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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?

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  • $\begingroup$ 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. $\endgroup$ – DanielC Oct 7 '17 at 16:31
  • $\begingroup$ @DanielC yes but then you can reformulate my question in "why is the photon its own antiparticle" if you want ! $\endgroup$ – StarBucK Oct 7 '17 at 17:16
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The equations of motion and gauge redundacies determine the physical number of degrees of freedom of a theory.

Whether a scalar is real or complex also affects the number of degrees of freedom and we know the photon has two.

Thus, in order to have the appropriate degrees of freedom the field is taken to be real, and the degrees of freedom in total off shell and on shell must be the same.

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  • $\begingroup$ If I understand well, we can choose to represent it as complex or real, but if we go from real to complex we will have 1 degree of freedom more that would need to be restricted in a way. But it would be possible in principle it just change the way we quantize things. Am I right ? Also what do you call by "off shell" and "on shell" degrees of freedom ? $\endgroup$ – StarBucK Oct 7 '17 at 16:03
  • $\begingroup$ @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. $\endgroup$ – JamalS Oct 7 '17 at 16:09
  • $\begingroup$ Ok thank you. So finally, if we want to have the appropriate degrees of freedom we are forced to have a real field, we have no choice on it $\endgroup$ – StarBucK Oct 7 '17 at 17:45

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