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 $I_0(2X)$. For the general case, I guess one can at least expand the exponent and use the Weingarten functions, and then perform a resummation. But I know too little about the properties of the Weingarten functions to organize the resummation into any simple, explicit form. Does someone know how to do this, or perhaps where formulae like this can be found?

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

$\begingroup$ @fedja Let's say one would like $U$ to be $N\times N$ unitary, orthogonal or symplectic matrix. $\endgroup$– JingYuan ChenDec 9, 2017 at 4:41

$\begingroup$ this is essentially a duplicate of mathoverflow.net/questions/256066/… $\endgroup$– Abdelmalek AbdesselamDec 14, 2017 at 15:24

1$\begingroup$ @AbdelmalekAbdesselam Thank you! The references are very useful! $\endgroup$– JingYuan 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 HarishChandraItzyksonZuber 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 ZinnJustin and Zuber, https://arxiv.org/pdf/mathph/0209019.pdf (they attribute the result to Balantekin and to GuhrWetting, although I guess one can trace it all the way back to HarishChandra). Now in your case $A=I$ and in particular the entries of $A$ are not distinct, but resolving this involves a straightforward 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.


$\begingroup$ @fedja I guess you mean (3.5). It’s the hermitian conjugate (what mathematicians call adjoint). $\endgroup$– lcvDec 9, 2017 at 18:28

$\begingroup$ 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

$\begingroup$ Actually following your reference I found hepth/0007161 link that has done this problem (in the unitary case) in Eq(5.5). Thank you! $\endgroup$ Dec 11, 2017 at 3:04