Solution for ”Quantum Computation and Quantum Information: 10th Anniversary Edition” by Nielsen and Chuang goropikari August 11, 2018
Copylight Notice: cbna
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Contents 0.1
Eratta list . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
Intro duction to quantum mechanics
3
8
Quantum noise and quantum operations
37
9
Distance measures for quantum information
39
1
2
0.1 0.1
CONTENTS
Era Eratta tta lis list
• p.101. eq (2.150) ρ (2.150) ρ = = m p( p(m)ρm should be ρ be ρ ′ = m p( p(m)ρm . • p.408. eq (9.49) i piD(ρi, σi) + D( pi, q i) should be i piD(ρi, σi) + 2 + 2D D( pi , q i ).
∑
∑
∑
� � ≤ � �
eqn (9. (9.48) =
pi Tr( Tr(P P ((ρi
i
pi Tr( Tr(P P ((ρi
i
=
− σi)) +
� � ∑
( pi
i
r(P σi ) − q i) Tr(P
( pi
i
− q i) ( i ( pi − q i )
∵
pi Tr( Tr(P P ((ρi
− σi)) + 2
pi Tr( Tr(P P ((ρi
− σi)) + 2D 2D( pi , q i )
i
=
− σi)) +
∑
i
Tr(P Tr(P σi )
≤ 1)
2
• p.409 p.409..
Exer Exerci cise se 9.12. 9.12. If ρ = σ , then D(ρ, σ ) = 0. Further urthermor moree trace distanc distancee is nonnonnegative. Therefore 0 D( D ( (ρ), (σ )) 0 D( D ( (ρ), (σ)) = 0. So I think the map is not strictly contractive. If p p = 1 and ρ and ρ = σ, σ , then D then D(( (ρ), (σ)) < )) < D(ρ, σ) is satisfied.
≤ E E ≤ ⇒ E E E ̸ ̸ E E Tr(A† B ) = ⟨m|A ⊗ B |m⟩ should be Tr(A r(AT B ) = ⟨m|A ⊗ B |m⟩. • p.411. Exercise 9.16. eqn(9.73) Tr(A
� � � � �− � � � �− �
Simple counter example is the case that A that A = A† B = Tr(A r(A† B ) = A
⊗
i 0 0 0
−
i 0 1 0 . B = , In this case, 0 0 0 0
1 0 = 0 0
i 0 , 0 0
i,
i 0 B= 0 0
0 0 0 0
0 0 0 0
0 0 0 0
⟨m|A ⊗ B|m⟩ = ( ⟨00| + ⟨11⟩)(A )(A ⊗ B )(|00⟩ + |11⟩) = i. ̸ ⟨m|A ⊗ B|m⟩. Thus Tr(A Tr(A† B ) = By using following relation, we can prove.
⊗ ⊗ A) |m⟩ = (AT ⊗ I ) |m⟩ Tr(A r(A) = ⟨m|I ⊗ ⊗ A|m⟩
(I
Tr(A r(AT B ) = Tr(BA Tr(BA T ) = m I
⟨ | ⊗ ⊗ BAT |m⟩ ⊗ B)(I ⊗ AT )|m⟩ = ⟨m|(I ⊗ )(I ⊗ ⊗ B)(A = ⟨m|(I ⊗ )(A ⊗ I )|m⟩ = ⟨m|A ⊗ B |m⟩ .
Chapter 2
Introduction to quantum mechanics 2.1
� � � �−� � � � 1 1 + 1 2
2 0 = 1 0
−
2.2
A 0 = A 11 0 + A21 1 = 1
|⟩ |⟩ | ⟩ | ⟩⇒A A |1⟩ = A |0⟩ + A |1⟩ = |0⟩ ⇒ A 12
∴
input:
A =
22
11
= 0, A21 = 1
12
= 1, A22 = 0
� � 0 1 1 0
{| 0⟩ , |1⟩}, output: {| 1⟩ , |0⟩} A 0 = A 11 1 + A21 0 = 1
|⟩ |⟩ | ⟩ | ⟩⇒A A |1⟩ = A |1⟩ + A |0⟩ = |0⟩ ⇒ A 12
A =
22
11
= 1, A21 = 0
12
= 0, A22 = 1
� � 1 0 0 1
2.3
From eq (2.12) A vi =
| ⟩
B w j =
| ⟩
� � j
k
3
A ji w j
| ⟩
Bkj xk
| ⟩
4
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Thus
� | ⟩ � |⟩ � |⟩ � � | � |⟩ �
BA vi = B
| ⟩
A ji w j
j
=
A ji B w j
j
=
A ji Bkj xk
j,k
=
Bkj A ji
j
k
=
xk
⟩
(BA)ki xk
k
(BA)ki =
∴
Bkj A ji
j
2.4
I v j =
| ⟩
�
I ij vi = v j , j.
| ⟩ | ⟩ ∀ ⇒ I ij = δ ij
i
2.5
Defined inner product on
Cn is ((y1 ,
·· · , yn), (z , · · · , zn)) = 1
�∗
yi zi .
i
Verify (1) of eq (2.13).
�
(y1 ,
· ·· , yn),
� i
λi (zi1 ,
� � ∗ � �∗ � �� ∗ � � ··· � · ··
· · · , zin)
=
yi
i
=
λ j z ji
j
yi λ j z ji
i,j
=
λ j
j
=
yi z ji
i
λ j ((y1 ,
, yn ), (z j 1 ,
λi ((y1 ,
, yn ), (zi1 ,
j
=
i
· · · , z jn ))
·· · , zin)) .
5 Verify (2) of eq (2.13), ((y1 ,
�� ∗ �∗ �� ∗� �� ∗ �
· · · , yn), (z , · ·· , zn))∗ =
yi zi
1
(2.1)
i
=
yi zi
(2.2)
zi yi
(2.3)
· · · , zn), (y , ·· · , yn))
(2.4)
i
=
i
= ((z1 ,
1
. Verify (3) of eq (2.13), ((y1 ,
· ·· , yn), (y , ·· · , yn)) = 1
=
�∗ �| |
yi yi
i
yi
2
i
Since yi 2 0 for all i. Thus i yi 2 = ((y1 , , yn ), (y1 , From now on, I will show the following statement,
∑| |
| | ≥
((y1 ,
| | ≥
·· · , yn)) ≥ 0.
· · · , yn), (y , · ·· , yn)) = 0 iff (y , · · · , yn) = 0. 1
( ) This is obvious. ( ) Suppose ((y1 , , yn ), (y1 , Since yi 2 0 for all i, if for all i. Thus,
⇐ ⇒
···
···
1
2
∑
·· · , yn)) = 0. Then i |yi| = 0. i |yi | = 0, then |yi | = 0 for all i. Therefore |yi | 2
∑
2
(y1 ,
2
=0
· · · , yn) = 0.
2.6
�� i
� � � �∗ |⟩ |⟩ | ⟩ �� �∗ | ⟩ | ⟩ � ∗ | ⟩ | ⟩∗ � ∗| ⟩ |⟩
λi wi , v
| ⟩
=
v ,
λi wi
i
=
λi ( v , wi )
i
=
λi ( v , wi )
i
=
λi ( wi , v )
i
2.7
(∵ linearlity in the 2nd arg.)
⇔ yi = 0
6
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
⟨w |v ⟩ = |w⟩ = ∥|w⟩∥ |v ⟩ = ∥|v⟩∥
�− � − � � | ⟩ √ ⟨ | ⟩ √ � � | ⟩ √ ⟨ | ⟩ √ − 1 = 1 1
1 1
1=0
w 1 1 = 2 1 ww v 1 1 = 1 2 vv
2.8
If k = 1,
|v ⟩ = ∥||ww ⟩⟩ −− ⟨⟨vv ||ww ⟩|⟩|vv ⟩⟩∥ ⟨v |v ⟩ = ⟨v | ∥||ww ⟩⟩ −− ⟨⟨vv ||ww ⟩|⟩|vv ⟩⟩∥ ⟨v |w ⟩ − ⟨v |w ⟩ ⟨v |v ⟩ = ∥|w ⟩ − ⟨v |w ⟩ |v ⟩∥ 2
1
2
1
2
1
2
1
2
1
2
1
�
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
�
1
= 0.
Suppose v1 ,
{ · ·· vn} (n ≤ d − 1) is a orthonormal basis. Then vi wn+1 vi vi wn+1 vi n v j wn+1 i=1 vi wn+1 v j vi = n wn+1 i=1 vi wn+1 vi n v j wn+1 i=1 vi wn+1 δ ij = n wn+1 i=1 vi wn+1 vi v j wn+1 v j wn+1 = n wn+1 i=1 vi wn+1 vi
⟨v j |vn ⟩ = +1
n i=1 n i=1
∑∑ ⟨ | ⟩| ⟩ � ∥| ∑ ⟨ | ⟩| ⟩∥ ⟨ | ⟩ − ∑ ⟨ | ⟩⟨ | ⟩ ∥| ⟩ − ∑ ⟨ | ⟩| ⟩∥ ⟨ | ⟩ −∑ ⟨ | ⟩ ∥| ⟩ − ⟨ | ⟩ | ⟩∥ ⟨ | ∑⟩ − ⟨ | ⟩ ∥| ⟩ − ⟨ | ⟩ | ⟩∥ � ⟨ | | v j
wn+1 wn+1
⟩− ⟩−
= 0.
