Quantum Computation and Quantum Information is a textbook about quantum information science written by Michael Nielsen and Isaac Chuang, regarded as a standard text on the subject.[1] It is informally known as 'Mike and Ike', after the candies of that name.[2] The book assumes minimal prior experience with quantum mechanics and with computer science, aiming instead to be a self-contained introduction to the relevant features of both. (Lov Grover recalls a postdoc disparaging it with the remark, 'The book is too elementary â it starts off with the assumption that the reader does not even know quantum mechanics.'[3]) The focus of the text is on theory, rather than the experimental implementations of quantum computers, which are discussed more briefly.[4]
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Nielsen, Isaac L. In this first comprehensive introduction to the main ideas and techniques of quantum computation and information, Michael Nielsen and Isaac Chuang ask the question: What are the ultimate physical limits to computation and communication? They detail such remarkable effects as fast quantum algorithms, quantum. QUANTUM SYSTEMS FOR INFORMATION TECHNOLOGY Organizer: PDDr. Nielsen and Isaac L. Landau and E. Lifshitz, Quantum Mechanics, Vol. 3, § 50, Problem 3. Lectures by Valery Pokrovsky âSemiclassical and Adiabatic Approxi.
As of January 2020, the book has been cited over 39,000 times on Google Scholar.[5] In 2019, Nielsen adapted parts of the book for his Quantum Country project.[6]
Table of Contents (Tenth Anniversary Edition)[edit]
Reviews[edit]
Peter Shor called the text 'an excellent book'. Lou Grover called it 'the bible of the quantum information field'. Scott Aaronson said about it, ''Mike and Ike' as it's affectionately called, remains the quantum computing textbook to which all others are compared.'[7]David DiVincenzo said, 'More than any of the previous attempts, this book has identified the essential foundations of quantum information theory with a clarity that has even, in a few cases, permitted the authors to obtain some original results and point toward new research directions.'[8] A review in the November 2001 edition of Foundations of Physics says, 'Among the handful of books that have been written on this new subject, the present volume is the most complete and comprehensive.'[9]
Editions[edit]
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Solution for âQuantum Computation and Quantum Information: 10th Anniversary Editionâ by Nielsen and Chuang goropikari August 11, 2018 Copylight Notice: c bna This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Contents 0.1 Eratta list . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Introduction to quantum mechanics 2 3 8 Quantum noise and quantum operations 37 9 Distance measures for quantum information 39 1 2 CONTENTS 0.1 Eratta list â â ⢠p.101. eq (2.150) Ï = m p(m)Ïm should be Ïâ² = m p(m)Ïm . â â ⢠p.408. eq (9.49) i pi D(Ïi , Ïi ) + D(pi , qi ) should be i pi D(Ïi , Ïi ) + 2D(pi , qi ). eqn (9.48) = â pi Tr(P (Ïi â Ïi )) + i ⤠â â pi Tr(P (Ïi â Ïi )) + â i = â = (pi â qi ) (âµ Tr(P Ïi ) ⤠1) i â pi Tr(P (Ïi â Ïi )) + 2 i â (pi â qi ) Tr(P Ïi ) i i (pi â qi ) 2 pi Tr(P (Ïi â Ïi )) + 2D(pi , qi ) i ⢠p.409. Exercise 9.12. If Ï = Ï, then D(Ï, Ï) = 0. Furthermore trace distance is nonnegative. Therefore 0 ⤠D(E(Ï), E(Ï)) ⤠0 â D(E(Ï), E(Ï)) = 0. So I think the map E is not strictly contractive. If p ̸= 1 and Ï Ì¸= Ï, then D(E(Ï), E(Ï)) < D(Ï, Ï) is satisfied. ⢠p.411. Exercise 9.16. eqn(9.73) Tr(Aâ B) = â¨m|A â B|mâ© should be Tr(AT B) = â¨m|A â B|mâ©. ] ] [ [ 1 0 i 0 , In this case, . B= Simple counter example is the case that A = 0 0 0 0 ] ] [ ][ [ âi 0 âi 0 1 0 , = AB= 0 0 0 0 0 0 â Tr(Aâ B) = âi,  i 0 AâB = 0 0 0 0 0 0 0 0 0 0  0 0  0 0 â¨m|A â B|mâ© = (â¨00| + â¨11â©)(A â B)(|00â© + |11â©) = i. Thus Tr(Aâ B) ̸= â¨m|A â B|mâ©. By using following relation, we can prove. (I â A) |mâ© = (AT â I) |mâ© Tr(A) = â¨m|I â A|mâ© Tr(AT B) = Tr(BAT ) = â¨m|I â BAT |mâ© = â¨m|(I â B)(I â AT )|mâ© = â¨m|(I â B)(A â I)|mâ© = â¨m|A â B|mâ© . Chapter 2 Introduction to quantum mechanics 2.1 [ ] [ ] [ ] [ ] 1 1 2 0 + â = â1 2 1 0 2.2 A |0â© = A11 |0â© + A21 |1â© = |1â© â A11 = 0, A21 = 1 A |1â© = A12 |0â© + A22 |1â© = |0â© â A12 = 1, A22 = 0 ] [ 0 1 â´A= 1 0 input: {|0â© , |1â©}, output: {|1â© , |0â©} A |0â© = A11 |1â© + A21 |0â© = |1â© â A11 = 1, A21 = 0 A |1â© = A12 |1â© + A22 |0â© = |0â© â A12 = 0, A22 = 1 [ ] 1 0 A= 0 1 2.3 From eq (2.12) A |vi â© = â Aji |wj â© j B |wj â© = â k 3 Bkj |xk â© 4 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS Thus   â BA |vi â© = B ï£ Aji |wj â©ï£¸ = â j Aji B |wj â© j = â j,k = â Aji Bkj |xk ⩠ ï£ â  Bkj Aji  |xk â© k j k â â = (BA)ki |xk â© â´(BA)ki = Bkj Aji j 2.4 I |vj â© = â Iij |vi â© = |vj â© , âj. i â Iij = δij 2.5 Defined inner product on C n is ((y1 , · · · , yn ), (z1 , · · · , zn )) = â yiâ zi . i Verify (1) of eq (2.13). ( (y1 , · · · , yn ), â ) λi (zi1 , · · · , zin ) = i â  yiâ ï£ i = â = j = â  λj zji  j yiâ λj zji i,j â â λj ( â ) yiâ zji i λj ((y1 , · · · , yn ), (zj1 , · · · , zjn )) j = â i λi ((y1 , · · · , yn ), (zi1 , · · · , zin )) . 