IAA – inversion about the average amplitude, M1,2 – mirrors, f1,2 – focal lengths
E(x)
Oracle
plate
pulse
F
in M o
1
Amsterdam’s experiment of Bhattacharya et al. (2002)
classical implementation of Grover’s search
435
Simon’s problem (1994)
it is an oracle problem closely related to Shor’s algorithm
GIVEN:
f : {0, 1}n → {0, 1}n – a 2-to-1 function such that
∀x 6= y : f (x) = f (y) ⇔ y = x ⊕ r,
r – a fixed n-bit string called the function’s period
TASK:
find period r.
How many oracle queries are required
to find period r?
exponential number classically
polynomial number quantumly
Simon’s algorithm - step by step
• initial state
|ψ0i = |0i⊗n|0i⊗n ≡ |0i|0i
|xi|0i.
|xi|f (x)i.
x∈{0,1}n
• apply n Hadamard gates to the first n qubits:
X
1
|ψ1i = √
2n
X
where |xi ≡ |x1, x2, · · · , xni
• apply quantum oracle:
1
|ψ2i = √
2n
x∈{0,1}n
437
438
• measure the last n qubits and obtain a certain f (x′) ∈ {0, 1}n,
which yields the (n-qubit) state:
1
|ψ3i = √ (|x′i + |x′ ⊕ ri).
2n
• apply n Hadamard gates to the n qubits:
X ′
′
1
|ψ4i = √
(−1)x ·y + (−1)(x ⊕r)·y |yi
2n+1
y∈{0,1}n
X
′
1
(−1)x ·y |yi.
= √
2n−1 r·y=0
• measure |ψ4i to get y such that r · y = 0.
• repeat the above steps a polynomial number of times
to get, with high probability,
439
440
n linearly-independent values y = {y1, y2, . . . , yn} such that
y · r = r, which determines r.
multiplicative order of an integer
k = ord(x, n) ≡ ord(x)
Mathematica: MultiplicativeOrder[x,n]
1 = xk mod n
multiplicative order of x is the smallest integer k for which
example
ord(8, 21) =?
81 mod 21 = 8
ord(8, 21) = 2
82 mod 21 = 64 mod 21 = 64 − 3 · 21 = 1
⇒
441
a1 +
1
1
a2 + K+a
N
1
≡ {a0, a1, a2, K, aN }
442
4+ 2
= 1 + 1 1 ≡ {1, 4, 2}
Mathematica: ContinuedFraction[11/9] ֒→ {1,4,2}
2
11 = 1 + 2 = 1 + 1
9
9
9
Example: CF for r=11/9
Note: The procedure will halt iff r is rational.
1. split r into integer part [[r]] and fractional part f = r − [[r]]
2. stop if f = 0
3. calculate 1/f and return to step 1.
How to calculate CF for number r ?
a0 +
Continued fractions (CF)
• polynomial time using quantum phase estimation
(a part of Shor’s algorithm)
• exponential time using known classical algorithms
complexity of algorithms for order calculation
orders of all elements modulo 21
x = 1 2 4 5 8 10 11 13 16 17 19 20
ord(x,21) = 1 6 3 6 2 6 6 2 3 6 6 2
25 mod 21 = 32 mod 21 = 11
26 mod 21 = 22 mod 21 = 1
24 mod 21 = 16
23 mod 21 = 8
22 mod 21 = 4
21 mod 21 = 2
another example
3.245
4.082
12.250
4.000
12+ 14
4+ 1
1
= {3, 4, 12, 4}
= 0.245 1 / 0.245 = 4.082
= 0.082 1 / 0.082 = 12.250
= 0.250 1 / 0.250 = 4.000
= 0.000
3.245 = 3 +
-3
-4
- 12
-4
443
⊗m
⊗n
H
⊗n
y
x
U
a x mod N
4. Measurement
3. Quantum Fourier transform
2. Modular exponentiation
1. Hadamard transform
x
QFT −1
444
p
q is a convergent of CF and GCD(p, q) = 1
Basic steps of Shor’s algorithm:
1
0
⇒
Shor's factorization algorithm
p
q − x < 2q12
Useful criterion for Shor’s algorithm
is a truncated CF x = {a0, a1, ..., an , ..., aN }.
The nth convergent, denoted by {a0, a1, ..., an },
A convergent of CF
Mathematica: ContinuedFraction[3245/1000] ֒→ {3, 4, 12, 4}
3
4
12
4
Example: CF for 3.245
Shor’s algorithm to factorize an integer N
1. If N is even then return factor f = 2.
2. Test classically whether N 6= ab.
3. Choose randomly an integer a (1 < a < N ) and apply
the Euclidean algorithm to check whether
a and N are coprime, i.e. GCD(a, N ) = 1
If not then choose another a.
4. Prepare two registers:
register #2 has n2 = ⌈log2 N ⌉ qubits
(this is the number of qubits to store N );
register #1 has n1 = 2n2 qubits
445
446
(in optimized versions of the algorithm, n1 can be smaller).
5. Initialize both registers:
|ψ1i = |0i⊗n1 |1i⊗n2
6. Apply Hadamard gates to register # 1:
i.e. create an equally-weighted superposition
x=0
N1−1
1 X
|ψ2i = Ĥ ⊗n1 |ψ1i = √
|x, 15i
N1
where N1 = 2n1 .
7. Apply modular exponential gate to register # 2:
|ψ3i = Ûmod.exp|ψ2i
aR1 mod 15
R1
|0i
|1i
|2i
...
|N1 − 1i
R2 |a0 mod N i |a1 mod N i |a2 mod N i ... |aN1−1 mod N i
8. Measure register #2 to get some state |x′i:
|ψ4i = N 2hx′|ψ3i
where N is an (unimportant) renormalization constant.
9. Apply QFT† on the register # 1:
|ψ5i = QF T †|ψ4i
and measure it.
10. Apply the classical method of continued fractions
to find period r.
11. If r is even and r2 6= −1( mod N )
then calculate f =GCD(ar/2 ± 1, N )
If f 6= 1 or f 6= N then return f .
Otherwise repeat the algorithm.
How to factorize N=15?
15=3*5
447
448
1. choose x such that
1<x<N-1, GCD(x,N)=1
e.g. x=11
2. find multiplicative order r=ord(x):
111 112 113 114 115 116
11 1
11 1
11 1 (mod N) so r=2
3. find y: y2=1 mod N
since xr=1 mod N then y=xr/2=11
4. calculate GCD(y+1,N)=GCD(12,15)=3
GCD(y-1,N) =GCD(10,15)=5
so
⇒ OK
⇒ OK
⇒ OK
x=0
|y, 15i
aR1 mod 15
R1
|0i
|1i
|2i
...
|255i
0
1
2
255
R2 |7 mod N i |7 mod N i |7 mod N i ... |7
mod N i
= |1i
= |7i
= |49i ≡ |4i ...
= |13i
7. apply modular exponential gate to register # 2:
where N1 = 2n1 = 256
1
|ψ2i = Ĥ ⊗n1 |ψ1i = √
N1
NX
1−1
i.e. create equally-weighted superposition
6. apply Hadamard gates to register # 1:
450
Mathematica: 2∧∧1111 ֒→ 15
|ψ1i = |0i⊗n1 |1i⊗n2 = |00000000i|1111i ≡ |0i|15i
5. initialize both registers:
Mathematica: Ceiling[Log[2, 15]] ֒→ 4
register # 2: n2 = ⌈log2 N ⌉ = ⌈3.906⌉ = 4
register # 1: n1 = 2n2 = 8
4. find the required dimension of registers (number qubits):
GCD(7,15) = 1
Euclidean algorithm to check whether a and N are coprime:
3. let’s, e.g., choose a = 7 (unlucky choice) and apply
2. N 6= ab
1. N is odd
Example 1: Factorize N = 15
449
451
|3i
|13i
|7i
|13i
|11i ...
|13i ...
r=4
⇒
12. final test 3 × 5 = 15
⇒ OK
these are the sought factors :-)
GCD(13r/2 − 1, 15) = GCD(168, 15) = 3
= GCD(169 + 1, 15) = GCD(17 × 2 × 5, 3 × 5) = 5
GCD(13r/2 + 1, 15) = GCD(132 + 1, 15)
⇒
11. r is even and r2 6= −1 mod N
64
1
256 = 4
(which reduces to a trivial case now) to find the period r
10. apply the classical continued fraction method
|ψ5i = QF T †|ψ4i = 21 (|0i1 − |64i1 + |128i1 − |192i1)
9. apply QFT† on the register # 1:
|ψ4i = N 2h13|ψ3i = N (|3i1 + |7i1 + |11i1 + ...)
where N is a renormalization constant
R1
R2
for example, we get the state |13i:
8. measure register #2:
452
so
Mathematica: PowerMod[aˆx, 15]
R1 |0i |1i |2i |3i |4i |5i |6i |7i |8i |9i |10i |11i ...
R2 |1i |7i |4i |13i |1i |7i |4i |13i |1i |7i |4i |13i ...
a^3
8
12
4
5
6
13
2
9
10
11
3
7
14
a^4
1
6
1
10
6
1
1
6
10
1
6
1
1
...
2 4
3 9
4 1
5 10
6 6
7 4
8 4
9 6
10 10
11 1
12 9
13 4
14 1
8 1
12 6
4 1
5 10
6 6
13 1
2 1
9 6
10 10
11 1
3 6
7 1
14 1
unlucky choice
lucky choice
bad choice
2 4 8 1
3 9 12 6
4 1 4 1
5 10 5 10
6 6 6 6
7 4 13 1
8 4 2 1
9 6 9 6
10 10 10 10
11 1 11 1
12 9 3 6
13 4 7 1
14 1 14 1
⇒ long period
⇒ short period
⇒ GCD(x, N ) 6= 1
⇒
⇒
⇒
lucky, unlucky and bad choices of a modulo N=15
a^2
4
9
1
10
6
4
4
6
10
1
9
4
1
condition: GCD(a,N)=1
a
2
3
4
5
6
7
8
9
10
11
12
13
14
a=2, 7, 8, 13
a=4, 11, 14
a=3,5,6,9,10,12
2
3
4
5
6
7
8
9
10
11
12
13
14
⇒
⇒
⇒
453
a^14(mod N)
4 unlucky
9 wrong
1 lucky
10 wrong
6 wrong
4 unlucky
4 ...
6
10
1
9
4
1
harder factorization
easier factorization
excluded
454
Example 2: Factorize again N = 15 but for a = 11
1–2. ditto
⇒ OK
3. we choose a = 11 (lucky choice)
GCD(11,15) = 1
R1 |0i |1i |2i |3i |4i |5i |6i |7i |8i ...
R2 |1i |11i |1i |11i |1i |11i |1i |11i |1i ...
8. measure register #2:
|1i
|11i
|3i
|11i
|5i ...
|11i ...
for example, we get state |11i:
R1
R2
|ψ4i = N 2h11|ψ3i = N (|1i1 + |3i1 + |5i1 + ...)
where N is a renormalization constant
2
9. apply QFT† on the register # 1:
|ψ5i = QF T †|ψ4i = √1 (|0i1 + |128i1)
r=2
⇒
10. apply the classical continued fraction method
⇒
to find period r
128 = 1
256
2
11. r is even and r2 6= −1 mod N
455
456
GCD(11r/2 + 1, 15) = GCD(11 + 1, 15) = GCD(3× 4, 3× 5) = 3
GCD(11r/2-1, 15) = GCD(10, 15) = 5
which are the sought factors.
Actually, Shor has found his algorithm by generalizing Simon’s