Thus Gram-Schmidt procedure produces an orthonormal basis. 2.9
σ0 = I = 0 0 + 1 1
| ⟩ ⟨ | | ⟩⟨ | σ = X = |0⟩⟨1| + |1⟩⟨0| σ = Y = −i |0⟩⟨1| + i |1⟩ ⟨0| σ = Z = |0⟩ ⟨0| − |1⟩⟨1| 1 2 3
2.10
( j
≤ n)
7
|v j ⟩⟨vk | = I V |v j ⟩ ⟨vk | I V = |v p⟩ ⟨v p| |v j ⟩ ⟨vk |
�� � �� � | ⟩⟨ | � | ⟩⟨ | ⟩ ⟨ | ⟩ ⟨ | � | ⟩⟨ | vq
p
=
v p v p v j
vk vq
vq
q
vq
p,q
=
δ pj δ kq v p vq
p,q
Thus ( v j
| ⟩ ⟨vk |) pq = δ pj δ kq
2.11
� �
0 1 X = , det(X 1 0 If λ =
− λI ) = det
��− �� λ 1
1 λ
−
= 0
⇒ λ = ±1
−1,
� �� � � � 1 1 1 1
c1 0 = c2 0
Thus
��
� � − √
1 c 1 = 1 = c2 2
|λ = − ⟩
1 1
If λ = 1
� � √
1 1 λ = 1 = 2 1
| X =
⟩
�− �
1 0 w.r.t. 0 1
{|λ = −1⟩ , |λ = 1⟩}
2.12
det
�� � − � 1 0 1 1
λI = (1
2
− λ)
=0
⇒ λ = 1
Therefore the eigenvector associated with eigenvalue λ = 1 is
|λ = 1⟩ = Because λ = 1 λ = 1 =
|
⟩⟨
|
� � � �̸
�� 0 1
0 0 , 0 1
� �
1 0 0 0 = c λ = 1 λ = 1 = 1 1 0 c
|
⟩⟨
|
8
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.13
Suppose ψ , φ are arbitrary vectors in V .
| ⟩ | ⟩
∗ (|ψ⟩ , (|w⟩⟨v |) |φ⟩)∗ = (|w⟩⟨v |)† |ψ⟩ , |φ⟩ = |φ⟩ , (|w⟩ ⟨v |)† |ψ⟩
� �
= φ ( w v )† ψ .
⟨ | | ⟩⟨ | | ⟩
� �
On the other hand, ( ψ , ( w v ) φ )∗ = ( ψ w v φ )∗
| ⟩ | ⟩⟨ | | ⟩
⟨ | ⟩⟨ | ⟩ = ⟨φ|v⟩ ⟨w|ψ ⟩ .
Thus
⟨φ| (|w⟩ ⟨v|)† |ψ⟩ = ⟨φ|v⟩⟨w|ψ⟩ for arbitrary vectors |ψ⟩ , |φ⟩ ( |w⟩ ⟨v |)† = |v ⟩⟨w| ∴
2.14
((ai Ai )† φ , ψ ) = ( φ , ai Ai ψ )
|⟩ | ⟩
∴
(ai Ai )† = a ∗i A†i
|⟩ | ⟩ = a i (|φ⟩ , Ai |ψ⟩) = a i (A†i |φ⟩ , |ψ⟩) = (a∗i A†i |φ⟩ , |ψ⟩)
2.15
((A† )† ψ , φ ) = ( ψ , A† φ )
|⟩ |⟩
∴
2.16
(A† )† = A
|⟩ |⟩ = (A† |φ⟩ , |ψ⟩)∗ = ( |φ⟩ , A |ψ⟩)∗ = (A |ψ⟩ , |φ⟩)
9
P =
�| ⟩⟨ | �� � � | ⟩⟨ | | ⟩⟨ | �| ⟩⟨ | ⟩⟨ | �| ⟩⟨ | �| ⟩⟨ | i i .
i
P 2 =
i i
j j
i
=
j
i i j j
i,j
=
i j δ ij
i,j
=
i i
i
= P
2.17
Proof. (
⇒) Suppose A is Hermitian. Then A = A†. Let |λ⟩ be eigenvectors of A with eigenvalues
λ, that is,
A
|⟩ = λ |λ⟩ .
Therefore
⟨λ|A|λ⟩ = λ ⟨λ|λ⟩ = λ. On the other hand, λ∗ = λ A λ ∗ = λ A† λ = λ A λ = λ λ λ = λ.
⟨| |⟩
⟨| |⟩ ⟨| |⟩
⟨|⟩
Hence eigenvalues of Hermitian matrix are real. ( ) Suppose eigenvalues of A are real. From spectral theorem, normal matrix A can be written by
⇐
A =
�
λi λi λi
i
| ⟩⟨ |
where λ i are real eigenvalues with eigenvectors λi . By taking adjoint, we get A† =
� � i
=
i
= A Thus A is Hermitian. 2.18
| ⟩ λ∗i |λi ⟩⟨λi | λi λi λi
| ⟩⟨ |
(∵ λ i are real)
(2.5)
10
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Suppose v is a eigenvector with corresponding eigenvalue λ.
| ⟩
U v = λ v .
|⟩
|⟩ 1 = ⟨v |v ⟩ = ⟨v | I |v ⟩ = ⟨v | U † U |v ⟩ = λλ ∗ ⟨v |v ⟩ = ∥λ∥ 2
∴
λ = eiθ
2.19
X 2 =
� �� � � � 0 1 1 0
0 1 1 0 = = I 1 0 0 1
2.20
U
≡
′
�|
wi vi
⟩⟨ |
i
Aij = vi A v j
⟨ | | ⟩ = ⟨vi |UU † AUU † |v j ⟩ ⟨vi|w p⟩ ⟨v p|vq ⟩ ⟨wq |A|wr ⟩ ⟨vr |vs⟩⟨ws|v j ⟩ =
� �⟨ | ⟩ � ⟨ | ⟩⟨
p,q,r,s
=
′′
vi w p δ pq Aqr δ rs ws v j
⟨ | ⟩
p,q,r,s
=
′′
vi w p wr v j A pr
p,r
| ⟩
2.21
Suppose M be Hermitian. Then M = M † . M = I MI = (P + Q)M (P + Q) = P MP + QMP + P MQ + QMQ
Now P MP = λP , QMP = 0, P MQ = P M † Q = (QMP )∗ = 0. Thus M = P MP + QM Q. Next prove QM Q is normal. QMQ(QMQ)† = QMQQM † Q = QM † QQMQ
(M = M † )
= (QM † Q)QMQ Therefore QMQ is normal. By induction, QMQ is diagonal ... (following is same as Box 2.2)
11
2.22
Suppose A is a Hermitian operator and vi are eigenvectors of A with eigenvalues λ i . Then
| ⟩ ⟨vi|A|v j ⟩ = λ j ⟨vi|v j ⟩ .
On the other hand,
⟨vi|A|v j ⟩ = ⟨vi|A†|v j ⟩ = ⟨v j |A|vi⟩∗ = λ∗i ⟨v j |vi⟩∗ = λ∗i ⟨vi|v j ⟩ = λi ⟨vi|v j ⟩ Thus (λi
− λ j ) ⟨vi|v j ⟩ = 0.
If λ i = λ j , then vi v j = 0.
̸
⟨ | ⟩
2.23
Suppose P is projector and λ are eigenvectors of P with eigenvalues λ. Then P 2 = P .
| ⟩ P |λ⟩ = λ |λ⟩ and P |λ⟩ = P |λ⟩ = λP |λ⟩ = λ |λ⟩ . 2
2
Therefore λ = λ 2 λ(λ 1) = 0 λ = 0 or 1.
−
2.24
Def of positive v A v 0 for all v . Suppose A is a positive operator. A can be decomposed as follows.
⟨ | | ⟩ ≥ A =
| ⟩
A + A† A A† +i 2 2i
−
= B + iC where B =
A + A† A A† , C = . 2 2i
−
Now operators B and C are Hermitian.
⟨v|A|v⟩ = ⟨v|B + iC |v⟩ = ⟨v |B |v ⟩ + i ⟨v |C |v ⟩
= α + iβ where α = v B v , β = v C v .
⟨| |⟩
⟨| |⟩
Since B and C are Hermitian, α, β R. From def of positive operator, β should be vanished because v A v is real. Hence β = v C v = 0 for all v , i.e. C = 0. Therefore A = A † .
⟨ | | ⟩
∈ ⟨| |⟩
| ⟩
12
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Reference: MIT 8.05 Lecture note by Prof. Barton Zwiebach. https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/ lecture-notes/MIT8_05F13_Chap_03.pdf Proposition. 2.0.1. Let T be a linear operator in a complex vector space V . If (u,Tv) = 0 for all u, v V , then T = 0 .
∈
Proof. Suppose u = T v. Then (T v , T v) = 0 for all v implies that T v = 0 for all v. Therefore
T = 0. Theorem. 2.0.1. If (v,Av) = 0 for all v
∈ V , then A = 0.
Proof. First, we show that (u,Tv) = 0 if (v,Av) = 0. Then apply proposition 2.0.1
Suppose u, v
∈ V . Then (u,Tv) is decomposed as
�
1 (u,Tv) = (u + v, T (u + v)) 4
− (u − v, T (u − v)) + 1i (u + iv,T (u + iv)) −
1 (u i
�
− iv,T (u − iv))
.
If (v , T v) = 0 for all v V , the right hand side of above eqn vanishes. Thus (u,Tv) = 0 for all u, v V . Then T = 0.
∈
∈
2.25
2
⟨ψ|A†A|ψ⟩ = ∥A |ψ⟩∥ ≥ 0 for all |ψ⟩ . Thus A† A is positive.
2.26
|ψ⟩⊗
2
=
√ 12 (|0⟩ + |1⟩) ⊗ √ 12 (|0⟩ + |1⟩
1 = ( 00 + 01 + 10 + 11 ) 2 1 1 1 = 2 1 1
| ⟩ | ⟩ | ⟩ | ⟩
13
|ψ⟩⊗
3
=
√ 12 (|0⟩ + |1⟩) ⊗ √ 12 (|0⟩ + |1⟩ ⊗ √ 12 (|0⟩ + |1⟩
=
1 √ (|000⟩ + |001⟩ + |010⟩ + |011⟩ + |100⟩ + |101⟩ + |110⟩ + |111⟩) 2 2
1 1 1 1 1 = 2 2 1 1 1 1
√
2.27
� �⊗� � − − − � �⊗� � � �⊗� � 0 1 1 0
X
⊗ Z =
0 0 = 1 0 I
⊗ X =
⊗ I =
0 0 0 1
1 0 0 0
1 0 0 1
0 1 = 0 0 X
1 0
1 0 0 0
0 0 0 1
0 1 0 0
0 1 1 0
0 0 0 1
0 0 1 0
0 1 1 0
0 0 = 1 0
0 1
1 0 0 1
1 0 0 0
0 1 0 0
In general, tensor product is not commutable. 2.28
(A
⊗ B)∗ = =
∗
·· ·
Am1 B
·· · AmnB ·· · A∗nB
A11 B ∗ .. .
