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Schur functions are irreducible characters of the unitary group $\mathcal{U}(N)$. This implies the integration formulae $$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)dU=\frac{|s_\lambda(A)|^2}{s_\lambda(1)}$$ and $$ \int_{\mathcal{U}(N)}s_\lambda(AU)\overline{s_\mu(A U)}dU=\frac{\delta_{\lambda\mu}s_\lambda(AA^{\dagger})}{s_\lambda(1)},$$ where the overline means complex conjugation.

My question is whether we can compute the integral $$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)\overline{s_\mu(AUA^\dagger U^\dagger)}dU$$

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As a first step towards a solution, using the identity $$\int_{\mathcal{U}(N)} U_{\alpha a}U_{\alpha' a'}\bar{U}_{\beta b}\bar{U}_{\beta' b'}\,dU=\frac{1}{N^{2}-1}\bigl( \delta_{\alpha\beta}\delta_{ab}\delta_{\alpha'\beta'}\delta_{a'b'}+ \delta_{\alpha\beta'}\delta_{ab'}\delta_{\alpha'\beta}\delta_{a'b}\bigr)$$ $$\qquad\qquad\mbox{}-\frac{1}{N(N^{2}-1)}\bigl( \delta_{\alpha\beta}\delta_{ab'}\delta_{\alpha'\beta'}\delta_{a'b}+ \delta_{\alpha\beta'}\delta_{ab}\delta_{\alpha'\beta}\delta_{a'b'}\bigr),$$

I computed this integral for the trace, $$\int_{\mathcal{U}(N)}{\rm tr}\, (AUA^\dagger U^\dagger)\,\overline{{\rm tr}\,(AUA^\dagger U^\dagger)}\,dU=\frac{1}{N^2-1}\left(|{\rm tr}\, A|^4+|{\rm tr}\,AA^\dagger|^2\right)-\frac{2}{N(N^2-1)}|{\rm tr}\,A|^2({\rm tr}\,AA^\dagger).$$

As a check, take $A=1$ and see that the right-hand-side reduces to $(N^4+N^2)/(N^2-1)-2N^3/(N^3-N)=N^2$.

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  • $\begingroup$ Interesting. Might I ask about the method? $\endgroup$
    – Marcel
    Nov 22 '18 at 17:15
  • $\begingroup$ Did you write the trace in terms of matrix elements and then used Weingarten calculus? $\endgroup$
    – Marcel
    Nov 22 '18 at 17:22
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    $\begingroup$ I added the identity that I applied to the integral over the product of traces. $\endgroup$ Nov 22 '18 at 17:46
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    $\begingroup$ @lcv that is what I called "Weingarten calculus". You can search for "Weingarten functions". For 8 U's, four of them being conjugated, you get a sum over all 24 elements of the permutation group $S_4$. $\endgroup$
    – Marcel
    Nov 22 '18 at 19:24
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    $\begingroup$ @lcv --- no, I have never seen a formula with all the Kronecker delta's explicitly printed out for more than 4 unitaries; such calculations quickly get tedious; there exists a Mathematica package $\endgroup$ Nov 22 '18 at 21:41

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