Least Action

Nontrivializing triviality..and vice versa.

Scalar Propagator

In this post, I want to discuss various aspects of the scalar field propagator. This is the first propagator one encounters in quantum field theory. In canonical quantization, there is quite a bit of spadework that goes into deriving the expression. However, as the more astute student will recognize, the expression is encountered well before one actually writes it. Those familiar with the theory of Fourier transforms will recognize that the propagator as normally written, is just the Green’s function of the wave equation, expressed as an integral over momentum space which is nothing but the inverse Fourier transform of the momentum space propagator. So, one should recognize straight out that a propagator is just a Green’s function. This doesn’t undermine the importance of the propagator of course (what would one do without it in field theory), but is merely meant to emphasize the crucial role played by Green’s functions in physics. Capturing the essence of the propagator, even if it means doing a lot of similar looking calculations repeatedly, is important in understanding both Feynman diagrams as well as Cluster Expansions.

Nevertheless, my reason for writing this post is that most of the ‘obvious’ calculational steps leading to the important expressions are often left to the imagination of readers in books, and sometimes its the steps and not the final result (as much) which give insight into the physics.

So, without further ado, let us begin with the expression for the propagator:

$D(x-y) = \int \frac{d^{4}p}{(2\pi)^4}\frac{e^{ip\cdot(x-y)}}{p^2-m^2 + i\epsilon}$

The $i\epsilon$ has been added to make the expression sensible at $p^2 = m^2$ (which is the “on-shell” condition, if you’re familiar with special relativity). Note that we have not imposed this condition in this integral, so p can actually run free over all possible values. Remember that the restriction comes into place only if we put in a “hand-made” delta function inside with its argument equal to $p^2 - m^2$. Actually the “$i\epsilon$ prescription” (as it is referred to in most QFT books) goes a bit deeper than this:- it is required for the path integral of the scalar field to converge at large values of the scalar field. That the same $\epsilon$ should help us regularize the denominator from blowing up on shell is not unexpected.

For simplicity, let me consider the case $y = 0$. In any event, I can always obtain $D(x-y)$ given a general expression for the propagator.

So,

$D(x) = \int \frac{d^{4}p}{(2\pi)^4}\frac{e^{i(p^0x^0-\vec{p}\cdot\vec{x})}}{p^2-m^2 + i\epsilon}$

which I write as

$D(x) = \int \frac{d^{3}p}{(2\pi)^3}e^{-i\vec{p}\cdot\vec{x}}\frac{1}{2\pi}\int dp^0\frac{e^{i p^0x^0}}{p^2-m^2 + i\epsilon}$

Also $p^2 = (p^0)^2-\vec{p}^2$, so

$p^2-m^2 + i\epsilon = (p^0)^2-\vec{p}^2-m^2 + i\epsilon$

The poles are located at

$\pm \sqrt{\vec{p}^2+m^2-i\epsilon} = \pm\sqrt{\omega_{p}^2-i\epsilon}$ where $\omega_{p} = \sqrt{\vec{p}^2+m^2}$

So, in the $\epsilon \rightarrow 0$ limit, the poles are at

$\omega_{p} - i\epsilon$
$-\omega_{p} + i\epsilon$

So the poles are in the second and fourth quadrant. This is important (why? If you loved the joke about why the mathematician named his dog Cauchy, you probably know why).

The integral

$I = \int dp^0\frac{e^{i p^0x^0}}{p^2-m^2 + i\epsilon} = \int dp^0\frac{e^{i p^0x^0}}{(p^0-(\omega_p - i\epsilon))(p^0-(-\omega_p+i\epsilon))}$

can be easily computed using the Residue Theorem. There are two cases to consider, depending on whether $x^0 > 0$ or $x^0 < 0$.

Case I: $x^0 > 0$

In this case, we must have Im($x^0$) > 0 on the contour of integration, for the integrand to be bounded (a necessary condition for the indefinite integral to converge). This is most easily seen if one expresses $p^0$ as a complex number:

$p^0 = a + ib$

Now, $ip^0 = ia - b$, so $e^{ip^0 x^0} = e^{-bx^0} e^{ia x^0}$. So, if $x^0 > 0$, we must have $b > 0$ or else the integrand will blow up at large $p^0$. I know that this is just the nifty trick known to contour integrators as Jordan’s lemma, and I probably don’t need to go into it in so much detail here, but the reason I do it is that it is easy to make sign errors subconsciously. So at least for the first few attempts, one must go through each and every step in detail before becoming an “expert propagator” 😉

Anyway, so much for that. The upper half plane contribution comes from the pole in the second quadrant, and the residue is simple to compute. It turns out to be

$\frac{e^{ip^0(-\omega_{p} + i\epsilon)}}{(-\omega_{p} + i\epsilon)-(\omega_{p} - i\epsilon)} = \frac{e^{-i\omega_p t}}{-2\omega_p}$

where $t = x^0$.

So,

$I = 2\pi i \frac{e^{-i\omega_p t}}{-2\omega_p}$

Case II: $x^0 < 0$

Obviously, I have to pick out the contribution from the pole in the 4th quadrant now. The residue is

$\frac{e^{ip^0(\omega_{p} - i\epsilon)}}{(\omega_{p} - i\epsilon)-(-\omega_{p} + i\epsilon)} = \frac{e^{i\omega_p t}}{2\omega_p}$

So that

$I = 2\pi i \frac{e^{i\omega_p t}}{2\omega_p}$

Plugging the results back into the original expression for $D(x)$, one gets

$D(x) = \int \frac{d^3 p}{(2\pi)^3} e^{-i\vec{p}\cdot\vec{x}}\left[-i\frac{e^{-i\omega_p t}}{2\omega_p}\theta(p^0) + i \frac{e^{i\omega_p t}}{2\omega_p}\theta(-p^0)\right]$

Now, the volume integral over the “3-momentum” variables is invariant under inversion in the 3-momentum space: $\vec{p} \rightarrow -\vec{p}$. So, I can write this as

$D(x) = \int \frac{d^3 p}{(2\pi)^3} \left[-i\frac{e^{-i\omega_p t}e^{i\vec{p}\cdot\vec{x}}}{2\omega_p}\theta(p^0) + i \frac{e^{i\omega_p t}e^{-i\vec{p}\cdot\vec{x}}}{2\omega_p}\theta(-p^0)\right]$

where I’ve performed the invariance operation of inversion in the first term. This is just

$D(x) = \int \frac{d^3 p}{(2\pi)^3} \left[-i\frac{e^{-i(\omega_p t-\vec{p}\cdot\vec{x})}}{2\omega_p}\theta(p^0) + i \frac{e^{i(\omega_p t-\vec{p}\cdot\vec{x})}}{2\omega_p}\theta(-p^0)\right]$

Under Construction