# Show that, averaging over uniformly random unitaries, $\mathbb{E}[UXU^\dagger]=\operatorname{Tr}(X)\frac{I}{d}$

As mentioned e.g. in this answer, if we compute the average $$\Phi(X)\equiv \mathbb{E}_U[UXU^\dagger] \equiv \int_{\mathbf U(d)} d\mu(U) UXU^\dagger,$$ where $$d\mu(U)$$ is the Haar measure over the unitary group $$\mathbf U(d)$$, we get $$\operatorname{Tr}(X) I/d$$.

This is relatively standard and intuitive, and a common argument to support it is in terms of how $$\Phi(X)$$ is invariant under the action of any unitary, meaning here that $$V\Phi(X)V^\dagger=\Phi(X)$$ for any $$V\in\mathbf U(d)$$. This kind of operation is also often referred to as twirling, discussed e.g. in this other answer here. Reasoning in terms of symmetries, the statement follows from the fact that $$[V,\Phi(X)]=0$$ for all $$V\in\mathbf U(d)$$, and the fact that multiples of the identity are the only matrices that commute with all unitaries.

I have the vague notion that this can also be seen via Schur's lemma. In such terms, I think I'd want to consider an irreducible representation of $$\mathbf U(d)$$ and presumably find that $$\Phi(X)$$ commutes with this action, for any $$X$$, hence must be a multiple of the identity. The most obvious such representation to consider would be the representation of $$\mathbf U(d)$$ on the space of linear operators $$\operatorname{Lin}(\mathbb{C}^d)$$ defined via $$V.X\equiv \rho(V)(X)\equiv VXV^\dagger$$ for $$V\in\mathbf U(d)$$, $$X\in\operatorname{Lin}(\mathbb{C}^d)$$, and $$\rho(V)\in\operatorname{Lin}(\operatorname{Lin}(\mathbb{C}^n))$$ the representation (in other words, we're representing the unitaries as "quantum maps"). Then, we clearly have $$\rho(V)(\Phi(X))= \Phi(\rho(V)(X))$$, i.e. $$[\rho(V),\Phi]=0$$ for all $$V$$. However, following this line of thought I would end up concluding that $$\Phi=c \operatorname{Id}$$, which is different than the statement at hand, where $$\Phi$$ is not the identity (as in, the identity map) but rather the map sending all states to the identity operator. I'm not sure, but this wrong conclusion is probably due to this representation not being irreducible, and thus Schur's lemma not being actually applicable.

Assuming I'm not mistaken in all this, is there a way to make some similar kind of group-theoretic reasoning work? Maybe considering subrepresentations of the above that are actually irreducible? It feels like it should work but I'm not sure how to formalise the precise steps.

You are right about all the ingredients here. First, the result can indeed be obtained from the Schur's lemma. Second, the naive argument fails because your representation is not irreducible. Finally, the argument can be fixed by decomposing your representation into irreps.

## Decomposing $$\rho$$ into irreps

Let $$\sigma_k\in\mathrm{Lin}(\mathbb{C}^d)$$ for $$k=0,\dots,d^2-1$$ be a basis of $$\mathrm{Lin}(\mathbb{C}^d)$$ where $$\sigma_0=I$$ is the identity and for $$k>0$$ the operators $$\sigma_k$$ are traceless with the same set of eigenvalues. For example, if $$d$$ is a power of two, then we could choose $$\sigma_k$$ to be the Pauli basis.

The key observation is that $$\rho(V)$$ is a similarity transformation for every $$V\in\mathbf{U}(d)$$. In particular, $$\rho(V)(\sigma_0)=\sigma_0$$ and $$\rho(V)$$ sends traceless matrices to traceless matrices for every $$V\in\mathbf{U}(d)$$. Therefore, $$\mathrm{span}(\sigma_0)\subset\mathrm{Lin}(\mathbb{C}^d)$$ is a one-dimensional irreducible subrepresentation of $$\rho$$. Moreover, since $$\sigma_k$$ for $$k>0$$ have the same eigenvalues, for any two $$\sigma_i$$ and $$\sigma_j$$ with $$i,j>0$$ there exists $$W\in\mathbf{U}(d)$$ such that $$\rho(W)(\sigma_i)=\sigma_j$$. Therefore, $$\mathrm{span}(\sigma_1,\dots,\sigma_{d^2-1})$$ is another irreducible subrepresentation of $$\rho$$.

Thus, we obtain decomposition of $$\rho$$ into irreps as $$\mathrm{Lin}(\mathbb{C}^d)=\mathrm{span}(\sigma_0)\oplus\mathrm{span}(\sigma_1,\dots,\sigma_{d^2-1}).\tag1$$ In other words, $$\rho$$ decomposes into the trace and the traceless matrices.

