How Do You Compute Gaussian Functional Integrals in Quantum Field Theory?

[psi*psi]

I am new to path integral and struggling with the computation involving Gaussian functional integrals. Could anyone show me the steps of computing the following integral?
$$\int D \phi e^{-S},$$
where
$$S = \int dx~d \tau [(\frac{\partial \phi}{\partial x})^2+2 i \frac{\partial \theta}{\partial \tau} \frac{\partial \phi}{\partial x} ]$$.
$\theta$ is a function that is not being integrated over.
I know I am supposed to use the formula
$$\int D v(x) \mathrm{exp}[-\frac{1}{2} \int d x d x'~v(x) A(x,x') v(x') + \int d x~j(x) v(x)] \propto \mathrm{exp} [\frac{1}{2} \int d x d x'~j(x) A^{-1}(x,x') j(x')],$$
where $A(x,x')$ is an operator. But I am having trouble in applying this result.

Thanks in advance.

 

[eljose79]

Hello,

I understand that you are new to path integral and are struggling with the computation involving Gaussian functional integrals. I will try my best to explain the steps involved in computing the given integral.

Firstly, let's rewrite the given integral in a more convenient form:

\int D \phi e^{-S} = \int D \phi \mathrm{exp} \left[ -\int dx~d \tau (\frac{\partial \phi}{\partial x})^2 - 2 i \int dx~d \tau \frac{\partial \theta}{\partial \tau} \frac{\partial \phi}{\partial x} \right]

Now, we can use the formula you mentioned:

\int D v(x) \mathrm{exp}[-\frac{1}{2} \int d x d x'~v(x) A(x,x') v(x') + \int d x~j(x) v(x)] \propto \mathrm{exp} [\frac{1}{2} \int d x d x'~j(x) A^{-1}(x,x') j(x')]

In this formula, we can identify v(x) as \phi and j(x) as -2i\frac{\partial \theta}{\partial \tau}. Also, A(x,x') is given by \delta(x-x')\frac{\partial}{\partial x}.

Substituting these values in the formula, we get:

\int D \phi e^{-S} \propto \mathrm{exp} \left[ \frac{1}{2} \int dx~d \tau \left( -2i \frac{\partial \theta}{\partial \tau} \right) \left( \delta(x-x') \frac{\partial}{\partial x} \right) \left( -2i \frac{\partial \theta}{\partial \tau} \right) \right]

Simplifying this further, we get:

\int D \phi e^{-S} \propto \mathrm{exp} \left[ \int dx~d \tau \frac{\partial \theta}{\partial \tau} \frac{\partial^2 \theta}{\partial x^2} \right]

Now, we can use the fact that \theta is a function that is not being integrated over, so it can be treated as a constant. Therefore, we can