A∗m1 B ∗
= A ∗
⊗ B∗.
..
A1n B ...
∗ ∗
A11 B .. .
.
..
1
.
·· ·
...
A∗mn B ∗
14
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
(A
⊗ B)T =
T
· ··
Am1 B
· ·· AmnB · · · Am BT
..
A1n B ...
A11 B .. .
.
A11 B T 1 . . .. .. ... = A1n B T Amn B T
··· ···
A11 B T A1m B T .. .. ... = . . An1 B T Anm B T = A T
(A
⊗ BT .
···
⊗ B)† = ((A ⊗ B)∗)T = (A∗ ⊗ B ∗ )T = (A∗ )T ⊗ (B ∗ )T = A† ⊗ B † .
2.29
Suppose U 1 and U 2 are unitary operators. Then (U 1
⊗ U )(U ⊗ U )† = U U † ⊗ U U † = I ⊗ I. 2
1
2
1
1
2
2
Similarly, (U 1
⊗ U )†(U ⊗ U ) = I ⊗ I. 2
1
2
2.30
Suppose A and B are Hermitian operators. Then (A Thus A
⊗ B)† = A† ⊗ B† = A ⊗ B.
(2.6)
⊗ B is Hermitian.
2.31
Suppose A and B are positive operators. Then
⟨ψ| ⊗ ⟨φ| (A ⊗ B) |ψ⟩ ⊗ |φ⟩ = ⟨ψ|A|ψ⟩ ⟨φ|B|φ⟩ . Since A and B are positive operators, ⟨ψ|A|ψ⟩ ≥ 0 and ⟨φ|B |φ⟩ ≥ 0 for all |ψ⟩ , |φ⟩. ⟨ψ|A|ψ⟩⟨φ|B|φ⟩ ≥ 0. Thus A ⊗ B is positive if A and B are positive. 2.32
Then
15 Suppose P 1 and P 2 are projectors. Then (P 1
2
= P 12
⊗ P ) 2
2 2
⊗ P = P ⊗ P . 1
Thus P 1
2
⊗ P . is also projector. 2
2.33
� � √
1 1 H = 2 1
H ⊗2 =
1 1
−
� � � � √ − ⊗ √ − 1 1 2 1
1 1 1 1 = 1 2 1 1
1 1 2 1
1 1
1 1 1 1
(2.7)
− −
1 1 1 1
1 1 1 1
− − − −
2.34
Suppose A =
� �
4 3 . 3 4 det(A
2
2
− λI ) = (4 − λ) − 3 = λ − 8λ + 7 = (λ − 1)(λ − 7) 2
Eigenvalues of A are λ = 1, 7. Corresponding eigenvectors are λ = 1 =
|
⟩
� � √ 1
2
1 . 1
Thus A = λ = 1 λ = 1 + 7 λ = 7 λ = 7 .
|
√
⟩⟨
|
|
⟩⟨
|
√ | √ 7 |λ = 7⟩⟨λ = 7| 7
A = λ = 1 λ = 1 +
|
1 = 2 =
1 2
⟩⟨
� −� � � �− √ √ − √ √ � 1 1
1 + 1 2
1+ 7 1+ 7
−
1 1 1 1
1+ 7 1+ 7
log(A) = log(1) λ = 1 λ = 1 + log(7) λ = 7 λ = 7
|
=
� �⟩⟨
log(7) 1 1 1 1 2
|
|
⟩⟨
|
� � √ | 1
2
1 , λ = 7 = 1
−
⟩
16
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.35
3
v⃗ σ⃗ =
·
� � � � −� � � − � � − vi σi
i=1
= v 1 =
0 1 0 + v2 1 0 i
i 1 + v3 0 0
0 1
v3 v1 iv2 v1 + iv2 v3
−
det(⃗ v σ⃗
· − λI ) = (v − λ)(−v − λ) − (v − iv )(v + iv ) = λ − (v + v + v ) = λ − 1 ( v|⃗ | = 1) Eigenvalues are λ = ±1. Let |λ± ⟩ be eigenvectors with eigenvalues ±1. Since · ⃗ is Hermitian, ⃗v σ· ⃗ is diagonalizable. Then ⃗v σ v⃗ σ · ⃗ = |λ ⟩⟨λ | − |λ− ⟩⟨λ− | 3
3
2
2 1
1
2 2
2
2
1
2
2 3
∵
1
1
1
1
1
Thus exp(iθ⃗ v σ⃗ ) = eiθ λ1 λ1 + e−iθ λ−1 λ−1
·
| ⟩⟨ |
| ⟩⟨ | = (cos θ + i sin θ) |λ ⟩⟨λ | + (cos θ − i sin θ) |λ− ⟩⟨λ− | = cos θ(|λ ⟩⟨λ | + |λ− ⟩⟨λ− |) + i sin θ(|λ ⟩⟨λ | − |λ− ⟩⟨λ− |) ·⃗ = cos(θ)I + i sin(θ)⃗ v σ. 1
1
∵
1
1
1
1
1
1
Since ⃗v σ⃗ is Hermitian, λ1 and λ−1 are orthogonal. Thus
·
| ⟩
| ⟩ |λ ⟩⟨λ | + |λ− ⟩⟨λ− | = I. 1
1
1
1
2.36
Tr(σ1 ) = Tr Tr(σ2 ) = Tr Tr(σ3 ) = Tr
2.37
�� �� �� − �� �� �� 0 1 1 0
= 0
0 i
i 0
= 0
1 0
0 1
= 1
−
−1=0
1
1
1
1
17
Tr(AB) =
�⟨ | | ⟩ �⟨ | | ⟩ �⟨ | | ⟩⟨ | | ⟩ � ⟨ | | ⟩⟨ | | ⟩ �⟨ | | ⟩ i AB i
i
=
i AIB i
i
=
i A j j B i
i,j
=
j B i i A j
i,j
=
j BA j
j
= Tr(BA) 2.38
Tr(A + B) =
�⟨ | | ⟩ �⟨| |⟩ ⟨| |⟩ �⟨ | | ⟩ �⟨ | | ⟩ iA+B i
i
=
( iAi + iBi )
i
=
iAi +
i
iBi
i
= Tr(A) + Tr(B).
Tr(zA) =
�⟨ | | ⟩ � ⟨| |⟩ �⟨ | | ⟩ i zA i
i
=
z iAi
i
= z
iAi
i
= z Tr(A). 2.39
(1) (A, B)
≡ Tr(A†B).
(i)
� � � � † �� �� λi Bi = Tr A
A,
i
λi Bi
i
= Tr(A† λ1 B1 ) + ·· · + Tr(A† λn Bn ) (∵ Execise 2.38) = λ 1 Tr(A† B1 ) + · ·· + λn Tr(A† Bn ) = λi Tr(A† Bi )
� i
18
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
(ii)
∗ (A, B)∗ = Tr(A† B)
� � � ∗ ⟨ | †| ⟩⟨ | | ⟩ � ⟨ | † | ⟩∗ ⟨ | | ⟩∗ � ⟨ | | ⟩∗ ⟨ | †| ⟩∗ � ⟨ | †| ⟩ ⟨ | | ⟩ �⟨ | † | ⟩
=
i A j j B i
i,j
=
i A j
j B i
j B i
i A j
i,j
=
i,j
=
i B j j A i
i,j
=
iB Ai
i
= Tr(B † A) = (B, A). (iii)
(A, A) = Tr(A† A) =
�⟨ | † | ⟩ iA Ai
i
Since A † A is positive, i A† A i 0 for all i . Let ai be i-th column of A. If i A† A i = 0, then
⟨ |
|⟩≥ ⟨ |
| ⟩ |⟩
⟨i|A†A|i⟩ = a†i ai = ∥ai∥
2
= 0 iff a i = 0 .
Therefore (A, A) = 0 iff A = 0 . (2) (3) 2.40
[X, Y ] = X Y = = =
Y X
� −�� − � − � − �� � � � − �− � � −� 0 1 1 0 i 0
2i 0
= 2iZ
0 i
0 i
0 2i
−
i 0
i 0 0 i
0 i
i 0
0 1 1 0
19
[Y, Z ] = =
� − �� � − � �� − � � � − − 0 i
i 0
1 0
0 1
1 0
0 1
0 i
i 0
0 2i 2i 0
= 2iX
[Z, X ] =
� �� � − � �� � − � −− � 1 0
0 1
0 i
= 2i
0 1 1 0
0 1 1 0
1 0
0 1
i 0
= 2iY
2.41
{σ , σ } = σ σ + σ σ 1
2
1
= =
2
0 1 1 0 i 0
2
3
0 i
i 0
0 1 1 0
i 0 0 i
� − �� � � �� − �
{σ , σ } = 1
i 0 + 0 i
−
0 i
i 0
1 0
=0
3
1
0 + i
=0
{σ , σ } =
2
� �� − � � − �� � � � �− � 0 1 + 1 0
−
0 1
0 i
−
i 0
� �� � � �� � 1 0
0 1
0 1 0 1 + 1 0 1 0
−
=0
σ02 = I 2 = I 2
� � � −� � �
0 1 σ = 1 0 2 1
0 σ = i
i 0
1 σ = 0
0 1
2 2
2 3
−
= I
2
= I 2
= I
1 0
0 1
−
20
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.42
[A, B] + A, B AB = 2
{
}
− BA + AB + BA = AB 2
2.43
From eq (2.75) and eq (2.76), σ j , σk = 2δ jk I . From eq (2.77),
{
}
[σ j , σk ] + σ j , σk 2 3 2i l=1 ϵ jkl σl + 2δ jk I = 2
{
σ j σk =
∑
}
3
= δ jk I + i
�
ϵ jkl σl
l=1
2.44
By assumption, [A, B] = 0 and A, B = 0, then AB = 0. Since A is invertible, multiply by A−1 from left, then
{
}
A−1 AB = 0 IB = 0 B = 0. 2.45
[A, B]† = (AB
− BA)† = B † A† − A† B †
� † †�
= B ,A 2.46
[A, B] = AB
− BA = −(BA − AB) = − [B, A]
2.47
(i [A, B])† =
−i [A, B]† = −i B † , A† = −i [B, A]
� �
= i [A, B]
21
2.48
(Positive ) Since P is positive, it is diagonalizable. Then P = J =
√ †
P P =
√
P P =
√
2
P =
∑
i
λi i i , (λi
| ⟩⟨ |
� � | ⟩⟨ | � λ2i i i =
i
i
≥ 0). λi |i⟩⟨i| = P.
