Lecture # 2, Quantum Computation 2: QEC Criteria

Lecture # 2, Quantum Computation 2: QEC Criteria Outline: 0. Review 1. Classical Coding 2. Q. Coding 3. Operator Measurement and Error Syndromes 4. Shor 9 Qubit Code 5. Quantum Error correction Codes Criteria (QEC criteria) 2 0. Review � χ ≡ χ∗ β(χ) = � k Ek χEk† where � k Ek Ek† = I 3 1. CLASSICAL CODING 1-p 0 0 p p 1 1-p 1 FIG. 1: a binary symmetric channel P = prob of error Definition: A Classical [n,k,d] code is a set of 2k n-bit strings which have a minimum Hamming distance d. Definition: A Hamming distance between two bit strings is d(x, y) = w(x ∃ y) where ∃ is the x-or operator, and w is an operation that counts the number of ones. Example: 0L(ogical) = 000, 1L = 111 is a [3,1,3] code could send could receive prob decode 0L = 000 000 (1 − p)3 0 001 p(1 − p)2 0 010 p(1 − p)2 0 100 p(1 − p)2 0 011 p2 (1 − p) 1 p2 (1 − p)+ 101 p2 (1 − p) 1 p2 (1 − p)+ 110 p2 (1 − p) 1 p2 (1 − p)+ 111 p3 1 p3 So, the total probability of error is 3p2 − 2p3 = O(p2 ) prob. of error 4 2. QUANTUM CODING 1995: Thought error correction to be impossible! 1. States collapse on measurement 2. Classically error occurs or does not occur. In Q. M., errors are continuous: �|0→ + � |1→ ≡ (� + β)|0→ + ... 3. No cloning Thm prohibits copying, so cannot create �|0→ + �|1→ ≡ (�|0→ + � |1→)(�|0→ + � |1→)(�|0→ + � |1→) The Solutions: 1. Measure only the effect of the environment, not the state (i.e. did an error occur?) 2. & 3. Orthogonalize errors using entanglement: the environment has done one thing, or another, in an entangled way. �|didsomething→ + � |didnothing→ Example: The Quantum Bit Flip Code: |0L → = |000→ |1L → = |111→ |�L → = �|0L → + � |1L → suppose β(χ) = (1 − P )χ + P XχX where P is the probability of error and X is the error operator. �(�) A≡ B Define: An [[n,k]] quantum code C is a k-qubit subspace of an n-qubit Hilbert space. So, for our example, k=3, n=1. 5 ��3 Input |�� = �|000� + � |111� 3. � Output prob decode prob. of error �|000� + � |111� (1 − p) 3 0 �|001� + � |110� p(1 − p)2 0 �|010� + � |101� p(1 − p)2 0 �|100� + � |011� p(1 − p)2 0 �|011� + � |100� p2 (1 − p) 1 p2 (1 − p)+ �|101� + � |010� p2 (1 − p) 1 p2 (1 − p)+ �|110� + � |001� p2 (1 − p) 1 p2 (1 − p)+ �|111� + � |000� p3 1 p3 OPERATOR MEASUREMENT Given U with eigenvalues ±1, eigenvectors |u± → Definition: Measuring U H H 0 C0 u+ U measurement result z � C1 u- Initially, the state is |0→(C0 |u+ → + C1 |u− →) �1 (|0→ + |1→)(C0 |u+ → + C1 |u− →) 2 Controlled-U gate, �12 (|0→(C0 |u+→ + C1 |u− →) + |1→(C0 |u+ → − C1 |u− →)) last Hadamard, �12 (|0→ + |1→)(C0 |u+ → + C1 |u− →) + (|0→ − |1→)(C0 |u+ → − After the first Hadamard, After the After the = |0→C0 |u+ → + |1→C1 |u− → If the measurement is z = 0, then |P si→ = |u+→. (With prob C02 , z = 0) If the measurement is z = 1, then |P si→ = |u− →. (With prob C12 , z = 1) 3.1. U1 = δz1 δz2 = δz δz I U2 = δz2 δz3 = Iδz δz Error Correction Syndrome Measurement C1 |u− →) 6 state U1 U2 �|000� + � |111� 0 0 �|001� + � |110� 0 1 �|010� + � |101� 1 1 �|100� + � |011� 1 0 TABLE I: 0 represents a +1 eigenstate of U i , and 1 represents a -1 eigenstate. Steps to Error Correction: 1. measure syndrome operators (here, U1 & U2 2. Apply recovery operator R (here, 00 ≡ I, 01 ≡ δx3 , 11 ≡ δx2 , 10 ≡ δx1 To create the initial state |P siL →: � 0 +� 1 � 000 0 + � 111 0 And then to error correct: 0 � H H αz Ψ αz R αz αz 0 H H note: the double lines indicate classical information. 