Compton Scattering amplitudes in Feynman Diagrams

Question

While reading "An introduction to Quantum Field Theory" by Peskin & Schroeder I came across the following expression for the amplitude of one of the Feynman Diagrams for Compton scattering: $$iM=\bar u(p')(-ie\gamma^\mu)\epsilon^*_\mu(k')\frac{i(\not{p}+\not{k}+m)}{(p+k)^2-m^2}\epsilon_\nu(k)(-ie\gamma^\nu)u(p)$$ for which the authors then write: $$iM=(ie)^2\epsilon^*_\mu(k')\epsilon_\nu(k)\bar u(p')\gamma^\mu\frac{i(\not{p}+\not{k}+m)}{(p+k)^2-m^2}\gamma^\nu u(p)$$

My question is: the photon polarisation are 4-vectors, then how can they be put on the left, while ignoring the $\gamma^\mu$ matrix? Is it because $\epsilon^*_\mu(k')\epsilon_\nu(k)$ is a number? Covariant 4-vectors are column vectors (is this the definition or just a convention?) and therefore $\epsilon^*_\mu(k')\epsilon_\nu(k)$ is the product of a $1\times4$ by a $4\times1$ matrix, giving a number?

Answer 1

In QED, there are two sorts of indices going around, which can be understood best by looking at structure of $\gamma^\mu$. Each $\gamma$ matrix ($\gamma^0$, $\gamma^1$, ...) is a $4 \times 4$ matrix acting on the internal degrees of freedom of the electron. On top of this matrix structure, the four matrices are packaged into a contravariant four-vector, $\gamma^\mu$. The photon polarization vectors are also four-vectors, but each of their elements are just numbers. Because they're just numbers, they can be moved past the $\gamma$ matrices.