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Consider a Lagrangian with a real scalar field $\varphi$ and massless vector field $A_\mu$ with field strength $F_{\mu\nu}$,

$$\mathcal{L} = -\frac{1}{2}\left(\partial\varphi\right)^2 - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \lambda\left(\varphi^2 - v^2\right)^2 - g\varphi^2 A_\mu A^\mu.$$

After SSB, $\varphi\left(x\right) \rightarrow \pm v + h\left(x\right)$, the scalar field has a non-zero VEV and the vector field acquires a mass.

Do we not now have one more degree of freedom than we started with, and if so, is it a problem? In the context of SSB in gauge theories, emphasis is put on the fact that the Goldstone d.o.f.s are eaten by the gauge bosons, and the total d.o.f. count remains the same.

tomdodd4598
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  • Sorry, maybe I am missing something, but can you please tell me what is the new degree of freedom? – Matteo Aug 30 '22 at 09:03
  • @Matteo I guess your idea is: Before SSB, we have one scalar (1 dof) and a massless vector (2 dof), after SSB we still have the scalar and a massive vector, so 1+3 dof? However, that doesn't work out because the Langrangean isn't even gauge invariant (as Kosm's answer points out), so this is not analogous to an Abelian Higgs model. – Toffomat Aug 30 '22 at 10:33
  • @Toffomat ah, so would it be correct to say that a massless vector field still has 3 d.o.f.s if we don't have gauge invariance, since that is what reduces the d.o.f. count to 2? – tomdodd4598 Aug 31 '22 at 07:53
  • Well, a massless vector field without gauge invariance contains negative-norm states and is non-unitary (see eg https://physics.stackexchange.com/questions/317028/why-do-negative-norm-states-break-unitarity and liks therein), so it is generally an inconsistent theory. So, while you may formally have more degrees of freedom, they are not dofs in the usual sense of the word. – Toffomat Aug 31 '22 at 09:31

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The only symmetry here is $\varphi\rightarrow\pm\varphi$ (the last term explicitly breaks the gauge symmetry of $A_m$). Also, the vector is not massless because of the last term. To see this you can just redefine the scalar as you did below the Lagrangian, treating $\upsilon$ as a constant parameter. In the abelian Higgs model, the non-invariance of the mass term of $A_m$ is compensated by the transformation of the Goldstone-dependent terms.

Kosm
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