# Prove the invariance upon change of variables in the definition of twirled channel

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}

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 $$U$$. $$V\mathcal{E}_T(\rho)V^\dagger=\int dW W^{\dagger}\mathcal{E}(W\tilde \rho W^\dagger)W=\mathcal{E}_T(\tilde\rho).$$
• 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? Mar 27, 2023 at 22:30
• 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 $dW$. Mar 28, 2023 at 6:36