## Twirling by Schur's lemma

Now that we know the decomposition of $$\rho$$ into irreps, we can justify the use of Schur's lemma along the lines suggested in the question. To that end, we consider $$\Phi:\mathrm{Lin}(\mathbb{C}^d)\to\mathrm{Lin}(\mathbb{C}^d)$$ restricted to the vector spaces associated with each of the two irreducible representations indicated by the right hand side of $$(1)$$ and apply Schur's lemma to each one. We obtain that $$\Phi$$ acts as a scalar multiple of the identity on the subspace spanned by $$\sigma_0$$ and as a possibly different scalar multiple of the identity on $$\mathrm{span}(\sigma_1,\dots,\sigma_{d^2-1})$$. In other words, $$\Phi=c_0 P_0 + c_1 P_1\tag2$$ where $$c_0,c_1\in\mathbb{C}$$, $$P_0$$ is the projector on $$\mathrm{span}(\sigma_0)$$ and $$P_1$$ is the projector on $$\mathrm{span}(\sigma_1,\dots,\sigma_{d^2-1})$$. Now, $$\sigma_0$$ is orthogonal$$^1$$ to all traceless matrices, so $$P_0P_1=P_1P_0=0$$. Evaluating $$(2)$$ on $$\sigma_0$$ we see that $$c_0=1$$. Similarly, evaluating $$(2)$$ on two distinct $$\sigma_i$$ and $$\sigma_j$$ with $$i,j>0$$ and exploiting the symmetry of $$\Phi$$ we have $$c_1\sigma_i=\Phi(\sigma_i)=\Phi(\sigma_j)=c_1\sigma_j$$. However, $$\sigma_i$$ and $$\sigma_j$$ are linearly independent so $$c_1=0$$. Finally, for any $$X\in\mathrm{Lin}(\mathbb{C}^d)$$ we have $$P_0(X)=\mathrm{tr}(X)\frac{I}{d}$$, so $$\Phi(X)=\mathrm{tr}(X)\frac{I}{d}\tag3$$ as expected.

$$^1$$ With respect to Hilbert-Schmidt inner product.

• very nice. I think the argument about matrices having the same set of eigenvalues being connected within the representation is the main ingredient I missed. It's kinda interesting: we're saying the action of $\mathbf U(d)$ partitions the set of Hermitian operators into orbits labeled by their set of eigenvalues? I don't think I've encountered this particular fact before (I know we probably don't need such a general statement here, but still).
– glS
Oct 25, 2022 at 17:03
• I do find a bit peculiar the fact that for this argument you need to assume operatorial bases with the same eigenvalues. I could in principle decompose states using more general bases (in 3dim Gell-Man matrices being a standard example). Would you then get more irreducible subrepresentations? Eg for Gell-Mann matrices used as basis, $\lambda_8$ would span a different 1d irrep? Then we'd have $\Phi=c_0 P_0+c_1 P_1+c_2 P_2$.
– glS
Oct 25, 2022 at 17:06
• @gIS Yes, orbits of Hermitian (or even just normal) matrices under the action of $\mathbf{U}(d)$ by conjugation are indeed labeled by sets of eigenvalues. Oct 25, 2022 at 17:49
• Remember that a subrepresentation has to satisfy two conditions: it must be preserved by the action and it must be a vector space. It's the combination of the two that gives $(1)$ its structure. Specifically, traceless matrices form a vector space invariant under conjugation. Similarly, the matrices with "flat" spectrum form a vector space invariant under conjugation that is necessarily one-dimensional. No more matrices are needed to generate arbitrary matrix by taking linear combinations, so the decomposition is done. Oct 25, 2022 at 17:50
• Thus, starting with a more general basis won't lead to more irreps. What will thwart this is the fact that taking linear combinations does not preserve the spectrum. However, showing irreducibility would be harder with such a basis. Oct 25, 2022 at 17:50

Another intuitive way of looking at it is to note that the Haar average of all pure states is a maximally mixed state, $$\int |\psi\rangle\langle\psi| d\psi=\frac{\mathbb{I}}{d}$$.

Considering $$|\psi\rangle{=}U|\alpha \rangle$$, with $$|\alpha \rangle$$ being a fixed state, and expressing the density matrix $$|\alpha\rangle\langle \alpha|$$ in a Hilbert-schmidt basis: $$\{\sigma_k\}$$, with $$k{\in}\{0,1,\dots,d^2-1\}$$, $$\sigma_0{=}\mathbb{I}$$, $$Tr(\sigma_k)=0$$, $$Tr(\sigma_k\sigma_l){=}d\delta_{kl}$$, we have \begin{align} &\int U|\alpha\rangle\langle \alpha| U^\dagger dU=\frac{\mathbb{I}}{d} \\ &\implies\frac{1}{d}\int U(I + \hat{n}.\vec{\sigma}) U^\dagger dU=\frac{\mathbb{I}}{d} \\ &\implies\int U \sigma_k U^\dagger dU = 0 \ \ \mathrm{for \ k{\neq}0} \end{align}.

This already gives the hint that no trace-less operator survives the integral. Note to derive the above equation, the purity of the fixed state is not important. However, it gives a good intuitive starting point.

Now expressing an operator $$X$$ in Hilbert-Schmidt basis, $$X{=}\sum_{k=0}^{d^2-1}a_k \sigma_k$$, with $$Tr(X)=a_0.d$$, we have

$$\int U X U^\dagger dU = \sum_{k=0}^{d^2-1}a_k\int U \sigma_k U^\dagger dU=a_0.\mathbb{I}=Tr(X). \frac{\mathbb{I}}{d}$$.

It seems like all the other answers are unnecessarily complicated. You already did the work of deducing $$\Phi = c I$$. Now you just need to acknowledge that $$c$$ depends on $$X$$ and you can calculate it by taking the trace on both sides

$$\mathrm{Tr} \int UXU^\dagger d\mu(U) = \int \mathrm{Tr} UXU^\dagger d\mu(U) = \int \mathrm{Tr} X d\mu(U) = \mathrm{Tr} X = \mathrm{Tr} (c I) = cd$$

which immediately gives $$c = \mathrm{Tr} X / d$$.