# How is the functional integral over momentum performed in the case of the real scalar field?

Let's follow Peskin and Schroeder section 9.2, page 282.

The Hamiltonian of a free real scalar field is

$$H=\int{}d^3x[\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+V(\phi)]$$

so the expression for the functional integral is

$$\langle\phi_b|e^{-iHT}|\phi_a\rangle=\int\cal{D}\phi\cal{D}\pi{}e^{i\int_ 0^Td^4x\,(\pi\partial_t\phi-\frac{1}{2}\pi^2-\frac{1}{2}(\nabla\phi)^2-V(\phi))}$$

then Peskin and Schroeder say that in order to integrate over the $\pi$ you just have to complete the square which gives us

$$\langle\phi_b|e^{-iHT}|\phi_a\rangle=\int\cal{D}\phi\cal{D}\pi{}e^{i\int_ 0^Td^4x\,(\frac{i}{\sqrt{2}}\pi-\frac{i}{\sqrt{2}}\partial_t\phi)^2}e^{i\int_0^Td^4x\,\cal{L}}$$

now, my question. How do you get rid of the first exponent? My teacher said something about Gaussian integrals but this doesn't convince me. This is not a regular integral, this is a functional integral, so we shouldn't use directly the formula for Gaussians here. How do you perform this integral without resorting to hand wavy arguments?

That's not a handwaving and I think that this particular question is covered in practically every textbook containing path integrals. First of all you should note that we can integrate by $\pi$ not touching the second exponent, i.e. $$\langle\phi_b\vert e^{-iHT}\vert\phi_a\rangle =\int \mathcal{D}\phi e^{i\int_0^T d^4x\mathcal{L}}\int\mathcal{D}\pi e^{i\int_0^Td^4x\Big(\frac{i}{\sqrt{2}}\pi-\frac{i}{\sqrt{2}}\partial_t\phi\Big)^2}$$
The path integral is actually a limit of a regularized integrals such as if we take the values of $\pi$ not on a continuum but only at a lattice. For start forget about spatial coordinates and calculate the following integral, \begin{eqnarray}\int \prod_{k=0}^{N-1}\frac{d\pi_k}{2\pi}\exp\left({i\sum_{k=0}^{N-1} \Big(\frac{i}{\sqrt{2}}\pi_k-\frac{i}{\sqrt{2}}\frac{\phi_{k+1}-\phi_k}{t_{k+1}-t_k}\Big)^2(t_{k+1}-t_k)}\right)\\ =\prod_{k=0}^{N-1}\int \frac{d\pi_k}{2\pi}\exp\left({i\Big(\frac{i}{\sqrt{2}}\pi_k-\frac{i}{\sqrt{2}}\frac{\phi_{k+1}-\phi_k}{t_{k+1}-t_k}\Big)^2(t_{k+1}-t_k)}\right)\end{eqnarray}, which is a product of usual Gaussian integrals.
I'd suggest you to calculate this integral and take the limit $N\rightarrow+\infty$. There is an important part - like usual integral, to be defined properly your path integral should not depend on the method you introduced the lattice. Also consider integrals with different quadratic operators like e.g. $$\int \mathcal{D}\phi e^{i\int_0^T dt \Big((\partial_t\phi)^2-m^2\phi^2\Big)}$$ which will give you a propagator for a Harmonic oscillator. Then you may apply your skills to QFT path integrals with some quadratic operator and discover that you will get very simple generalization of the usual Gaussian integral formula.