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I'm confused with the gauge fixing in the Higgs mechanism.

So if we have an action like $$S=\int |D\phi|-\frac{1}{4}F^2 -V(\phi) ~ ,\tag{1}$$ then expand around some non-trivial vacuum, then we have terms like $$\int \partial_\mu \phi\partial^\mu \phi -2 m A_\mu \partial^\mu \phi +m^2 A_\mu A^\mu +\dots ~,\tag{2}$$ which can be written as $$\int m^2(A_\mu -\frac{1}{m}\partial_\mu \phi)(A^\mu -\frac{1}{m}\partial^\mu \phi)+\dots ~.\tag{3}$$ Then since the original action (1) has a gauge symmetry $A\rightarrow A-\partial \alpha$ for any function $\alpha$, we can fix the gauge (by choosing some $\alpha$), in this case $\alpha=\frac{1}{m}\phi$. Then above actions looks like $$\int m^2 B_\mu B^\mu+\dots ~, $$ where $B\equiv A-\frac{1}{m}\partial \phi$. I think this is how Higgs mechanism works and essencially you are removing the "unphysical degree of freedom" (in this case $\phi$)

My question is what if we take $m=0$? Then Eq.(2) looks like $$\int \partial_\mu \phi\partial^\mu \phi -2 m A_\mu \partial^\mu \phi +\dots ~.$$ So we can't rewrite the action in nice forms like Eq.(3). Then what kind of gauge should I choose to see the physical degree of freedom? I'm guessing I can't choose a gauge like $A\rightarrow A-\partial \alpha A$ where $\partial \alpha = A$.

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  • $\begingroup$ What does your standard summary say? $\endgroup$ – Cosmas Zachos Aug 12 at 16:48
  • $\begingroup$ When $m=0$ then you don't have a Higgs particle. Also, why is there a $m$ in the last term in the last equation- there aren't $2$ different masses - the last term should vanishes too. $\endgroup$ – Cinaed Simson Aug 12 at 18:42
  • $\begingroup$ @CinaedSimson Very true. Sorry did't even notice. Thanks $\endgroup$ – user239970 Aug 13 at 10:03

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