As we know, the number of physical degrees of freedom(DoF) for a photon is 2.(adsbygoogle = window.adsbygoogle || []).push({});

I can understand this by gauging away redundant DoF's by gauge fixing.

For example, in QED, by fixing the Lorentz gauge [tex] \partial_\mu A^\mu = 0 [/tex],

we could get rid of one DoF, moreover, the residual gauge symmetry, which is

[tex] A^\mu \rightarrow A^\mu + \partial^\mu f(x) [/tex]

with [tex] \partial^2 f = 0 [/tex] could allow us to remove another DoF.

This meansthe physical DoF of a photon is 2.

------

However, on the other hand, we know that the virtual photon

which appearing in the internal legs of Feynman diagrams could have some longitudinal component.

And this longitudinal DoF could interact with other particles in a Feynman diagram.

However, this means the above symmetry argument in the first part of my post could NOT apply

to virtual photons. I don't know why. We could always gauge away two DoF's,

however, consideration of Feynman diagrams says that we could only gauge away 1 DoF of

virtual particles, why is that?

Thanks!

**Physics Forums - The Fusion of Science and Community**

# DoF of a gauge boson

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: DoF of a gauge boson

Loading...

**Physics Forums - The Fusion of Science and Community**