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The twirled operation of a quantum channel $\varepsilon$ is defined as \begin{align} \varepsilon_T(\rho) &= \int dU U^\dagger \varepsilon(U \rho U^\dagger)U, \end{align} where the integral is over the normalized Haar measure $dU$. Then for any unitary $V$, \begin{align} V \varepsilon_T(\rho) V^\dagger &= \int dU V U^\dagger \varepsilon(U \rho U^\dagger)U V^\dagger. \end{align} Making the change of variables $W \equiv UV^\dagger$, how can we obtain the following result: \begin{align} V \varepsilon_T(\rho) V^\dagger &= \varepsilon_T(V \rho V^\dagger). \end{align}

This result was in this article by Nielsen.

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If $W=UV^\dagger$ then $U=WV$. This means that you can rewrite $$ U\rho U^\dagger=W(V\rho V^\dagger)W^\dagger. $$ So, let me write $\tilde\rho=V\rho V^\dagger$. Your integral becomes $$ V\mathcal{E}_T(\rho)V^\dagger=\int dU W^{\dagger}\mathcal{E}(W\tilde \rho W^\dagger)W. $$ But we might as well be integrating over $W$ rather than $U$. $$ V\mathcal{E}_T(\rho)V^\dagger=\int dW W^{\dagger}\mathcal{E}(W\tilde \rho W^\dagger)W=\mathcal{E}_T(\tilde\rho). $$

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  • $\begingroup$ Thank you! Why integrating over $W$ is the same as integrating over $U$, i.e., $dW$ is the same as $dU$, which is different from elementary calculus? $\endgroup$
    – Michael.Andy
    Mar 27 at 22:30
  • $\begingroup$ Well... hidden within the $\int dU$ are implicitly all the different integration elements to make everything uniform over the Haar measure. Unless you can see all of that, you cannot simply apply the elementary calculus rules. The way I think about it (I don't remember how to formalise it off the top of my head) is that $dU$ is sampling every $U$ with equal likelihood. That's just the same as sampling every $UV^\dagger$ with equal likelihood (because the $V^\dagger$ is a rotation, and doesn't contain any scale factors etc) and hence we can replace $dU$ with $dW$. $\endgroup$
    – DaftWullie
    Mar 28 at 6:36
  • $\begingroup$ It's probably worthwhile attempting a couple of example calculations for one or two qubits to convince yourself. $\endgroup$
    – DaftWullie
    Mar 28 at 6:36

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