In field theory, one typically encounters integrals of the form:

$$ \mathcal{Z}[J] = \int \mathcal{D}[\phi] \exp \left( - \frac{1}{2} \int d^Dx d^Dx' \ \phi(x)A(x,x')\phi(x')+ \int d^Dx \phi(x) J(x)\right). $$

This can easily be done by discretizing space, and assuming that $A(x,x')$ is real and symmetric so that it is diagonalizable. Suppose however that there are multiple fields $\phi_a(x)$, and a path integral of the form

$$\mathcal{Z}[J] = \int \mathcal{D}[\phi] \exp \left( - \frac{1}{2}\int d^Dx d^Dx' \ \phi_i(x)A^{ij}(x,x')\phi_j(x')+ \int d^Dx \phi(x) J(x) \right).$$

What is the condition on $A^{ij}(x,x')$ such that I can write

$$\mathcal{Z}[J] \propto \exp \left( \frac{1}{2}\int d^Dx d^Dx' \ J_i(x)[A^{-1}]^{ij}(x,x')J_j(x') \right)?$$ Any help will be appreciated.

My ideas: Fixing two $x$ and $x'$, and requiring $A^{ij}(x,x')=A^{ji}(x,x')$, we could write $A(x,x') = O^T D(x,x') O$, diagonalizing the $3 \times 3$ matrix. Then, performing the transformation $\tilde{\phi}_a = O_{ab} \phi_b$, would break the integral into three independent components, one for each $\tilde{\phi}_a$. Relabelling $\tilde{J} = O^T J$, yields

$$\mathcal{Z}[J] \propto \exp \left( \frac{1}{2}\int d^Dx d^Dx' \ J(x)[A^{-1}]^{ij}(x,x')J_j(x') \right).$$

However, is it sufficient if $A^{ij}(x,x')=A^{ji}(x',x)$ instead of just $A^{ij}(x,x')=A^{ji}(x,x')$?


1 Answer 1

  1. We may assume w.l.o.g. that the integral kernel $$A(x,x^{\prime})_{ik}=A(x^{\prime},x)_{ki}\tag{1}$$ is symmetric (because an antisymmetric part would not be visible under the integral/summation).

  2. Assuming that the fields $\phi^i\in\mathbb{R}$ are real, in order for the Gaussian functional integral to formally be convergent, the real integral kernel $${\rm Re}A(x,x^{\prime})_{ik}\tag{2} $$ should correspond to a positive definite (and hence invertible) operator$^1$.

  3. One may show that the condition (2) in principle implies that the complex integral kernel $A(x,x^{\prime})_{ik}$ corresponds to an invertible operator, so that the Gaussian formula is well-defined.

  4. For the corresponding finite-dimensional Gaussian integral, see e.g. this related Math.SE post.


$^1$ There are subtleties with unbounded operators, domains, selfadjoint extensions, etc., that we ignore in this answer.


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