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?

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


1 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.

  • $\begingroup$ I don’t see the three dof in the $\xi=0$ case. The gauge propagator has the transversal projector in its numerator, which tells me that there are 2 dof? $\endgroup$
    – ersbygre1
    Oct 24, 2018 at 0:58
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    $\begingroup$ The object $P^{\mu\nu}:=\eta^{\mu\nu}-p^\mu p^\nu/p^2$ is a projector into the direction transverse to $p^\mu$. It is not a projector into the $\epsilon_\pm$ subspace, $P^{\mu\nu}\neq\sum_{\sigma=\pm}\epsilon^\mu_\sigma(\boldsymbol p)\epsilon^{\nu*}_\sigma(\boldsymbol p):=\tilde P^{\mu\nu}$ (for one thing, $\tilde P$ is not Lorentz covariant and $P$ is). The object $P$ is actually given by the sum of the polarisation vectors for $\sigma=\pm,0,L$, and so it includes three physical d.o.f.s, and one unphysical one. In short: $\tilde P$ corresponds to two d.o.f.s., and $P$ to four. $\endgroup$ Oct 24, 2018 at 2:23
  • $\begingroup$ Thank you, now I understand the $\xi=0$ case, but the other two cases not anymore. (1) Does the equation in my footnote correspond to the sum over $\sigma=\{\pm, L\}$? (2) How do I see the dof‘s in the $\xi=1$ case? Is it correct to say that due to the opposite sign in the spin-1 propagator, the $\mu\nu=00$ component has the wrong sign $\to$ unphysical and the others represent 3 massive scalar propagators (?) $\endgroup$
    – ersbygre1
    Oct 24, 2018 at 2:53
  • $\begingroup$ 1) The equation corresponds to $\sigma=\pm,0$ (the three physical, spin 1 states). 2) Yes, that is a fine way to think about it. Recall that the physical polarisation states are spacelike, $\epsilon_\sigma^2=+1$ for $\sigma=\pm,0$, and the unphysical one is timelike, $\epsilon_L^2\propto p^2/m^2=-1$. $\endgroup$ Oct 24, 2018 at 2:57
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    $\begingroup$ @Stephan Recall that the physical polarisation states for a massive spin 1 particle are $\epsilon_+,\epsilon_0,\epsilon_-$. The state $\epsilon_0$ is not longitudinal (but it satisfies $\epsilon^\mu_0=(p^\mu/m)+\mathcal O(1)$ as $m\to0$, and so it becomes longitudinal in the massless limit). On the other hand, the unphysical state is $\epsilon_L^\mu\propto p^\mu$, which is exactly longitudinal. In short: for massive particles, none of the physical states is longitudinal; but one of them is approximately longitudinal if the mass is very small (or, equivalently, the momentum very high). $\endgroup$ Oct 24, 2018 at 13:37

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