5 Verify (2) of eq (2.13), ( ((y1 , · · · , yn ), (z1 , · · · , zn ))â = ( = ( = â )â yiâ zi i â ) yi ziâ i â (2.1) (2.2) ) ziâ yi (2.3) i = ((z1 , · · · , zn ), (y1 , · · · , yn )) (2.4) . Verify (3) of eq (2.13), ((y1 , · · · , yn ), (y1 , · · · , yn )) = â yiâ yi i = â |yi |2 i â Since |yi |2 ⥠0 for all i. Thus i |yi |2 = ((y1 , · · · , yn ), (y1 , · · · , yn )) ⥠0. From now on, I will show the following statement, ((y1 , · · · , yn ), (y1 , · · · , yn )) = 0 iï¬ (y1 , · · · , yn ) = 0. (â) This is obvious. â (â) Suppose ((y1 , · · · , yn ), (y1 , · · · , yn )) = 0. Then i |yi |2 = 0. â Since |yi |2 ⥠0 for all i, if i |yi |2 = 0, then |yi |2 = 0 for all i. Therefore |yi |2 = 0 â yi = 0 for all i. Thus, (y1 , · · · , yn ) = 0. 2.6 ( â ) λi |wi â© , |vâ© ( |vâ© , = i [ = â )â λi |wi â© i â λi (|vâ© , |wi â©) i = â λâi (|vâ© , |wi â©)â i = â i 2.7 ]â λâi (|wi â© , |vâ©) (âµ linearlity in the 2nd arg.) 6 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS ] 1 =1â1=0 â1 [ ] |wâ© |wâ© 1 1 â â = = â¥|wâ©â¥ 2 1 â¨w|wâ© [ ] |vâ© |vâ© 1 1 =â =â â1 â¥|vâ©â¥ 2 â¨v|vâ© [ ] â¨w|vâ© = 1 1 [ 2.8 If k = 1, |w2 â© â â¨v1 |w2 â© |v1 â© â¥|w2 â© â â¨v1 |w2 â© |v1 â©â¥ ( ) |w2 â© â â¨v1 |w2 â© |v1 â© â¨v1 |v2 â© = â¨v1 | â¥|w2 â© â â¨v1 |w2 â© |v1 â©â¥ â¨v1 |w2 â© â â¨v1 |w2 â© â¨v1 |v1 â© = â¥|w2 â© â â¨v1 |w2 â© |v1 â©â¥ = 0. |v2 â© = Suppose {v1 , · · · vn } (n ⤠d â 1) is a orthonormal basis. Then â ) |wn+1 â© â ni=1 â¨vi |wn+1 â© |vi â© â (j ⤠n) â¨vj |vn+1 â© = â¨vj | â¥|wn+1 â© â ni=1 â¨vi |wn+1 â© |vi â©â¥ ân â¨vj |wn+1 â© â i=1 â¨vi |wn+1 â© â¨vj |vi â© â = â¥|wn+1 â© â ni=1 â¨vi |wn+1 â© |vi â©â¥ â â¨vj |wn+1 â© â ni=1 â¨vi |wn+1 ⩠δij â = â¥|wn+1 â© â ni=1 â¨vi |wn+1 â© |vi â©â¥ â¨vj |wn+1 â© â â¨vj |wn+1 â© â = â¥|wn+1 â© â ni=1 â¨vi |wn+1 â© |vi â©â¥ ( = 0. Thus Gram-Schmidt procedure produces an orthonormal basis. 2.9 Ï0 = I = |0â© â¨0| + |1â© â¨1| Ï1 = X = |0â© â¨1| + |1â© â¨0| Ï2 = Y = âi |0â© â¨1| + i |1â© â¨0| Ï3 = Z = |0â© â¨0| â |1â© â¨1| 2.10 7 |vj â© â¨vk | = IV |vj â© â¨vk | IV ) ( ) ( â â |vq â© â¨vq | = |vp â© â¨vp | |vj â© â¨vk | p = â q |vp â© â¨vp |vj â© â¨vk |vq â© â¨vq | p,q = â δpj δkq |vp â© â¨vq | p,q Thus (|vj â© â¨vk |)pq = δpj δkq 2.11 [ ] ([ ]) 0 1 âλ 1 X= , det(X â λI) = det = 0 â λ = ±1 1 0 1 âλ If λ = â1, [ Thus 1 1 1 1 ][ ] [ ] c1 0 = c2 0 [ ] [ ] 1 â1 c1 |λ = â1â© = =â c2 2 1 If λ = 1 [ ] 1 1 |λ = 1â© = â 2 1 [ â1 0 X= 0 1 ] w.r.t. {|λ = â1â© , |λ = 1â©} 2.12 ([ ] ) 1 0 â λI = (1 â λ)2 = 0 â λ = 1 det 1 1 Therefore the eigenvector associated with eigenvalue λ = 1 is [ ] 0 |λ = 1â© = 1 [ ] 0 0 Because |λ = 1â© â¨Î» = 1| = , 0 1 [ ] [ ] 1 0 0 0 ̸= c |λ = 1â© â¨Î» = 1| = 1 1 0 c 8 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.13 Suppose |Ïâ© , |Ïâ© are arbitrary vectors in V . ( )â (|Ïâ© , (|wâ© â¨v|) |Ïâ©)â = (|wâ© â¨v|)â |Ïâ© , |Ïâ© ( ) = |Ïâ© , (|wâ© â¨v|)â |Ïâ© = â¨Ï| (|wâ© â¨v|)â |Ïâ© . On the other hand, (|Ïâ© , (|wâ© â¨v|) |Ïâ©)â = (â¨Ï|wâ© â¨v|Ïâ©)â = â¨Ï|vâ© â¨w|Ïâ© . Thus â¨Ï| (|wâ© â¨v|)â |Ïâ© = â¨Ï|vâ© â¨w|Ïâ© for arbitrary vectors |Ïâ© , |Ïâ© â´ (|wâ© â¨v|)â = |vâ© â¨w| 2.14 ((ai Ai )â |Ïâ© , |Ïâ©) = (|Ïâ© , ai Ai |Ïâ©) = ai (|Ïâ© , Ai |Ïâ©) = ai (Aâ i |Ïâ© , |Ïâ©) = (aâi Aâ i |Ïâ© , |Ïâ©) â´ (ai Ai )â = aâi Aâ i 2.15 ((Aâ )â |Ïâ© , |Ïâ©) = (|Ïâ© , Aâ |Ïâ©) = (Aâ |Ïâ© , |Ïâ©)â = (|Ïâ© , A |Ïâ©)â = (A |Ïâ© , |Ïâ©) â´ (Aâ )â = A 2.16 9 P = â |iâ© â¨i| . i P2 = ( â ) |iâ© â¨i| ï£ i = â â  |jâ© â¨j| j |iâ© â¨i|jâ© â¨j| i,j = â |iâ© â¨j| δij i,j = â |iâ© â¨i| i =P 2.17 Proof. (â) Suppose A is Hermitian. Then A = Aâ . Let |λ⩠be eigenvectors of A with eigenvalues λ, that is, A |â© = λ |λ⩠. Therefore â¨Î»|A|λ⩠= λ â¨Î»|λ⩠= λ. On the other hand, λâ = â¨Î»|A|λâ©â = â¨Î»|Aâ |λ⩠= â¨Î»|A|λ⩠= λ â¨Î»|λ⩠= λ. Hence eigenvalues of Hermitian matrix are real. (â) Suppose eigenvalues of A are real. From spectral theorem, normal matrix A can be written by â A= (2.5) λi |λi â©â¨Î»i | i where λi are real eigenvalues with eigenvectors |λi â©. By taking adjoint, we get â Aâ = λâi |λi â©â¨Î»i | i = â i =A Thus A is Hermitian. 2.18 λi |λi â©â¨Î»i | (ⵠλi are real) 10 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS Suppose |vâ© is a eigenvector with corresponding eigenvalue λ. U |vâ© = λ |vâ© . 1 = â¨v|vâ© = â¨v| I |vâ© = â¨v| U â U |vâ© = λλâ â¨v|vâ© = â¥Î»â¥2 ⴠλ = eiθ 2.19 [ ][ ] [ ] 0 1 0 1 1 0 X = = =I 1 0 1 0 0 1 2 2.20 Uâ¡ â |wi â© â¨vi | i â² Aij = â¨vi |A|vj â© = â¨vi |U U â AU U â |vj â© â = â¨vi |wp â© â¨vp |vq â© â¨wq |A|wr â© â¨vr |vs â© â¨ws |vj â© p,q,r,s = â â²â² â¨vi |wp ⩠δpq Aqr δrs â¨ws |vj â© p,q,r,s = â â²â² â¨vi |wp â© â¨wr |vj â© Apr p,r 2.21 Suppose M be Hermitian. Then M = M â . M = IM I = (P + Q)M (P + Q) = P M P + QM P + P M Q + QM Q Now P M P = λP , QM P = 0, P M Q = P M â Q = (QM P )â = 0. Thus M = P M P + QM Q. Next prove QM Q is normal. QM Q(QM Q)â = QM QQM â Q = QM â QQM Q (M = M â ) = (QM â Q)QM Q Therefore QM Q is normal. By induction, QM Q is diagonal ... (following is same as Box 2.2) 11 2.22 Suppose A is a Hermitian operator and |vi â© are eigenvectors of A with eigenvalues λi . Then â¨vi |A|vj â© = λj â¨vi |vj â© . On the other hand, â¨vi |A|vj â© = â¨vi |Aâ |vj â© = â¨vj |A|vi â©â = λâi â¨vj |vi â©â = λâi â¨vi |vj â© = λi â¨vi |vj â© Thus (λi â λj ) â¨vi |vj â© = 0. If λi ̸= λj , then â¨vi |vj â© = 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 2 |λ⩠= λP |λ⩠= λ2 |λ⩠. Therefore λ = λ2 λ(λ â 1) = 0 λ = 0 or 1. 2.24 Def of positive â¨v|A|v⩠⥠0 for all |vâ©. Suppose A is a positive operator. A can be decomposed as follows. A= A + Aâ A â Aâ +i 2 2i = B + iC where B = A + Aâ A â Aâ , C= . 