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

Question

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.

Answer 1

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.

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

Answer 2

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

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

Answer 3

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

$$\mathrm{Tr} X = \mathrm{Tr} \Phi(X) = \mathrm{Tr} (c I) = cd $$

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

Edit: The first equality is fully general, namely you can just always use $\mathrm{Tr} \Phi = \mathrm{Tr}$ (where more explicitly I mean $\mathrm{Tr} \Phi \equiv \mathrm{Tr} \circ \Phi$ as function composition) and any map satisfying such condition is known as being "trace preserving". Any twirl will satisfy this because

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