algorithm, i.e. by replacing Simon’s Hadamard transforms (Fourier
transform over Z2n) by Fourier transform over ZN .
Shor’s algorithm resembles Simon’s algorithm.
Note 2:
Note 1:
= |11i
4–6. ditto
= |121i ≡ |1i ...
7. apply modular exponential gate to register # 2:
= |11i
To find period r, one usually has to apply
the complete classical continued fraction method.
aR1 mod 15
= |1i
R1
|0i
|1i
|2i
...
|255i
R2 |110 mod N i |111 mod N i |112 mod N i ... |11255 mod N i
so
ÛÛ
Û
Û
Û
Ú
Û
Û
Û
256
;
Ý
Ü
Û
Û
Ü
ã
á
à
Û
ÛÛ
Û
Û
Ú
â
â
Ý
ä
åæ
é
é
é
é
é
é
é
éé
é
1
3
FromContinuedFraction
⇒
Õ
ÙÙ
Ù
Ù
Ù
criterion 1, 3, x
Ø
⇒
PERIOD r 6= 3
False
çè
0, 3
1
a
1
65 1
= 0.079 · · · <
/ 2=
= 0.055 · · ·
− x = −
b
3 256 2b
18
1 1 a
= ≡
3 3 b
458
↑
|ψ1i
/
H
|0i
H
..
H
H
|1i⊗n
PERIOD r = 4
0, 3, 1
TRUE
460
459
↑
|ψ2i
Uf1
•
Uf2
•
Uf3
•
...
...
...
Ufn
•
↑
|ψ3i
nm
↑
|ψ4i
QF T †
Quantum circuit for Shor’s factorization algorithm
⇒
True
|0i
|0i
|0i
î
1st convergent of the continued fraction
î
criterion 1, 4, x
êë
{0, 3} ≡ 0 +
Û
65
256
1
4
⇒
1
1 a
= ≡
3 + 11 4 b
1
a
1
65 1
1
=
<
=
− x = −
b
4 256 256 2b2 32
{0, 3, 1} ≡ 0 +
2nd convergent of the continued fraction
î
0, 3, 1, 15, 4
=
;
457
ê
65
256
Û
í
Û
15+ 14
Û
îî
3 + 1+ 1 1
Û
1
Û
{0, 3, 1, 15, 4} ≡ 0 +
Û
2 b2
Û
0, 3, 1, 15, 4
1
ÛÛ
ContinuedFraction x
a
65
256
ß
x
b
InputForm
Þ
criterion a_, b_, x_ : Abs
x
65
Mathematica:
Example: Find period r from
Û
ìí
×Ø
ÕÖ
↑
|ψ5i
nm
nm
nm
..
nm
2:
1:
H
H
H
•
89:;
?>=<
•
89:;
?>=<
B
•
89:;
?>=<
C
•
89:;
?>=<
•
D
•
89:;
?>=<
E
•
89:;
?>=<
B
89:;
?>=<
•
F
•
89:;
?>=<
•
D
•
•
89:;
?>=<
89:;
?>=<
•
G H
•
•
89:;
?>=<
G
QF T †
QF T †
461
nm
nm
nm
462
nm
nm
nm
7-qubit circuit for Shor’s factorization of N = 15
3:
4:
5:
6:
7:
A
89:;
?>=<
•
optimized version of the 7-qubit circuit
for Shor’s factorization of N = 15
H
(gates C, E, F, H are removed)
1:
H
H
2:
3:
4:
5:
6:
7:
A
H
R2
•
•
R2
R3
•
H
H
•
R3
R2
•
•
R2
H
H
|x2i
|x1i
|x0i
|y2i
|y1i
|y0i
Quantum Fourier transform and its inverse for 3 qubits
QFT
|x2i
|x1i
|x0i
H
QFT−1=QFT†
|y2i
|y1i
|y0i
where the rotations are R2 = R(900), R3 = R(450 )
Outline
1. Generalized projective measurements
2. Positive operator valued measure (POVM)
3. Kraus representation
4. Damping channels for a single qubit
5. Imperfect photocount detectors
6. Bell-state and GHZ-state analyzers
463
464
7. Fidelity and other measures of quality of the state generation
8. Entanglement measures
466
where
m,n,M,N
X
(|µihµ|)mM,nN (ρsys)mn(ρaux)MN
M,N
X
(|µihµ|)mM,nN (ρaux)MN
Pµ = Tr(Aµρsys)
(Aµ)mn =
which be rewritten as
Pµ = Tr[|µihµ|(ρsys ⊗ ρaux)] =
• probability of outcome µ:
Different outcomes correspond to orthogonal and complete projectors |µihµ|
A maximal test is then performed in the combined Hilbert space K.
The combined, uncorrelated state of the original quantum system and
the ancilla is
(ρsys ⊗ ρaux)mM,nN = (ρsys)mn (ρaux)MN
• Let’s prepare an auxiliary quantum system (ancilla) in a known
state ρ̂aux
Generalized projective measurement
do not commute and are therefore not simultaneously observable
• measurement operators, corresponding to nonorthogonal states,
also called the projection valued (PV) measure
|µihµ| – measurement operator or projection operator
= probability of the measurement outcome µ
Pµ – probability that the system is in state |µi
where
Pµ = Tr{|µihµ|ρsys}
is represented by complete, orthonormal set of states |µi
von Neumann-type projective measurement
465
|ψi = α|ui + β|vi
general state
Can you distinguish whether qubit is in state |ui or |vi?
where θ the angle between polarization vectors
hu|vi = cos θ
non-orthogonal linear polarization states |ui and |vi
Cryptographic example of POVM
(at leat sometimes)
How to distinguish conclusively between
two non-orthogonal states?
468
• POVM allow extraction of more mutual information that can the
usual von Neumann-type projective measurement
nonorthogonal states.
• POVM allow the possibility of measurement outcomes associated with
• the number of available outcomes may differ from the number of available preparations (of a given state) and the dimension of the Hilbert space.
Advantages of POVM over PV measures
Pµ = Tr(Aµρsys)
• probability that a quantum system is in a particular state is given by the
expectation value of the POVM operator corresponding to that state
µ
which are positive Hermitian operators acting on original Hilbert space
X
Aµ = 1
positive operator valued measures (POVM)
= set of Aµ’s
467
⇒
471
Neumark’s theorem
469
POVM operators
1 − |vihv|
1 + hu|vi
inconclusive measurement
Av =
are not projective measurements
1 − |uihu|
,
1 + hu|vi
⇒
detector Dv will not click
470
= |α + β|2 cos θ
⇒
detector Du will not click
Du
M
Dv
analogy
a pure state of the bipartite system AB may behave like
a mixed states when we observe subsystem A alone,
similarly
an orthonormal measurement of the system AB may be
a nonorthonormal POVM on A alone
Kraus representation
= operator sum representation
basic idea
an elegant representation to describe open system dynamics
=
TrE
{ρSE (t)}
• reduced density matrix
ρS (t)
X
h{ki}|ρSE (t)|{ki}i
{ki }
where |{ki}i ≡ |k1 · · · ki · · · i = Πi ⊗ |kii
=
ρ(0) = ρS (0) ⊗ ρE (0) −→ ρSE (t) = U (t)ρS (0) ⊗ ρE (0)U †(t)
• So let’s analyze evolution of both the system and environment:
• The final state of an open system cannot be described by a unitary transformation of the initial state.
472
It is also generally thought that a POVM belongs to the most general test to
which a quantum system may be subjected.
most general measurement operator
There always exists an experimentally realizable procedure generating any
desired POVM represented by given matrices Aµ
physical implication
One can extend the Hilbert space H of states, in which the Aµ are defined, in
such a way that there exists, in the extended space K, a set of orthogonal and
complete projectors |µihµ| such that Aµ is the result of projecting |µihµ| from
K into H.
Au =
A? = 1 − Au − Av
Pi = hψ|Ai|ψi
= |β|2(1 − cos θ),
Pv = 0
⇒
P?
probabilities of measurements
= |α|2(1 − cos θ),
Pv
⇒
Pu = 0
Pu
|ψi = |ui
⇒
special input states
|ψi = |vi
an optical implementation of POVM
D?
BS
HWP
4
(NOT)
π
optical implementation of POVM
BS
PBS
π
HWP
8
(H)
ψ
is an orthonormal basis of the environment Hilbert space
{ki }
†
′
i ki
=k
USE
p
A0 = 1 − p,
Aµ =
E hµ|USE |0iE
A1 = A2 = A3 =
so
474
473
p
3
r h
i
p
σx|ψiS ⊗|1iE +σy |ψiS ⊗|2iE +σz |ψiS ⊗|3iE
3
Kraus operators
p
: |ψiS ⊗|0iE → 1 − p|ψiS ⊗|0iE +
evolution of single qubit in depolarizing channel
and error (bit flip and/or phase flip) occurs with probability p
i.e. the qubit remains intact with probability 1 − p
depolarizing channel
Example of Kraus representation
it seems extremely difficult to use K.r. if the environment is at T > 0
• drawback
k
A†k (t)Ak (t) = I
P
h{ki}|U (t)|{0}i
Ak (t)ρS (0)Ak (t)
stands for summation under the condition
• completeness relation
X
where
P′
Ak (t) =
k
X
k=0
in terms of Kraus operators
ρS (t) =
∞
X
• alternatively, the evolution can be given in
the so-called Kraus representation
|1i → −|1i
σ̂z ≡
|1i → −i|0i
|ψi → σ̂y |ψi
|0i → i|1i,
|ψi → σ̂z |ψi
|0i → |0i,
|1i → |0i
|ψi → σ̂x|ψi
|0i → |1i,
where the σk are the Pauli operators
0 1
0 −i
σ̂x ≡
, σ̂y ≡
,
1 0
i 0
3. both errors
†
p
Fµ
1 0
0 −1
and Aµ are Kraus operators
2. phase flip error
1. bit flip error
Fµρs
where
µ
Xp
Fµ = A µ A µ
ρs →
modifies a density matrix as follows
three types of errors
POVM
POVM from Kraus representation
476
475
+
√
p|0iS |1iE
damping channels for a single qubit
1. amplitude-damping channel
→
p|1iS |0iE
|0iS |0iE → |0iS |0iE
p
1−
|1iS |0iE
2. phase-damping channel
p
√
|0iS |0iE → 1 − p|0iS |0iE + p|0iS |1iE
p
√
1 − p|1iS |0iE + p|1iS |2iE
|1iS |0iE →
^
^
b2
r
b^ 3
b^ 4
N4 photons
N3 photons
N2 photons
Φ
b̂1
b̂
2
b̂3
b̂4
ρˆ 1
N4
477
photons
N3 photons
N2 photons
478
i
ph