7 Claim: This scheme also corrects for a small continuous rotation error! We will do this on one bit to demonstrate. β(χ) = eiεδx χeiεδx e−εδx = Rx (2ε) Rx1 (2ε)|�→ ∝ = |�→ − iεδ 1 |P si→ ∈ |�∗ → x The fidelity is F = |√�|�∗ →|2 ∝ = 1−ε Syndrome measurement collapses error into either I or δx1 F (R(β(χ)), |�→) ∝ =? ∝ = 1 − ε2 Example: The Phase Flip Code βphasef lip(χ) = (1 − P )χ + P δz χδz Recall Hδx H = δz , Hδz H = δx So, Hβphasef lip(HχH)H = βbitf lip Explicitly, � 0 +� 1 H 0 H 0 H 0L = |+++ 1L = | - - ­ Ψ H H R H + = 0 + 1 / sqrt(2) _ = 0 - 1 / sqrt(2) For the bit flip: Uo = δz δz I and U1 = Iδz δz For the phase flip: Uo = δx δx I and U1 = Iδx δx Claim: arbitrary errors can be described as δx , δz , and δx δz errors Proof Argument: � β(χ) = k Ek χEk† where we are guaranteed � k Ek Ek† = I Recall pauli matrices δj = I, δx , δy , δz , and that δy = −iδx δz � Since δj is a basis for all 2x2 hermitian matrices, let Ek = j Ckj δj . 8 � Then, β(χ) = kjj � Ckj Ck�j � δj χδj � � β(χ) jj � �jj � δj χδj � is the “Chi representation or OSR” Example: recall ∝ |P si→ − iεδx |P si→ Rx (2ε)|�→ = β(χ) = χ − iεδx χ − iεχδx + ε2 δx χδx The −iεδx χ − iεχδx term disappears in the syndrome measurement, and the χ + ε2 δx χδx term remains. The result is that the syndrome measurement projects the environment into a definite error state. 9 4. SHOR 9 QUBIT CODE ⊕ |0L → = (|000→ + |111→)�3 / 8 ⊕ |1L → = (|000→ − |111→)�3 / 8 this code will correct ANY single qubit error. Syndrome Measurements: for a bit flip: δz1 δz2 , δz2 δz3 , δz4 δz5 , δz5 δz6 , δz7 δz8 , δz8 δz9 for a phase flip: δx1 δx2 , δx3 δx4 , δx4 δx5 , δx5 δx6 , δx6 δx7 , δx8 δx9 , 10 5. Channel: E(χ) = � k QEC CRITERIA/CONDITIONS Ek χEk† Thm: Let C be a quantum Code defined by the orthonormal states { |�l → } � a quantum recovery operation R correction β on C iff: 1. Orthogonality: if I have 2 errors j and k, √�l |Ej† Ek |�l → = 0 2. Nondeformation criteria: √�l |Ek† Ek |�l → = dk �l this is so you cannot distinguish shrinking on different code words, all shrinking is the same. C Ψ C C C No Overlap or Unique Deformations C C C note that dk implies probability loss, but not information loss, � † k Ek Ek =1 � k dk = 1 since Proof: (≡) � Let P = l |�l →√�l | (project onto C) note P Ej† Ek P = dk αjk P (*) note by Polar decomposition (extracting rotation and shrinkage) Ek P = Uk ⊕ ⊕ dk Uk P where dk is the shrinkage and Uk P is the rotation. 1. Syndrome measurement: � P Ek† Ek P = 11 let Pk = Uk P Uk† = Ek P Uk† � dk = Uk P Ek† � dk By (*), the Pk s are orthogonal: �k ∼= j, Pk Pj ∀ Uk P Ek† Ej P Uj† = 0 measure Pk output k syndrome. 2. Apply Recovery R � R(χ) = k Uk† Pk χPk Uk note for |�→ � C , Uk† Pk Ej |�→ = Uk† Uk P Ek† � Ej P |�→ dk = �jk dk P � |�→ dk = ⊕ dk αjk |�→ Thus: R(β(|�→√�|) = R( � j Ej |�→√�|EjT ) = � jk Uk† Pk Ej† Pk Uk = � jk dk αjk P = |�→√�|