Therefore polar decomposition of P is P = U P for all P . Thus U = I , then P = P . (Unitary) Suppose unitary U is decomposed by U = W J where W is unitary and J is positive, J = U † U .
√
J =
√ †
U U =
√
I = I
Since unitary operators are invertible, W = U J −1 = U I −1 = U I = U . Thus polar decomposition of U is U = U . (Hermitian) Suppose H = U J . J =
√
H † H =
√
HH =
√
H 2 .
√
Thus H = U H 2 .
√
In general, H = H 2 . From spectral decomposition, H =
̸
√
2
H =
� � i
2
∑
i
λi i i , λ i
| ⟩⟨ | ∈ R.
λi i i =
| ⟩⟨ |
� � | ⟩⟨ | � | | | ⟩⟨ | ̸ λ2i i i =
λi i i = H
i
i
2.49
Normal matrix is diagonalizable, A =
∑
λi i i .
| ⟩⟨ | √ |λi||i⟩⟨i| . J = A† A = i
� i
U =
� | ⟩⟨ | � | || ⟩⟨ | ei i
i
A = U J =
λi ei i .
i
2.50
� �
� �
1 0 2 1 Define A = . A† A = . 1 1 1 1 Characteristic equation of A † A is det(A† A
√ λ± = ±2 5 3
− λI ) = λ − 3λ + 1 = 0. Eigenvalues of A† A are 2 √ and associated eigenvectors are |λ± ⟩ = √ √ −1 ± 5 . ∓ 2
1
10
2
5
�
�
22
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
A† A = λ + λ+ λ+ + λ− λ− λ− .
| ⟩⟨ |
J =
√ †
A A = =
| ⟩⟨ |
√ | ⟩⟨ | √ − | −⟩⟨ −| � √ − √ � √ − � � − √ √ √ · · λ+ λ+ λ+ +
λ
λ
λ
3 + 5 5 5 4 2 5 2 40
J −1 =
−
1 λ
| ⟩⟨ |
√
5 5 + 5 2 40
−
1 λ+ λ+ + λ+
√
3
2 5 2 + 2 6 2 5
λ
λ
√ − | −⟩⟨ −|
�
−2√ 5√ −2
4 2 5
− √ − 2
6+2 5
.
U = AJ −1 I’m tired. 2.51
H † H =
� √ � ��† √ � � 1 1 2 1
1 1 2 1
1 1
−
� � � � � � √ √
1 1 1 = 1 2 1
−
1 1
−
1 1 2 1
1 2 0 1 = = I. 1 2 0 2
−
2.52
H † =
� √ � ��† 1 1 2 1
1 1
−
� � √
1 1 = 2 1
1 = H. 1
−
Thus H 2 = I . 2.53
det(H
� √ − ��− √ − � −
1 λ 2 1 1 = λ 2 2 2 2 = λ 1
− λI ) =
1 2
λ
1 2
− − −
Eigenvalues are λ± = 2.54
1
±1 and associated eigenvectors are |λ±⟩ = √ ∓ 4
� √
2
2
1
−1 ±
� √ 2
.
�
23 Since [A, B] = 0, A and B are simultaneously diagonalize, A = exp(A) exp(B) =
�� � � �
��� | ⟩⟨ |
exp(ai ) i i
i
=
|i⟩⟨i|, B =
exp(bi ) i i
�
∑
i bi
|i⟩⟨i|.
| ⟩⟨ |
exp(ai + b j ) i i j j
| ⟩ ⟨ | ⟩⟨ |
exp(ai + b j ) i j δ i,j
| ⟩⟨ |
i,j
=
i ai
i
i,j
=
∑
exp(ai + bi ) i i
| ⟩⟨ |
i
= exp(A + B)
2.55
H =
�
E E E
| ⟩⟨ |
E
�− − � � − � � � �− − � | ⟩⟨ |�� �− ′ � � �− − ′ − � | ⟩⟨ ′| � � | ⟩⟨ | � | ⟩⟨ | iH (t2
U (t2 − t1 )U † (t2 − t1 ) = exp =
t1 )
ℏ
iE (t2
exp
′
i(E
exp
E E t1 )
iE (t2
exp
�
1
E E δ E,E
exp(0) E E
E
=
E E
E
= I Similarly, U † (t2
− t )U (t − t ) = I . 1
2
1
2.56
U =
∑
i λi
|λi⟩⟨λi| (|λi| = 1). log(U ) =
�
log(λ j ) λ j λ j =
| ⟩⟨ |
j
K =
−i log(U ) =
K † = ( i log U )† =
−
� j
�
�
iθ j λ j λ j where θ j = arg(λ j )
| ⟩⟨ |
j
θ j λ j λ j .
j
| ⟩⟨ |
θ j λ j λ j
| ⟩⟨
† � ∗ | | =
θ j λ j λ j =
j
⟩⟨ |
� j
�
− t ) |E ′⟩⟨E ′| ℏ
′
ℏ
E,E
=
t1 )
E )(t2
′
t1 )
ℏ
ℏ
E,E
=
iH (t2
exp
θ j λ j λ j = K
| ⟩⟨ |
24
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.57
|φ⟩ ≡
Ll ψ
� ⟨ | |† ⟩ | ⟩ ψ Ll Ll ψ
† M L |ψ⟩ ⟨ ψ|L†l M m m l † ⟨φ|M mM m|φ⟩ = † ⟨ψ|L L |ψ⟩ l
l
M m φ
� ⟨ |
|⟩
† M φ φ M m m
|⟩
=
� ⟨ | †
ψ Ll Ll ψ
M m Ll ψ
| ⟩
� ⟨ | † | |⟩ ⟩ · � ⟨ | †
† M L ψ ψ Ll M m m l
ψ Ll Ll ψ
=
| ⟩
M m Ll ψ
| ⟩
� ⟨ | †
† M L ψ ψ Ll M m m l
|⟩
2.58
⟨M ⟩ = ⟨ψ|M |ψ⟩ = ⟨ψ|m|ψ⟩ = m ⟨ψ|ψ⟩ = m ⟨M ⟩ = ⟨ψ|M |ψ⟩ = ⟨ψ|m |ψ⟩ = m ⟨ψ|ψ⟩ = m deviation = ⟨M ⟩ − ⟨M ⟩ = m − m = 0. 2
2
2
2
2
2
2
2
2
2.59
⟨X ⟩ = ⟨0|X |0⟩ = ⟨0|1⟩ = 0 ⟨X ⟩ = ⟨0|X |0⟩ = ⟨0|X |1⟩ = ⟨0|0⟩ = 1 standard deviation = ⟨X ⟩ − ⟨X ⟩ = 1 2
2
�
2
2
2.60 3
v⃗ σ⃗ =
·
� � � � −� � � − � � − vi σi
i=1
= v 1 =
0 1 0 + v2 1 0 i
i 1 + v3 0 0
0 1
v3 v1 iv2 v1 + iv2 v3
−
det(⃗ v σ⃗
· − λI ) = (v − λ)(−v − λ) − (v − iv )(v + iv ) = λ − (v + v + v ) = λ − 1 ( v|⃗ | = 1) 3
2
3
2 1
2
Eigenvalues are λ = (i) if λ = 1
2 2
1
2
2 3
∵
±1. v⃗ σ⃗
· − λI = ⃗v σ· ⃗ − I v −1 =
�
3
v1 + iv2
v1 iv2 v3 1
− − −
�
1
2
=
N lm ψ
� ⟨ |
|⟩
† N ψ ψ N lm lm
| ⟩
25
� � � | ⟩ − − � � − | ⟩⟨ | − �− − � − � − � � � − − �� 1
1+v3 2
Normalized eigenvector is λ1 =
λ1 λ1 = =
.
v3 v1 iv2 1
1 + v3 2
1
1
v3 iv2
1
v1
v1 iv2 1+v3 1 v3 1+v3
1
1 + v3 2
1 v3 v1 +iv2
v1 +iv2 1+v3
1 1 + v3 v1 iv2 2 v1 + iv2 1 v3 1 v3 v1 iv2 = I + v1 + iv2 v3 2 1 = (I +v⃗ σ⃗ ) 2 =
−
·
(ii) If λ =
−1. v⃗ σ⃗
· − λI = ⃗v ·σ⃗ + I
� � − �
v3 + 1 v1 iv2 = v1 + iv2 v3 + 1
Normalized eigenvalue is λ−1 =
| ⟩
2
|λ− ⟩⟨λ− | = 1
v3
1
1
1
=
−
1
− v −viv
1+ 3 1
�
�
.
� � − − − � � − − − − − − � − − −− � �− − � − �� −v
1
3
1+v3
2
1
2
−
1+v3 v1 +iv2
1
v1 iv2
1
v3
v1 +iv2 1 v3
2
v1 iv2 1 v3 1+v3 1 v3
1 1 v3 (v1 iv2 ) (v1 + iv2 ) 1 + v3 2 1 v3 v1 iv2 = I (v1 + iv2 v3 2 1 = (I v⃗ σ⃗ ). 2 =
−
− ·
While I review my proof, I notice that my proof has a defect. The case (v1 , v2 , v3 ) = (0, 0, 1), v3 second component of eigenstate, v11− −iv2 , diverges. So I implicitly assume v1 iv2 = 0. Hence my proof is incomplete. Since the exercise doesn’t require explicit form of projector, we should prove the problem more abstractly. In order to prove, we use the following properties of ⃗v σ⃗
− ̸
·
•v⃗ σ· ⃗ is Hermitian • v(⃗ σ· ⃗ ) = I where ⃗v 2
is a real unit vector.