σx|ψiS |1iE + σy |ψiS |2iE + σz |ψiS |3iE
3
• it is a „caricature” model of decoherence in real systems
• no bit flip in system!
p
1 − p|ψiS |0iE +
3. depolarizing channel
|ψiS |0iE →
a^ 2
a^3
b1
ρˆ 1
2) (b3) (b4 )
Tr(b2,b3,b4) Π̂(b
N2 Π̂N3 Π̂N4 |ΦihΦ|
POVM for the kth imperfect photocounter detecting Nk photons
four-mode state before the measurements
single-mode state after the measurements
renormalization constant
ρ̂1 = N
a^ 4
conditional measurements using imperfect photocounters
a^ 1
(b )
Π̂Nkk
|Φi
ρ̂1
N
•
479
POVM for photocount detectors (I)
Perfectly-Resolving Photon Counter
m=0 n=0
( No clicks )
480
η − inefficiency
ν − dark count rate
( N−n)
∞ N
−ν ν
ˆ
η n (1−η )m−n C mn m m
Π
=
e
∑∑
(N − n)!
N
∞
m =0
ˆ c = 1− Π
ˆc
Π
1
0
single-photon counter
( 1 click )
POVM for photocount detectors (II)
• Dark count rate ~ 100 − 1000 s − 1
( Click )
ˆ c = ∑ e −ν (1 − η ) m m m
Π
0
• Conventional Photon Counter (CPC)
•
m=0
∞
ˆ vl = ∑ e −ν (1 − η ) m m m ( no clicks )
Π
0
( 2 clicks )
10 4 s −1
(1−n)
∞ 1
−ν ν
ˆ vl
Π
ηn (1−η )m−n C mn m m
1 = ∑∑ e
(1− n)!
m=0 n=0
ˆ vl = 1 − Π
ˆ vl − Π
ˆ vl
Π
2
0
1
• high dark count rate ~
for detection of single photon
η < 25%
⇒
⇒
basically noise free
useless for QC
for detection of single photon
⇒
small resolution time (for detection of two photons)
482
•
•
•
2
|00i ± |11i
√
|Φ±i =
How to distinguish experimentally the GHZ states?
Is it possible to unambiguously discriminate between
all the Bell states using only linear optics?
Do they exist perfect linear Bell-state analyzers?
2
|01i ± |10i
√
|Ψ±i =
How to discriminate between the optical Bell states?
Bell-state and GHZ-state analyzers
τ ∼ 2ns
η ∼ 47% for detection of two photons
η ∼ 88%
with noise free avalanche photomultiplication
• VLPC = visible light photon counters
Fnoise = 1
η ∼ 70 − 80%
• SSPM = solid state photomultiplier
η ∼ 6% for detection of two photons
• PMT = photomultiplier tube
relatively high noise
⇒
Fnoise ≥ 2
η ∼ 70 − 80%
• Si APD = Si avalanche photodiode (working in Geiger mode)
Photocounters (photon count detectors)
481
A
A
PBSAB
HWP
PBS 1
BS
PBS 1
B
B
D3
D4
Ψ + ← (1, 3) o r ( 2 , 4 )
D1
PBS 2
D3
D4
Φ − ← (1, 3) o r ( 2 , 4 )
Ψ ± ← (1,1), ( 2, 2 ), (3, 3), o r ( 4 , 4 )
HWP
D2
Φ + ← (1, 4 ) o r ( 2 , 3)
484
Φ ± ← (1,1), ( 2 , 2 ), (3, 3), o r ( 4 , 4 )
PBS 2
D2
Ψ − ← (1, 4 ) o r ( 2 , 3)
Bell-state analyzer (2)
D1
Bell-state analyzer (1)
483
485
What Bell states can uniquely be distinguished in
the (Pan-Zeilinger) analyzer?
notation: particles A,B; modes 1,2
1. general input state
2. state after PBSAB
if horizontal (vertical) polarization component is transmitted (reflected) then
PBS1
|ψini = α|HAi|HB i + β|HAi|VB i + γ|VA i|HB i + δ|VAi|VB i
|ψi = α|HA2i|HB1i + β|HA2i|VB2i + γ|VA1 i|HB1i + δ|VA1i|VB2i
A
B
D3
HWP
D4
D5
PBS3
GHZ-state analyzer
D1
D2
PBS2
HWP
PBSBC
C
polarization GHZ states
1 |Φ±i = √ |Hi|Hi|Hi + |V i|V i|V i
2
1 |Ψ1±i = √ |V i|Hi|Hi + |Hi|V i|V i
2
1 |Ψ2±i = √ |Hi|V i|Hi + |V i|Hi|V i
2
1 |Ψ3±i = √ |Hi|Hi|V i + |V i|V i|Hi
2
• the Pan-Zeilinger analyzer discriminates between
only two (|Φ±i) among 2N GHZ states
PBSAB
HWP
3. indistinguishability of photons
implies that we can omit subscripts A, B:
D4 ≡ DH2
486
|ψi = α|H1i|H2i + β|H2i|V2i + γ|V1i|H1i + δ|V1i|V2i
1
1
1
1
+
−
+
−
= √ (α + δ)|Φtwo
i + √ (α − δ)|Φtwo
i + √ (β + γ)|Ψone
i + √ (β − γ)|Ψone
i
2
2
2
2
where
1 1 ±
±
i = √ |H1i|V1i ± |V2i|H2i
|Φtwo
i = √ |H1i|H2i ± |V1i|V2i , |Ψone
2
2
4. state after quarter-wave plates and detection
D3 ≡ DV 2 ,
1
|Vii → √ (|Hii − |Vii)
2
• quarter-wave plates change polarizations as follows
D2 ≡ DV 1 ,
1
|Hii → √ (|Hii + |Vii),
2
• notation: D1 ≡ DH1,
• case 1:
1 +
+
|Φtwo
i → |Φtwo
i = √ |H1i|H2i + |V1i|V2i
2
then photons are detected in D1 and D4 or D2 and D3
1
|H1i|H1i ± |H2i|H2i − |V1i|V1i ∓ |V2i|V2i
2
• case 2:
1 −
+
|Φtwo
i
→
|Ψ
|H1i|V2i + |V1i|H2i
twoi = √
2
then photons are detected in D1 and D3 or D2 and D4
• cases 3,4:
+
|Ψone
i→
then both photons are detected in either D1, or D2, or D3, or D4
487
488
D6
Dn
…
490
Linear optical elements can provide any arbitrary unitary mapping
only over creation operators, not over a general input state.
Why no-go?
Within this subclass it is not possible to discriminate unambiguously four
equiprobable Bell states with a probability higher than 50%.
Calsamiglia-Lütkenhaus theorem:
There is no perfect Bell state linear analyzer on two qubits in polarization
entanglement without conditional measurements.
No-go theorem:
Do they exist perfect Bell-state linear analyzers?
[Calsamiglia i Lütkenhaus, 2000]
Uˆ2(N1)
N1
Uˆ3(N2)
N2
…
…
…
N3
QND
HWP
on
(P)BS – (polarization) beam splitter.
HWP – half-wave plate,
CCL – classical coincidence & logic,
Key: QND – quantum non-demolition detector,
PBS
off
BS
492
off
on
Bell state QND CCL
|Ψ+i
off off
|Ψ−i
off
on
|Φ+i
on
off
|Φ−i
on
on
CCL
nonlinear Mach-Zehnder interferometer
based on
[Paris et al. (2000)]
Complete nonlinear Bell-state analyzer
Bell state
source
491
1
→1
1 + f (Nph, Ncm)
where Ncm – number of conditional measurements
Nph – number of entangled photons in auxiliary modes
auxiliary
modes
Psuccess = 1 −
…
1
2
…
auxiliary
modes
Û1
…
…
Psuccess ≤
auxiliary
modes
input
modes
- based on conditional measurements
Complete linear Bell-state analyzer
…
probability to discriminate unambiguously between all the Bell states is
bˆn
…
vacuum
auxiliary
aˆn
modes
Not complete linear Bell-state analyzer
â1
b̂1
D1
â2
input
b̂
2
modes â3
D2
â4
Û
â5
489
…
signal
α
0
BS
QND – Fock filter
Kerr
PS
BS
Key: Kerr nonlinear medium described by 3rd order susceptibility χ(3)
|αi – strong coherent field
PS – phase shifter
'
%6
0
'
493
494
QND measurement does not destroy coherence of the signal
state, but only adds extra phase, which can easily be corrected.
.HUU
PHGLXP
Kerr effect and QND
%6
0
θ
YDFXXP
SUREH
VLJQDO
on
off
Measures of quality of the state generation
1. Fidelity
= Uhlmann’s transition probability for mixed states
q
2
p
p
Tr
ρ̂thρ̂exp ρ̂th
F (ρ̂exp, ρ̂th) =
ρ̂exp – density matrix for the experimentally generated state
ρ̂th – theoretically predicted density matrix
2. Bures distance
a measure of discrepancy between ρ̂exp and ρ̂th:
q
DB (ρ̂exp||ρ̂th) = 2 − 2 F (ρ̂exp, ρ̂th)
• it satisfies the usual metric properties including symmetry
DB (ρ̂exp||ρ̂th) = DB (ρ̂th||ρ̂exp)
495
496
3. Quantum relative entropy = Kullback-Leibler ‘distance’
S(σ||ρ) = Tr (σ lg σ − σ lg ρ)
it is not a true metric since
S(σ||ρ) 6= S(ρ||σ)
4. MaxLik parameters
- used by us in the tomographic reconstruction of physical density matrices
5. Relative entropy of entanglement
minimum of the quantum relative entropy
over set D of all separable states ρ:
E (σ ) = minρ∈D S (σ||ρ) = S (σ||ρ̄)
ρ̄ is the separable state closest to σ.
i pi E(ρi )
≥ E(
i p i ρi )
P
entanglement of distillation ED
•
= EC = ER = ED
EF ≥ EC ≥ ER ≥ ED
for mixed states
EF
for pure states
• ···
relative entropy of entanglement ER
entanglement cost EC = limn→∞ EF (ρn
entanglement of formation EF
•
•
•
⊗n
Measures of entanglement
)
498
C5.∗ E(ρ) should reduce to the entropy of entanglement for a pure state
i.e., mixing cannot increase E(ρ)
P
C4. Entanglement is convex under discarding information:
C3. LOCC operations cannot increase E(ρ)
C2. local unitary transformations UA ⊗ UB do not change E(ρ)
E(ρ) = 1 for a Bell state
E(ρ) = 0 for an unentangled state
C1. E(ρ) ≥ 0
ρ – the density matrix of a given state
Criteria for a good entanglement measure
497
1
2 [1 +
√
λ1 −
λ2 −
maximally
entangled pairs
distillation
LOCC
LOCC
formation
partially
entangled pairs
asymptotic conversion ratios for
entanglement of formation and distillation
500
“C(ρ) is a measure of the entanglement of formation in its own right.” [Wootters]
where λi’s are in nonincreasing order the eigenvalues of
ρ(σy ⊗ σy )ρ∗(σy ⊗ σy )
C(ρ) = max{0,
√
p
2
1 − C (ρ)]
in terms of the concurrence
EF (ρ) = H
√
√
