# Show that ${\cal C}(\rho)={\rm Tr}[U(\rho\otimes|0\rangle\!\langle0|)U^\dagger]$ can be corrected iff $t=0$, where $U=I+(e^{it}-1)|\Psi^-〉〈\Psi^-|$

Let $$\mathcal{C}$$ be the quantum channel defined by $$\mathcal{C}(\rho)=\operatorname{Tr}_{2}\left[U(\rho \otimes|0\rangle\langle 0|) U^{\dagger}\right],$$ where $$U$$ is the two-qubit unitary gate $$U=I \otimes I+\left(e^{i t}-1\right)\left|\Psi^{-}\right\rangle\left\langle\Psi^{-}\right|, \quad t \in \mathbb{R}$$ $$\left|\Psi^{-}\right\rangle=(|0\rangle \otimes|1\rangle-|1\rangle \otimes|0\rangle) / \sqrt{2}$$, and $$\operatorname{Tr}_{2}$$ denotes the partial trace on the second qubit. Show that $$\mathcal{C}$$ can be corrected if and only if $$t=0$$.

I wonder if there are any other ways to do this without expanding all the terms.

• What are the Kraus operators of the channel? Dec 8 '21 at 7:51