Here we study some issues with tensorial propagators that are often encountered in the study of quantum field theory. We will use the photon propagator from electromagnetism as an example to guide us through the troublesome calculations.

We start with the electromagnetic Lagrangian and, as usual, we write out the path integral and attempt to simplify the expression in the exponent. The Lagrangian for electromagnetism is

$latex \displaystyle \mathcal{L}_{EM} =-\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + J^\mu A_\mu$

where $latex F_{\mu\nu}$ is the electromagnetic tensor which encapsulates the behavior of the electric and magnetic field, $latex A_\mu$ is the vector potential on which $latex F_{\mu\nu}$ depends, in fact $latex F_{\mu\nu} = \partial_{\mu}A_{\nu} – \partial_{\nu}A_{\mu}$, and $latex J^{\mu}$ is the source for the electromagnetic fields. Replacing the electromagnetic tensor with its definition, we obtain

$latex \displaystyle \begin{aligned} i\int d^2x \mathcal{L}_{EM} &= i\int d^2x \bigg[-\frac{1} {4}F_{\mu\nu}F^{\mu\nu}+J_\mu A^\mu\bigg] \\ &= i\int d^2x \bigg[-\frac{1}{4}(\partial_\mu A_\nu-\partial_\nu A_\mu)(\partial^\mu A^\nu-\partial^\nu A^\mu)+J_\mu A^\mu\bigg] \\ &= i\int d^2x \bigg[-\frac{1}{2}A_\mu(\partial^2g^{\mu\nu} – \partial^\mu\partial^\nu)A_\nu+J_\mu A^\mu\bigg], \end{aligned}$

where we have integrated the second line and applied our boundary conditions in order to have the cross terms disappear. **Note:** the term $latex g_{\mu\nu}$ is the metric tensor $latex -$ an integral part of Einstein’s general theory of relativity. This term encodes the curvature of our background spacetime. Now, if we attempt to find the photon propagator, $latex S_{\nu\lambda}(x)$, using the method outlined above, we find that

$latex \displaystyle \begin{aligned} (\partial^2g^{\mu\nu} – \partial^\mu\partial^\nu)S_{\nu\lambda}(x) &= (\partial^2g^{\mu\nu}-\partial^\mu\partial^\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} + k^\mu k^\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). \end{aligned} $

We may *naively* conclude that $latex (-k^2 g^{\mu\nu} + k^\mu k^\nu)S_{\nu\lambda}(k)=\delta_\lambda^\mu$, and attempt to solve the equation by dividing the $latex (-k^2 g^{\mu\nu} + k^\mu k^\nu)$ through. However, in general, $latex M=-k^2 g^{\mu\nu} + k^\mu k^\nu$ is not invertible. One way to see this is through a direct check using the Minkowski metric, $latex \eta_{\mu\nu}$. The reason why we are allowed to switch $latex g_{\mu\nu}$ with the Minkowski metric arises from a heuristic argument. Since the curvature on the quantum scales is extremely negligible, we may take the limit as $latex g_{\mu\nu}\to\eta_{\mu\nu}$ without worrying too much.

**Example:** Using the Minkowski metric with the signature $latex (+,-)$, we see that

$latex -k^2 \eta^{\mu\nu} + k^\mu k^\nu = \left( \begin{array}{cc} -k^2+k^0 k^0 & k^0k^1 \\ k^1 k^0 & k^2 + k^1k^1 \end{array} \right)$.

Then since $latex k^2 = (k^0)^2-(k^1)^2,$ we find

$latex -k^2 \eta^{\mu\nu} + k^\mu k^\nu =\left( \begin{array}{cc} (k^1)^2 & k^0k^1 \\ k^1 k^0 & (k^0)^2 \end{array} \right)$.

Looking at the determinant of the above expression, we can immediately write that

$latex \det[M] = (k^1)^2(k^0)^2-(k^0k^1)(k^1k^0)=0$

Notice that, while we have assumed that $latex k^0$ and $latex k^1$ commute, the last line remains true even if we throw out that assumption. Since $latex (k^i)^2 \in \mathbb{R}$ is a scalar, which commute, and since $latex k^0k^1k^1k^0 = k^0(k^1)^2k^0 = (k^1)^2k^0k^0 = (k^1)^2(k^0)^2,$ the above line is satisfied.

Well, we have found that the determinant vanishes $latex -$ which is terrible. A vanishing determinant implies a non-invertible matrix which prevents the term on the left hand side to be divided over. To show how to obtain a tensorial propagator, we have to introduce gauge transformations $latex -$ the importance and application of which will be discussed in Part II.

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