Implementation of transversal Hadamard in Knill's $C_6$ code

Implementation of transversal Hadamard in Knill's C 6 𝐶 6 code Ask Question Asked 30 days ago Modified today Viewed 93 times 1 I have been reading Knill's paper on the C 4 / C 6 𝐶 4 / 𝐶 6 architecture and I am getting confused by how the Hadamard can be implemented. It is claimed in the paper that the Hadamard operation is transversal (up to a swap operation in the case of C 4 𝐶 4 , see figure 12). My algebra seems to show that this is the case for C 4 𝐶 4 but not C 6 𝐶 6 , so I am hoping someone can point out a flaw in my reasoning. First let me start with C 4 𝐶 4 because I understand this case. The C 4 𝐶 4 code is a [[4,2,2]] [ [ 4 , 2 , 2 ] ] code with stabilizers ZZZZ 𝑍 𝑍 𝑍 𝑍 and XXXX 𝑋 𝑋 𝑋 𝑋 , in addition to logical operators X 1 =XXII, Z 1 =ZIZI, X 2 =IXIX, Z 2 =IIZZ 𝑋 1 = 𝑋 𝑋 𝐼 𝐼 , 𝑍 1 = 𝑍 𝐼 𝑍 𝐼 , 𝑋 2 = 𝐼 𝑋 𝐼 𝑋 , 𝑍 2 = 𝐼 𝐼 𝑍 𝑍 . It is then easy to check that H ⊗4 X 1 H ⊗4 =ZZII=IIZZ= Z 2 𝐻 ⊗ 4 𝑋 1 𝐻 ⊗ 4 = 𝑍 𝑍 𝐼 𝐼 = 𝐼 𝐼 𝑍 𝑍 = 𝑍 2 H ⊗4 X 2 H ⊗4 =IZIZ=ZIZI= Z 1 𝐻 ⊗ 4 𝑋 2 𝐻 ⊗ 4 = 𝐼 𝑍 𝐼 𝑍 = 𝑍 𝐼 𝑍 𝐼 = 𝑍 1 where the last equality in both cases follows from applying stabilizers. Hence we can see that applying Hadamard transversely and swapping the logical qubits is equivalent to the product of logical Hadamard operations. Now let us consider the case of C 6 𝐶 6 . In this case the stabilizers are XIIXXX,XXXIIX,ZIIZZZ,ZZZIIZ 𝑋 𝐼 𝐼 𝑋 𝑋 𝑋 , 𝑋 𝑋 𝑋 𝐼 𝐼 𝑋 , 𝑍 𝐼 𝐼 𝑍 𝑍 𝑍 , 𝑍 𝑍 𝑍 𝐼 𝐼 𝑍 and the logical operators are X 1 =IXXIII, Z 1 =IIZZIZ, X 2 =XIXXII, Z 2 =IIIZZI 𝑋 1 = 𝐼 𝑋 𝑋 𝐼 𝐼 𝐼 , 𝑍 1 = 𝐼 𝐼 𝑍 𝑍 𝐼 𝑍 , 𝑋 2 = 𝑋 𝐼 𝑋 𝑋 𝐼 𝐼 , 𝑍 2 = 𝐼 𝐼 𝐼 𝑍 𝑍 𝐼 . Now let us see what the action of transversal Hadamard is in this case. For X 1 𝑋 1 , we have H ⊗6 X 1 H ⊗6 =IZZIII=IIIZZI= Z 2 𝐻 ⊗ 6 𝑋 1 𝐻 ⊗ 6 = 𝐼 𝑍 𝑍 𝐼 𝐼 𝐼 = 𝐼 𝐼 𝐼 𝑍 𝑍 𝐼 = 𝑍 2 where we have applied stabilizers again. This is already strange - Knill's paper does not mention the need for a swap operation in C 6 𝐶 6 , but we seem to have gotten Z 2 𝑍 2 instead of Z 1 𝑍 1 as we would expect. Things are even stranger for X 2 𝑋 2 : H ⊗6 X 2 H ⊗6 =ZIZZII≠ Z 1 𝐻 ⊗ 6 𝑋 2 𝐻 ⊗ 6 = 𝑍 𝐼 𝑍 𝑍 𝐼 𝐼 ≠ 𝑍 1 That is, there is no stabilizer relating Z 1 𝑍 1 to H ⊗6 X 2 H ⊗6 𝐻 ⊗ 6 𝑋 2 𝐻 ⊗ 6 . I'm either misunderstanding what Knill is saying or I am making some kind of error. If you can see how the implementation should go please let me know, thanks! quantum-gateerror-correctionstabilizer-codelogical-gates Share Improve this question Follow asked Feb 22 at 17:38 miggle 2151 1 silver badge 