# Spin 1 quantization $A_\mu$ (E.M field): why is the field real?

In my QFT course, when we quantize the spin 1 field corresponding to the E.M potential vector $A_\mu$ we consider it as a real one. I would like to know why it is the case

To explain what the teacher is doing : He prove from the Lagrangian written in Feynman Gauge, using Gupta Bleuler quantization (I could explain more if necessary) that $A_\mu$ follows the Klein Gordon equation: $\Box A_\mu=0$

And then it is like if we would have 4 scalar fields. Then he writes it in the form:

$$A_\mu(x)= \int d^3k ~ e^{ikx}a_\mu(\vec{k})+e^{-ikx}a_\mu^{\dagger}(\vec{k})$$

And we can see that when we do it we implicitly consider that it is a real field.

Why is it so? Why would be $A_\mu$ a real field and not a complex one when we study the electromagnetic field? Is there a physical reason behind?

• The photon in the Standard Model is its own antiparticle. This forces the gauge field to be a real function in the classical sense, or self-adjoint in the quantum one. – DanielC Oct 7 '17 at 16:31
• @DanielC yes but then you can reformulate my question in "why is the photon its own antiparticle" if you want ! – StarBucK Oct 7 '17 at 17:16

• @StarBucK For example, a field $F$ with Lagrangian $\mathcal L = F^* F$ has equations of motion $F=0$. Thus on shell the theory has no degrees of freedom to contribute but off shell it does not need to obey the equations of motion. – JamalS Oct 7 '17 at 16:09