λ3 − λ4 }
i piE(|ψi ihψi |)
P
For two qubits [Wootters, PRL’98]
Special case
EF (ρ) = minpi,ψi
It is the minimized
P average entanglement of any ensemble of pure states |ψii
realizing ρ = i pi|ψiihψi|:
[Bennett, DiVincenzo, Smolin, and Wootters, PRA’96]
Entanglement of formation
499
Negativity
501
[Życzkowski et al. PRA’98, Eisert, Plenio JMO’99, Vidal, Werner, PRA’02]
a quantitative version of the Peres-Horodecki entanglement
criterion [Peres, PRL’96, Horodecki et al., PLA’96]
N (ρ) = max{0, −2 mini µi}
P
N (ρ) = max{0, −2 µi}
where µi eigenvalues of the partial transpose of ρ
Logarithmic negativity
EN (ρ) = log2[N (ρ) + 1]
a measure of the PPT entanglement cost
[Audenaert et al. PRL’03, Ishizaka PRA’04]
502
EN gives upper bounds on the teleportation capacity and the
entanglement of distillation ED [Vidal, Werner PRA’02]
σ
*
S (σ || ρ sep
)
geometric measure of entanglement
entangled states
ρ
*
sep
separable states
Relative entropy of entanglement (REE)
[Vedral, Plenio, Jacobs, Knight, PRA’1997]
E (σ ) = inf ρ∈D S (σ||ρ) = S (σ||ρ∗)
ρ∗ – the closest separable state to σ
quantum relative entropy
or a quantum Kullback-Leibler distance
S (σ||ρ) = Tr (σ log σ − σ log ρ)
NOTE:
S(σ||ρ) is a “distance” between σ and ρ
• but it is not a true metric:
it is neither symmetric nor satisfies the triangle inequality
• it is not a unique measure of the distance:
503
504
see also Bures measure with the Uhlmann transition probability (or fidelity)
p
√ √
S ′(σ||ρ) = 2 − 2 F (σ, ρ) with F (σ, ρ) = [T r( ρσ ρ)1/2]2
Bell-inequality violation
3
P
n,m=1
I⊗ I + r · σ ⊗ I + I ⊗ s · σ +
tnm σn ⊗ σm
!
For two qubits the Bell inequality due to Clauser, Horne, Shimony and Holt (CHSH) [PRL’69]:
|Tr (ρ BCHSH)| ≤ 2
where Bell operator is
BCHSH = a · σ ⊗ (b + b′) · σ + a′ · σ ⊗ (b − b′) · σ
1
4
and arbitrary ρ in Hilbert-Schmidt basis is
ρ=
with tnm = Tr (ρ σn ⊗ σm)
Horodecki et al. [PLA’95] showed that
p
M (ρ)
maxBCHSH Tr (ρ BCHSH) = 2
where M (ρ) = maxj<k {uj + uk };
uj are eigenvalues of Uρ = TρT Tρ; Tρ = [tnm]3×3
the maximal violation of Bell inequality
state ρ admits local hidden variable model
max {0, M (ρ) − 1 }
⇒
⇒
p
−
506
for
⇒ Entanglement without Bell inequality violation for p ∈
states are entangled iff 1/3 < p ≤ 1.
1 √1
3, 2
E
0≤p≤1
n
o
C(ρW ) = N (ρW ) = max 0, 12 (3p − 1)
3. Concurrence & negativity
thus the
√ Werner state violates the Bell inequality
iff 1/ 2 < p ≤ 1
B(ρW ) = max{0, 2p2 − 1}1/2
2. Degree of Bell-inequality violation
ρW = p|Ψ−ihΨ−| + 1−p
4 I ⊗I
Mixture of the maximally entangled state (singlet) |Ψ i and the (separable) maximally mixed state [Werner, PRA’89]:
1. Definition
Two-qubit Werner states
C (Ψ) = N (Ψ) = B (Ψ) = 2|c00c11 − c01c10|
Entanglement measures for two-qubit pure states
|Ψi = c00|00i + c01|01i + c10|10i + c11|11i
B(ρ) = 1
B(ρ) = 0
then
B(ρ) ≡
• useful parameter
Horodecki theorem: the Bell inequality is violated iff M (ρ) > 1
Degree of the Bell-inequality violation (BIV)
505
508
No, if we want to define entanglement measures for
examining the problems of how to prepare and use the entanglement.
Can we avoid the state-ordering ambiguity?
Yes, as these incomparable states cannot be
transformed to each other with unit efficiency by LOCC.
Is it physically reasonable?
It is implied by the requirements of equivalence and continuity of the measures
on pure states
Why is it so?
all good asymptotic entanglement measures,
which reduce to the entropy of entanglement for pure states,
are either equivalent or do not impose the same state ordering
Virmani-Plenio theorem:
ordering of states can depend on the applied measures of entanglement
Eisert-Plenio conclusion from Monte Carlo simulations:
Is it a necessary condition for consistency
of entanglement measures?
Strange, but no.
Eisert and Plenio [J. Mod. Opt. 1999] - numerical example
Can this condition be violated?
Yes.
QUESTIONS:
E ′(ρ1) < E ′(ρ2) ⇔ E ′′(ρ1) < E ′′(ρ2)
Entanglement measures should impose
the same ordering of states
CONJECTURE
507
Questions
509
1. Can we find analytical examples of two-qubit states for
which entanglement measures impose different orderings ?
2. Are there mixed states more entangled than pure states ???
3
Bob
ABCD
U
EPR source
out
510
3. Can we find a physical process manifesting the different
orderings ?
ideal teleportation
Alice
ABCD
2
Bennett, Brassard, Crepeau, Jozsa, Peres, Wootters (1993)
BSM
1
in
teleportation fidelity F = 〈 in | ρ out | in 〉 = 1
imperfect BSM
1
in
Bob
ABCD
out
511
512
ρ
imperfect U
imperfect EPR state
3
real teleportation
Alice
ABCD
2
?
E (ρ ') > E (ρ '') ⇒ F (ρ ') > F (ρ '')
1
0.8
0.6
0.4
0.2
0
0
0.8
pure & Werner states
Horodecki states
0.4
0.6
concurrence
Numerical simulations of 5 × 104 states.
0.2
1
Upper & lower bounds for negativity vs concurrence
negativity
= C (ρ) ⇐⇒
Examples of the minimum-negativity states
states have the minimum negativity for a given concurrence
⇒
N (ρW ) = C (ρW ) = max{0, 3p−1
2 }
2(a) Werner state ρW (p)
2. Bell diagonal states
= p|ψ−ihψ−| + 1−p
4 I ⊗I
Examples of the maximum-negativity states
1. pure states
•
concurrence C(ρH ) = p
p
negativity N (ρH ) = (1 − p)2 + p2 − (1 − p)
⇒
1. Horodecki state ρH (p) = p|ψ−ihψ−| + (1 − p)|00ih00|
2. Horodecki-like state ρ′H (p) = p|ψ−ihψ−| + (1 − p)|ψ ′ihψ ′ |
where |ψ ′i = 12 (|00i + |01i + |10i + |11i)
•
⇒
two eigenvalues are vanishing and
the other two correspond to eigenvectors, which are
a Bell state and separable state orthogonal to it
= fC (ρ) ⇐⇒
states have the maximum negativity for a given concurrence
2. N (ρ)
⇒
the eigenvector corresponds to
the negative eigenvalue of ρTA is a Bell state
1. N (ρ)
514
[1 − C (ρ)]2 + C 2(ρ) + C (ρ) − 1 ≡ fC (ρ)
Structure of the extremal states:
C (ρ) ≥ N (ρ) ≥
q
[Verstraete et al. JPA’01]
Upper & lower bounds for negativity vs concurrence
513
0.2
Y
0.4
X
0.6
concurrence
V
Z
P
R
0.8
T
O
S
1
515
and the Werner states:
√
ρY = ρW ( 2/3),
ρZ = ρW (2/3),
√
ρV = ρW (1/3 + 2/6)
√
√
or pure state |Ψ(p)i = p |01i + 1 − p |10i:
p
√
ρY = |Ψ(p)ihΨ(p)| for p = 1/2 ± 1 + 2 2/4,
√
ρZ for p = 1/2 ± 3/4,
√
ρV for p = 1/2 ± 14/8
Let us choose the Horodecki state:
ρX = ρH (1/2)
Explicit examples of states extremely violating
the ordering condition
516
Two states, when one corresponds to a point O (or X) and the other to any
other point in yellow regions, exhibit different state orderings for N and C.
0
0
0.2
0.4
0.6
0.8
1
Regions of different state orderings
for negativity N and concurrence C
negativity
0.6
0.8
1
1
0.8
0.6
0.4
0.2
0
0
(b)
0.2
0.6
concurrence
0.4
0.8
1
1
0.8
0.6
0.4
0.2
0
0
(c)
0.2
0.6
concurrence
0.4
0.8
517
518
1
States specifically violating the ordering condition
1
(a)
0.4
(a) states with constant negativity
(b) states with constant concurrence
(c) states for which C(ρ1) − C(ρ2) = −[N (ρ1) − N (ρ2)]
concurrence
negativity
0.8
0.6
0.2
negativity
ρ′ = ρ̄(p, q ′) for q ′ =
ρ′′ =
ρ′′′ = ρ̄(p, q ′′′) for q ′′′ = 1 +
[N (ρ)+C(ρ)+1−2p]2−1
2p(1−p)
3. states giving exactly opposite predictions:
ρ̄(C0, q)
2. states with the same concurrence C0:
N0 [N0+2(1−p)]−p2
2p(1−p)
• Three classes of states:
1. states with the same negativity N0:
then
p
N (ρ̄(p, q)) = 1 − 2p(1 − p)(1 − q) − (1 − p)
C(ρ̄(p, q)) = p
States specifically violating the ordering condition
ρ̄(p, q ) = p|ψ ihψ | + (1 − p)|ψq ihψq |
−
−
p
√
1 − q|00i + q|01i
with |ψq i =
0
0
0.2
0.4
negativity
519
REE vs concurrence and negativity
REE for Bell diagonal (including Werner) states
pi|ψAiihψAi| ⊗ |ψBiihψBi|
520
EW (C) = EW (N ) = 12 [(1 + C) log(1 + C) + (1 − C) log(1 − C) ]
REE for pure states
√
EP (C) = EP (N ) = H 12 [1 + 1 − C 2]
REE for Horodecki states
σH = C|ψ−ihψ−| + (1 − C)|00ih00|
EH (C) = (C − 2) log(1 − C/2) + (1 − C) log(1 − C)
p
2N (1 + N ) − 1)
EH (C =
Numerics for two-qubit REE:
[Vedral and Plenio, PRA’1998]
Caratheodory’s theorem:
Any state in D can be decomposed into a sum of