2 2i Now operators B and C are Hermitian. â¨v|A|vâ© = â¨v|B + iC|vâ© = â¨v|B|vâ© + i â¨v|C|vâ© = α + iβ where α = â¨v|B|vâ© , β = â¨v|C|vâ© . Since B and C are Hermitian, α, β â R. From def of positive operator, β should be vanished because â¨v|A|vâ© is real. Hence β = â¨v|C|vâ© = 0 for all |vâ©, i.e. C = 0. Therefore A = Aâ . 12 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 0013 0010 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, T v) = 0 for all u, v â 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 â V , then A = 0. Proof. First, we show that (u, T v) = 0 if (v, Av) = 0. Then apply proposition 2.0.1 Suppose u, v â V . Then (u, T v) is decomposed as [ 1 1 (u, T v) = (u + v, T (u + v)) â (u â v, T (u â v)) + (u + iv, T (u + iv)) 4 i ] 1 â (u â iv, T (u â iv)) . i If (v, T v) = 0 for all v â V , the right hand side of above eqn vanishes. Thus (u, T v) = 0 for all u, v â V . Then T = 0. 0012 2.25 â¨Ï|Aâ A|Ïâ© = â¥A |Ïâ©â¥2 ⥠0 for all |Ïâ© . Thus Aâ A is positive. 2.26 1 1 |Ïâ©â2 = â (|0â© + |1â©) â â (|0â© + |1â© 2 2 1 = (|00â© + |01â© + |10â© + |11â©) 2  1 1 1  =  2 1 1 0011 13 1 1 1 |Ïâ©â3 = â (|0â© + |1â©) â â (|0â© + |1â© â â (|0â© + |1â© 2 2 2 1 = â (|000â© + |001â© + |010â© + |011â© + |100â© + |101â© + |110â© + |111â©) 2 2   1 1   1    1  1 â =  2 2 1  1   1 1 2.27 [ ] [ ] 0 1 1 0 â 1 0 0 â1   0 0 1 0 0 0 0 â1  = 1 0 0 0  0 â1 0 0 X âZ = [ 1 I âX = 0  0 1 = 0 0 ] ] [ 0 1 0 â 1 0 1  1 0 0 0 0 0  0 0 1 0 1 0 [ 0 X âI = 1  0 0 = 1 0 ] ] [ 1 0 1 â 0 1 0  0 1 0 0 0 1  0 0 0 1 0 0 In general, tensor product is not commutable. 2.28  â A11 B · · · A1n B  ..  .. (A â B)â =  ... . .  Am1 B · · · Amn B  â â  A11 B · · · Aâ1n B â  ..  .. =  ... . .  Aâm1 B â · · · Aâmn B â = Aâ â B â . 14 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS  T A11 B · · · A1n B  ..  .. (A â B)T =  ... . .  Am1 B · · · Amn B   A11 B T · · · Am1 B T   .. .. =  ...  . . T T A1n B · · · Amn B   T A11 B · · · A1m B T   .. .. =  ...  . . · · · Anm B T An1 B T = AT â B T . (A â B)â = ((A â B)â )T = (Aâ â B â )T = (Aâ )T â (B â )T = Aâ â B â . 2.29 Suppose U1 and U2 are unitary operators. Then (U1 â U2 )(U1 â U2 )â = U1 U1â â U2 U2â = I â I. Similarly, (U1 â U2 )â (U1 â U2 ) = I â I. 2.30 Suppose A and B are Hermitian operators. Then (A â B)â = Aâ â B â = A â B. (2.6) Thus A â B is Hermitian. 2.31 Suppose A and B are positive operators. Then â¨Ï| â â¨Ï| (A â B) |Ïâ© â |Ïâ© = â¨Ï|A|Ïâ© â¨Ï|B|Ïâ© . Since A and B are positive operators, â¨Ï|A|Ï⩠⥠0 and â¨Ï|B|Ï⩠⥠0 for all |Ïâ© , |Ïâ©. Then â¨Ï|A|Ïâ© â¨Ï|B|Ï⩠⥠0. Thus A â B is positive if A and B are positive. 2.32 15 Suppose P1 and P2 are projectors. Then (P1 â P2 )2 = P12 â P22 = P1 â P2 . Thus P1 â P2 . is also projector. 2.33 [ ] 1 1 1 H=â 2 1 â1 H â2 (2.7)   1 1 1 1 [ ] [ ] 1 1 1 1 1 â1 1 â1 1 1 1  =â ââ =  2 1 1 â1 â1 2 1 â1 2 1 â1 1 â1 â1 1 2.34 ] 4 3 . Suppose A = 3 4 [ det(A â λI) = (4 â λ)2 â 32 = λ2 â 8λ + 7 = (λ â 1)(λ â 7) ] 1 , |λ = 7â© = â1 [ Eigenvalues of A are λ = 1, 7. Corresponding eigenvectors are |λ = 1â© = [ ] 1 â1 . 2 1 Thus A = |λ = 1â©â¨Î» = 1| + 7 |λ = 7â©â¨Î» = 7| . â â A = |λ = 1â©â¨Î» = 1| + 7 |λ = 7â©â¨Î» = 7| [ ] â [ ] 1 1 â1 7 1 1 = + 2 â1 1 2 1 1 â â ] [ 1 1 + 7 â1 + 7 â â = 2 â1 + 7 1 + 7 log(A) = log(1) |λ = 1â©â¨Î» = 1| + log(7) |λ = 7â©â¨Î» = 7| [ ] log(7) 1 1 = 1 1 2 â1 2 16 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.35 âv · âÏ = 3 â vi Ïi i=1 [ ] [ ] [ ] 0 1 0 âi 1 0 = v1 + v2 + v3 1 0 i 0 0 â1 [ ] v3 v1 â iv2 = v1 + iv2 âv3 det(âv · âÏ â λI) = (v3 â λ)(âv3 â λ) â (v1 â iv2 )(v1 + iv2 ) = λ2 â (v12 + v22 + v32 ) = λ2 â 1 (âµ |âv | = 1) Eigenvalues are λ = ±1. Let |λ±1 â© be eigenvectors with eigenvalues ±1. Since âv · âÏ is Hermitian, âv · âÏ is diagonalizable. Then âv · âÏ = |λ1 â©â¨Î»1 | â |λâ1 â©â¨Î»â1 | Thus exp (iθâv · âÏ ) = eiθ |λ1 â©â¨Î»1 | + eâiθ |λâ1 â©â¨Î»â1 | = (cos θ + i sin θ) |λ1 â©â¨Î»1 | + (cos θ â i sin θ) |λâ1 â©â¨Î»â1 | = cos θ(|λ1 â©â¨Î»1 | + |λâ1 â©â¨Î»â1 |) + i sin θ(|λ1 â©â¨Î»1 | â |λâ1 â©â¨Î»â1 |) = cos(θ)I + i sin(θ)âv · âÏ . âµ Since âv · âÏ is Hermitian, |λ1 â© and |λâ1 â© are orthogonal. Thus |λ1 â©â¨Î»1 | + |λâ1 â©â¨Î»â1 | = I. 2.36 ([ 0 Tr(Ï1 ) = Tr 1 ([ 0 Tr(Ï2 ) = Tr i ([ 1 Tr(Ï3 ) = Tr 0 2.37 1 0 ]) =0 ]) âi =0 0 ]) 0 =1â1=0 â1 17 Tr(AB) = â â¨i|AB|iâ© i = â â¨i|AIB|iâ© i = â â¨i|A|jâ© â¨j|B|iâ© i,j = â â¨j|B|iâ© â¨i|A|jâ© i,j = â â¨j|BA|jâ© j = Tr(BA) 2.38 Tr(A + B) = â â¨i|A + B|iâ© i = â (â¨i|A|iâ© + â¨i|B|iâ©) i = â â¨i|A|iâ© + i â â¨i|B|iâ© i = Tr(A) + Tr(B). Tr(zA) = â â¨i|zA|iâ© i = â z â¨i|A|iâ© i =z â â¨i|A|iâ© i = z Tr(A). 2.39 (1) (A, B) â¡ Tr(Aâ B). (i) ( A, â ) λi Bi [ ( = Tr Aâ i â )] λ i Bi i = Tr(Aâ λ1 B1 ) + · · · + Tr(Aâ λn Bn ) â â = λ1 Tr(A B1 ) + · · · + λn Tr(A Bn ) â = λi Tr(Aâ Bi ) i (âµ Execise 2.38) 18 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS (ii) ( )â (A, B)â = Tr(Aâ B)  â â =ï£ â¨i|Aâ |jâ© â¨j|B|iâ©ï£¸ i,j = â â â¨i|Aâ |jâ© â¨j|B|iâ©â i,j = â â â¨j|B|iâ©â â¨i|Aâ |jâ© i,j = â â¨i|B â |jâ© â¨j|A|iâ© i,j = â â¨i|B â A|iâ© i = Tr(B â A) = (B, A). (iii) (A, A) = Tr(Aâ A) â = â¨i|Aâ A|iâ© i Since Aâ A is positive, â¨i|Aâ A|i⩠⥠0 for all |iâ©. Let ai be i-th column of A. If â¨i|Aâ A|iâ© = 0, then â¨i|Aâ A|iâ© = aâ i ai = â¥ai â¥2 = 0 iï¬ ai = 0. Therefore (A, A) = 0 iï¬ A = 0. (2) (3) 2.40 [X, Y ] = XY [ 0 = 1 [ i = 0 [ 2i = 0 = 2iZ âYX ][ ] [ ][ ] 1 0 âi 0 âi 0 1 â 0 i 0 i 0 1 0 ] [ ] 0 âi 0 â âi 0 i ] 0 â2i 19 ] [ ][ ] [ ][ 1 0 0 âi 0 âi 1 0 â [Y, Z] = i 