3
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ I presume a factor $({\rm tr}\,X)$ is missing in front of $(1-p)$ $\endgroup$ Jan 29, 2015 at 21:36
  • $\begingroup$ Yes, you are right! I made the correction. I still don't understand why $p$ becomes greater than 1. $\endgroup$
    – Atnap
    Jan 29, 2015 at 22:11
  • $\begingroup$ For the case that $X$ is a positive-definite matrix, the integral is positive, but the second expression (with $p$>1) may give negative results. $\endgroup$
    – Atnap
    Jan 29, 2015 at 23:22

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you, I really enjoyed your comprehensive solution! $\endgroup$
    – Atnap
    Jan 30, 2015 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.