Consider a controlled-NOT (CX) gate between the two qubits, implemented with an interaction of the form

$ \widehat{H}_{\mathrm{CX}}=V\left[\left(\frac{\hat{I}_1+\widehat{Z}_1}{2}\right) \otimes \hat{I}_2+\left(\frac{\hat{I}_1-\widehat{Z}_1}{2}\right) \otimes \widehat{X}_2\right] $

with the evolution unitary $\widehat{U}(t)=\exp \left(-i \widehat{H}_{\mathrm{CX}} t\right)$. Here, the qubits labeled 1 and 2 are the control (data qubit) and target (ancilla qubit), respectively. A CX gate is realized at time $T$ when $V T=\pi / 2$ and

$ \widehat{U}(T)=\left[\left(\frac{\hat{I}_1+\widehat{Z}_1}{2}\right) \otimes \hat{I}_2+\left(\frac{\hat{I}_1-\widehat{Z}_1}{2}\right) \otimes \widehat{X}_2\right] $

where we have ignored an overall phase. In this case, a phase-flip error in the target qubit at time $0 \leq \tau \leq T$ modifies the evolution to

$\widehat{U}_{\mathrm{e}}(T) =\widehat{U}(T-\tau) \hat{I}_1 \otimes \widehat{Z}_2 \widehat{U}(\tau) =\hat{I}_1 \otimes \widehat{Z}_2 e^{i V(T-\tau)\left(\hat{I}_1-\hat{Z}_1\right) \otimes \hat{X}_2} \widehat{U}(T)$

Apparently, this error on qubit 2 propagates as an uncorrectable high-weight error on the first qubit. How do I prove the last identity?



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