5 5 bronze badges "Knill's paper does not mention the need for a swap operation in C 6 𝐶 6 ". It does. In the "Clifford gates for C4 and C6" section, it says. "The codes C4, C6 and their concatenations have the property that encoded CNOTs, HADs and measurements that act in parallel on both encoded qubits in a pair can be implemented “transversally” with physical qubit relabeling". The relabeling are SWAPs here. Anyway, I'm also confused that Fig 12 explicitly calls out relabeling for C4 but not for C6. –  Victory Omole Commented Feb 23 at 0:57 @VictoryOmole My point was that this was worded ambiguously - given that the qubit relabeling was explicitly provided for C 4 𝐶 4 , the absence of such a scheme for C 6 𝐶 6 makes it seem that relabeling is not necessary in that case. However it doesn't really matter, I have found a method to implement Hadamard which I have written up as an answer. –  miggle Commented Feb 23 at 15:00 Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 After some experimentation, I figured out how to implement the product of logical Hadamards in C 6 𝐶 6 . It doesn't seem to be consistent with what Knill says but I have confirmed that this works in a few ways. Here is my solution: First, we apply the Hadamard gate transversely. Next, we apply what Knill calls the ∗ u 2 ∗ 𝑢 2 operation to neighboring pairs of qubits, which is equivalent to a CNOT gate followed by a swap operation. Let's verify that this implements the expected logic. For the case of X 1 ¯ ¯ ¯ ¯ ¯ ¯ 𝑋 1 ¯ , transversal Hadamard produces the Pauli string IZZIII 𝐼 𝑍 𝑍 𝐼 𝐼 𝐼 . Putting neighboring qubits through the ∗ u 2 ∗ 𝑢 2 gate produces ZZIZII=IIZZIZ= Z 2 ¯ ¯ ¯ ¯ ¯ ¯ 𝑍 𝑍 𝐼 𝑍 𝐼 𝐼 = 𝐼 𝐼 𝑍 𝑍 𝐼 𝑍 = 𝑍 2 ¯ , where equality follows by applying stabilizers. Next consider X 2 ¯ ¯ ¯ ¯ ¯ ¯ 𝑋 2 ¯ . Transversal Hadamard in this case produces ZIZZII 𝑍 𝐼 𝑍 𝑍 𝐼 𝐼 which is then mapped to IZZIII 𝐼 𝑍 𝑍 𝐼 𝐼 𝐼 by ∗ u 2 ∗ 𝑢 2 . Stabilizers can then be used to map this to IIZZII= Z 2 ¯ ¯ ¯ ¯ ¯ ¯ 𝐼 𝐼 𝑍 𝑍 𝐼 𝐼 = 𝑍 2 ¯ . Maybe Knill just has an error, I'm not 100% sure because I have not read the entire paper. In any case, hopefully this can be a useful reference for others in the future. Share Improve this answer Follow answered Feb 23 at 14:58 miggle 2151 1 silver badge 5 5 bronze badges Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions quantum-gateerror-correctionstabilizer-codelogical-gates See similar questions with these tags. 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