Counting degrees of freedom in the Higgs mechanism for different gauges

Question

I am wondering how to count the degrees of freedom (dof) for a massive gauge field in different gauges. I've been reading some other answers, but haven't found a solution yet.

I am looking at the model of a complex scalar field $\phi$ with spontaneously broken symmetry ($\to$ Higgs mechanism) and a $U(1)$ gauge invariance, which means one gauge field $A$ (scalar QED). (see e.g. in Peskin & Schroeder's QFT book ch.20/21)

For a general $R_\xi$ gauge, the gauge field propagator is $$ \Delta^{\mu\nu}_F(p) = \frac{\eta^{\mu\nu}-\frac{p^\mu p^\nu}{m_A^2}}{p^2-m_A^2+\text{i}\epsilon}+ \frac{\frac{p^\mu p^\nu}{m_A^2}}{p^2-\xi m_A^2+\text{i}\epsilon} $$ This term should theoretically have 4 dof, three physical(=actual) ones from the first part (this includes the spin sum over all physical states${}^1$ = 3 dof) and one unphysical(=fictitious) from the second part (proportional to a scalar propagator $\to$ 1 dof).

So how does this hold up in certain gauges?

1. $\xi=0$, which leads to the Lorenz/Landau gauge. The gauge and Goldstone propagator looks like $$ \Delta^{\mu\nu}_F(p) = \frac{\eta^{\mu\nu}-\frac{p^\mu p^\nu}{p^2}}{p^2-m_A^2+\text{i}\epsilon},\qquad S_{gold}(p)=\frac{1}{p^2+\text{i}\epsilon}\tag{21.27} $$ The Goldstone boson is an unphysical field, since its mass depends on the gauge parameter. Therefore it contributes 1 unphysical dof. The gauge propagator contains the transversal projection operator in its numerator, so this should represent 2 dof?

2. $\xi=1$, which leads to the Feynman/'t Hooft gauge. Again, the propagators are $$ \Delta^{\mu\nu}_F(p) = \frac{\eta^{\mu\nu}}{p^2-m_A^2+\text{i}\epsilon},\qquad S_{gold}(p)=\frac{1}{p^2-m_A^2+\text{i}\epsilon}\tag{21.28} $$ Below Eq.(21.1), P&S write that this form of the gauge propagator has 4 components (2 transv., 1 long., 1 timelike) and the Goldstone propagator also has 1 dof.

3. $\xi\to\infty$, which leads to the unitary gauge. In a unitary gauge, only physical degrees of freedom remain. $$ \Delta^{\mu\nu}_F(p) = \frac{\eta^{\mu\nu}-\frac{p^\mu p^\nu}{m_A^2}}{p^2-m_A^2+\text{i}\epsilon},\qquad S_{gold}(p)=0\tag{21.29} $$ Here we have again the sum over all physical states in the numerator ($\to$ 3 physical dof) and the Goldstone propagator decouples from the theory.

How can I make sense of this dof-counting? Is it even reasonable to count dof's, since the unphysical terms cancel anyway and I should only focus on the physical states?

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${}^1$See Eq.(21.26) in P&S's book: $$ \sum\limits_{\epsilon^\mu q_\mu=0}\epsilon^\mu (\epsilon^\nu)^* = -\left( \eta^{\mu\nu}-\frac{q^\mu q^\nu}{m_A^2} \right)\quad \leftarrow \text{physical polarization states} $$

Answer 1

For any (finite) value of $\xi$, the propagator $$ \Delta^{\mu\nu}_F(p) = \frac{\eta^{\mu\nu}-\frac{p^\mu p^\nu}{m_A^2}}{p^2-m_A^2+\text{i}\epsilon}+ \frac{\frac{p^\mu p^\nu}{m_A^2}}{p^2-\xi m_A^2+\text{i}\epsilon} $$ contains three physical and one unphysical d.o.f.s. The Goldstone propagator $$ \Delta_\varphi(p)=\frac{1}{p^2-\xi m_A^2+i\epsilon} $$ contains one unphysical d.o.f. that cancels the unphysical d.o.f. in the gauge propagator.

This counting is essentially valid for any $\xi$ (including $\xi=0$). The only exception is the limit $\xi\to\infty$, where the Goldstone bosons become infinitely massive and freeze up. In this gauge, the gauge propagator contains only three d.o.f.s, and they are all physical.

See also this PSE post.