I am a physicist and I am wondering whether the following integral over Haar measure （edit: say $U$ is unitary, orthogonal or symplectic matrix) \begin{align} \int dU \: \exp\left( \mathrm{tr}(UX) + \mathrm{tr}(X^\dagger U^\dagger) \right) \end{align} have an explicit expression in terms of the matrix X. For example, if the group is $U(1)$, then the result would be the modified Bessel function $I_0(2|X|)$. For the general case, I guess one can at least expand the exponent and use the Weingarten functions, and then perform a re-summation. But I know too little about the properties of the Weingarten functions to organize the re-summation into any simple, explicit form. Does someone know how to do this, or perhaps where formulae like this can be found?
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2$\begingroup$ For example, if the group is U(1), then And in your case the group is ??? $\endgroup$– fedjaDec 9, 2017 at 4:10
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$\begingroup$ @fedja Let's say one would like $U$ to be $N\times N$ unitary, orthogonal or symplectic matrix. $\endgroup$– Jing-Yuan ChenDec 9, 2017 at 4:41
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$\begingroup$ this is essentially a duplicate of mathoverflow.net/questions/256066/… $\endgroup$– Abdelmalek AbdesselamDec 14, 2017 at 15:24
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1$\begingroup$ @AbdelmalekAbdesselam Thank you! The references are very useful! $\endgroup$– Jing-Yuan ChenDec 15, 2017 at 8:32
1 Answer
Depending on what you mean by "explicit", in the unitary case this can be read off from a generalization of the Harish--Chandra-Itzykson-Zuber formula. To see that, note that your integral can be rewritten as $$J=\int_{U_N}\int_{U_N} \exp(\Re (\mbox{tr} V YU)) dU dV,$$ where $Y$ is a diagonal real matrix whose entries are the singular values of $X$. Now, for fixed diagonal $A,B$ consider the integral $$J(A,B)=\int_{U_N} \int_{U_N} \exp(\Re (\mbox{tr} A V BU)) dU dV.$$ Then $J=J(I,Y)$. For $A,B$ with distinct entries, $J(A,B)$ has an explicit formula involving determinants in Bessel functions, see for example formula (3.6) in the review paper of Zinn-Justin and Zuber, https://arxiv.org/pdf/math-ph/0209019.pdf (they attribute the result to Balantekin and to Guhr--Wetting, although I guess one can trace it all the way back to Harish-Chandra). Now in your case $A=I$ and in particular the entries of $A$ are not distinct, but resolving this involves a straight-forward limit, replacing $I$ by $A=I+\epsilon \Delta$ where $\Delta$ has distinct real entries, and taking $\epsilon \to 0$.
I suspect that the case of $A=I$ has an even simpler formula, but I don't see it. Maybe somebody else, more versed in representations than me, can comment on that.
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$\begingroup$ @fedja I guess you mean (3.5). It’s the hermitian conjugate (what mathematicians call adjoint). $\endgroup$– lcvDec 9, 2017 at 18:28
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$\begingroup$ Thank you very much Ofer, this is very helpful. If the matrix $X$ in question is not full rank, but say of rank $n<N$, I guess the original $SU(N)$ integral can be reduced to over $SU(n)$ without causing other changes. Is this correct? I also would hope there is a discussion of $SO(N)$ somewhere. $\endgroup$ Dec 11, 2017 at 2:44
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$\begingroup$ Actually following your reference I found hep-th/0007161 link that has done this problem (in the unitary case) in Eq(5.5). Thank you! $\endgroup$ Dec 11, 2017 at 3:04