We can easily check above conditions. v(⃗ σ⃗ )† = (v1 σ1 + v2 σ2 + v3 σ3 )†
·
= v 1 σ1† + v2 σ2† + v3 σ3† = v 1 σ1 + v2 σ2 + v3 σ3 = ⃗v σ⃗
·
(∵ Pauli matrices are Hermitian.)
26
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
3 2
v(⃗ σ⃗ ) =
·
� � � � � � � � � � (v j σ j )(vk σk )
j,k=1 3
=
v j vk σ j σk
j,k=1 3
=
3
v j vk
δ jk I + i
ϵ jkl σl
j,k=1
l=1
3
=
(∵ eqn(2.78) page78)
3
v j vk δ jk I + i
j,k=1
ϵ jkl v j vk σl
j,k,l =1
3
v j2 I
=
j =1
= I
v j2 = 1
∵
j
Proof. Suppose λ is an eigenstate of ⃗v σ⃗ with eigenvalue λ. Then
| ⟩
·
v⃗ σ⃗ λ = λ λ
· |⟩ |⟩ v(⃗ σ · ⃗ ) |λ⟩ = λ |λ⟩ 2
2
On the other hand v(⃗ σ⃗ )2 = I ,
·
v(⃗ σ⃗ )2 λ = I λ = λ
·
|⟩
∴
λ 2
|⟩ |⟩ |λ⟩ = |λ⟩ .
Thus λ2 = 1
⇒ λ = ±1. Therefore ⃗v σ· ⃗ has eigenvalues ±1. Let |λ ⟩ and |λ− ⟩ are eigenvectors with eigenvalues 1 and −1, respectively. I will prove that P ± = |λ± ⟩⟨λ± |. 1
1
1
1
In order to prove above equation, all we have to do is prove following condition. (see Theorem 2.0.1)
1
2
⟨ψ|(P ± − |λ± ⟩⟨λ± |)|ψ⟩ = 0 for all |ψ⟩ ∈ C . 1
(2.8)
Sincev⃗σ⃗ is Hermitian, λ1 and λ−1 are orthonormal vector (∵ Exercise 2.22). Let ψ be an arbitrary state. ψ can be written as
·
| ⟩
| ⟩
| ⟩
| ⟩∈C
2
|ψ⟩ = α |λ ⟩ + β |λ± ⟩ (|α| + |β | 1
1
2
= 1, α , β
∈ C).
2
27
⟨ψ|(P ± − |λ±⟩⟨λ±|)|ψ⟩ = ⟨ψ|P ±|ψ⟩ − ⟨ψ|λ±⟩ ⟨λ± |ψ⟩ . ⟨ψ|P ±|ψ⟩ = ⟨ψ| 12 (I ±v⃗ σ· ⃗ )|ψ⟩ 1 1 = ± ⟨ψv|⃗ σ · ⃗ )|ψ⟩ 2 2 1 1 = ± (|α| − |β | ) 2 2 1 1 = ± (2|α| − 1) ( |α| + |β | = 1) 2 2 ⟨ψ|λ ⟩ ⟨λ |ψ⟩ = |α| ⟨ψ|λ− ⟩ ⟨λ− |ψ⟩ = |β | = 1 − |α| Therefore ⟨ψ|(P ± − |λ± ⟩⟨λ± |)|ψ⟩ = 0 for all |ψ⟩ ∈ C . Thus P ± = |λ± ⟩⟨λ± |. 2
2
2
1
2
1
1
2
2
2
1
1
∵
2
2
1
1
1
2.61
⟨λ |0⟩ ⟨0|λ ⟩ = ⟨0|λ ⟩ ⟨λ |0⟩ 1 · ⃗ )|0⟩ = ⟨0| (I +v⃗ σ 2 1
1
1
1
1 = (1 + v3 ) 2
Post-measurement state is
|λ ⟩⟨λ |0⟩ = ⟨0|λ ⟩⟨λ |0⟩ 1
√
1
1
1
= =
�
1
1 1 + v3 v + iv2 1 (1 + v3 ) 2 1 2
� · � � � � � � 1 (1 + v3 ) 2 1 + v3 2
= λ1 .
| ⟩
�
1
v1 +iv2 1+v3
1
−v v −iv 1
1
3
2
2.62
† M = M . Suppose M m is a measurement operator. From the assumption, E m = M m m m Then
⟨ψ|E m|ψ⟩ = ⟨ψ|M m|ψ⟩ ≥ 0. for all ψ . Since M m is positive operator, M m is Hermitian. Therefore,
| ⟩
† M m = M m M m = M 2 = M m . E m = M m m Thus the measurement is a projective measurement.
28
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.63
† M m = M m =
√ † √ √ √ E m U m U m
E m
E m I E m
= E m .
† M is POVM. Since E m is POVM, for arbitrary unitary U , M m m 2.64
Read following paper: Lu-Ming Duan, Guang-Can Guo. Probabilistic cloning and identification of linearly independent quantum states. Phys. Rev. Lett.,80:4999-5002, 1998. arXiv eprint quant-ph/9804064. https://arxiv.org/abs/quant-ph/9804064 2.65
+ |1⟩ |0⟩ − |1⟩ |+⟩ ≡ |0⟩√ , |−⟩≡ √ 2 2 2.66
X 1 Z 2
⟨X Z ⟩ = 1
2
�⟨
00 + 11 2
|√ ⟨ |
�|
00 + 11 2
⟩√ | ⟩
� �| X 1 Z 2
�
=
00 + 11 2
⟩√ | ⟩
|10⟩√ − |01⟩ 2
�
=
⟨00|√ + ⟨11| |10⟩ − |01⟩ · √ = 0 2
2
2.67
Unsolved W V V = W W ⊥ . U : W V , U ′ : V V . ′ U w = U w U ′ (V ) U (W ) ′ U = U I ???
⊂ → ⊕ → → | ⟩ | ⟩ ∈ L ∈ L ⊕
2.68
|ψ⟩ = |
⟩√ | ⟩ . 2
00 + 11
Suppose a = a 0 0 + a1 1 and b = b 0 0 + b1 1 .
| ⟩
| ⟩ | ⟩ | ⟩ | ⟩ | ⟩ |a⟩ |b⟩ = a b |00⟩ + a b |01⟩ + a b |10⟩ + a b |11⟩ . If |ψ ⟩ = |a⟩ |b⟩, then a b = 1, a b = 0, a b = 0, a b = 1 since {|ij ⟩} is an orthonormal 0 0
0 0
basis.
0 1
0 1
1 0
1 0
1 1
1 1
29 If a0 b1 = 0, then a 0 = 0 or b 1 = 0. When a0 = 0 , this is contradiction to a0 b0 = 1. When b1 = 0 , this is contradiction to a1 b1 = 1. Thus ψ = a b .
| ⟩ ̸ | ⟩ | ⟩
2.69
Define Bell states as follows.
√ √ − √ √ −
1 00 + 11 1 0 = 2 2 0 1
|ψ ⟩ ≡ | ⟩√ | ⟩ 1
− |11⟩ = |ψ ⟩ ≡ |00⟩√ 2 2
1 0 0 1
1 2
0 01 + 10 1 1 = 2 2 1 0
|ψ ⟩ ≡ | ⟩√ | ⟩ 3
− |10⟩ = |ψ ⟩ ≡ |01⟩√ 2 4
0 1 1 0
1 2
First, we prove
{|ψi⟩} is a linearly independent basis. a |ψ ⟩ + a |ψ ⟩ + a |ψ ⟩ + a |ψ ⟩ = 0 1
∴
1
2
√
2
3
− − − −
3
4
4
a1 + a2 1 a3 + a4 =0 2 a3 a4 a1 a2
a1 + a2 = 0 a3 + a4 = 0
∴
∴
a3
a4 = 0
a1
a2 = 0
a 1 = a 2 = a 3 = a 4 = 0
Thus ψi is a linearly independent basis. Moreover ψi = 1 and ψi ψ j = δ ij for i, j = 1, 2, 3, 4. Therefore thonormal basis.
{| ⟩}
∥| ⟩∥
⟨ | ⟩
{|ψi⟩} forms an or-
2.70
For any Bell states we get ψi E I ψi = 21 ( 0 E 0 + 1 E 1 ). Suppose Eve measures the qubit Alice sent by measurement operators M m . The probability † M I ψ . Since M m† M is positive, p (m) are same that Eve gets result m is p i (m) = ψi M m m i m i values for all ψi . Thus Eve can’t distinguish Bell states.
| ⟩
⟨ | ⊗ | ⟩ ⟨ | | ⟩ ⟨ | | ⟩ ⟨ | ⊗ | ⟩
30
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
2.71
From spectral decomposition, ρ =
� � � � i
2
ρ =
2
| ⟩ ⟨ | ⟩⟨ | | ⟩⟨ |
p2i i i
Tr(ρ ) = Tr
� � | ⟩⟨ | ∑
2
pi i i
i
i
pi p j i j δ ij
i
��
pi = 1.
pi p j i i j j
i,j
=
≥ 0,
| ⟩⟨ |
i,j
=
�
pi ψi ψi , pi
=
| ⟩⟨ |
2
pi Tr( i i ) =
| ⟩⟨ |
i
�
2
pi i i =
⟨ | ⟩
i
� ≤ � p2i
pi = 1
i
(∵ p2i
i
≤ pi)
Suppose Tr(ρ2 ) = 1. Then i p2i = 1. If 0 pi < 1, then p 2i < pi . Thus only one p i = 1 and otherwise are 0. Therefore ρ = ψi ψi is pure state. Conversely if ρ is pure, then ρ = ψ ψ .
