# How $X$ noise propagates through controlled-$S$ gate

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

Here`s the circuit.

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:

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.

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):

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.

• May you share the code on wolframalpha.com? Oct 9, 2022 at 11:19
• 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. Oct 9, 2022 at 11:24
• Thanks!! I sent you a request. Oct 9, 2022 at 13:12