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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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1 Answers1

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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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