≤
| ⟩⟨ | | ⟩⟨ | Tr(ρ ) = Tr(|ψ⟩ ⟨ψ |ψ⟩ ⟨ψ|) = Tr(|ψ⟩⟨ψ|) = ⟨ψ|ψ⟩ = 1. 2
2.72
a b , a, d b∗ d C w.r.t. standard basis. Because ρ is density matrix, Tr(ρ) = a + d = 1. Define a = (1 + r3 )/2, d = (1 r3 )/2 and b = (r1 ir2 )/2, (ri R). In this case, (1) Since density matrix is Hermitian, matrix representation is ρ =
b
∈
� �
−
−
� � �
1 1 + r3 r1 a b ρ = ∗ = b d 2 r1 + ir2 1
∈
− ir −r
2
3
�
=
1 (I +r⃗ σ⃗ ). 2
·
Thus for arbitrary density matrix ρ can be written as ρ = 21 (I +r ⃗ σ⃗ ). Next, we derive the condition that ρ is positive. If ρ is positive, all eigenvalues of ρ should be non-negative.
·
det(ρ
2
2
2
− λI ) = (a − λ)(b − λ) − |b| = λ − (a + d)λ + ad − |b | = 0 (a + d) ± (a + d) − 4(ad − |b| ) λ =
√ � � � ± − − − √ ± − − − − ± √ | | 2
2
1 = = =
1
4
1
r32
r12 +r22
4
4
2
1
1
r12
(1
2
1
r⃗
2 1 r⃗ = 2
±| |
2
r22
r32 )
2
∈ R and
31 Since ρ is positive, 1−|2r⃗| 0 r⃗ 1. Therefore an arbitrary density matrix for a mixed state qubit is written as ρ = 21 (I +r ⃗ σ⃗ ).
≥ →| |≤
(2) ρ = I /2
·
→r⃗ = 0. Thus ρ = I /2 corresponds to the origin of Bloch sphere.
(3) 1 1 ρ2 = (I +r⃗ σ⃗ ) (I +r ⃗ σ⃗ ) 2 2
·
=
·
1 I + r2⃗ σ⃗ + 4
·
� � || r j rk
δ jk I + i
j,k
· ||
ϵ jkl σl
l=1
1 I + r2⃗ σ⃗ +r⃗ 2 I 4 1 Tr(ρ2 ) = (2 + 2r⃗ 2 ) 4 =
� � 3
If ρ is pure, then Tr(ρ2 ) = 1. 1 1 = Tr(ρ2 ) = (2 + 2r ⃗ 2 ) 4 ∴ r ⃗ = 1.
||
||
Conversely, if r ⃗ = 1, then Tr(ρ2 ) = 41 (2 + 2r⃗ 2 ) = 1. Therefore ρ is pure.
| |
||
2.73
Theorem 2.6
ρ =
�
pi ψi ψi =
i
| ⟩⟨ |
�| i
ψ˜i ψ˜i =
⟩⟨ |
�| j
where u is unitary. Transformation in theorem 2.6, ψ˜i =
ϕ˜ j ϕ˜ j =
⟩⟨ |
�
q j ϕ j ϕ j
| ⟩⟨ | ⇔ | ⟩
j
∑ |⟩ �| ⟩ ·· · | ⟩� � | ⟩ ·· · | ⟩ � j u ij
| ⟩
ψ˜1
where k = rank(ρ).
ψ˜i =
� j
uij ϕ˜ j
| ⟩
ϕ˜ j , corresponds to
ψ˜k =
ϕ˜1
ϕ˜k
U T
d k=1 λk
∑
From spectral theorem, density matrix ρ is decomposed as ρ = k k where d = dim . Without loss of generality, we can assume pk > 0 for k = 1 , l where l = rank(ρ) and pk = 0 for k = l + 1, , d. Thus ρ = lk=1 pk k k = lk=1 k˜ k˜ , where k˜ = λk k . Suppose ψi is a state in support ρ. Then
H
···
| ⟩
∑ �
∑
| ⟩⟨ |
| ⟩⟨ |
l
|ψi⟩ = cik |k⟩ , k √ picik and uik = √ . λ =1
Define p i =
1 cik 2 λk
∑| k
|
k
�|
cik
k
2
|
= 1.
·· ·
| ⟩⟨ | | ⟩ √ | ⟩
32
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Now
�|
uik
k
2
|
=
� k
pi cik λk
2
| |
= pi
cik 2 = 1. λk
�| k
|
Next prepare an unitary operator 1 such that ith row of U is [ui1 define another ensemble such that
�|
ψ˜1
| ⟩ √ | ⟩
ψ˜i
⟩ ·· · |
� � ⟩ ·· · | ⟩ | ψ˜l
=
k˜1
where ψ˜i = pi ψi . From theorem 2.6, ρ =
k˜ k˜ =
� | ⟩⟨ | � | k
� ⟩ ·· · | ⟩ k˜l
· ·· uik · · · uil]. Then we can
U T
ψ˜k ψ˜k .
k
⟩⟨ |
Therefore we can obtain a minimal ensemble for ρ that contains ψi . Moreover since ρ −1 = k λ1k k k ,
∑
| ⟩⟨ |
⟨ψi|ρ− |ψi⟩ = Hence,
| ⟩
1
� k
1 ψi k k ψi = λk
⟨ | ⟩⟨ | ⟩
cik 2 1 = . λk pi
�| k
|
1 ⟨ψ |ρ |ψ ⟩ = p i . i
1
−
i
2.74
ρAB = a a A
| ⟩⟨ | ⊗ |b⟩⟨b|B ρA = TrB ρAB = |a⟩⟨a| Tr(|b⟩⟨b|) = |a⟩⟨a|
Tr(ρ2A ) = 1 Thus ρA is pure. 2.75
Define Φ± =
| ⟩
√ 12 (|00⟩ ± |11⟩) and |Ψ± ⟩ = √ 12 (|01⟩ ± |10⟩).
|Φ±⟩⟨Φ±|AB = 21 (|00⟩⟨00| ± |00⟩⟨11| ± |11⟩⟨00| + |11⟩⟨11|) 1 I TrB (|Φ± ⟩⟨Φ± |AB ) = (|0⟩⟨0| + |1⟩⟨1|) = 2 2 |Ψ±⟩⟨Ψ± | = 21 (|01⟩⟨01| ± |01⟩⟨10| ± |10⟩⟨01| + |10⟩⟨10|) 1 I TrB (|Ψ± ⟩⟨Ψ± |) = (|0⟩⟨0| + |1⟩⟨1|) = 2 2 2.76 1
By Gram-Schmidt procedure construct an orthonormal basis { uj } (row vector) with
... . Then define unitary U = . . u1
ui
.
ul
ui
= [ui1 · · · uik · · · uil ].
33 Unsolved. I think the polar decomposition can only apply to square matrix A, not arbitrary linear operators. Suppose A is m n matrix. Then size of A† A is n n. Thus the size of U should be m n. Maybe U is isometry, but I think it is not unitary. I misunderstand linear operator.
×
×
×
Quoted from ”Advanced Liner Algebra” by Steven Roman, ISBN 0387247661. A linear transformation τ : V
→ V is called a linear operator on V .
2
Thus coordinate matrices of linear operator are square matrices. And Nielsen and Chaung say at Theorem 2.3, ”Let A be a linear operator on a vector space V .” Therefore A is a linear transformation such that A : V V .
→
2.77
|ψ⟩ = |0⟩ |Φ ⟩ 1 = |0⟩ √ (|00⟩ + |11⟩) 2 +
�
�
�⟩ √ |
= (α φ0 + β φ1 )
| ⟩
|
�⟩
1 ( φ0 φ0 + φ1 φ1 ) 2
⟩ |
where φi are arbitrary orthonormal states and α, β C. We cannot vanish cross term. Therefore ψ cannot be written as ψ = i λi i A i B i C .
| ⟩ | ⟩
| ⟩
2.78
∑
∈ |⟩ |⟩ |⟩
Proof. Former part.
If ψ is product, then there exist a state φA for system A, and a state φB for system B such that ψ = φA φB . Obviously, this Schmidt number is 1. Conversely, if Schmidt number is 1, the state is written as ψ = φA φB . Hence this is a product state.
| ⟩
| ⟩
| ⟩ | ⟩| ⟩
| ⟩
| ⟩ | ⟩ | ⟩
Proof. Later part.
( ) Proved by exercise 2.74. ( ) Let a pure state be ψ = i λi iA iB . Then ρA = TrB ( ψ ψ ) = i λ2i i i . If ρA is a pure state, then λ j = 1 and otherwise 0 for some j. It follows that ψ j = jA jB . Thus ψ is a product state.
⇒ ⇐
|⟩
| ⟩
∑
| ⟩| ⟩
∑
| ⟩⟨ | | ⟩⟨ | | ⟩ | ⟩| ⟩
2.79
2
According to Roman, some authors use the term linear operator for any linear transformation from V to W .
34
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Procedure of Schmidt decomposition. Goal: ψ = i λi iA iB
| ⟩
∑ √ |
⟩| ⟩
• Diagonalize reduced density matrix ρ A = ⟨ iA | ) | ψ ⟩ • Derive |iB ⟩, |iB ⟩ = (I ⊗√
∑
i
λi iA iA .
| ⟩⟨ |
λi
• Construct |ψ⟩.
(i)
√ 12 (|00⟩ + |11⟩) This is already decomposed. (ii)
�
� �
|00⟩ + |01⟩ + |10⟩ + |11⟩ = |0⟩√ |0⟩√ + |1⟩ + |1⟩ ⊗ 2 2 2
�
+ |1⟩ | ⟩ |0⟩√ 2
= ψ ψ where ψ =
| ⟩| ⟩
(iii)
|ψ⟩AB = √ 13 (|00⟩ + |01⟩ + |10⟩) ρAB = |ψ⟩⟨ψ|AB det(ρA
−
1 ρA = TrB (ρAB ) = (2 0 0 + 0 1 + 1 0 + 1 1 ) 3 2 1 1 λI ) = λ λ =0 3 3 9 1 λ2 λ + = 0 9 1 5/3 3 5 λ = = 2 6
| ⟩⟨ | | ⟩⟨ | | ⟩⟨ | | ⟩⟨ | − −
� − �� �
−
± √
Eigenvector with eigenvalue λ 0
≡
± √
√ 3+ 5 6
is λ0
|
√ 5 − Eigenvector with eigenvalue λ ≡ is |λ 6
1
1+ 5 2
1
1
1
5
5
ρA = λ 0 λ0 λ0 + λ1 λ1 λ1 .
| ⟩⟨ |
| ⟩⟨ |
|a ⟩ ≡ (I ⊗⟨√ λλ |) |ψ⟩ |a ⟩ ≡ (I ⊗⟨√ λ |) |ψ⟩ 0
0
0
1
λ1
Then 1
|ψ⟩ =
� √ | ⟩| ⟩
λi ai λi .
i=0
.
