Quantum channels that commute with any unitary channel

Question

Consider a quantum channel $\Phi$ that maps from density operators $\mathcal{S}(\mathcal{H}_A)$ to itself, that commutes with any unitary channel $\mathcal{U}$ on $\mathcal{S}(\mathcal{H}_A)$, i.e. $\Phi(\mathcal{U}(\rho))=\mathcal{U}(\Phi(\rho)),\,\,\forall\rho\in\mathcal{S}(\mathcal{H}_A)$.

Is there a general characterisation of such channels? (As far as I can think of it, $\Phi$ can either be a unitary channel or a depolarization channel. What are the other possibilities?)

Answer 1

Yes, there is, you're asking about the commutant of the representation $U\mapsto U(\cdot)U^\dagger$ of the unitary group $U(d)$. From a representation-theoretic point of view, the superoperators in the commutant are easy to characterize: By Schur's lemma, such an operator has to be proportional to the identity on every irrep of $U\mapsto U(\cdot)U^\dagger$.

These irreps are well-known and are given by the following orthogonal decomposition of the vector space of linear operators $L(\mathbb C^d)$: $$ L(\mathbb C^d) = \langle \mathbb I \rangle \oplus \{\text{traceless operators}\}. $$ The projection on $\langle \mathbb I \rangle$ is the completely depolarizing channel: $$ \mathcal{D}(X) := \mathrm{tr}(X) \frac{\mathbb I}{d}. $$ The projection on the traceless operators is simply $\mathrm{id}-\mathcal{D}$. Hence, every element $X$ in the commutant can be written as $$ X = a \mathcal{D} + b (\mathrm{id}-\mathcal{D}) = a' \mathcal{D} + b' \mathrm{id}. $$ The latter decomposition is not orthogonal, but has to advantage to answer your question: The only quantum channels which commute with unitary channels are the completely depolarizing channel and the identity channel, and of course any convex combination of those.

For more details on the rep theory, see my answer here: Why does the twirl of a quantum channel give a depolarizing channel?