0 0 â1 0 â1 i 0 [ ] 0 2i = 2i 0 = 2iX [ ][ ] [ ][ ] 1 0 0 1 0 1 1 0 [Z, X] = â 0 â1 1 0 1 0 0 â1 [ ] 0 âi = 2i i 0 = 2iY 2.41 {Ï1 , Ï2 } = Ï1 Ï2 + Ï2 Ï1 ] ][ ] [ ][ [ 0 âi 0 1 0 1 0 âi + = 1 0 i 0 1 0 i 0 ] ] [ [ âi 0 i 0 + = 0 i 0 âi =0 ] ][ ] [ ][ [ 0 âi 1 0 0 âi 1 0 + {Ï2 , Ï3 } = 0 â1 i 0 0 â1 i 0 =0 ] ][ ] [ ][ 0 1 1 0 0 1 1 0 {Ï3 , Ï1 } = + 1 0 0 â1 0 â1 1 0 [ =0 Ï02 = I 2 = I [ ]2 0 1 2 Ï1 = =I 1 0 [ ]2 0 âi 2 Ï2 = =I i 0 [ ]2 1 0 2 Ï3 = =I 0 â1 20 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.42 [A, B] + {A, B} AB â BA + AB + BA = = AB 2 2 2.43 From eq (2.75) and eq (2.76), {Ïj , Ïk } = 2δjk I. From eq (2.77), [Ïj , Ïk ] + {Ïj , Ïk } 2 â3 2i l=1 ϵjkl Ïl + 2δjk I = 2 3 â = δjk I + i ϵjkl Ïl Ïj Ïk = l=1 2.44 By assumption, [A, B] = 0 and {A, B} = 0, then AB = 0. Since A is invertible, multiply by from left, then Aâ1 Aâ1 AB = 0 IB = 0 B = 0. 2.45 [A, B]â = (AB â BA)â = B â Aâ â Aâ B â ] [ = B â , Aâ 2.46 [A, B] = AB â BA = â(BA â AB) = â [B, A] 2.47 (i [A, B])â = âi [A, B]â [ ] = âi B â , Aâ = âi [B, A] = i [A, B] 21 2.48 (Positive ) â Since P is positive, it is diagonalizable. Then P = i λi |iâ©â¨i|, (λi ⥠0). â â ââ â â λi |iâ©â¨i| = P. J = P â P = P P = P 2 = λ2i |iâ©â¨i| = i i 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, â â J = U U. â â J = U â U = I = I Since unitary operators are invertible, W = U J â1 = U I â1 = U I = U . Thus polar decomposition of U is U = U . (Hermitian) Suppose H = U J. J= â H â H = â HH = â H 2. â Thus H = U H 2 . 0013 0010 â In general, H ̸= H 2 . â From spectral decomposition, H = i λi |iâ©â¨i|, λi â R. ââ â ââ â H2 = λ2i |iâ©â¨i| = |λi | |iâ©â¨i| ̸= H λ2i |iâ©â¨i| = i i i 0012 0011 2.49 â Normal matrix is diagonalizable, A = i λi |iâ©â¨i|. â â J = Aâ A = |λi | |iâ©â¨i| . U= â i |ei â©â¨i| i A = UJ = â |λi | |ei â©â¨i| . i 2.50 [ ] [ ] 1 0 2 1 â Define A = . A A= . 1 1 1 1 Characteristic equation of Aâ A is det(Aâ A â λI) = λ2 â 3λ + 1 = 0.] Eigenvalues of Aâ A are [ â 2â . λ± = 3±2 5 and associated eigenvectors are |λ± â© = â 1 â 10â2 5 â1 ± 5 22 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS Aâ A = λ+ |λ+ â©â¨Î»+ | + λâ |λâ â©â¨Î»â | . J= â â â λ+ |λ+ â©â¨Î»+ | + λâ |λâ â©â¨Î»â | â â â â [ â â [ â â ] ] 3+ 5 5â 5 3â 5 5+ 5 4 2 5 â 2 4 â2 5 â 2 â â + â â = · · 2 5â2 6â2 5 â2 5 â 2 6 + 2 5 2 40 2 40 Aâ A = 1 1 |λ+ â©â¨Î»+ | + â |λâ â©â¨Î»â | . J â1 = â λ+ λâ U = AJ â1 Iâm tired. 2.51 â H H= ( ])â ] ] ] ] [ [ [ [ [ 1 1 1 1 1 1 1 1 1 1 2 0 1 1 1 â â â =â = = I. 2 0 2 2 1 â1 2 1 â1 2 1 â1 2 1 â1 2.52 â H = ( ])â ] [ [ 1 1 1 1 1 1 â =â = H. 2 1 â1 2 1 â1 Thus H 2 = I. 2.53 ( )( ) 1 1 1 det (H â λI) = â â λ ââ â λ â 2 2 2 1 1 = λ2 â â 2 2 2 =λ â1 ] 1â . â1 ± 2 [ Eigenvalues are λ± = ±1 and associated eigenvectors are |λ± â© = â 2.54 1 â 4â2 2 23 Since [A, B] = 0, A and B are simultaneously diagonalize, A = ( exp(A) exp(B) = â )( exp(ai ) |iâ©â¨i| i = â â â i ai |iâ©â¨i|, B= â i bi |iâ©â¨i|. ) exp(bi ) |iâ©â¨i| i exp(ai + bj ) |iâ© â¨i|jâ© â¨j| i,j = â exp(ai + bj ) |iâ©â¨j| δi,j i,j = â exp(ai + bi ) |iâ©â¨i| i = exp(A + B) 2.55 H= â E |Eâ©â¨E| E ) ( ) ( iH(t2 â t1 ) iH(t2 â t1 ) exp U (t2 â t1 )U (t2 â t1 ) = exp â â â ( ) )( ( ) ) â( iE(t2 â t1 ) iE â² (t2 â t1 ) = exp â |Eâ©â¨E| exp â |E â² â©â¨E â² | â â E,E â² ( ) ) â( i(E â E â² )(t2 â t1 ) = exp â |Eâ©â¨E â² | δE,E â² â E,E â² â = exp(0) |Eâ©â¨E| â E = â |Eâ©â¨E| E =I Similarly, U â (t2 â t1 )U (t2 â t1 ) = I. 2.56 U= â i λi |λi â©â¨Î»i | log(U ) = (|λi | = 1). â log(λj ) |λj â©â¨Î»j | = j K = âi log(U ) = â â iθj |λj â©â¨Î»j | where θj = arg(λj ) j θj |λj â©â¨Î»j | . j  K â = (âi log U )â = ï£ â j â θj |λj â©â¨Î»j | = â j θjâ |λj â©â¨Î»j | = â j θj |λj â©â¨Î»j | = K 24 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.57 Ll |Ïâ© |Ïâ© â¡ â â¨Ï|Lâ l Ll |Ïâ© â â¨Ï|Mm Mm |Ïâ© = â â¨Ï|Lâ l Mm Mm Ll |Ïâ© â¨Ï|Lâ l Ll |Ïâ© â â¨Ï|Lâ l Ll |Ïâ© Mm Ll |Ïâ© Mm Ll |Ïâ© Nlm |Ïâ© â =â ·â =â =â â â â â â¨Ï|Lâ l Ll |Ïâ© â¨Ï|Lâ l Mm Mm Ll |Ïâ© â¨Ï|Lâ l Mm Mm Ll |Ïâ© â¨Ï|Nlm Nlm |Ïâ© â¨Ï|Mm Mm |Ïâ© Mm |Ïâ© 2.58 â¨M â© = â¨Ï|M |Ïâ© = â¨Ï|m|Ïâ© = m â¨Ï|Ïâ© = m â¨M 2 â© = â¨Ï|M 2 |Ïâ© = â¨Ï|m2 |Ïâ© = m2 â¨Ï|Ïâ© = m2 deviation = â¨M 2 â© â â¨M â©2 = m2 â m2 = 0. 2.59 â¨Xâ© = â¨0|X|0â© = â¨0|1â© = 0 â¨X 2 â© = â¨0|X 2 |0â© = â¨0|X|1â© = â¨0|0â© = 1 â standard deviation = â¨X 2 â© â â¨Xâ©2 = 1 2.60 âv · âÏ = 3 â vi Ïi i=1 [ ] [ ] [ ] 0 1 0 âi 1 0 = v1 + v2 + v3 1 0 i 0 0 â1 [ ] v3 v1 â iv2 = v1 + iv2 âv3 det(âv · âÏ â λI) = (v3 â λ)(âv3 â λ) â (v1 â iv2 )(v1 + iv2 ) = λ2 â (v12 + v22 + v32 ) = λ2 â 1 (âµ |âv | = 1) Eigenvalues are λ = ±1. (i) if λ = 1 âv · âÏ â λI = âv · âÏ â I [ ] v3 â 1 v1 â iv2 = v1 + iv2 âv3 â 1 25 Normalized eigenvector is |λ1 â© = â [ 1+v3 2 1 + v3 |λ1 â©â¨Î»1 | = 2 = = = = 1 ] 1âv3 v1 âiv2 . [ ] 1 1âv3 v1 âiv2 [ 1 1âv3 v1 +iv2 ] ] [ v1 âiv2 1 1 + v3 1+v3 1âv3 v1 +iv2 2 1+v3 1+v3 [ ] 1 1 + v3 v1 â iv2 2 v1 + iv2 1 â v3 ( [ ]) 1 v3 v1 â iv2 I+ v1 + iv2 âv3 2 1 (I + âv · âÏ ) 2 (ii) If λ = â1. âv · âÏ â λI = âv · âÏ + I [ ] v3 + 1 v1 â iv2 = v1 + iv2 âv3 + 1 [ ] â 1 1âv3 Normalized eigenvalue is |λâ1 â© = . 3 2 â v1+v 1 âiv2 [ ] ] [ 1 â v3 1 3 1 â v1+v |λâ1 â©â¨Î»â1 | = 1+v3 +iv2 1 â 2 v âiv ] [ 1 2 1 âiv2 1 â v1âv 1 â v3 3 = 1+v3 1 +iv2 â v1âv 2 1âv3 3 ] [ 1 1 â v3 â(v1 â iv2 ) = 1 + v3 2 â(v1 + iv2 ) ( [ ]) 1 v3 v1 â iv2 = Iâ (v1 + iv2 âv3 2 1 = (I â âv · âÏ ). 