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I am struggling to find how an $X$ noise propagates through a controlled-$S$ gate.

Here`s the circuit.

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2 Answers 2

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In Quirk, the trick to finding the equivalent operation after an intermediate operation is to check that their phase kickback exactly cancels even when operating on entangled qubits:

enter image description here

Note that the bottom qubit ends up OFF. This is a proof that X before CS is the inverse of $S \cdot X \cdot CZ$ after CS.

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If I understand you correctly, you would like to find how an $X$ error before a $cS$ (controlled-S, where $S=diag(1,i)$) gate behave after a $cS$ gate.

What you should solve is simply the equation (I assume an $X$ error on the control as in your circuit):

$$cS X_1 = E cS \Leftrightarrow E=cS X_1 cS^{\dagger}$$

Solving it on mathematica gives the following two-qubit matrix for $E$ (in the two-qubit computational basis):

enter image description here

In practice if you decompose $E$ on the Pauli-matrices you will have a non-trivial expression (linear combination of tensor products of $I$, $X$, $Y$ and $Z$ operators) as $cS$ is non-Clifford. Finally, given the shape of the $E$ matrix above, I think in that case the error propagates on the two registers after the gate.

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  • $\begingroup$ May you share the code on wolframalpha.com? $\endgroup$ Oct 9, 2022 at 11:19
  • $\begingroup$ I am not sure of how I could give it to you through wolfram alpha website. However, you can contact me on linkedin and I can provide you the code. $\endgroup$ Oct 9, 2022 at 11:24
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    $\begingroup$ Thanks!! I sent you a request. $\endgroup$ Oct 9, 2022 at 13:12

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