Path Integral Formulation of Quantum Electrodynamics

Abstract

Let us consider a pure Abelian gauge theory given by the Lagrangian

$$\displaystyle\begin{array}{rcl} \mathcal{L}_{\text{photon}}& =& -\frac{1} {4}F_{\mu \nu }F^{\mu \nu } \\ & =& -\frac{1} {4}\left (\partial _{\mu }A_{\nu } - \partial _{\nu }A_{\mu }\right )\left (\partial ^{\mu }A^{\nu } - \partial ^{\nu }A^{\mu }\right ){}\end{array}$$

(36.1)

or, after integration by parts,

$$\displaystyle\begin{array}{rcl} \mathcal{L}_{\text{photon}}& =& -\frac{1} {2}\left [-\left (\partial _{\mu }\partial ^{\mu }A_{\nu }\right )A^{\nu } + \left (\partial ^{\mu }\partial ^{\nu }A_{\mu }\right )A_{\nu }\right ] \\ & =& \frac{1} {2}A_{\mu }\left [g^{\mu \nu }\square - \partial ^{\mu }\partial ^{\nu }\right ]A_{\nu }\quad, {}\end{array}$$

(36.2)

and therefore, the corresponding action is given by

$$\displaystyle\begin{array}{rcl} S\left [A_{\mu }\right ]& =& \frac{1} {2}\int (dx)A_{\mu }(x)\left [g^{\mu \nu }\square - \partial ^{\mu }\partial ^{\nu }\right ]A_{\nu }(x) \\ & =& -\frac{1} {2}\int \frac{(dk)} {(2\pi )^{4}} \tilde{A}_{\mu }(k)\left [g^{\mu \nu }k^{2} - k^{\mu }k^{\nu }\right ]\tilde{A}_{\nu }(-k)\quad.{}\end{array}$$

(36.3)

The operator $M^{\mu \nu }(k) = \left (g^{\mu \nu }k^{2} - k^{\mu }k^{\nu }\right )$ has no inverse, because it has eigenfunctions k _ν with eigenvalues zero:

$$\displaystyle{ M^{\mu \nu }(k)k_{\nu } = \left (k^{\mu }k^{2} - k^{\mu }k^{2}\right ) = 0\quad. }$$

(36.4)