2 While I review my proof, I notice that my proof has a defect. The case (v1 , v2 , v3 ) = (0, 0, 1), 3 second component of eigenstate, v1âv , diverges. So I implicitly assume v1 â iv2 ̸= 0. Hence 1 âiv2 my proof is incomplete. Since the exercise doesnât require explicit form of projector, we should prove the problem more abstractly. In order to prove, we use the following properties of âv · âÏ â¢ âv · âÏ is Hermitian ⢠(âv · âÏ )2 = I where âv is a real unit vector. We can easily check above conditions. (âv · âÏ )â = (v1 Ï1 + v2 Ï2 + v3 Ï3 )â = v1 Ï1â + v2 Ï2â + v3 Ï3â = v1 Ï1 + v2 Ï2 + v3 Ï3 = âv · âÏ (âµ Pauli matrices are Hermitian.) 26 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS (âv · âÏ )2 = 3 â (vj Ïj )(vk Ïk ) j,k=1 = 3 â vj vk Ï j Ï k j,k=1 = 3 â ( vj vk δjk I + i j,k=1 = 3 â 3 â j=1 =I ) ϵjkl Ïl (âµ eqn(2.78) page78) l=1 vj vk δjk I + i j,k=1 = 3 â 3 â ϵjkl vj vk Ïl j,k,l=1 vj2 I  ï£âµ â  vj2 = 1 j Proof. Suppose |λ⩠is an eigenstate of âv · âÏ with eigenvalue λ. Then âv · âÏ |λ⩠= λ |λ⩠(âv · âÏ )2 |λ⩠= λ2 |λ⩠On the other hand (âv · âÏ )2 = I, (âv · âÏ )2 |λ⩠= I |λ⩠= |λ⩠ⴠλ2 |λ⩠= |λ⩠. Thus λ2 = 1 â λ = ±1. Therefore âv · âÏ has eigenvalues ±1. Let |λ1 â© and |λâ1 â© are eigenvectors with eigenvalues 1 and â1, respectively. I will prove that P± = |λ±1 â©â¨Î»Â±1 |. In order to prove above equation, all we have to do is prove following condition. (see Theorem 2.0.1) 0013 0010 â¨Ï|(P± â |λ±1 â©â¨Î»Â±1 |)|Ïâ© = 0 for all |Ïâ© â C2 . (2.8) 0012 0011 Since âv ·âÏ is Hermitian, |λ1 â© and |λâ1 â© are orthonormal vector (âµ Exercise 2.22). Let |Ïâ© â be an arbitrary state. |Ïâ© can be written as |Ïâ© = α |λ1 â© + β |λ±1 â© (|α|2 + |β|2 = 1, α, β â C). C2 27 â¨Ï|(P± â |λ± â©â¨Î»Â± |)|Ïâ© = â¨Ï|P± |Ïâ© â â¨Ï|λ± â© â¨Î»Â± |Ïâ© . 1 â¨Ï|P± |Ïâ© = â¨Ï| (I ± âv · âÏ )|Ïâ© 2 1 1 = ± â¨Ï|âv · âÏ )|Ïâ© 2 2 1 1 = ± (|α|2 â |β|2 ) 2 2 1 1 = ± (2|α|2 â 1) (âµ |α|2 + |β|2 = 1) 2 2 â¨Ï|λ1 â© â¨Î»1 |Ïâ© = |α|2 â¨Ï|λâ1 â© â¨Î»â1 |Ïâ© = |β|2 = 1 â |α|2 Therefore â¨Ï|(P± â |λ±1 â©â¨Î»Â±1 |)|Ïâ© = 0 for all |Ïâ© â C2 . Thus P± = |λ±1 â©â¨Î»Â±1 |. 2.61 â¨Î»1 |0â© â¨0|λ1 â© = â¨0|λ1 â© â¨Î»1 |0â© 1 = â¨0| (I + âv · âÏ )|0â© 2 1 = (1 + v3 ) 2 Post-measurement state is [ ] |λ1 â© â¨Î»1 |0â© 1 1 1 + v3 â =â · 2 v1 + iv2 1 â¨0|λ1 â© â¨Î»1 |0â© (1 + v ) 3 2 â [ ] 1 1 (1 + v3 ) v1 +iv2 = 2 1+v3 â [ ] 1 + v3 1 = 1âv3 2 v1 âiv2 = |λ1 â© . 2.62 â Suppose Mm is a measurement operator. From the assumption, Em = Mm Mm = Mm . Then â¨Ï|Em |Ïâ© = â¨Ï|Mm |Ï⩠⥠0. for all |Ïâ©. Since Mm is positive operator, Mm is Hermitian. Therefore, â 2 Em = Mm Mm = Mm Mm = M m = Mm . Thus the measurement is a projective measurement. 28 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.63 â â Em Um Um Em â â = Em I Em â Mm Mm = â = Em . â Since Em is POVM, for arbitrary unitary U , Mm Mm is POVM. 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 |+â© â¡ |0â© + |1â© â , 2 |ââ© â¡ |0â© â |1â© â 2 2.66 ( X1 Z2 ( â¨X1 Z2 â© = â¨00| + â¨11| â 2 |00â© + |11â© â 2 ) ( X1 Z2 ) = |00â© + |11â© â 2 |10â© â |01â© â 2 ) = â¨00| + â¨11| |10â© â |01â© â â · =0 2 2 2.67 Unsolved W â V â V = W â W â¥. U : W â V , Uâ² : V â V . U â² |wâ© = U |wâ© U â² â L(V ) U â L(W ) U â² = U â I ??? 2.68 â |Ïâ© = |00â©+|11â© . 2 Suppose |aâ© = a0 |0â© + a1 |1â© and |bâ© = b0 |0â© + b1 |1â©. |aâ© |bâ© = a0 b0 |00â© + a0 b1 |01â© + a1 b0 |10â© + a1 b1 |11â© . If |Ïâ© = |aâ© |bâ©, then a0 b0 = 1, a0 b1 = 0, a1 b0 = 0, a1 b1 = 1 since {|ijâ©} is an orthonormal basis. 29 If a0 b1 = 0, then a0 = 0 or b1 = 0. When a0 = 0 , this is contradiction to a0 b0 = 1. When b1 = 0 , this is contradiction to a1 b1 = 1. Thus |Ï⩠̸= |aâ© |bâ©. 2.69 Define Bell states as follows. |Ï1 â© â¡ |00â© + |11â© â = 2 |Ï2 â© â¡ |00â© â |11â© â = 2 |Ï3 â© â¡ |01â© + |10â© â = 2 |Ï4 â© â¡ |01â© â |10â© â = 2   1 1  0  â   2 0 1   1 1  0  â   2 0 â1   0  1 1  â  2 1 0   0 1  1  â   2 â1 0 First, we prove {|Ïi â©} is a linearly independent basis. a1 |Ï1 â© + a2 |Ï2 â© + a3 |Ï3 â© + a4 |Ï4 â© = 0   a1 + a2 1 a3 + a4  =0 â´â  2 a3 â a4  a1 â a2  a1 + a2 = 0    a +a =0 3 4 ⴠ a3 â a4 = 0    a1 â a2 = 0 â´ a1 = a2 = a3 = a4 = 0 Thus {|Ïi â©} is a linearly independent basis. Moreover â¥|Ïi â©â¥ = 1 and â¨Ïi |Ïj â© = δij for i, j = 1, 2, 3, 4. Therefore {|Ïi â©} forms an orthonormal basis. 2.70 For any Bell states we get â¨Ïi |E â I|Ïi â© = 12 (â¨0|E|0â© + â¨1|E|1â©). Suppose Eve measures the qubit Alice sent by measurement operators Mm . The probability â â that Eve gets result m is pi (m) = â¨Ïi |Mm Mm â I|Ïi â©. Since Mm Mm is positive, pi (m) are same values for all |Ïi â©. Thus Eve canât distinguish Bell states. 30 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS 2.71 From spectral decomposition, â â Ï= pi |Ïi â©â¨Ïi | , pi ⥠0, pi = 1. i 2 Ï = â i pi pj |iâ© â¨i|jâ© â¨j| i,j = â pi pj |iâ©â¨j| δij i,j = â p2i |iâ©â¨i| i ( 2 Tr(Ï ) = Tr â i ) p2i |iâ©â¨i| = â i â p2i Tr(|iâ©â¨i|) = â i p2i â¨i|iâ© = â p2i ⤠i â pi = 1 (âµ p2i ⤠pi ) i Suppose Tr(Ï2 ) = 1. Then i p2i = 1. If 0 ⤠pi < 1, then p2i < pi . Thus only one pi = 1 and otherwise are 0. Therefore Ï = |Ïi â©â¨Ïi | is pure state. Conversely if Ï is pure, then Ï = |Ïâ©â¨Ï|. Tr(Ï2 ) = Tr(|Ïâ© â¨Ï|Ïâ© â¨Ï|) = Tr(|Ïâ©â¨Ï|) = â¨Ï|Ïâ© = 1. 