at most (dim(HA) × dim(HB ))2 products of pure states.
16
X
i=1
there are at most 15 + 16 × 4 = 79 real parameters
ρ=
Thus, any disentangled 2 qubit state can be given by
⇒
How to minimize S (σ||ρ) over 79 parameters?
S(σ||ρ) is a convex function
D is a convex set (convex hull) of its pure states
a convex function over a convex set can only have a global minimum
0.8
1
0.2
0.4
0.6
concurrence
Horodecki states
pure states
Werner states
0.8
1
Relative entropy vs concurrence:
Numerical simulations of 5 × 104 states
0
0
0.2
0.4
0.6
0.8
1
522
0.2
0.4
0.6
0.8
entanglement of formation
Horodecki states
pure states
Werner states
1
524
523
0
0
0.5
1
0.2
0.4
N
0.6
0.8
1
1
0.8
0.6
0.4
C
0.2
0
How to find different state orderings imposed by E, C & N
Relative entropy vs entropy of formation:
Numerical simulations of 5 × 104 states
0
0
0.4
0.6
concurrence
0
0
0.4
0.6
0.8
1
0.2
0.2
Horodecki states
pure states
Werner states
521
0.2
0.4
0.6
0.8
1
Relative entropy vs concurrence:
Epure(C ) > EW(C ) > EH(C ) for C ∈ (0, 1)
relative entropy
relative entropy
relative entropy
E
E’’-E’
N’’-N’
1.
2.
All cases of different state orderings
C’’-C’
5.
8.
4.
7.
3.
6.
11.
14.
10.
13.
9.
12.
Concurrences
C(σ ′) < C(σ ′′)
C(σ ′) < C(σ ′′)
C(σ ′) > C(σ ′′)
C(σ ′) < C(σ ′′)
C(σ ′) = C(σ ′′)
C(σ ′) < C(σ ′′)
C(σ ′) = C(σ ′′)
C(σ ′) < C(σ ′′)
C(σ ′) = C(σ ′′)
C(σ ′) < C(σ ′′)
C(σ ′) = C(σ ′′)
C(σ ′) > C(σ ′′)
C(σ ′) = C(σ ′′)
C(σ ′) < C(σ ′′)
Negativities
N (σ ′) < N (σ ′′)
N (σ ′) > N (σ ′′)
N (σ ′) < N (σ ′′)
N (σ ′) < N (σ ′′)
N (σ ′) = N (σ ′′)
N (σ ′) = N (σ ′′)
N (σ ′) < N (σ ′′)
N (σ ′) < N (σ ′′)
N (σ ′) = N (σ ′′)
N (σ ′) = N (σ ′′)
N (σ ′) < N (σ ′′)
N (σ ′) = N (σ ′′)
N (σ ′) > N (σ ′′)
N (σ ′) > N (σ ′′)
REEs
E(σ ′) < E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) > E(σ ′′)
E(σ ′) = E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) = E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) = E(σ ′′)
E(σ ′) = E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) < E(σ ′′)
E(σ ′) = E(σ ′′)
Table 1: All cases of different state orderings by E, C & N.
Class
1
2
3
4
5
6
7
8
9
10
11
12
13
14
525
526
527
528
Quantum information processing
with quantum dots
1. based on electron spins of quantum dots
using optical methods
2. based on nuclear spins of quantum dots
using NMR methods
What are the quantum dots?
Quantum dots (q-dots or artificial atoms) are
small metal or semiconductor boxes that hold a well-defined
number of electrons
useful property
number of electrons in a q-dot can be adjusted by changing
the dot electrostatic environment
parameters
number of electrons: from 0 to hundreds
size of q-dots: from 30 nm to 1 micron
energy
energy
1D
0D
530
energy
microdisk
lasers
microcavity
Quantum-information processing based on q-dots
and cavity QED [Imamoǧlu et al., PRL’99]
energy
529
quantum well quantum wire quantum dot
2D
DOS
bulk system
DOS
3D
DOS
Energy spectra of
q-dots vs other nano-structures
DOS
531
microdisk
microcavity
laser
Identical quantum dots in microcavity
microdisk
microcavity
lasers
532
Scalable system of quantum dot interactions and cavity QED
|1>n
ωcav
(1)
∆ωn
|v>n
|0>n
(0)
∆ωn
(L)
ωn
conduction band
∆n
valence band
533
Hamiltonian for
three-level q-dots interacting with N + 1 fields
534
Level structure of quantum dots in V configuration
N
n
X
X
n
(L)
†
(L) † (L)
) ân ,
h̄gnv1(âcav σ̂n1v + âcavσ̂nv1),
(L)
(L)
h̄gnv0[ân σ̂n0v + (ân )†σ̂nv0]
†
âcav +
= h̄ωcavâcav
=
h̄ωn (ân
Ĥ = ĤQD + ĤF + Ĥint,
X
=
(En(0)σ̂n00 + En(1)σ̂n11 + En(v)σ̂nvv ),
ĤQD
ĤF
Ĥint
nX
+
n
where the nth dot operator is σ̂nxy = |xinnhy|
Note: Q-dots are coupled only indirectly via the
cavity and laser fields
Derivation of q-dot interaction Hamiltonian
(L)
3-level Hamiltonian for |gn >, |en >, |vn >, an , acav
(L)
effective 2-level Hamiltonian for |gn >, |en >, an , acav
535
536
effective spin-spin interaction Hamiltonian for |gn >, |en >
equivalent-neighbor (SVW) Hamiltonian
Hamiltonian after adiabatic elimination
where
gn(t)gm(t)
∆
n
P
− ei(∆n−∆m)t + σ̂ −σ̂ + e−i(∆n−∆m)t]
Ĥeff = h̄2 n6=m κnm(t)[σ̂n+σ̂m
n m
κnm(t) =
1
(0)
∆ωn
+
1
(1)
∆ωn
gn(t) = gnv0gnv1|En
(k = e, g)
(L)
(t)|
∆ωk(n) = ωk(n) − ωV(n)B − ωcav
∆n = ωe(n) − ωg(n) + ωL(n) − ωcav = ∆ωe(n) − ∆ωg(n)
Adiabatic elimination requires
(1)
(0)
1. coupling strength, cavity decay rate, and thermal fluctuations
(x)
≪ h̄∆n, h̄∆ωn , En − En
2. valence-band levels |vin are far off resonance.
i
h
+σ̂ − + σ̂ −σ̂ + + γ(σ̂ +σ̂ + + σ̂ −σ̂ − )
σ̂
T
n m
n m
n m
n6=m nm n m
P
x x
n6=m σ̂n σ̂m
P
538
[Koashi,Bužek,Imoto, PRA’00]
Tight bound for symmetric sharing of entanglement
Cij ≤ 2/N
Entangled webs in
spin van der Waals (SVW) model
Ĥint = 4κ
SVW model for γ = 1
Non-conservative SVW model (NCE model)
SVW model for γ = 0
P
− + σ̂ −σ̂ +
Ĥint = κ n6=m σ̂n+σ̂m
n m
Conservative SVW model (CE model)
Frenkel model for Tnm=const
P
− + σ̂ −σ̂ + + γ(σ̂ +σ̂ + + σ̂ −σ̂ − )]
Ĥint = κ n6=m[σ̂n+σ̂m
n m
n m
n m
Spin van der Waals (SVW) model
Ĥint = h̄
Frenkel-type model or Heisenberg model
Models
537
6. tomography of nuclear spins
5. quantum algorithms
4. quantum gates
3. pseudo-pure states
2. control of nuclear spins
1. nuclear qubits and qudits
540
N-M
Quantum Computing Based on
Nuclear Magnetic Resonance
(NMR)
M
Q-dot bipartite entanglement in
SVW model
539
541
Quantum computer in a cup of coffee?
(3)
(1) (2)
(1) (3)
542
3,7-Dihydro-1,3,7-trimethyl-1H-purine-2,6-dione
C8H10N4O2
caffeine molecule
- qubits encoded in nonequivalent C-nuclei
CHEMICAL NAME :
(2)
(2) (3)
three qubits encoded in
alanine molecule
13C-labeled
(1)
J-coupling Hamiltonian for carbon-13 nuclei
Ĥint = ν1σ̂z + ν2σ̂z + ν3σ̂z + 21 (J12σ̂z σ̂z + J23σ̂z σ̂z + J13σ̂z σ̂z )
antenna gate
543
544
3 1
,−
2 2
3 1
,
2 2
3 3
,
2 2
3 3
,−
2 2
NMR quantum computing in nanostructures
oscillating
magnetic-field
high-frequency current
insulator
ω
/
01
ω12/
ω 23/
point contact channel
split Schottky gates
barrier
GaAs
barrier
GaAs backgate
ω12 = ω 0
0
ω = ω −ω
01
quadrupole
interaction
q
ω 23 = ω 0 + ω q
energy levels of spin I=3/2
ω0
ω0
ω0
Zeeman
interaction
2nd order
shifts
(usually neglected)
ω 01
ω12
ω 23
ω 02
ω13
ω 03
Mz – longitudinal component of M static along B0
0 ≡ 00
3 1
,−
2 2
3 3
,−
2 2
546
3 1
,
2 2
≡ 3,3
2 2
1 ≡ 01 ≡
2 ≡ 10 ≡
3 ≡ 11 ≡
Mxy – transverse component of M precessing in the x–y plane
(Larmor precession)
B1 – magnetic field component of the r.f. field
B0 – static magnetic field
M – magnetization vector of an ensemble of nuclear spins
Larmor precession
quartit = ququart = 4-level qudit
ω0
ω0
ω0
quartit is equivalent to 2 virtual qubits
545
Ĥ0 = 21h̄ω0σ̂z
ih̄
h1
e−iω0tp/2
0
,
0
eiω0tp/2
h̄ωnut
(σ̂x cos φp + σ̂y sin φp)
2
where ωnut = γBrf is the nutation frequency
′
Ĥrf,rot
=
• Hamiltonian in the rotating frame
ωLarmor = ωref
spin Hamiltonian during the r.f. pulse set on resonance
– ωref is the spectrometer reference frequency
– Brf is the amplitude of the oscillating r.f. field
– φp is the phase of the pulse
Brf (t) = Brf cos(ωref + φp)ex
coil is along x-axis generating an r.f. pulse field
• assumptions
X - and Y -rotations
i
Ĥ0 tp =
which is equal to Ẑ(θ) for θ = ω0tp.
R̂z (ω0tp) = exp
in terms of the propagator
|ψ(tp)i = R̂z (ω0tp)|ψ(0)i
• solution of Schödinger equation
• Hamiltonian
corresponds to free evolution of a spin-1/2 system without r.f. fields
Z -rotation
rotations via NMR techniques
548
547
• solution of Schödinger equation
549
550
for a pulse at resonant frequency ω of duration tp, which corresponds to the
nutation angle θp = ωnuttp, and general phase φp can be given as
|ψ(tp)i = χ̂(φp, θp)|ψ(0)i
where the propagator is
X̂(θ) = χ̂(0, θ)
= Ẑ(φp)X̂(θp)Ẑ(−φp)
i
h1
′
Ĥrf,rot
t
χ̂(φ , θ ) = exp
p
p
p
i
h ih̄ i
= exp − θp(σ̂x cos φp + σ̂y sin φp)
2
"
#
θ
θ
cos p
−i exp(−iφp) sin 2p
2
=
θ
θ
−i exp(iφp) sin 2p
cos 2p
• special cases
Ŷ (θ) = χ̂(π/2, θ)
selective rotations in a quartit




