# Least Action

Nontrivializing triviality..and vice versa.

## Klein Gordon Field Part II

In the last post, we looked at how a general solution to the sourced KG equation can be written in terms of a source-free KG solution and the Green’s function of the KG field. The important thing to note here is that the Green’s function depends only on the structure of the equation, specifically, on the differential operator $(\Box + m^2)$, and not on the source $J(x)$. Note also, that $x$ is short for $x_{\mu}$, the position 4 vector. In our notation $\bf{x}$ will denote the position 3 vector.

The first step in our analysis is to solve for the propagator. To do, we write $G(x-x')$ in terms of its Fourier transform $G(p)$:

$G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')} G(p)$

where the different components of $p^{\mu}$ should be treated as independent, unrelated by the energy momentum relation. Right now, we’re only at the level of mathematics, so we do not enforce $p^{\mu}p_{\mu} - m^2 = 0$. Let us operate on the above equation by $(\Box_{x} + m^2)$. This gives us

$(\Box_{x} + m^2) G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')}(-p^2 + m^2) G(p)$

Using the integral representation of the Dirac delta distribution, we get

$G(p) = \frac{1}{p^2-m^2} = \frac{1}{(p^{0})^2-|{\bf{p}}|^2-m^2} = \frac{1}{(p^{0})^2-E_{p}}$

where we have used the energy momentum relation in the last two equalities. Putting this back into the analysis equation, we get

$G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')}G(p)$

Do you see a problem with this expression?

The integration over $p^0$ cannot be performed as the integrand has poles at $p^{0} = \pm E_{p}$. This is where Feynman’s genius comes in…we’ll take a look at it in the next blog post!