Consider the theory of scalar QED with the Lagrangian
$$\mathcal{L} = – \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \phi)^* (D_\mu \phi) – m^2 \phi^* \phi \tag{1}$$
where $\phi$ is a complex scalar field with mass $m$. Counting the degrees of freedom, we have
- two massless real degrees of freedom from $A_\mu$
- two massive real degrees of freedom from $\phi$
Now, even though there's no symmetry breaking going on here we can still choose to go to unitary gauge, i.e. fixing the gauge so that $\phi$ is real. We now have the gauge-fixed Lagrangian
$$\mathcal{L} = – \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \varphi) (D_\mu \varphi) – \frac12 m^2 \varphi^2\tag{2}$$
where $\varphi$ is a canonically normalized real scalar field, and there is no gauge symmetry. Then we have
- three real degrees of freedom from $A_\mu$
- one massive real degree of freedom from $\phi$
where I know there are three real degrees of freedom in $A_\mu$, because gauge fixing always removes one and we have no gauge fixing here.
What confuses me is that there must be two massive degrees of freedom, just as there were in the original theory. So that somehow means that one of the degrees of freedom in $A_\mu$ is massive while the other two aren't — but how can that be? There's no mass term for $A_\mu$ in sight.
Best Answer
What you can remove with a field redefinition (of the form of a $U(1)$ gauge transformation) is the phase of $\phi$. But the total number of degrees of freedom (dof's) is not going to change since the resulting lagrangian is not longer gauge invariant. The new counting is $3+1$ where the $3$ dof's come from $A_\mu$ (with a covariant constraint, see below, so that really $3=4-1$), and the 1 dof comes instead from the scalar radial mode of $\phi$.
More explicitly, write $$\phi(x)=e^{ie\pi(x)}\frac{h(x)}{\sqrt{2}}$$ (with both $\pi$ and $h(x)$ real scalar fields) and the covariant derivative $D_\mu \phi=(\partial_\mu-i e A_\mu)\phi$ becomes $$ D_\mu \phi=e^{ie\pi}\left[ie(\partial_\mu\pi-A_\mu)+\frac{1}{\sqrt{2}}\partial_\mu h\right] $$ so that the lagrangian reads $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{e^2}{2}h^2(\partial_\mu\pi-A_\mu)^2+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2\,. $$ Now, defining the new variable $A_\mu^\prime=A_\mu-\partial_\mu\pi$ (which is the gauge invariant combination), the lagrangian becomes $$ \mathcal{L}=-\frac{1}{4}F^\prime_{\mu\nu}F^{\prime \mu\nu}+\frac{e^2}{2}h^2 A_\mu^{\prime 2}+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2 $$ in full analogy to the abelian Higgs mechanism except that $h$ has vanishing vacuum expectation value. Now, this last lagrangian depends on the fields $A_\mu^\prime$ and $h$: how many dof's are there? Well, the $h(x)$ certainly counts 1. The $A_\mu^\prime$ on the other hand counts $3$ and neither $4$ (even though $\mu=0,1,2,3$) nor $2$. Indeed, from the equations of motion $\partial_\mu F^{^\prime\mu\nu}=-e^2h^2 A^{\prime \nu}$ we see the covariant constrain $$ \partial_\mu (A^{\prime\mu} h^2)=0 $$ which sends us from $4$ to $3$. But notice that there is no gauge invariance for $A_\mu^\prime$ in its lagrangian above (the $e^2 h^2 A_\mu^{\prime 2}$-term breaks it) and no longitudinal mode can be therefore removed in that way: $A_\mu^\prime$ is a different physical configuration than say $A^\prime_\mu-\partial_\mu \Omega$ (in particular one solves the equation of motion, the other doesn't). So the counting ends here and it matches that for massive spin-1 plus a real scalar, that is $3+1$ as expected. Notice however, that the spin-1 particle is not massive and all this gymnastic is just artificial as you wanted to move around one scalar dof $\pi$ inside $A_\mu$.