√1
1 0
0
0
− √i2 0 0
i
1
 2



0 0
 − √i √1
 0 √2 − √2 0 
2
2
X̂01( π2 ) = 
,
 , X̂12( π2 ) = 
0
 0 − √i2 √12 0 
0
1 0
0 0
0
1
0
0
0 1




√1
√1 − √1 0 0
0 − √i2 0
2
2

 2


√1 √1
1 0
0
0 0

0

X̂02( π2 ) =  √i
 , Ŷ01( π2 ) =  2 2
,
 − 2 0 √12 0 
0 0
1 0
0 0
0 1
0
0 0
1




0 −i 0 0
0 −1 0 0




 −i 0 0 0  ,
 1 0 0 0 , ···
X̂01(π) = 
Ŷ01(π) = 

0
0
1
0
0 0 1 0
0 0 0 1
0 0 0 1
Who has introduced NMR quantum computing?
the method has been developed independently by:
551
1. Cory, Fahmy, Havel in the article on "NMR spectroscopy: an experimentally accessible paradigm for quantum computing" (1996)
552
2. Gershenfeld and Chuang in "Bulk quantum computation" (1996).
How to generate pseudo-pure states?
pseudo-pure states (PPS) or effective pure states
can be obtained via
1. spatial averaging
2. temporal averaging
3. logical labeling
Here, we describe only a version of spatial averaging.
How to obtain pseudo-pure states?
population of the spin energy level |ki
Nk = A + (4 − k)∆
where
N 5N h̄ω
N h̄ω0
0
A =
−
, ∆=
,
4
8kB T
4kB T
N – total umber of spins
ω0 – Larmor frequency
Note: A does not contribute to the observed NMR signal
Assumption: quadrupole interaction ≪ Zeeman interaction
••↔ ∆ – relative occupation of the corresponding quantum level
|3i ••
|2i ••••
|1i ••••••
|0i ••••••••
thermal
or equivalently
••
••••’
••••••’
••••’
X̂23( π2 )
−→ ••••
••••’
••••••••
••••••••
|0i
••••
••••••••’
X̂23(π)
••••’
−→
••••
|2i
••
••••••’
X̂23(π)
•••••• −→
••••••’
−|3i
••••••’
••’
X̂12(π)
•••••• −→
••••••
−|2i
• states (−|ki)
••••
••••
X̂12(π)
••••••••’ −→
••••’
|1i
|3i ••
|2i ••••
X̂02( π2 )
|1i ••••••
−→
|0i ••••••••
thermal
|3i ••••
|2i ••••
X̂01(π)
|1i ••••
−→
|0i ••••••••
|0i
• states |ki (k = 1, 2, 3):
554
••••••
••••••’
••’
••••••
−|1i
••••••••’
••••’
••••
••••
|3i
How to obtain pseudo-pure states in a quartit
|3i ••
••••’
|2i ••••
••••
X̂13( π2 )
−→ ••••’
|1i ••••••
|0i ••••••••
••••••••
thermal
|0i
thermal
|3i ••
|2i ••••
X̂12(π)
−→
|1i ••••••
|0i ••••••••
How to obtain pseudo-pure state |0i in a quartit
553
556
1
2
P
k
|ki
••
•••••’
•••••
••••
••••
•••••’
X̂23( π2 )
X̂03( π2 )
−→ •••••• −→ •••••’
••••••
••••••••
•••••’
•••••
1 P |ki
thermal
k
2
|3i
|2i
|1i
|0i
Ĥ A ⊗ Ĥ B |0i =
Apply gate equivalent to two-qubit Hadamard
gate.
How to prepare equally-weighted superposition of
pseudo-pure states in a quartit?
Whether |11i or |00i is the most populated depends on
our labeling and direction of external magnetic field.
Note: here |11i – is the most populated state in equilibrium.
Initial state
Interchanging pseudo-pure states in a quartit
555
(a)
(b)
(c)
(d)
(e)
1
2
3
4
1
01
2/3
01
2
01
2/3
01
2/3
01
PPS 0 1
1
4
0
11
2
00
2/3
11
2
00
2/3
11
2/3
00
2/3
11
2/3
00
2
11
2/3
00
1
10
3
2
2/3
10
2/3
10
2
10
2/3
10
Eq
j00i pps
j01i pps
j10i pps
j11i pps
PPS − 1 0
557
PPS − 1 1
ω 01 ω12 ω 23
ω
23
ω 01
ω12
558
PPS - pseudo-pure states, Eq - equilibrium [Mahesh et al.’03]
ω 01 ω12 ω 23
PPS 0 0
NMR spectra for a quartit
population
,
energy level
NMR spectra for pseudo-pure states of two spin-1/2 systems
Key:
equilibrium
ω 01 ω12 ω 23
ω 01 ω12 ω 23 ω 01 ω12 ω 23
3
2
1
0
[Hirayama et al.’06]
≡
3
!ω 23 Rk (θ )
2
1
0
≡
Rk (θ )
Rk (θ )
experimental generation of pseudo-pure states
in solid-state systems of 69Ga nuclei
Rk (θ )
How to rotate “qubit” B in a quartit?
qubit A
qubit B
!ω 01 Rk (θ )
559
560
ω 01
ω12
ω 23
Rk (θ )
≡
0
1
!ω 02 Rk (θ )
2
!ω13 Rk (θ )
≡
≡
Y (−π )
Rk (θ k )
Rk (θ )
Y (π )
0
1
2
3
≡
Y (−π )
Rk (θ k )
Rk (θ )
ω 01
ω13
ω 02
ω 23
ω12
ω 01
ω 23
Y (π ) ω12
562
Rk (θ )
Rk (θ )
Another method to rotate “qubit” A in a quartit
qubit B
qubit A
3
How to rotate “qubit” A in a quartit?
561
01
00
01
00
pulses are applied simultaneously
10
11
10
11
RB
564
RC
000
000
RA
simultaneous application of pulses
001
001
110
101
111
100
011
010
RB
100
011
010
110
101
111
000
001
100
011
010
101
110
111
rotations of a virtual qubit in 8D qudit
RA
rotations of virtual qubit in quartit
563
1
0
0
0
0
0
0
1
= iX̂(π)

0

1
A
= ÛNOT
= iX̂02(π)X̂13(π)
0
0
0 1
1 0
NOT gates via rotations
• NOT gate for a qubit
ÛNOT = σ̂x =
0
0
0
1
• NOT gate for qubit A in a quartit

