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$$\begin{bmatrix} XZZXI\\IXZZX\\XIXZZ\\ZXIXZ\end{bmatrix}$$

The above is the generator of the Hamming code scheme [[5,1,3]].

I'd like to know step by step, how should I encode a single physical qubit $|\varphi\rangle = \alpha|0\rangle + \beta|1\rangle$ into 5.

EDIT: Also, consider a transversal $CNOT$ with target over the code. What is the Clifford transformation to make it logical?

Daniele Cuomo
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    This previuos question should help https://quantumcomputing.stackexchange.com/questions/21534/how-to-perform-encoding-and-syndrome-measurement-in-stim – unknown Jun 06 '22 at 17:58
  • BTW do you mean the stabilizer matrix? The $[[5,1,3]]$ is a fairly well known qecc but I've never seen it referred to as "Hamming" code. Also your matrix ix 5x5; if its the list stabilizers and they're independent then that defines a 0 dimensional code (i.e. a stabilizer state). – unknown Jun 06 '22 at 20:01
  • Thanks, I fixed the matrix. I call it Hamming code because it saturates the Hamming bound. – Daniele Cuomo Jun 07 '22 at 10:21
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    Do you want to have explicit representations of the code words, or circuits that prepare them? – Nikita Nemkov Jun 07 '22 at 11:17
  • I'd like to read the circuit preparation, and how to perform logical operations over the code. I edited the question, to make clearer what I need! – Daniele Cuomo Jun 08 '22 at 15:03
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    @DanieleCuomo I'm pretty sure $[[5,1,3]]$ does not admit a transversal CNOT. It does admit transversal hadamard and phase gates. I think the Steane code $[[7,1,3]]$ is the smallest code with transversal CNOT. – unknown Jun 08 '22 at 16:56
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    @DanieleCuomo see https://quantumcomputing.stackexchange.com/questions/15432/css-code-in-disguise for proof – unknown Jun 08 '22 at 17:53
  • @unknown Thanks a lot! May you give me some reference to the encoding for [[7,1,3]]? – Daniele Cuomo Jun 09 '22 at 11:52
  • @DanieleCuomo try this https://quantumcomputing.stackexchange.com/questions/13033/how-to-create-the-logical-0-l-rangle-state-for-the-steanes-7-qubit-code ...google should bring out many others – unknown Jun 09 '22 at 15:48
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    [[4,2,2]] code has transversal $ CNOT^{\otimes 4} $ implements logical $ CNOT $. I think 4 qubit code is the smallest you can go with transversal CNOT. Obviously downside of this is that [[4,2,2]] just detects 1 qubit errors doesn't correct them like the [[7,1,3]] code. – Ian Gershon Teixeira Jul 29 '22 at 13:29

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