Some Background: In mathematical physics (matrix quantum mechanics in particular), one often runs into path integrals of the form: $$ Z = \int \prod_{ab} DM^1_{ab}(\tau) DM^2_{ab}(\tau) e^{-NS(M^1,M^2,\dot M^1,\dot M^2)}$$ Where $M^1$ and $M^2$ are $N$ by $N$ Hermitian matrices and $S$ is a function invariant under simultaneous conjugation of $M^1$ and $M^2$ by $U(N)$ group action. To understand properties of this integral, physicists often look at a simpler version without time dependence: $$ Z' = \int \prod_{ab} dM^1_{ab} dM^2_{ab} e^{-NV(M^1,M^2)}$$ Where $V$ is the potential energy in the action $S$ which is usually taken to be a trace of powers of $M^1$ and $M^2$. The usual trick is to make a change of coordinates and write the measure as: $$ \int \prod_{ab} dM^1_{ab} = \int \prod_{a} d \lambda^1_a dU^1 \cdot \Delta(\lambda^1)^2$$ Where $\lambda^1_a$'s are the eigenvalues of $M^1$ and $dU^1$ schematically means integral over the group manifold $U(N)/U(1)^N$. The $\Delta(\lambda^1)^2$ term is just a Jacobian factor. After this change of coordinates, we get (for example) integrals of the form: $$ \int dU e^{tr(M^1UM^2U)} $$ This integral is hard but there is a nice Itzykson-Zuber formula for it. For details see Terry Tao's blog post.

My question: is there a generalization of the Itzykson-Zuber formula for path integrals? In particular, I wonder if there is in general a nice formula for the integral: $$ \int \mathcal{D}U(\tau) e^{tr(A(\tau)U(\tau)B(\tau)U(\tau)} $$ Written in terms of $A(\tau),B(\tau)$.

  • $\begingroup$ Sure, if $IZ(\tau)$ denotes the usual Itzykson-Zuber integral with time parameter $\tau$, then the corresponding path integral trivially factorizes in a formal product $\prod_{\tau}IZ(\tau) $ over time . $\endgroup$
    – Qmechanic
    Jun 25, 2017 at 18:16
  • $\begingroup$ So would the answer be something like the following? $$\prod_{\tau} IZ(\tau) = e^{\sum_{i=0}^N \log IZ(i\frac{\tau}{N}) \cdot \frac{\tau}{N} \cdot \frac{N}{\tau}} $$ In the end as I take a large N limit (small step size), it seems like I get something divergent: $$ e^{\int_0^{\tau} \log IZ(\tau) \cdot \frac{N}{\tau}} = IZ(\tau)^N $$ That is confusing to me. Could you help me understand that better? $\endgroup$ Jun 29, 2017 at 1:20

1 Answer 1


The answer is positive, modulo the inherent non-rigor in the path integral quantization. There is a series of papers by Antti Niemi and collaborators where this problem was considered. I'll give you here an introduction to their main reasoning and some references.

The Itzykson-Zuber formula belongs to a special case of integrals where the WKB semiclassical approximation is exact due to the Duistermaat-Heckmann theorem (which is also described in Terry Tao's lecture).

Physically, the integrals satisfying this property can be expressed as classical partition functions: $$ Z = \int_{\mathcal{M}} e^{-\beta H(x)} d{\mu}(x)$$ Where the integration is over a classical phase space $\mathcal{M}$ which is a symplectic manifold, and $ d{\mu}(x)$ is the Liouville measure on the manifold, which can be locally written as: $$d{\mu}(x) = det(\omega) d_L(x)$$ Where $ \omega$ is the symplectic form and $d_L$ is the Lebesgue measure.

In the Itzykson-Zuber formula, the integrand is invariant under the action of the maximal torus $T^n$ of $U(n)$. Thus the integration can be effectively performed on the flag manifold $U(n)/T^n$, which is a symplectic manifold.

In physics, the exactness of the WKB approximation is attributed to the existence of supersymmetry.

This property is explained by A Keski-Vakkuri, Niemi, Semenoff and Tirkkonen in the following work. I'll repeat the main arguments (which I'll bring here their main arguments heuristically)

This supersymmetry is obtained by exponentiating the determinant factor of the symplectic form by means of a Berezin Grassmann integral: $$ Z = \int_{\mathcal{M}} e^{-\beta ( H(x) + c^a \omega_{ab} c^b)} d_L(x)$$ One can easily check that the integrand is invariant under the following supersymmetry operator: $$ Q = c^a \partial_a + \omega^{ab}\partial_aH \frac{\partial}{\partial c^b}$$, i.e., $$[Q, H(x) + c^a \omega_{ab} c^b] = 0$$ Thus: $$[Q, e^{-\beta ( H(x) + c^a \omega_{ab} c^b)}] = 0$$ This operator squares to: $$Q^2 = c^a \partial_a( \omega^{bc} \partial_c H )\frac{\partial}{\partial c^b}$$

Thus this operator vanishes in the stationary points $\partial_a H = 0$.

The Berezin integral picks the top form, the action of Q on the top form has two components, one which reduces the ghost number and the second increases the ghost number, which acts as an exterior derivative. Thus the top form is exact in all points where $Q$ is not zero, that is when $\partial_a H \neq 0$. Thus only points where the Hamiltonian is stationary contribute to the integral. Thus the semiclassical approximation is exact.

Niemi generalized the same reasoning for path integrals in 0+1 dimensions.

Final remarks

  1. The above heuristic reasoning seems to apply to any Hamiltonian, but a deeper analysis shows that it applies only to Hamiltonians which are perfect Morse functions on the phase space.

  2. Witten applied the same reasoning even in field theory computations for example $N=2$ Susy Yang-Mills theory in $4D$.


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.