2.72 ] a b , a, d â R and (1) Since density matrix is Hermitian, matrix representation is Ï = â b d b â C w.r.t. standard basis. Because Ï is density matrix, Tr(Ï) = a + d = 1. Define a = (1 + r3 )/2, d = (1 â r3 )/2 and b = (r1 â ir2 )/2, (ri â R). In this case, ] [ ] [ 1 1 + r3 r1 â ir2 1 a b = = (I + âr · âÏ ). Ï= â b d 2 r1 + ir2 1 â r3 2 [ Thus for arbitrary density matrix Ï can be written as Ï = 21 (I + âr · âÏ ). Next, we derive the condition that Ï is positive. If Ï is positive, all eigenvalues of Ï should be non-negative. det(Ï â λI) = (a â λ)(b â λ) â |b|2 = λ2 â (a + d)λ + ad â |b2 | = 0 â (a + d) ± (a + d)2 â 4(ad â |b|2 ) λ= 2 â ( 2 ) 1âr3 r12 +r22 1± 1â4 â 4 4 = 2 â 1 ± 1 â (1 â r12 â r22 â r32 ) = 2 â 1 ± |âr|2 = 2 1 ± |âr| = 2 31 r| Since Ï is positive, 1â|â r| ⤠1. 2 ⥠0 â |â Therefore an arbitrary density matrix for a mixed state qubit is written as Ï = 12 (I + âr · âÏ ). (2) Ï = I/2 â âr = 0. Thus Ï = I/2 corresponds to the origin of Bloch sphere. (3) 1 1 Ï2 = (I + âr · âÏ ) (I + âr · âÏ ) 2 2 ) ( 3 â â 1 ϵjkl Ïl  rj rk δjk I + i = I + 2âr · âÏ + 4 l=1 j,k ) 1( = I + 2âr · âÏ + |âr|2 I 4 1 2 Tr(Ï ) = (2 + 2|âr|2 ) 4 If Ï is pure, then Tr(Ï2 ) = 1. 1 1 = Tr(Ï2 ) = (2 + 2|âr|2 ) 4 â´ |âr| = 1. Conversely, if |âr| = 1, then Tr(Ï2 ) = 14 (2 + 2|âr|2 ) = 1. Therefore Ï is pure. 2.73 0013 0010 Theorem 2.6 Ï= â pi |Ïi â©â¨Ïi | = i â |ÏËi â©â¨ÏËi | = i â |ÏËj â©â¨ÏËj | = â j qj |Ïj â©â¨Ïj | â |ÏËi â© = j â uij |ÏËj â© j where u is unitary. â Transformation in theorem 2.6, |ÏËi â© = j uij |ÏËj â©, corresponds to [ ] [ ] |ÏË1 ⩠· · · |ÏËk â© = |ÏË1 ⩠· · · |ÏËk â© U T where k = rank(Ï). 0012 0011 â From spectral theorem, density matrix Ï is decomposed as Ï = dk=1 λk |kâ©â¨k| where d = dim H. Without loss of generality, we â can assume pk > â 0 for k = 1 · · · , l where l â = rank(Ï) and Ë k|, Ë where |kâ© Ë = λk |kâ©. pk = 0 for k = l + 1, · · · , d. Thus Ï = lk=1 pk |kâ©â¨k| = lk=1 |kâ©â¨ Suppose |Ïi â© is a state in support Ï. Then |Ïi â© = l â k=1 Define pi = â 1 |cik |2 k λk and uik cik |kâ© , â pi cik = â . λk â k |cik |2 = 1. 32 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS Now â |uik |2 = â pi |cik |2 k k λk = pi â |cik |2 λk k = 1. Next prepare an unitary operator 1 such that ith row of U is [ui1 · · · uik · · · uil ]. Then we can define another ensemble such that [ ] [ ] |ÏË1 ⩠· · · |ÏËi ⩠· · · |ÏËl â© = |kË1 ⩠· · · |kËl â© U T â where |ÏËi â© = pi |Ïi â©. From theorem 2.6, â â Ë k| Ë = Ï= |kâ©â¨ |ÏËk â©â¨ÏËk | . k k Therefore we can obtain â a 1minimal ensemble for Ï that contains |Ïi â©. â1 Moreover since Ï = k λk |kâ©â¨k|, â¨Ïi |Ïâ1 |Ïi â© = â 1 â |cik |2 1 â¨Ïi |kâ©â¨k|Ïi â© = = . λk λk pi k Hence, 1 â¨Ïi |Ïâ1 |Ïi â© k = pi . 2.74 ÏAB = |aâ©â¨a|A â |bâ©â¨b|B ÏA = TrB ÏAB = |aâ©â¨a| Tr(|bâ©â¨b|) = |aâ©â¨a| Tr(Ï2A ) = 1 Thus ÏA is pure. 2.75 Define |Φ± â© = â1 (|00â© 2 ± |11â©) and |Ψ± â© = â1 (|01â© 2 ± |10â©). 1 |Φ± â©â¨Î¦Â± |AB = (|00â©â¨00| ± |00â©â¨11| ± |11â©â¨00| + |11â©â¨11|) 2 1 I TrB (|Φ± â©â¨Î¦Â± |AB ) = (|0â©â¨0| + |1â©â¨1|) = 2 2 1 |Ψ± â©â¨Î¨Â± | = (|01â©â¨01| ± |01â©â¨10| ± |10â©â¨01| + |10â©â¨10|) 2 1 I TrB (|Ψ± â©â¨Î¨Â± |) = (|0â©â¨0| + |1â©â¨1|) = 2 2 2.76 By Gram-Schmidt procedure construct an orthonormal basis {uj } (row vector) with ui = [ui1 · · · uik · · · uil ].   u1  ..   .     Then define unitary U =   ui .  .   ..  1 ul 33 Unsolved. I think the polar decomposition can only apply to square matrix A, not arbitrary linear operators. Suppose A is m à n matrix. Then size of Aâ A is n à n. Thus the size of U should be m à n. Maybe U is isometry, but I think it is not unitary. I misunderstand linear operator. Quoted from âAdvanced Liner Algebraâ by Steven Roman, ISBN 0387247661. A linear transformation Ï : V â 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, âLet A be a linear operator on a vector space V .â Therefore A is a linear transformation such that A : V â V . 2.77 |Ïâ© = |0â© |Φ+ â© [ ] 1 = |0â© â (|00â© + |11â©) 2 ] [ 1 = (α |Ï0 â© + β |Ï1 â©) â (|Ï0 Ï0 â© + |Ï1 Ï1 â©) 2 where |Ïi â© are arbitrary orthonormalâstates and α, β â C. We cannot vanish cross term. Therefore |Ïâ© cannot be written as |Ïâ© = i λi |iâ©A |iâ©B |iâ©C . 2.78 Proof. Former part. If |Ïâ© is product, then there exist a state |ÏA â© for system A, and a state |ÏB â© for system B such that |Ïâ© = |ÏA â© |ÏB â©. Obviously, this Schmidt number is 1. Conversely, if Schmidt number is 1, the state is written as |Ïâ© = |ÏA â© |ÏB â©. Hence this is a product state. Proof. Later part. (â) Proved by exercise 2.74. â â (â) Let a pure state be |Ïâ© = i λi |iA â© |iB â©. Then ÏA = TrB (|Ïâ©â¨Ï|) = i λ2i |iâ©â¨i|. If ÏA is a pure state, then λj = 1 and otherwise 0 for some j. It follows that |Ïj â© = |jA â© |jB â©. Thus |Ïâ© 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 0013 0010 Procedure ofâ Schmidt decomposition. â Goal: |Ïâ© = i λi |iA â© |iB ⩠⢠Diagonalize reduced density matrix ÏA = ⢠Derive |iB â©, |iB â© = 0012 â i λi |iA â©â¨iA |. (I â â¨iA |) |Ïâ© â λi ⢠Construct |Ïâ©. 0011 (i) 1 â (|00â© + |11â©) This is already decomposed. 2 (ii) |00â© + |01â© + |10â© + |11â© = 2 ( |0â© + |1â© â 2 ) ( â |0â© + |1â© â 2 ) = |Ïâ© |Ïâ© where |Ïâ© = (iii) 1 |Ïâ©AB = â (|00â© + |01â© + |10â©) 3 ÏAB = |Ïâ©â¨Ï|AB 1 ÏA = TrB (ÏAB ) = (2 |0â©â¨0| + |0â©â¨1| + |1â©â¨0| + |1â©â¨1|) ( )( 3 ) 2 1 1 det(ÏA â λI) = âλ âλ â =0 3 3 9 1 λ2 â λ + = 0 9â â 1 ± 5/3 3± 5 λ= = 2 6 [ â ] â 1+ 5 3+ 5 1 2 Eigenvector with eigenvalue λ0 â¡ . is |λ0 â© â¡ â â 6 1 5+ 5 2 [ â ] â 1â 5 3â 5 1 2 Eigenvector with eigenvalue λ1 â¡ is |λ1 â© â¡ â â . 