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As the Shor's code is a CSS code, it admits a transversal implementation of logical CNOT.

An immediate implementation may perform 9 (reversed) CNOT, by respecting the order of the qubits.

However. Considering the answer to this question, I'm led to think that I may define a better implementation, involving just 3 CNOTs.

Is this possible?

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    $\begingroup$ All CSS codes admit transversal CNOT; Shor's code is CSS so it does. What do you mean by the 1-1 map...? $\endgroup$
    – unknown
    Jun 18 at 15:58
  • $\begingroup$ @unknown Thank you. I expanded my question! $\endgroup$ Jun 18 at 17:05
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    $\begingroup$ What is your standard sequence of CNOT? you have two copies of the code (so this is an 18 qubit system). You connect each of the 9 "bottom" qubits to a corresponding "top" qubit...so you have 9 CNOT's between the two code blocks. Is this what you tried and failed in your experiment? $\endgroup$
    – unknown
    Jun 18 at 19:21
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    $\begingroup$ The previous questions says you can take $Z_L=X_1X_2X_3$ and $X_L=Z_1Z_4Z_7$; (there are other combinations but let's work with this one). Did you try connecting qubits 1,2,3,4,7 from code block 1 to 1,2,3,4,7 from code block 2?..so 5 CNOT's instead of 9. $\endgroup$
    – unknown
    Jun 29 at 3:29
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    $\begingroup$ With "5 cnot" connection, you should get $X_L \otimes I \to X_L \otimes X_L$, $I \otimes Z_L \to Z_L \otimes Z_L$, $I \otimes X_L \to I \otimes X_L$, $Z_L\otimes I \to Z_L\otimes I$. The "3 cnot" connection won't give you that. $\endgroup$
    – unknown
    Jun 29 at 14:48

1 Answer 1

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For 5-CNOT connection (and 9-CNOT connection)

$$X_L \otimes I = Z_1Z_4Z_7 \otimes I \to Z_1 Z_4 Z_7 \otimes I = X_L \otimes I$$ $$Z_L \otimes I = X_1X_2X_3 \otimes I \to X_1 X_2 X_3 \otimes X_1 X_2 X_3 = Z_L \otimes Z_L$$ $$I \otimes X_L = I \otimes Z_1Z_4Z_7 \to Z_1 Z_4 Z_7 \otimes Z_1 Z_4 Z_7 = X_L \otimes X_L$$ $$I \otimes Z_L = I \otimes X_1X_2X_3 \to I \otimes X_1 X_2 X_3 = I \otimes Z_L$$

this is logical CNOT (with the second block as control logical).

For 3-CNOT connection $$Z_L \otimes I = X_1X_2X_3 \otimes I \to X_1 X_2 X_3 \otimes X_1 \neq Z_L \otimes Z_L$$ Since there is no connection between qubits 2,3 of block1 and 2,3 of block 2.

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  • $\begingroup$ Thanks. Did you have the opportunity to verify this experimentally? $\endgroup$ Jun 29 at 18:01
  • $\begingroup$ Here's why I think it doesn't work: your proof doesn't catch the whole logic here. Actually, as there are 5 CNOTs, an operation becomes, say, $X_1 X_2 X_3 X_4 X_7$ which may not be a $Z_L$. $\endgroup$ Jun 29 at 18:07
  • $\begingroup$ I don't see the issue; $X_1X_2X_3X_4X_7$ is not a logical to begin with. $\endgroup$
    – unknown
    Jun 29 at 19:02
  • $\begingroup$ Why not? 5 CNOTs propagates that. $\endgroup$ Jun 29 at 19:35
  • $\begingroup$ I'm afraid I don't know what you mean by "propogate" here. In what I write the $\to$ refers to conjugation by the 5-CNOT gate...Besides the action on the logicals there could be other restrictions for fault tolerance...it might help if you describe what you do in more detail. How do you show that the 9-CNOT circuit works and the 5 and 3 versions fail? $\endgroup$
    – unknown
    Jun 30 at 0:06

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