0
0
A
ÛNOT
= ÛNOT ⊗ Iˆ = 

1
0
1
0
0
0

0

0
B
= ÛNOT
= iX̂01(π)X̂23(π)
1
0
• NOT gate for qubit B in a quartit

0

B
1
ÛNOT
= Iˆ ⊗ ÛNOT = 
0
0
(a) NOP
(t) NOT(I2)+
XOR1
(n) SWAP+
XNOR2
(h) XNOR2
(b) NOT(I1)
(u) NOT(I1)+
XNOR2
(d) NOT(I1,I2)
(e) XOR1
(f) XOR2
(j) SWAP+NOT (k) SWAP+XOR1(l) SWAP+XOR2
(v) NOT(I2)+
XNOR1
(w) SWAP+
NOT(I1)
(x) SWAP+
NOT(I2)
(o) SWAP+NOT (p) SWAP+NOT (q) SWAP+NOT (r) SWAP+NOT
+XOR1
+XOR2
+XNOR1
+XNOR2
(i) SWAP
(c) NOT(I2)
NMR spectra for classical gates in a spin-1/2 system
(g) XNOR1
(m) SWAP+
XNOR1
(s) NOT(I1)+
XOR2
[Mahesh et al.’03]
565
566
Hadamard gates via rotations
• Hadamard gate for a qubit
0
1
0
1
1
0
−1
0
0
0
1
1

0

0 
= iX̂01(π)Ŷ01( π2 )X̂23(π)Ŷ23( π2 )
1 
−1
567
568

0

1 
= iŶ12(π)X̂01(π)Ŷ01( π2 )X̂23(−π)Ŷ23(− π2 )Ŷ12(−π)
0 
−1
for qubit A in a quartit
1 1 1
= iX̂(π)Ŷ ( π2 ) = iŶ ( π2 )Ẑ(π)
Ĥ = √
2 1 −1
• Hadamard gate

1
1 0
Ĥ A = Ĥ⊗Iˆ = √ 
2 1
0
1
−1
0
0
• Hadamard gate for qubit B in a quartit

1
1 1
Ĥ B = Iˆ ⊗ Ĥ = √ 
2 0
0
X (π )
X (π )
H
≡
≡
ω 01
ω12
Z (π )
Y ( π2 )
Y ( π2 )
ω 23
Z (π )
HB
Equivalent realizations of Hadamard gate H B (up to factor i)
qubit A
Y ( π2 )
qubit B
≅
Y ( π2 )
≡
0
1
0
0
0
0
0
1
CNOT1
in | out
-------00 | 00
01 | 01
10 | 11
11 | 10
ÛCNOT2
ÛCNOT1

Y (− π2 )
Y ( π2 )
≅
Y (π )
Y (−π )
0
0
0
1
0
0
1
0
≡
qubit B
qubit A
H
CNOT2
in | out
-------00 | 00
01 | 11
10 | 10
11 | 01
SWAP
in | out
-------00 | 00
01 | 10
10 | 01
11 | 11
570
Y (π )

0
0
 = diag([1, 1, −1, 1]) · Ŷ23(π),
1
0

0
1
 = diag([1, 1, 1, −1]) · Y12(π)Ŷ23(π)Ŷ12(−π)
0
0
truth tables
X (π )
Y (− π2 ) X (−π )
Y ( π2 )
CNOT gates via rotations
Z (π )
Z (π )
HA
1
0
= 
0
0

1
0
= 
0
0
Y (−π )
ω 23
ω12
ω 01
569
A
Equivalent realizations of Hadamard gate H (up to factor i)
00
00
01
10
10
01
11
11
CNOT21
572
100
011
010
001
000
001
000
110
101
111
100
011
010
110
101
111
CNOT32
CNOT23
CNOT gates of 3 virtual qubits #1
CNOT12
CNOT gates of 2 virtual qubits
571
CNOT12
573
CNOT gates of 3 virtual qubits #2
111
110
101
111
010
100
011
110
101
111
010
100
011
110
101
111
CNOT31
110
101
100
011
010
001
CNOT13
100
011
010
001
001
000
CNOT21
001
000
0
1
0
0

0

0
= ÛCNOT1ÛCNOT2ÛCNOT1
0
1
574
000
000
0
0
1
0
SWAP gates via rotations

1
0
ÛSWAP = 

0
0
0
0
1
0
0
−1
0
0

1

′′
0
ÛSWAP
=
0
0
similar gates

0

0
= Ŷ12(π),
0
1
relation between them
0
0
−i
0
0
−i
0
0

0

0
= X̂12(π)
0
1
which acts as follows
ÛSWAP c0|00i + c1|01i + c2|10i + c3|11i = c0|00i + c2|01i + c1|10i + c3|11i

1

′
0
ÛSWAP
=
0
0
ÛSWAP
′
′
= ÛSWAP
· diag([1, 1, −1, 1]) = diag([1, −1, 1, 1]) · ÛSWAP
′′
′′
= ÛSWAP
· diag([1, i, i, 1]) = diag([1, i, i, 1]) · ÛSWAP
NMR implementations of two-qubit algorithms
575
[Leung et al.’99]
[Fu et al.’99]
[Jones et al.’98]
1. Deutsch-Jozsa algorithm [Chuang et al.’98, Jones et al.’98]
2. Grover search algorithm
3. quantum Fourier transform
4. quantum error detection
[Fang et al.’99]
[Somaroo et al.’99]
6. dense coding
[Zhu et al.’01]
5. quantum simulations
7. Hogg algorithm
[Teklemarian et al.’02]
576
8. quantum erasers
Hogg algorithm – a highly structured search algorithm
What can be done with two-virtual qubits?
NMR implementations of two-qubit algorithms on spin-3/2 nuclei:
1. demonstration of classical gates
[Khitrin et al.’00, Sinha et al.’01, Kumar et al.’02]
2. demonstration of quantum gates
[Sarthour et al.’03, Kampermann et al.’02, Steffen’03,
3. generation of Bell states
[Sarthour et al.’03, Kampermann et al.’02]
4. quantum tomography
[Bonk et al.’04, Steffen’03, Kampermann et al.’05]
577
(∗) - not implemented yet
ω 23
ω12
ω 01
H
qubit B 1
y
x
Uf
y ⊕ f ( x)
x
H
HA HB
Uf
HA
Deutsch's algorithm for a quartit
H
qubit A 0
Deutsch's algorithm for two qubits 578
12. ∗ quantum erasers
11. ∗ Hogg algorithm
10. ∗ dense coding
9. ∗ quantum simulations
8. ∗ quantum error detection
[Kampermann et al.’05]
[Das et al.’03, Kampermann et al.’05]
6. Deutsch-Jozsa algorithm
[Ermakov et al.’02, Steffen’03, Kampermann et al.’05]
5. Grover search algorithm
2
π
Uf
2
π
Uf
580
1

0

0
0

0
1
0
0
0
0
1
0

0

0

0
1
0

1

0
0

1
0
0
0
0
0
0
1
1
1
0
0

0

0

1
0
1

0

0
0

0
1
0
0
0
0
0
1
0
1
f3

0

0

1
0
Pulse no pulse X̂01(π)X̂23(π) X̂23(π)
Uf
0
1
f2
f1
1
0
0
0
0
0
1
0
3 1
,
2 2
3 3
,
2 2
1 ≡ 01 ≡
0 ≡ 00 ≡
3
1
,−
2
2
3
3
,−
2
2
2 ≡ 10 ≡
3 ≡ 11 ≡

0

0

0
1
X̂01(π )
0

1

0
0

1
0
f4
NMR pulses for the oracle in Deutsch’s algorithm
Constant
Balanced
ω 23
ω12
ω 01
experimental Deutsch's algorithm for a quartit
qubit B 0
qubit A 0
579
experimental Deutsch's algorithm for two qubits
ω ω
1
Û1
2
4
Û 2
3
ω ω
ω
1
4
Û 3
ω
ω2
3
|10〉
ω1 ω2 ω3 ω4
|11〉
| 00〉
Û 3
581
Û 4
Û 4
582
ω1 ω2 ω3 ω4
exemplary NMR spectra for
Deutsch algorithm in 2 qubits
ω ω
3
ω1 ω2
| 01〉
ω4
Û 2
ω01 ω12 ω23
ω01 ω12 ω23
exemplary NMR spectra for
Deutsch algorithm in a quartit
Û1
ω01 ω12 ω23
|1〉
| 2〉
| 3〉
ω01 ω12 ω23
ω
23
ω12
ω01
| 0〉
HA HB
step II
Z ( π2 )
Z (−π )
Z ( 32π )
III.1
HA HB
III.2
a circuit for Grover’s search in a quartit
ω 01
ω12
ω 23
H
H
step II
f Xo
III.1
H
H
III.2
a circuit for Grover search in two qubits
qubit A
qubit B
III.3
Z ( π2 )
Z (−π )
Z ( 32π )
III.3
f0
NMR detections of magnetization of a spin-3/2
III.4
HA HB
III.4
H
H
583
584
• Mxy -detection of a two spin-1/2 system enables determination of the offdiagonal elements marked in boxes:


ρ00 ρ01 ρ02 ρ03
ρ
ρ 
 10 ρ11 ρ12 13 
ρ̂ = 
ρ20 ρ21 ρ22 ρ23 
ρ30 ρ31 ρ32 ρ33

ρ03

ρ13 
ρ23 
ρ33
• Mxy -detection of a spin-3/2 system gives the other off-diagonal elements:


ρ00 ρ01 ρ02 ρ03
 10 ρ11 ρ12 ρ13 
ρ

ρ̂ = 

ρ20 ρ21 ρ22 ρ23 
ρ30 ρ31 ρ32 ρ33
• Mz -detection of a spin-3/2 system gives:

ρ00 ρ01 ρ02

ρ
 ρ10 11 ρ12
ρ̂ = 
ρ
ρ
20 ρ21 22
ρ30 ρ31 ρ32
≡ Ŷ
( π2 )
R̂3 = Iˆ ⊗ Ŷ ,
R̂6 = X̂ ⊗ Ŷ ,
R̂9 = Ŷ ⊗ Ŷ
586
⇒ XXXXXX, IIIIII, YYYYYY, IIXXYY, XXYYII, YYIIXX, IXIYXY, XYXIYI,
YIYXIX, IIXYYX, XXYIIY, YYIXXI, IIYIXI, XXIXYX, YYXYIY, IYYXIX, XIIYXY, YXXIYI,
IXIIIY, IYYXXI, XIIYYX, YIXXIY, YXXIXX, XYXYII, XYYIYY, YXYYYI, IIIXYI, IYIYXX,
XIYXXX, XYXIIY, IIXIXI, IXXYYX, XXIXIX, YIIYXY, YXYIYY, IYYIIX, YYIXXY, IXYYXY,
YXYYII, YYIIYX, IIIXII, XIXXXX, IYXXYY, XXXXXI, XYYYYY, IIXYIX, YIYXIY,
IXIIXY, XIIIII, IXXXYY, IYIYXI, XYYIYI, YXXYXX, XXXIIY, YIIIII, YIYXYX, XYIIXX, YYXYII, YYYXXI, IXXYIX, YXIXIX, YXIYYI, IIIXYY, XIXYXY, XYXXXX, IIYIXY, IXYIYX, XIYYII, XXIXYY, XYYYYX, YIXIYI, YYYIIX, IXIYIY, IYIIYX, XXYXXX,
XYIXIY, YIXIYY, IXIIXI, IXXXII, IXYXII, IYIYYI, IYXIXI, XIIIYX, XIIXYI, YIYYXX,
YXYIXY, YYXXYX, IIIXXX, IIXIIX, IIYYII, XXIIXI, XXXYYX, XYYXIY, IIXYXI, IIYXYY,
IYIYIX, IYXIXY, XIXXXI, YIIYXX, YXXXYY, IIIIXX, IIXYYI, IIYIIX, IIYIYX, IIYYIY,
IXIYXX, IXYXYI, IYIIIY, IYXXYX, IYYYXI
n=6
⇒ XXXXX, IIIII, YYYYY, IIIXX, XXYYY, YYXII, IIXIY, XYXXI, YXIYY,
XXYIX, IYIXX, IIXYI, YIYXI, XXIYY, YXYXI, YYXIX, IYYYY, XIXIY, IIIYI, IXYIX,
XIYYX, IXXXY, XYIXI, YXIIY, XIYIX, XYYXI, YIIXX, YYXYX, IIXXY, IXXYI, XYIII,
IYYIY, YYXYY, IXIXX, XIXXY, XXXIX, XYIYI, YXYXX, IIYIY, IYIXY
n=5
YYXY, XYXI, IIYX, IXIY, IIXI, IYIY
2. other sets of rotations for tomography of n-qubits
n=1 ⇒ X, I
n=2 ⇒ XX, II, IX, IY;
n=3 ⇒ XXX, III, IIY, XYX, YII, XXY, IYY
n=4 ⇒ XXXX, IIII, IIYY, YYXX, IIIY, XYXX, YXYI, IXYI, IIIX, XIYY, YXII,
X̂( π2 ), Ŷ
R̂2 = Iˆ ⊗ X̂,
R̂5 = X̂ ⊗ X̂,
R̂8 = Ŷ ⊗ X̂,
where X̂ ≡
ˆ
R̂1 = Iˆ ⊗ I,
ˆ
R̂4 = X̂ ⊗ I,
ˆ
R̂7 = Ŷ ⊗ I,
1. Chuang’s set of rotations for Mxy tomography of 2 qubits
ρ̂(k) = R̂k ρ̂R̂k†
• The remaining matrix elements can be obtained by rotating the original
density matrix ρ̂ through properly chosen rotational operations R̂k :
• Mxy and Mz detections give only some elements of ρ̂
is a method for complete reconstruction of a given density
matrix ρ̂ in a serious of NMR measurements.
Principle ideas
NMR quantum state tomography
585
X 02
X 13
X 03
00
01
10
•
•
•
R̂11
= X̂03( π2 ),
three-photon transitions
= X̂02( π2 ),
R̂9 = X̂13( π2 ),
R̂7
two-photon transitions
R̂1
= X̂01( π2 ),
R̂3 = X̂12( π2 ),
R̂5 = X̂23( π2 ),
single-photon transitions
R̂12 = Ŷ03( π2 )
R̂10 = Ŷ13( π2 )
R̂8 = Ŷ02( π2 )
= Ŷ01( π2 )
R̂4 = Ŷ12( π2 )
R̂6 = Ŷ23( π2 )
R̂2
a natural set of rotations for tomography of a quartit state
(3+2+1)=6
so 12 readouts (including Y-rotations)
X 01
X 12
X 23
11
588
tomography of 4-level system
587
rotations in Mz -based tomography
e.g.
ρ̂ after R̂1 = X̂01( π2 )
...
...
ρ22
...

...

... 
... 
ρ33
ρ̂ after R̂5 = X̂23( π2 )

...
...
...

ρ11
...
...


... 1 (ρ − ρ23 − ρ32 + ρ33)
...
2 22
1
...
...
2 (ρ22 + ρ23 + ρ32 + ρ33 )

1
...
2 (ρ00 − ρ01 − ρ10 + ρ11)
1

...
2 (ρ00 + ρ01 + ρ10 + ρ11)

ρ̂′ = 
...
...
...
...

ρ00

 ...
ρ̂′′ = 
...
...
so let us apply both rotation before read-outs
Ŷ13(π)X̂01(θ)Ŷ13(−π)
Ŷ02(π)X̂23(−θ)Ŷ02(−π)
Ŷ01(π)X̂13(−θ)Ŷ01(−π)
Ŷ23(π)X̂02(θ)Ŷ23(−π)
Ŷ01(π)Ŷ23(π)X̂12(−θ)Ŷ23(−π)Ŷ01(−π)
···
tomography without three-photon rotations
X̂03(θ) =
=
=
=
=
=
Ŷ23(π)X̂12(θ)Ŷ23(−π)
Ŷ12(π)X̂23(−θ)Ŷ12(−π),
Ŷ23(π)Ŷ12(θ)Ŷ23(−π)
Ŷ12(π)Ŷ23(−θ)Ŷ12(−π),
Ŷ12(π)X̂01(θ)Ŷ12(−π)
Ŷ01(π)X̂12(−θ)Ŷ01(−π),
Ŷ12(π)Ŷ01(θ)Ŷ12(−π)
Ŷ01(π)Ŷ12(−θ)Ŷ01(−π).
tomography without two-photon rotations
X̂13(θ)
Ŷ13(θ)
X̂02(θ)
Ŷ02(θ)
=
=
=
=
=
=
=
=
589
590
7
591
+5
+4
+3
+2
+1 = 28
so 56 readouts (including X and Y-rotations)
[Tseng et al.’00]
000
001
100
011
010
110
101
111
tomography of 8-level system
+6
592
[Laflamme et al.’97]
NMR implementations of 3-qubit algorithms on spin-1/2 nuclei
1. generation of GHZ states
[Cory et al.’98]
[Nielsen et al.’98]
2. quantum error correction
3. quantum teleportation
[Linden et al.’98]
[Kim et al.’00]
5. refined Deutsch-Jozsa algorithm for entangled qubits
(a meaningful test of quantum parallelism)
6. Grover algorithm (with cancellation of systematic errors)
7. quantum simulation
[Chang et al.’01]
[Vandersypen et al.’00]
8. Schulman-Vazirani algorithm (a cooling scheme)
[Viola et al.’01]
[Weinstein et al.’01]
9. noiseless subsystems
[Peng et al.’02]
14. Hogg algorithm
[Steffen et al.’03]
[Khitrin et al.’01]
10. logical labeling
9. quantum teleportation
8. quantum error correction
7. generation of GHZ states
6. quantum tomography
• not implemented yet
3. test of phase coherence in EIT
[Murali et al.’04]
2. half-adder and subtractor operations [Murali et al.’02, Kumar et al.’02]
1. quantum simulation
• implemented (in liquid NMR systems)
596
595
2018 - estimated year for reaching the fundamental limits
• the end of Moore’s law?
transistors
2007 - Dual-Core Intel Itanium 2 chip contains 1.7 billion
1965 - approximately 60 devices on a chip.
The number doubles every 24 months.
• modified Moore’s law (1975)
The number of transistors and resistors on a chip
doubles every 18 months.
• original Moore’s law (1965)
Gordon E. Moore - a co-founder of Intel.
possible QIP applications of spin-7/2 nuclei
21. adiabatic quantum optimization algorithm
20. Hogg algorithm
empirical Moore’s law
594
19. phase estimation algorithm
18. quantum baker’s map
17. quantum erasers
15. noiseless subsystems
14. Schulman-Vazirani algorithm (a cooling scheme)
13. Grover algorithm (with cancellation of systematic errors)
(a meaningful test of quantum parallelism)
12. refined Deutsch-Jozsa algorithm for entangled qubits
What can be done with three virtual qubits?
Shor’s factorization algorithm
NMR implementation of a seven-qubit algorithm
[Murali et al.’04]
18. test of phase coherence in electromagnetically induced transparency (EIT)
17. quantum state and process tomography
16. adiabatic quantum optimization algorithm
15. half-adder and subtractor operations [Murali et al.’02, Kumar et al.’02]
[Lee et al.’02]
[Weinstein et al.’02]
593
[Teklemarian et al.’01]
13. phase estimation algorithm
12. quantum chaotic map (baker’s map)
11. quantum erasers
[source: Wikipedia]
end of Moore’s Law?
• how to provide energy to a chip?
• how to cool down a chip?
As power-driven heat can cause major malfunctions.
597
598
“Chip with 3nm-length gates would overheat itself.” [Gargini]
• when the gate length < 5 nm quantum effects become important
gate lengths
37 nm in 2007
∼5 nm in 2015-2018
quantum tunneling
source & drain are so close that the electrons will tunnel
even if voltage is not applied to the gate
⇒ Heisenberg uncertainty becomes important
⇒ transistor becomes unreliable
Physical principles of
quantum computing
1. Superposition
2. Interference
3. Entanglement
4. Non-cloning
5. Uncertainty
599
600
purely quantum applications of QC
1. Quantum cryptography
2. Quantum teleportation
advantages of QC
• superfast (Shor) and fast (Grover) algorithms
• understanding new aspects of measurement theory
• improvement of precision spectroscopy
• understanding dissipation in mesoscopic system
• partial control of decoherence
• quantum state engineering
• quantum simulations
"Quantum computers must be a
component of any world view that
seeks to be fundamental"
David Deutsch
• Information Science
• Cryptography
• Quantum physics
including
• Quantum Optics
• Nanotechnology
Quantum Computing is the frontier of
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Strony 201 – 220