5
2
2
1
1
5+ 5 2
3
1
� √ � ⟩ ≡ � √ � −√ � ⟩ ≡ � −√ 1
.
35 (It’s too tiresome to calculate ai )
| ⟩
2.80
∑∑ | ⟩ | ⟩ | ⟩⟨ |
Let ψ = Define U = Then
| ⟩
i
∑ ∑
λi ψi A ψi B and ϕ = i λi ϕi i ψ j ϕ j A and V = j ψ j ϕ j .
| ⟩A |ϕi⟩B . | ⟩⟨ |
| ⟩
(U
⊗ V ) |ϕ⟩ = =
� �
λi U ϕi
| ⟩A V |ϕi⟩B
i
λi ψi
| ⟩A |ψi⟩B
i
= ψ .
|⟩
2.81
Suppose ρA = TrR AR2 AR2 =
|
⟩⟨
|
∑
i
λi i i . Define AR1 = (I A
TrR ( AR1 AR1 ) = TrR
|
⟩⟨
|
= TrR
| ⟩⟨ |
� �|
|
⟩
2
� �
(I A ⊗ U R ) |AR2 ⟩⟨AR2 | (I A ⊗ U R† ) AR2 ⟩⟨AR2 | (I ⊗ U † )(I ⊗ U ) A
R
= TrR ( AR2 AR2 ) = ρ A .
⊗ U R) |AR ⟩.
|
⟩⟨
|
A
R
Thus AR1 is also a purification of ρ A .
|
⟩
2.82
(1) Let ψ =
| ⟩
∑ √ | i
pi ψi i .
⟩| ⟩
TrR ( ψ ψ ) =
| ⟩⟨ |
�
pi ψi ψi
| ⟩⟨ |
i
Thus ψ is a purification of ρ.
| ⟩
(2) Probability Tr [(I
⊗|i⟩⟨i|) |ψ⟩⟨ψ|] = ⟨ψ|(I ⊗ |i⟩⟨i|)|ψ⟩ = pi ⟨ψi|ψi⟩ = pi.
Post-measurement state
√ pi |ψi⟩ i⟩⟨i| |ψ⟩) ⊗|√ = √ = |ψi ⟩ .
(I
pi
(3)
pi
36
CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS
Suppose AR is a purification of ρ such that AR = others purification is written as (I U ) AR .
| ⟩
| ⟩
∑ √ | i
pi ψi ri . By exercise 2.81, the
⊗ | ⟩ √ pi |ψi⟩|ri⟩ (I ⊗ U ) |AR⟩ = (I ⊗ U ) i √ = pi |ψi ⟩ U |ri ⟩
⟩| ⟩
�
� � √ | i
=
pi ψi i
i
∑ | ⟩⟨ |
⟩| ⟩
where U = i i ri . By (2), if we measure the system R w.r.t i , post-measurement state for system A is ψi with probability p i , which prove the assertion.
| ⟩
| ⟩
Problem 2.1
From Exercise 2.35, ⃗n σ⃗ is decomposed as
·
n⃗ σ⃗ = λ1 λ1
| ⟩⟨ | − |λ− ⟩⟨λ− | σ · ⃗ with eigenvalues ±1. ·
where λ±1 are eigenvector of ⃗n Thus
| ⟩
1
1
⃗n σ⃗ ) = f (θ) λ1 λ1 + f ( θ) λ−1 λ−1 f (θ f (θ) + f ( θ) f (θ) f ( θ) f (θ) + f ( θ) f (θ) f ( θ) = + λ1 λ1 + 2 2 2 2 f (θ) + f ( θ) f (θ) f ( θ) = ( λ1 λ1 + λ−1 λ−1 ) + ( λ1 λ1 λ−1 λ−1 ) 2 2 f (θ) + f ( θ) f (θ) f ( θ) = I + n⃗ σ⃗ 2 2
·
�
Problem 2.2
Unsolved Problem 2.3
Unsolved
| ⟩⟨ | −
− | ⟩⟨ | − − | ⟩⟨ |
�
− | ⟩⟨ | | ⟩⟨ | − − − ·
�
− −
− −
− − | ⟩⟨ | − | ⟩⟨ |
�|
λ−1 λ−1
⟩⟨ |
Chapter 8
Quantum noise and quantum operations 8.1
Density operator of initial state is written by ψ ψ and final state is written by U ψ ψ U † . Thus time development of ρ = ψ ψ can be written by (ρ) = U ρU † .
| ⟩⟨ |
| ⟩⟨ |
| ⟩⟨ |
E
8.2
From eqn (2.147) (on page 100),
† M m ρM m
† M m ρM m
E (ρ)
m = = . † † Tr E m (ρ) Tr(M m M m ρ) Tr(M m ρM m ) † M ρ) = Tr(M ρM m† ) = Tr E (ρ). And from eqn (2.143) (on page 99), p(m) = Tr(M m m m m
ρm =
8.3
8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13
37
38
8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 8.29 8.30 8.31 8.32 8.33 8.34 8.35
CHAPTER 8. QUANTUM NOISE AND QUANTUM OPERATIONS
Chapter 9
Distance measures for quantum information 9.1
1 D((1, 0), (1/2, 1/2)) = ( 1 1/2 + 0 2 1 1 1 = + 2 2 2 1 = 2
| − | | − 1/2|)
� �
1 D ((1/2, 1/3, 1/6), (3/4, 1/8, 1/8)) = ( 1/2 3/4 + 1/3 1/8 + 1/6 2 1 = (1/4 + 5/24 + 1/24) 2 1 = 4
| − | | − | | − 1/8|)
9.2
D (( p, 1
− p), (q, 1 − q )) = 21 (| p − q | + |(1 − p) − (1 − q )|) 1 = ( | p − q | + | − p + q |) 2 = | p − q |
9.3
F ((1, 0), (1/2, 1/2)) =
√ · √ · √ · √
F ((1/2, 1/3, 1/6), (3/4, 1/8, 1/8)) = =
1 1/2 +
0 1/2 =
1/2 3/4 +
√ √ 4 6+ 3 12
39
√ 12
1/3 1/8 +
·
√
1/6 1/8
·
40
CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION
9.4
Define rx = p x
− q x. Let U be the whole index set. max p(S )
|
S
− q (S )| = max S = max S
S
∑∈
x S rx is
rx =
x S rx
( px rx
∈
rx +
x S rx 0
∈
q x
x S
� � �
x S
∑ ∈
x S
x S
written as
∈ − � −
px
x S
= max Since
� ∈ � ∈ �
q x )
rx ,
(9.1)
x S rx <0
∈ ≥
∈
is maximized when S = x U rx Define S + = x U rx 0 and S − = x Now the sum of all rx is 0,
{ ∈ | ≥ 0} or S = {x ∈ U |rx < 0}. { ∈ | ≥ } { ∈ U |rx < 0}.
� � � �∈ −∈ � ∈ rx =
rx +
x U ∴
x S +
rx = 0
x S
−
rx =
rx .
x S +
x S
∈
∈
−
Thus max S
On the other hand,
� � � ∈ ∈ − ∈ �| − | �∈ | | �∈ | | � | | �∈ − �∈ ∈ ∈ � � ∈ �∈ ∈ � rx =
x S
D( px , q x ) = = = = =
1 2 1 2 1 2 1 2 1 2
rx =
x S +
px
rx .
x S
−
q x
x U
rx
x U
rx +
x S +
1 2
x S +
x S
rx
x S
−
rx +
x S +
1 2
rx
x S +
= max S
rx
−
rx
=
1 2
rx .
x S
∈
rx
x S +
(∵ eqn(9.2))
(9.2)
41 Therefore D( px , q x ) = maxS 9.5
∑ ∈ − ∑ ∈ ∑ ∈ x S p x
From eqn (9.1) and (9.2), maximizing Hence
D( px , q x ) = max( p(S ) S
x S q x
x S rx
= maxS p(S )
|
− q (S )|.
is equivalent to maximizing
− q (S )) = max S
∑∈
x S r x .
�� � � − px
q x .
x S
x S
∈
∈
9.6
Define ρ =
3 4
1 4
2 3
1 3
|0⟩⟨0| + |1⟩⟨1|, σ = |1⟩⟨1| + |1⟩⟨1|. 1 D(ρ, σ) = Tr ρ σ 2 = D((3/4, 1/4), (2/3, 1/3)) 1 3 2 1 1 = + 2 4 3 4 3 1 1 1 = + 2 12 12 1 = 12
| − |
� − �
Define ρ =
3 4
1 4
2 3
− �
�
1 3
|0⟩⟨0| + |1⟩⟨1|, σ = |+⟩⟨+| + |−⟩⟨−|. |+⟩⟨+| = 21 (|0⟩⟨0| + |0⟩⟨1| + |1⟩⟨0| + |1⟩⟨1|) |−⟩⟨−| = 21 (|0⟩⟨0| − |0⟩⟨1| − |1⟩⟨0| + |1⟩⟨1|) ρ
− σ = =
(ρ
� − � | ⟩⟨ | − 3 4
1 2
1 0 0 4
1 0 0 ( 0 1 + 1 0 )+ 6 1 1 (0 1 + 1 0) 1 1 6 4
| ⟩⟨ | | ⟩⟨ |
� − � | ⟩⟨ | 1 4
1 2
1 1
| ⟩⟨ | − | ⟩⟨ | | ⟩⟨ | − | ⟩⟨ |
− σ)†(ρ − σ) = 41 |0⟩⟨0| − 4 1· 6 |0⟩⟨1| + 61 |0⟩⟨0| + 6 1· 4 |0⟩⟨1| − 4 1· 6 |1⟩⟨0| + 61 |1⟩⟨1| + 4 1· 6 |1⟩⟨0| + 41 |1⟩⟨1| 1 1 = + (|0⟩⟨0| + |1⟩⟨1|) 4 6 2
�
2
2
2
2
�
1 D(ρ, σ) = Tr ρ σ 2 1 1 = + 42 62
� | − |
9.7
2
42
CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION
Since ρ
− σ is Hermitian, we can apply spectral decomposition. Then ρ − σ is written as ρ
− σ =
k
n
�
�
λi i i +
| ⟩⟨ |
i=1
where λi are positive eigenvalues for i = 1, Define Q = ki=1 λi i i and S = Therefore ρ σ = P S .
