In Part I, we discovered that tensorial propagators in quantum field theory are often non-invertible which poses an issue when one is trying to determine the form of the propagator. Then, in Part II, we discussed possible techniques involving both symmetries and invariants of the Lagrangian. Here, we aim to reveal how invariant quantities may be useful for propagators, and gauge fields specifically, and we will ultimately derive the form of the photon propagator.

First we return to the Lagrangian for electromagnetism, given by $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + J_\mu A^\mu,$ where the electromagnetic tensor is defined as $F_{\mu\nu} = \partial_\mu A_\nu – \partial_\nu A_\mu$. Much as we had done in the previous examples showcasing the invariance of a vector’s length, and of the Lagrangian, under a continuous transformation, we aim to determine how to vary the gauge fields contained within the Lagrangian density to keep it invariant.

Recall that from classical electromagnetism, the most common method of gauge fixing is to modify the vector potential by $A_\mu \to A^\prime_\mu = A_\mu + \partial_\mu \Lambda$, where $\Lambda$ is some scalar field. Recalling that partial derivates commute, we can clearly see how this preserves the electromagnetic tensor $F_{\mu\nu}$ since

$$\begin{align*} F_{\mu\nu} \to F^\prime_{\mu\nu} &= \partial_\mu(A_\nu + \partial_\nu \Lambda) – \partial_\nu(A_\mu + \partial_\mu \Lambda) \\ &= \partial_\mu A_\nu – \partial_\nu A_\mu +\partial_\mu\partial_\nu \Lambda -\partial_\nu\partial_\mu \Lambda \\ &= F_{\mu\nu}.\end{align*}$$

Fixing the gauge in this way gives rise to a few different gauge conditions, namely the Coulomb gauge and the Lorenz gauge. While both may be used to determine the form of the photon propagator, we will make a key argument here for the Lorenz gauge. Not only is the Lorenz gauge easier to work with when performing the calculations, the Lorenz gauge also has the fortune of naturally bringing in Lorentz^{1} symmetry^{2}. Lorentz symmetry is one of more crucial aspects of physics, and definitely of general relativity, which states that experimental results are the same for all inertial observers. This symmetry holds true, especially for the speed of light, regardless of whether or not we are dealing with the physics of the quantum realm.

Now, using Lorenz gauge, we move forward! The Lorenz gauge states that $\partial_\mu A^{\mu} = 0$. Now that we have this condition, the Lagrangian density, after replacing the electromagnetic tensor with its definition and integrating^{3}, becomes $$\mathcal{L} = – \tfrac{1}{2} A^\mu(\partial^2g_{\mu\nu}-\partial_\mu\partial_\nu)A^\nu + J_\mu A^\mu =- \tfrac{1}{2}A^\mu(\partial^2g_{\mu\nu})A^\nu + J_\mu A^\mu,$$

where we have imposed the Lorenz gauge. Now finding the photon propagator, $S^{\nu\lambda}(x)$, is quite straightforward, we simply follow the same steps as outlined before in Part I without the term $\partial_\mu\partial_\nu$;

$$\begin{align*}(\partial^2g_{\mu\nu}) S^{\nu\lambda}(x) &= (\partial^2g_{\mu\nu})\int\frac{d^2k}{(2\pi)^2}e^{ik_\mu x^\mu}S^{\nu\lambda}(k) \\ &= \int \frac{d^2k}{(2\pi)^2}(-k^2 g_{\mu\nu} )e^{ik_\mu x^\mu}S^{\nu\lambda}(k) \\ &= \int\frac{d^2k}{(2\pi)^2} \delta^\lambda_\mu e^{ik_\mu x^\mu} =\delta^\lambda_\mu \delta^{(2)}(x)\\&\implies \boxed{S^{\nu\lambda}(k)=-\delta^\lambda_\mu \frac{g^{\mu\nu}}{k^2}}.\end{align*}$$

Simplifying for the photon propagator, and recasting back to the original frame with $k^2 \to k^2 +i\epsilon$, we immediately have that

$$\boxed{iS^{\mu\nu}(x-y) =\text{ } i\int \frac{d^2k}{(2\pi)^2}\frac{- g^{\mu\nu}}{k^2 +i\epsilon} e^{ik_\mu(x^\mu-y^\mu)}.}$$

And so, we have finally derived the form of the photon propagator!

**Credit**: Image found on Wikipedia. Author is Sanpaz.

#### Footnotes

- Lorenz and Lorentz are two very different physicists. Of the two, however, Lorentz is the more renown physicist.
- In physics, the rule of thumb is the more the symmetry – the better.
- Proof Shown Here:
$$\begin{align*}-\frac{1}{4}\int d^2 x F_{\mu\nu}F^{\mu\nu} &=-\frac{1}{4}\int d^2 x(\partial_\mu A_\nu – \partial_\nu A_\mu)(\partial^\mu A^\nu – \partial^\nu A^\mu) \\ &=-\frac{1}{4}\int d^2 x (\partial_\mu A_\nu\partial^\mu A^\nu -\partial_\mu A_\nu\partial^\nu A^\mu \\ &\hspace{1in} -\partial_\nu A_\mu\partial^\mu A^\nu +\partial_\nu A_\mu\partial^\nu A^\mu) \\ &=-\frac{1}{2}\int d^2 x (\partial_\mu A_\nu\partial^\mu A^\nu -\partial_\mu A_\nu\partial^\nu A^\mu), \\&\text{then by integrating each term by part}, \\&=\text{B.C.S } – \frac{1}{2}\int d^2 x (A^\mu\partial^2A^\mu – A^{\mu}\partial_{\mu}\partial_{\nu} A^{\nu} ), \\&\text{the B.C.S disappear accordingly to obtain},\\&=- \frac{1}{2}\int d^2 x A^\mu(\partial^2g_{\mu\nu}-\partial_\mu\partial_\nu)A^\nu.\end{align*} $$

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