An expectation of the product of random unitaries

Question

I want to find the answer of $$\int dU \ U^m X \ U^{\dagger m}$$

Where $m\in\mathbb{N}$ and $U$'s are unitary matrices in $U(n)$ and $dU$ is a normalized Haar measure. $X$ is a given self-adjoint matrix.

I used Schur lemma and found that the answer is of the form

(Edit: the correct form is ) $$pX+(1-p)tr(X)\frac{I}{n}$$

with $p\in[0,1]$, but if this is true then by comparing above expressions and using the results obtained in my last question, $p$ can be found as $$p=\frac{m-1}{n^2-1}$$ which is greater than 1 for $m>n^2$.

I don't know what is going wrong here. Does anyone have an idea about my solution or another way for solving this integral?

Answer 1

Let me try to work this out, along the lines of a similar calculation in the orthogonal (rather than unitary) group.

We need the fourth-order tensor $$\int_{{\rm U}(n)}(U^m)_{ij}(\bar{U}^m)_{kl}\,dU=a_{m}(n)\delta_{ij}\delta_{kl}+b_{m}(n)\delta_{ik}\delta_{jl}+c_{m}(n)\delta_{il}\delta_{jk}$$ so that the required integral takes the form $$\int_{{\rm U}(n)}U^m X(U^\dagger)^m\,dU=a_{m}(n)X+b_{m}(n)\mathbb{1}\,{\rm tr}\,X+c_{m}(n)X^{\rm t} $$ [note: The OP does not have the transpose $X^{\rm t}$, but I don't see a priori why this term will not appear.]

Substitution $X=\mathbb{1}$ gives a first relation $$a_m(n)+nb_m(n)+c_m(n)=1$$ one more relation follows from application of theorem 2.1.b of Diaconis and Evans: $$n^2 a_m(n)+nb_m(n)+nc_m(n)=\int_{{\rm U}(n)}\,({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)\,dU={\rm min}\,(n,m)$$ [note: in a related MO posting I had $m$ instead of ${\rm min}\,(n,m)$, I have now corrected this oversight and apologize for the confusion it may have caused]

I need a third relation $$na_m(n)+nb_m(n)+n^2 c_m(n)=\int_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU$$

To evaluate this integral I write $U^m=V\Lambda^m V^\dagger$, with $\Lambda$ the diagonal matrix of eigenvalues of $U$ and $V\in{\rm U}(n)$ independent of $\Lambda$. I then first average over the $V$ matrices, which is easy because there are just four of them: $$\int_{{\rm U}(n)}\,{\rm tr}\,(V\Lambda^m V^\dagger\overline{V\Lambda^m V^\dagger})\,dV=\frac{1}{n+1}\left[({\rm tr}\,\Lambda^m)({\rm tr}\,\bar{\Lambda}^m)+{\rm tr}\,(\Lambda^m\bar{\Lambda}^m)\right]=\frac{1}{n+1}\left[({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)+n\right].$$ and then the remaining average can be evaluated using Diaconis and Evans:

$$\int_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU =\frac{n+{\rm min}\,(n,m)}{n+1}$$

so now we have three equations with three unknowns and we're done:

$$a_{m}(n)= \frac{\min(n,m) -1}{n^2-1}=1-nb_{m}(n),\;\;c_m(n)= 0$$

and $c_m(n)$ does in fact turn out to be equal to zero.