∑
·· · ∑ −
| ⟩⟨ |
− − Proof of |ρ − σ | = Q + S .
λi i i
i=k+1
| ⟩⟨ |
, k and negative eigenvalues for i = k + 1, , n. i i . Then P and S are positive operator.
n i=k+1 λi
·· ·
| ⟩⟨ |
|ρ − σ| = |Q − S | = (Q − S )† (Q − S ) = (Q − S ) = Q − QS − SQ + S
� √ √ √ � � | ⟩⟨ | � | | | ⟩⟨ | 2
2
= =
2
Q2 + S 2
λ2i i i
i
=
λi i i
i
= Q + S
9.8
Suppose σ = σ i . Then σ = D
∑
i pi σi .
�� � �� � � � ≤ � pi ρi , σ = D
pi ρi ,
i
i
pi σi
(9.3)
i
pi D(ρi , σi )
(∵ eqn(9.50))
(9.4)
i
=
pi D(ρi , σ). (∵ assumption).
(9.5)
i
9.9 9.10 9.11 9.12
Suppose ρ = 21 (I +r ⃗ σ⃗ ) and σ = 21 (I +s⃗ σ⃗ ) where ⃗v and ⃗s are real vectors s.t. v⃗ ,s⃗
·
·
E (ρ) = p I 2 + (1 − p)ρ, E (σ) = p I 2 + (1 − p)σ.
| | | | ≤ 1.
43
1 D( (ρ), (σ)) = Tr (ρ) (σ) 2 1 = Tr (1 p)(ρ σ) 2 1 = (1 p) Tr ρ σ 2 = (1 p)D(ρ, σ) r ⃗ s⃗ = (1 p) 2
E E
|E − E | | − − | − | − | − − | −|
Is this strictly contractive? 9.13
√
Bit flip channel E 0 = pI , E 1 =
√ 1 − pσ . x
E (ρ) = E ρE † + E ρE † = pρ + (1 − p)σx ρσx . 0
Since σ x σx σx = σ x , σx σy σx = Thus
0
1
1
−σy and σ xσz σx = −σz , then σxr(⃗ σ· ⃗ ) = r σx − r σy − r σ . 1
2
3
3
1 D( (ρ), (σ)) = Tr (ρ) (σ) 2 1 = Tr p(ρ σ) + (1 p)(σx ρσx σx σσ x ) 2 1 1 p Tr ρ σ + (1 p) Tr σx (ρ σ)σx 2 2 = pD(ρ, σ) + (1 p)D(σx ρσx , σx σσx )
E E
|E − E | | − − | − | − −
≤
−
|
| |
−
= D(ρ, σ) (∵ eqn(9.21)). Suppose ρ0 = 21 (I +r ⃗ σ⃗ ) is a fixed point. Then
·
ρ0 = (ρ0 ) = pρ 0 + (1 p)σx ρ0 σx p)ρ0 (1 p)σx ρ0 σx = 0 ∴ (1
E
∴
−
− − − (1 − p)(ρ − σx ρ σx ) = 0 0
ρ 0 = σ x ρ0 σx 1 1 (I + r1 σx + r2 σy + r3 σz ) (I + r1 σx ∴ 2 2 ∴
− r σy − r σz ) 2
3
Since I, σx , σy , σz are linearly independent, r 2 = r2 and r 3 = r3 . Thus r 2 = r 3 = 0. Therefore the set of fixed points for the bit flip channel is ρ ρ = 21 (I + rσx ), r 1, r R
{
9.14
}
−
{ |
−
| | ≤ ∈ }
44
CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION
� � � � �
F (U ρU † , U σ U † ) = Tr (U ρU † )1/2 σ(U ρU † ) = Tr
Uρ1/2 σρ 1/2 U †
= Tr(U ρ1/2 σρ 1/2 U † ) = Tr( ρ1/2 σρ1/2 U † U ) = Tr
ρ1/2 σρ1/2
= F (ρ, σ)
√
√ U AU † = U AU † is not restricted for positive operator.
I think the fact Suppose A is a normal matrix. From spectral theorem, it is decomposed as
�
ai i i .
� � �
| ⟩⟨ |
A =
i
| ⟩⟨ |
Let f be a function. Then f (U AU † ) = f (
ai U i i U † )
i
=
f (ai )U i i U †
| ⟩⟨ |
i
= U (
f (ai )U i i U † )U †
i
| ⟩⟨ |
= U f (A)U †
9.15
√ |ψ⟩ = (U R ⊗ √ √ ρU Q ) |m⟩ is any fixed purification of ρ, and |φ⟩ = (V R ⊗ σV Q ) |m⟩ is purification √ √ √ √ √ of σ. Suppose ρ σ = | ρ σ |V is the polar decomposition of ρ σ. Then |⟨ψ|φ⟩ | = ⟨m| U R† V R ⊗ U Q† √ ρ√ σV Q |m⟩ √ √ = Tr (U R† V R )T U Q† ρ σV Q † √ ρ√ σV = Tr V RT U R∗ U Q Q √ √ = Tr V Q V RT U R∗ U Q† ρ σ √ √ = Tr V Q V RT U R∗ U Q† | ρ σ |V √ √ = Tr V V Q V RT U R∗ U Q† | ρ σ| ≤ Tr |√ ρ√ σ|
� � � � � �
= F (ρ, σ)
� � � � � �
Choosing V Q = V † , V RT = (U Q∗ U R† )† we see that equality is attained. 9.16
45 I think eq (9.73) has a typo. Tr(A† B) = m A B m should be Tr(AT B) = m A errata list. In order to show that this exercise, I will prove following two properties,
⟨ | ⊗ B|m⟩. See
⟨ | ⊗ | ⟩
⟨ | ⊗ A)|m⟩ , (I ⊗ A) |m⟩ = (AT ⊗ I ) |m⟩ where A is a linear operator and |m⟩ is unnormalized maximally entangled state, |m⟩ = Tr(A) = m (I
⟨m|I ⊗ A|m⟩ = =
� ⟨ | ⊗ | ⟩ �⟨ | | ⟩⟨ | | ⟩ � ⟨| | ⟩ �⟨ | | ⟩ ii (I
A) jj
ij
i I j i A j
ij
=
δ ij i A j
ij
=
iAi
i
= Tr(A) Suppose A =
∑
ij aij
|i⟩⟨ j |. (I
⊗ A) |m⟩ = =
� � ⊗ | ⟩⟨ | | ⟩ � | ⟩ ⊗ | ⟩⟨ | ⟩ � | ⟩⊗| ⟩ � | ⟩⊗| ⟩ � | ⟩⊗| ⟩ � � | ⟩⟨ | ⊗ | ⟩ � | ⟩⟨ | ⟩ ⊗ | ⟩ � | ⟩ ⊗| ⟩ � |⟩ I
aij i j
kk
ij
k
aij k
i j k
aij k
i δ jk
aij j
i
a ji i
j
ijk
=
ijk
=
ij
=
ij
(AT
⊗ I ) |m⟩ =
a ji i j
I
ij
=
kk
k
a ji i j k
k
ij
=
a ji i δ jk
k
ij
=
a ji ij
ij
= (I
⊗ A) |m⟩
Thus Tr(AT B) = Tr(BA T ) = m I
⟨ | ⊗ BAT |m⟩ = ⟨m|(I ⊗ B)(I ⊗ AT )|m⟩ = ⟨m|(I ⊗ B)(A ⊗ I )|m⟩ = ⟨m|A ⊗ B |m⟩ .
∑| ⟩ i
ii .
46
CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION
9.17
If ρ = σ, then F (ρ, σ) = 1. Thus A(ρ, σ) = arccos F (ρ, σ) = arccos 1 = 0. If A(ρ, σ) = 0, then arccos F (ρ, σ) = 0 cos(arccos F (ρ, σ)) = cos(0) p.411, the fifth line from bottom).
⇒
⇒ F (ρ, σ) = 1 (
∵
text
9.18
For 0 1,
≤ x ≤ y ≤ 1, arccos(x) ≥ arccos(y). From F (E (ρ), E (σ)) ≥ F (ρ, σ) and 0 ≤ F (E (ρ), E (σ)), F (ρ, σ) ≤ arccos F ( (ρ), (σ)) ∴
E E ≥ arccos F (ρ, σ) A(E (ρ), E (σ)) ≥ A(ρ, σ)
9.19
From eq (9.92) F
�� � � � √ ≥ � pi ρi ,
pi σi
i
pi pi F (ρi , σi )
i
i
=
pi F (ρi , σi ).
i
9.20
Suppose σ i = σ. Then F
�� � �� � � �� � � � ≥ � pi ρi , σ = F
pi ρi ,
i
i
= F
pi σ
i
pi ρi ,
i
pi σi
i
pi F (ρi , σi ) (∵ Exercise9.19)
i
=
pi F (ρi , σ)
i
9.21
1
− F (|ψ⟩ , σ)
2
=1
− ⟨ψ|σ|ψ⟩
D( ψ , σ) = max Tr(P (ρ
| ⟩
− σ)) (where P is projector.) ≥ Tr (|ψ⟩⟨ψ| (ρ − σ)) = ⟨ψ |(|ψ⟩⟨ψ| − σ)|ψ⟩ = 1 − ⟨ψ|σ |ψ⟩ = 1 − F (|ψ⟩ , σ) . P
2
9.22
(∵ eq(9.60))