6 1 5â 5 2 ÏA = λ0 |λ0 â©â¨Î»0 | + λ1 |λ1 â©â¨Î»1 | . (I â â¨Î»0 |) |Ïâ© â λ0 (I â â¨Î»1 |) |Ïâ© â |a1 ⩠⡠λ1 |a0 â© â¡ Then |Ïâ© = 1 â â λi |ai â© |λi â© . i=0 |0â© + |1â© â 2 35 (Itâs too tiresome to calculate |ai â©) 2.80 â â Let |Ïâ© = i λi |Ïi â©A |Ïi â©B and |Ïâ© = i λi |Ïi â©A |Ïi â©B . â â Define U = i |Ïj â©â¨Ïj |A and V = j |Ïj â©â¨Ïj |. Then (U â V ) |Ïâ© = â λi U |Ïi â©A V |Ïi â©B i = â λi |Ïi â©A |Ïi â©B i = |Ïâ© . 2.81 Suppose ÏA = TrR |AR2 â©â¨AR2 | = â i λi |iâ©â¨i|. Define |AR1 â© = (IA â UR ) |AR2 â©. ) ( TrR (|AR1 â©â¨AR1 |) = TrR (IA â UR ) |AR2 â©â¨AR2 | (IA â URâ ) ) ( = TrR |AR2 â©â¨AR2 | (IA â URâ )(IA â UR ) = TrR (|AR2 â©â¨AR2 |) = ÏA . Thus |AR1 â© is also a purification of ÏA . 2.82 (1) â â Let |Ïâ© = i pi |Ïi â© |iâ©. TrR (|Ïâ©â¨Ï|) = â pi |Ïi â©â¨Ïi | i Thus |Ïâ© is a purification of Ï. (2) Probability Tr [(I â |iâ©â¨i|) |Ïâ©â¨Ï|] = â¨Ï|(I â |iâ©â¨i|)|Ïâ© = pi â¨Ïi |Ïi â© = pi . Post-measurement state (I â |iâ©â¨i| |Ïâ©) = â pi (3) â pi |Ïi â© = |Ïi â© . â pi 36 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS â â Suppose |ARâ© is a purification of Ï such that |ARâ© = i pi |Ïi â© |ri â©. By exercise 2.81, the others purification is written as (I â U ) |ARâ©. ââ (I â U ) |ARâ© = (I â U ) pi |Ïi â© |ri â© = ââ i pi |Ïi â© U |ri â© i = ââ pi |Ïi â© |iâ© i â where U = i |iâ©â¨ri |. By (2), if we measure the system R w.r.t |iâ©, post-measurement state for system A is |Ïi â© with probability pi , which prove the assertion. Problem 2.1 From Exercise 2.35, ân · âÏ is decomposed as ân · âÏ = |λ1 â©â¨Î»1 | â |λâ1 â©â¨Î»â1 | where |λ±1 â© are eigenvector of ân · âÏ with eigenvalues ±1. Thus f (θân · âÏ ) = f (θ) |λ1 â©â¨Î»1 | + f (âθ) |λâ1 â©â¨Î»â1 | ( ) ( ) f (θ) + f (âθ) f (θ) â f (âθ) f (θ) + f (âθ) f (θ) â f (âθ) = + |λ1 â©â¨Î»1 | + â |λâ1 â©â¨Î»â1 | 2 2 2 2 f (θ) + f (âθ) f (θ) â f (âθ) = (|λ1 â©â¨Î»1 | + |λâ1 â©â¨Î»â1 |) + (|λ1 â©â¨Î»1 | â |λâ1 â©â¨Î»â1 |) 2 2 f (θ) â f (âθ) f (θ) + f (âθ) I+ ân · âÏ = 2 2 Problem 2.2 Unsolved Problem 2.3 Unsolved Chapter 8 Quantum noise and quantum operations 8.1 Density operator of initial state is written by |Ïâ©â¨Ï| and final state is written by U |Ïâ©â¨Ï| U â . Thus time development of Ï = |Ïâ©â¨Ï| can be written by E(Ï) = U ÏU â . 8.2 From eqn (2.147) (on page 100), Ïm = â Mm ÏMm â Tr(Mm Mm Ï) = â Mm ÏMm â Tr(Mm ÏMm ) = Em (Ï) . Tr Em (Ï) â â And from eqn (2.143) (on page 99), p(m) = Tr(Mm Mm Ï) = Tr(Mm ÏMm ) = Tr Em (Ï). 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 (|1 â 1/2| + |0 â 1/2|) 2( ) 1 1 1 = + 2 2 2 1 = 2 D((1, 0), (1/2, 1/2)) = 1 (|1/2 â 3/4| + |1/3 â 1/8| + |1/6 â 1/8|) 2 1 = (1/4 + 5/24 + 1/24) 2 1 = 4 D ((1/2, 1/3, 1/6), (3/4, 1/8, 1/8)) = 9.2 1 (|p â q| + |(1 â p) â (1 â q)|) 2 1 = (|p â q| + | â p + q|) 2 = |p â q| D ((p, 1 â p), (q, 1 â q)) = 9.3 F ((1, 0), (1/2, 1/2)) = â â 1 1 · 1/2 + 0 · 1/2 = â 2 â â â 1/2 · 3/4 + 1/3 · 1/8 + 1/6 · 1/8 â â 4 6+ 3 = 12 F ((1/2, 1/3, 1/6), (3/4, 1/8, 1/8)) = 39 40 CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION 9.4 Define rx = px â qx . Let U be the whole index set. â â max |p(S) â q(S)| = max px â qx S S xâS xâS â = max (px â qx ) S xâS â = max rx S Since â xâS xâS rx is written as â rx = xâS â â â rx + xâS rx â¥0 rx , (9.1) xâS rx <0 is maximized when S = {x â U |rx ⥠0} or S = {x â U |rx < 0}. r x xâS Define S+ = {x â U |rx ⥠0} and Sâ = {x â U |rx < 0}. Now the sum of all rx is 0, â â â rx = rx + rx = 0 xâU â´ â xâS+ rx = â xâS+ Thus xâSâ â rx . xâSâ â â â max rx = rx = â rx . S xâS xâS+ xâSâ On the other hand, 1â |px â qx | 2 xâU 1â |rx | = 2 xâU 1 â 1 â = |rx | + |rx | 2 2 D(px , qx ) = xâS+ xâSâ 1 â 1 â rx â rx = 2 2 xâS+ xâSâ 1 â 1 â = rx + rx 2 2 xâS+ xâS+ â = rx xâS+ â = max rx . S xâS (âµ eqn(9.2)) (9.2) 41 â â Therefore D(px , qx ) = maxS xâS px â xâS qx = maxS |p(S) â q(S)|. 9.5 â â From eqn (9.1) and (9.2), maximizing xâS rx is equivalent to maximizing xâS rx . Hence ( ) â â D(px , qx ) = max(p(S) â q(S)) = max px â qx . S S xâS xâS 9.6 Define Ï = 3 4 |0â©â¨0| + 14 |1â©â¨1|, Ï = 2 3 |1â©â¨1| + 31 |1â©â¨1|. 1 Tr |Ï â Ï| 2 = D((3/4, 1/4), (2/3, 1/3)) ) ( 1 3 2 1 1 = â + â 2 4 3 4 3 ( ) 1 1 1 = + 2 12 12 1 = 12 D(Ï, Ï) = Define Ï = 3 4 |0â©â¨0| + 14 |1â©â¨1|, Ï = 2 3 |+â©â¨+| + 13 |ââ©â¨â|. 1 |+â©â¨+| = (|0â©â¨0| + |0â©â¨1| + |1â©â¨0| + |1â©â¨1|) 2 1 |ââ©â¨â| = (|0â©â¨0| â |0â©â¨1| â |1â©â¨0| + |1â©â¨1|) 2 ( 3 1 â 4 2 ) 1 ÏâÏ = |0â©â¨0| â (|0â©â¨1| + |1â©â¨0|) + 6 1 1 1 = |0â©â¨0| â (|0â©â¨1| + |1â©â¨0|) â |1â©â¨1| 4 6 4 ( 1 1 â 4 2 ) |1â©â¨1| 1 1 1 1 1 1 1 1 |0â©â¨0| â |0â©â¨1| + 2 |0â©â¨0| + |0â©â¨1| â |1â©â¨0| + 2 |1â©â¨1| + |1â©â¨0| + 2 |1â©â¨1| 2 4 6 6·4 4·6 6 4·6 4 ( )4 · 6 1 1 = + (|0â©â¨0| + |1â©â¨1|) 42 62 (Ï â Ï)â (Ï â Ï) = 1 Tr |Ï â Ï| 2 â 1 1 = + 2 2 4 6 D(Ï, Ï) = 9.7 42 CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION Since Ï â Ï is Hermitian, we can apply spectral decomposition. Then Ï â Ï is written as ÏâÏ = k â n â λi |iâ©â¨i| + i=1 λi |iâ©â¨i| i=k+1 where λi are positive â eigenvalues for i = 1, â· · · , k and negative eigenvalues for i = k + 1, · · · , n. Define Q = ki=1 λi |iâ©â¨i| and S = â ni=k+1 λi |iâ©â¨i|. Then P and S are positive operator. Therefore Ï â Ï = P â S. 0013 0010 Proof of |Ï â Ï| = Q + S. |Ï â Ï| = |Q â S| â = (Q â S)â (Q â S) â = (Q â S)2 â = Q2 â QS â SQ + S 2 â = Q2 + S 2 ââ = λ2i |iâ©â¨i| = â i |λi | |iâ©â¨i| i =Q+S 0012 0011 9.8 â Suppose Ï = Ïi . Then Ï = i pi Ïi . ( ) ( ) â â â D pi Ïi , Ï = D pi Ïi , pi Ïi i ⤠â i (9.3) i pi D(Ïi , Ïi ) (âµ eqn(9.50)) (9.4) i = â pi D(Ïi , Ï). (âµ assumption). (9.5) i 9.9 9.10 9.11 9.12 Suppose Ï = 21 (I + âr · âÏ ) and Ï = 12 (I + âs · âÏ ) where âv and âs are real vectors s.t. |âv |, |âs| ⤠1. I E(Ï) = p + (1 â p)Ï, 2 I E(Ï) = p + (1 â p)Ï. 2 43 1 Tr |E(Ï) â E(Ï)| 2 1 = Tr |(1 â p)(Ï â Ï)| 2 1 = (1 â p) Tr |Ï â Ï| 2 = (1 â p)D(Ï, Ï) |âr â âs| = (1 â p) 2 D(E(Ï), E(Ï)) = Is this strictly contractive? 