## nutshell notes: free propagators

In section I.3 of QFT in a Nutshell, Zee examines the path integral for the Lagrangian $$\mathcal{L}(\phi) = \frac{1}{2}[(\partial \phi)^2-m^2\phi^2]$$ (the “free”/”Gaussian” Lagrangian) with a source $$J$$. This looks like
\begin{align} Z(J)&=\int D\phi\,e^{i\int d^4x\,\{\frac{1}{2}[(\partial \phi)^2-m^2\phi^2]+J\phi\}}\\ &=\int D\phi\,e^{i\int d^4x\,\{-\frac{1}{2}\phi(\partial^2+m^2)\phi+J\phi\}}. \end{align}
(Integration by parts allows us to replace $$(\partial\phi)^2$$ with $$\phi(\partial^2\phi)$$ in the exponent’s integrand.)

The exponent here is really $$\frac{i}{2}\phi\cdot A\cdot\phi+iJ\cdot\phi$$, where we’re thinking of $$\phi$$ and $$J$$ as vectors in the space of field configurations (something like “$$\mathbb{R}^{1,3}\rightarrow\mathbb{R}$$”) and $$A=-(\partial^2+m^2)$$ is an operator on this space. Our “path integral” is really just an integral over all field configurations (or at least those which vanish at infinity), so we are in the same situation as Appendix 1 of I.2 — computing a Gaussian integral over a vector space.

And we have a solution to this integral:
$$Z(J)=\left(\frac{(2\pi i)^N}{\det[A]}\right)^{1/2}e^{-(i/2)J\cdot A^{-1}\cdot J}.$$
The coefficient on the left is kind of mysterious: $$N$$ is the “dimension” of the vector space, which should be infinity. Perhaps some clever limit arguments could make sense of it (make a lattice out of space), but we don’t really care; we’ll just call it $$Z(J=0)$$. Then we have:
$$Z(J)=Z(J=0)e^{i W(J)}$$
where
$$W(J)=-\frac{1}{2}\int d^4x\,d^4y\,J(x)A^{-1}(x,y)J(y).$$
By translation invariance, $$A^{-1}(x,y)$$ must be some function $$D(x-y)$$. Since it’s the inverse of $$A=-(\partial^2+m^2)$$, we must have $$-(\partial^2+m^2)D(x-y)=\delta^{(4)}(x-y)$$ (the Dirac delta is the identity operator on continuous space). So $$D(x)$$, called the free propagator, is the Green’s function for the Klein-Gordon equation.

That was all prelude. The real topic of this blog post is: how do we determine $$D(x)$$? Zee doesn’t really motivate his work here, so I thought I would give it a try.

The Klein-Gordon equation $$-(\partial^2+m^2)\phi=0$$ has a solution $$\phi_k(x)=e^{ikx}$$ for every 4-vector $$k$$ with $$k^2=m^2$$. (You are probably more familiar thinking of this as $$\phi_{\omega,\vec{k}}=e^{i(\omega t – \vec{k}\cdot\vec{x})}$$ for $$\omega^2 = \vec{k}^2+m^2$$.) That’s great, but we have a source term (that Dirac delta) on the right-hand side, so we want $$\phi$$s with $$-(\partial^2+m^2)\phi\neq0$$. One way to do this is to take wave solutions with $$k^2\neq m^2$$:
\begin{align} -(\partial^2+m^2)e^{ikx} &= (k^2-m^2)e^{ikx} \\ -(\partial^2+m^2)\frac{e^{ikx}}{k^2-m^2} &= e^{ikx}. \end{align}
Huh; so we have the power to create any wave on the right-hand side, as long as it has $$k^2\neq m^2$$. Using linearity & Fourier analysis, we can combine these to form (almost) any source term. In particular, since
$$\delta^{(4)}(x) = \int\frac{d^4k}{(2\pi)^4}e^{ikx},$$
we will find:
$$-(\partial^2+m^2)\int\frac{d^4k}{(2\pi)^4}\,\frac{e^{ikx}}{k^2-m^2} = \delta^{(4)}(x).$$
So we have a solution for the free propagator:
$$D(x)=\int\frac{d^4k}{(2\pi)^4}\,\frac{e^{ikx}}{k^2-m^2}.$$

But there’s a flaw with this line of thought: When $$k^2=m^2$$ (a “Minkowski hyperboloid” of vectors), the integrand is undefined. Is the integral as a whole still defined? I would like to think that it is, that this is simply a measure-0 set we can integrate around. For their part, QFT people like to compute the integral by pushing the denominator away from zero in the complex plane (important, since the denominator can be either positive or negative!) and then taking this push to zero. That is:
$$D(x)=\lim_{\epsilon\rightarrow0}\int\frac{d^4k}{(2\pi)^4}\,\frac{e^{ikx}}{k^2-m^2+i\epsilon}.$$

Taking the $$\lim$$ as implied whenever we have an $$\epsilon$$, and separating $$k$$ into space and time components, we have:
\begin{align} D(x) &= \int\frac{d^3\vec{k}}{(2\pi)^3}\int\frac{d\omega}{2\pi}\,\frac{e^{i\omega t}e^{-i\vec{k}\vec{x}}}{\omega^2-\vec{k}^2-m^2+i\epsilon} \\ &= \int\frac{d^3\vec{k}}{(2\pi)^3}e^{-i\vec{k}\vec{x}}\int\frac{d\omega}{2\pi}\,\frac{e^{i\omega t}}{\omega^2-\omega_k^2+i\epsilon} \end{align}
(where $$\omega_k^2 = \vec{k}^2-m^2$$, as before.) That inside integral looks like fun; let’s do it.

[I finally figured out how to actually do contour integration, and it's pretty awesome. I'm too lazy to write it up here, though.]

And a conceptual note which doesn’t deserve its own post: I keep on getting confused while playing with path integrals, thinking I’m accidentally doing quantum mechanics because I’m connecting classical field configurations to classical field configurations. There are some pretty easy mnemonic tricks to convince myself that this is wrong, such as remembering that these classical field configurations could be made of real numbers for all I care, or some far more exotic species. Meanwhile, a QM wave function really has to made of complex numbers.

But there’s something deeper going on here, which is that the path integral formulation really is fundamentally quasi-classical. It operates on a single, privileged basis of the quantum field space — the classical field space. And to clarify, this has nothing to do with “position”. Even when we go to momentum space we are simply reparameterizing the basis of classical fields. In table form:

Physical space Classical field space Quantum field space
A $$X$$-valued field on Minkowski space $$\mathbb{R^{3}}$$ $$\mathbb{R^{3}}\rightarrow X$$ $$(\mathbb{R^{3}}\rightarrow X)\rightarrow \mathbb{C}$$
QM in 3 dimensions $$\mathbb{R^{0}}$$ $$\mathbb{R^{0}}\rightarrow \mathbb{R}^3$$ $$(\mathbb{R^{0}}\rightarrow \mathbb{R}^3)\rightarrow \mathbb{C}$$

I’m a little unclear as to the best way to work time into this picture… Especially for relativistic fields, it makes sense to think of the field as being defined over all of spacetime, instead of just on a single space-slice. But this breaks the analogy with quantum mechanics, and it makes the concept of a “path” more amorphous.