Funcional Integral in QFT

Question

I am reading Peskin and Schroeder's chapter on functional methods. They propose the amplitude:

$$ \langle \phi_b(\vec{x})|e^{-iHT}| \phi_a(\vec{x})\rangle = \int \mathcal{D}\phi\mathcal{D}\pi \exp \bigg[ i\int_0^T \left( \pi \dot\phi - \frac{1}{2}\pi^2 - \frac{1}{2} (\nabla\phi)^2-V(\phi) \right) \bigg] $$ They then said that we can complete the square and integrate over $\mathcal{D}\pi$ integral. How is this done?

I was able to complete the square: $\pi\dot\phi - \frac{1}{2} \pi^2=-\frac{1}{2}(\pi-\dot\phi)^2+\frac{1}{2}\dot\phi^2$ and rewrite the amplitude as:
$$ \langle \phi_b(\vec{x})|e^{-iHT}| \phi_a(\vec{x})\rangle = \int \mathcal{D}\phi \exp \bigg[ i\int_0^T \left( \frac{1}{2} \partial_\mu\phi\partial^\mu\phi-V(\phi) \right) \bigg] \int \mathcal{D}\pi \bigg[ -\frac{1}{2}i \int^T_0 (\pi-\dot\phi^2) \bigg] $$ I am only confused about the functional integral (mathematically), how is it done?

Answer 1

You do the following

$$\int D\pi \exp[ -\frac12 i \int_0^T (\pi-\dot \phi)^2] =\int D\pi \exp[-\frac12 i\int_0^T \pi^2]= constant$$

where I've shifted $\pi\to \pi+\dot \phi$ which leaves the measure $D\pi$ invariant. The result is a $\phi$ independent multiplicative constant. It's not quite well-defined; you'll need to determine this constant by other means, if necessary.

Answer 2

well functional integral can be thought as vector integral

$$I=\int d\vec{v}e^{-\frac{1}{2}\vec{v}^tA\vec{v}}=\prod_{a}\int dv_ae^{-1/2u_a^2(\lambda_{a})}=\sqrt{\det(2\pi A^{-1})}$$

where $\lambda_a(u_a)$ is eigenvalues(vectors) of $A$ the above calculation is very easy gaussian vector integral. Now instead of discrete vector we have continuous "vectors"

$$I=\int dv(x)e^{-\frac{1}{2}\int dx\int dy\ v(x)A(x,y)v(y)}=\prod_{x}\int du(x)e^{-1/2u(x)^2(\lambda(x))}\propto\sqrt{\det( A^{-1})}$$

in your case,

$$I=\int dv(x)e^{-\frac{1}{2}\int dx\int dy\ v(x)A(x,y)v(y)+\int dxj(x)v(x)}\propto\sqrt{\det( A^{-1})}\int d v(x)e^{\frac{1}{2}\int d x\int dy v(x)A^{-1}(x,y)v(y)}$$ so this last equation is automatically square completed version, just apply $\pi$ integral to your case in your case: $A=i$ and$j=i\dot{\theta}$ so if we apply our formula we get, $$\text{constant}\times e^{i1/2\int dx\dot{\theta}^2}$$