9.13 Bit flip channel E0 = â pI, E1 = â 1 â pÏx . E(Ï) = E0 ÏE0â + E1 ÏE1â = pÏ + (1 â p)Ïx ÏÏx . Since Ïx Ïx Ïx = Ïx , Ïx Ïy Ïx = âÏy and Ïx Ïz Ïx = âÏz , then Ïx (âr · âÏ ) = r1 Ïx â r2 Ïy â r3 Ï3 . Thus 1 Tr |E(Ï) â E(Ï)| 2 1 = Tr |p(Ï â Ï) + (1 â p)(Ïx ÏÏx â Ïx ÏÏx )| 2 1 1 ⤠p Tr |Ï â Ï| + (1 â p) Tr |Ïx (Ï â Ï)Ïx | 2 2 = pD(Ï, Ï) + (1 â p)D(Ïx ÏÏx , Ïx ÏÏx ) D(E(Ï), E(Ï)) = = D(Ï, Ï) (âµ eqn(9.21)). Suppose Ï0 = 12 (I + âr · âÏ ) is a fixed point. Then Ï0 = E(Ï0 ) = pÏ0 + (1 â p)Ïx Ï0 Ïx â´ (1 â p)Ï0 â (1 â p)Ïx Ï0 Ïx = 0 â´ (1 â p)(Ï â Ïx Ï0 Ïx ) = 0 â´ Ï0 = Ïx Ï0 Ïx 1 1 â´ (I + r1 Ïx + r2 Ïy + r3 Ïz ) (I + r1 Ïx â r2 Ïy â r3 Ïz ) 2 2 Since {I, Ïx , Ïy , Ïz } are linearly independent, r2 = âr2 and r3 = âr3 . Thus r2 = r3 = 0. Therefore the set of fixed points for the bit flip channel is {Ï | Ï = 21 (I + rÏx ), |r| ⤠1, r â R} 9.14 44 CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION â F (U ÏU , U ÏU ) = Tr (U ÏU â )1/2 Ï(U ÏU â ) â = Tr U Ï1/2 ÏÏ1/2 U â â = Tr(U Ï1/2 ÏÏ1/2 U â ) â = Tr( Ï1/2 ÏÏ1/2 U â U ) â = Tr Ï1/2 ÏÏ1/2 â â = F (Ï, Ï) 0013 â 0010 â I think the fact U AU â = U AU â is not restricted for positive operator. Suppose A is a normal matrix. From spectral theorem, it is decomposed as â A= ai |iâ©â¨i| . i Let f be a function. Then â f (U AU â ) = f ( ai U |iâ©â¨i| U â ) = â i = U( i f (ai )U |iâ©â¨i| U â â f (ai )U |iâ©â¨i| U â )U â i = U f (A)U â 0012 0011 9.15 â â |Ïâ© = (UR â ÏUQ ) |mâ© is any fixed purification of Ï, and |Ïâ© = (VR â ÏVQ ) |mâ© is purification â â â â â â of Ï. Suppose Ï Ï = | Ï Ï|V is the polar decomposition of Ï Ï. Then ( ) â â â | â¨Ï|Ïâ© | = â¨m| URâ VR â UQ Ï ÏVQ |mâ© ( ) â â â = Tr (URâ VR )T UQ Ï ÏVQ ( ) â â â = Tr VRT URâ UQ Ï ÏVQ ( ) â â â Ï Ï = Tr VQ VRT URâ UQ ( ) â â â = Tr VQ VRT URâ UQ | Ï Ï|V ( ) â â â = Tr V VQ VRT URâ UQ | Ï Ï| â â ⤠Tr | Ï Ï| = F (Ï, Ï) â U â )â we see that equality is attained. Choosing VQ = V â , VRT = (UQ R 9.16 45 I think eq (9.73) has a typo. Tr(Aâ B) = â¨m|A â B|mâ© should be Tr(AT B) = â¨m|A â B|mâ©. See errata list. In order to show that this exercise, I will prove following two properties, Tr(A) = â¨m|(I â A)|mâ© , (I â A) |mâ© = (AT â I) |mâ© where A is a linear operator and |mâ© is unnormalized maximally entangled state, |mâ© = â¨m|I â A|mâ© = â â¨ii|(I â A)|jjâ© ij = â â¨i|I|jâ© â¨i|A|jâ© ij = â δij â¨i|A|jâ© ij = â â¨i|A|iâ© i Suppose A = = Tr(A) â ij aij |iâ©â¨j|.  (I â A) |mâ© = ï£I â = â â  aij |iâ©â¨j| â ij |kkâ© k aij |kâ© â |iâ© â¨j|kâ© ijk = â aij |kâ© â |i⩠δjk ijk = â aij |jâ© â |iâ© ij = â aji |iâ© â |jâ© ij   â â (AT â I) |mâ© = ï£ aji |iâ©â¨j| â I  |kkâ© ij = â k aji |iâ© â¨j|kâ© â |kâ© ij = â aji |i⩠δjk â |kâ© ij = â aji |ijâ© ij = (I â A) |mâ© Thus Tr(AT B) = Tr(BAT ) = â¨m|I â BAT |mâ© = â¨m|(I â B)(I â AT )|mâ© = â¨m|(I â B)(A â I)|mâ© = â¨m|A â B|mâ© . â i |iiâ©. 46 CHAPTER 9. DISTANCE MEASURES FOR QUANTUM INFORMATION 9.17 If Ï = Ï, then F (Ï, Ï) = 1. Thus A(Ï, Ï) = arccos F (Ï, Ï) = arccos 1 = 0. If A(Ï, Ï) = 0, then arccos F (Ï, Ï) = 0 â cos(arccos F (Ï, Ï)) = cos(0) â F (Ï, Ï) = 1 (âµ text p.411, the fifth line from bottom). 9.18 For 0 ⤠x ⤠y ⤠1, arccos(x) ⥠arccos(y). From F (E(Ï), E(Ï)) ⥠F (Ï, Ï) and 0 ⤠F (E(Ï), E(Ï)), F (Ï, Ï) ⤠1, arccos F (E(Ï), E(Ï)) ⥠arccos F (Ï, Ï) â´ A(E(Ï), E(Ï)) ⥠A(Ï, Ï) 9.19 From eq (9.92) F ( â pi Ïi , i â ) ⥠pi Ïi i ââ pi pi F (Ïi , Ïi ) i = â pi F (Ïi , Ïi ). i 9.20 Suppose Ïi = Ï. Then F ( â ) pi Ïi , Ï =F i ( â pi Ïi , â i =F ⥠( â â ) pi Ï i pi Ïi , â i ) pi Ïi i pi F (Ïi , Ïi ) (âµ Exercise9.19) i = â pi F (Ïi , Ï) i 9.21 1 â F (|Ïâ© , Ï)2 = 1 â â¨Ï|Ï|Ïâ© (âµ eq(9.60)) D(|Ïâ© , Ï) = max Tr(P (Ï â Ï)) (where P is projector.) P ⥠Tr (|Ïâ©â¨Ï| (Ï â Ï)) = â¨Ï|(|Ïâ©â¨Ï| â Ï)|Ïâ© = 1 â â¨Ï|Ï|Ïâ© = 1 â F (|Ïâ© , Ï)2 . 9.22 47 (ref: QCQI Exercise Solutions (Chapter 9) - ãããã http://enakai00.hatenablog.com/entry/2018/04/12/134722) For all Ï, following inequality is satisfied, d(V U ÏU â V â , F ⦠E(Ï)) ⤠d(V U ÏU â V â , F(U ÏU â )) + d(F(U ÏU â ), F ⦠E(Ï)) ⤠d(V U ÏU â V â ) + d(U ÏU â , E(Ï)) ⤠E(V, F) + E(U, E). First inequality is triangular inequality, second is contractivity of the metric1 and third is from definition of E. Above inequality is hold for all Ï. Thus E(V U, F ⦠E) ⤠E(V, F) + E(U, E). 9.23 (â) If E(Ïj ) = Ïj for all j such that pj > 0, then â â â â F¯ = pj F (Ïj , E(Ïj ))2 = pj F (Ïj , Ïj )2 = pj 12 = pj = 1. j j j j (â) Suppose E(Ïj ) ̸= Ïj . Then F (Ïj , E(Ïj )) < 1 (âµ text p.411, the fifth line from bottom ). Thus â â F¯ = pj F (Ïj , E(Ïj ))2 < pj = 1. j j Therefore if F¯ = 1, then E(Ïj ) = Ïj . Problem 1 Problem 2 Problem 3 Theorem 5.3 of âTheory of Quantum Error Correction for General Noiseâ, Emanuel Knill, Raymond Laflamme, and Lorenza Viola, Phys. Rev. Lett. 84, 2525 â Published 13 March 2000. arXiv:quant-ph/9604034 https://arxiv.org/abs/quant-ph/9604034 1 Trace distance and angle are satisfied with contractive (eq (9.35), eq (9.91)), but I donât assure that arbitrary metric satisfied with contractive. ![]() Comments are closed.
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