# A Function integration of a Gaussian integral

1. Oct 11, 2016

### spaghetti3451

Consider the partition function $Z[J]$ of the Klein-Gordon theory

$Z[J] =\int \mathcal{D}\phi\ e^{i\int d^{4}x\ [\frac{1}{2}(\partial\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}+J\phi]} =\int \mathcal{D}\phi\ e^{-i\int d^{4}x\ [\frac{1}{2}\phi(\partial^{2}+m^{2})\phi]}\ e^{i\int d^{4}x\ J\phi}$

If the operator $\partial^{2}+m^{2}$ is symmetric and positive-definite, then

$Z[J]=\int \mathcal{D}\phi\ e^{i\int d^{4}x\ [\frac{1}{2}\phi(-\partial^{2}-m^{2})\phi]}\ e^{-\frac{i}{2}\int d^{4}x\ J(-\partial^{2}-m^{2})^{-1}J(x)}.$

The assumption of symmetry and positive-definiteness of the operator $\partial^{2}+m^{2}$ is crucial in the derivation of the Feynman diagrammatic rules of the scalar theory from the partition function of the scalar theory.

What does it mean for an operator to be symmetric and positive-definite? Does it mean that its matrix representation in any basis is symmetric and positive-definite?

2. Oct 12, 2016

### andrewkirk

For an operator $T$ on a vector space $V$ to be symmetric means that, for all $\vec u,v\in V$ we have $\vec u\cdot T\vec v=T\vec u\cdot\vec v$.

For it to be positive-definite means that, for all $\vec v\in V$ we have $\vec v\cdot T\vec v>0$, which of course requires that the field over which the vector space is defined have an ordered subfield, and that all $\vec v\cdot T\vec v$ are in that subfield. That is the case automatically if $V$ is over $\mathbb R$, and if the field is $\mathbb C$ then the subfield is the isomorphic copy of $\mathbb R$ that is embedded in $\mathbb C$.

It would follow from those definitions that the matrix representation in any basis be symmetric and positive-definite. In infinite-dimensional vector spaces, the converse may not necessarily follow.

I see from this link that there is another definition of positive-definite operator (definition (i) in the link), that is not necessarily the same as the above (definition (ii) in the link) for infinite-dimensional vector spaces.