How to count the number of degrees of freedom of an arbitrary field (vector or tensor)? In other words, what is the mathematical procedure of gauge fixing?
[Physics] Gauge-fixing of an arbitrary field: off-shell & on-shell degrees of freedom
degrees of freedomfield-theorygaugegauge-theoryquantum-field-theory
Related Solutions
If $\phi$ is non-zero, fixing the phase of $\phi$ is a perfectly valid gauge condition. It's used frequently in Standard Model calcuations involving the Higgs field, where it goes by the name unitarity gauge. This is a nice gauge in some ways, because it makes manifest the fact that there's a massive vector field in the system.
Edit: Some caution is required with unitary gauge. It's a complete gauge when you can reasonably treat $\phi$ as non zero, because it uses every degree of freedom in the gauge transformation. This means for example that it's ok to use in perturbative calculations around a Higgs condensate. But when $\phi$ can vanish, the phase function isn't uniquely defined, which means the gauge transformation is not invertible. This gauge isn't quite a gauge.
You really should split your question. I will answer the part where you do not understand how counting of degrees of freedom work.
Basically we count the number of propagating (physical) degrees of freedom per point of spacetime. Of course, the total number of degrees of freedom is infinite because spacetime is continuous and has an infinite number of points, but to ask for the number of degrees of freedom per spacetime point is a reasonable demand to make. Bear in mind that we only care about physical degrees of freedom by which we mean those that can be properly normalized.
You correctly state that photons can be off-shell but they are only those involved in internal processes. External photons are always on-shell. Moreover, gauge invariance is a physical property. External fields which you measure in your laboratory should be independent of your chosen gauge. In other words, the S-matrix should be gauge-invariant. On the other hand, there is nothing that stops me from having gauge-broken internal processes if ultimately I can make the S-matrix gauge-invariant. Therefore, the word "physical" should almost always give you a picture of external on-shell gauge-invariant quantities.
So yes, gauge redundancy kills one degree of freedom, and when we are talking about propagating physical degrees of freedom, one more is killed on-shell. You have to understand how that happens. It is not that every time you see an equation of motion, a degree of freedom is killed. Killing of degrees of freedom requires an elaborate process of imposing constraints on the equation of motion known as gauge-fixing. And this has to be done on a case by case basis.
For example, consider the four equations of motion (separated into temporal and spatial sets) for the massless photon $A^\mu = (\phi, \vec A)$ describing four on-shell degrees of freedom as follows.
\begin{align*} -\Delta \phi + \partial_t \vec\nabla\cdot\vec A = 0\,,\\ \square \vec A - \vec\nabla(\partial_t\phi-\vec\nabla\cdot\vec A) = 0\,.\\ \end{align*}
Since these equations exhibit a gauge symmetry $A_\mu \to A'_\mu := A_\mu + \partial_\mu \alpha_1(x)$, we can try to fix the gauge by choosing $\alpha_1$ such that, for instance, it is a solution of $\square \alpha_1 = -\vec\nabla\cdot\vec A$, giving us
\begin{align*} \Delta \phi' = 0\,, \\ \square \vec A' - \vec\nabla\partial_t\phi' = 0\,. \\ \\ \vec\nabla\cdot\vec{A}'=0\,.\\ \end{align*}
We have selected a divergence-free field, the so-called Coulomb gauge. Under this choice, the electric potential becomes non-propagating, that is there are no kinetic terms in the Lagrangian for it (observe that $\Delta \phi' = 0$ does not have any time derivatives).
In momentum space, this gauge condition reads $\vec p \cdot \vec \epsilon = 0$ where $ \vec \epsilon$ is the polarisation vector (Fourier transform of the magnetic potential). There are three solutions to this constraint. Choosing a frame in which $p^\mu = (E,0,0,E)$, we find that the three polarisation vectors are
$$ \epsilon^\mu_1 = (0,1,0,0), \qquad \epsilon_2^\mu=(0,0,1,0), \qquad \epsilon_t^\mu = (1,0,0,0) $$
The third polarisation is time-like and therefore cannot be normalized. It is unphysical, and we have to get rid of it. Luckily, the gauge symmetry is not exhausted. There are more available choices of gauge transformations which preserve the Coulomb gauge $\vec p \cdot \vec \epsilon = 0$. For example, we could go from $A'_\mu \to A_\mu:= A'_\mu + \partial_\mu \alpha_2(x)$ such that $\Delta \alpha_2 = 0,\ \partial_t \alpha_2 = - \phi'$ which preserves the divergence and sets $\phi = 0$.
Note that this time we have to make sure that this gauge transformation happens on-shell, namely that $\Delta \phi = 0$, otherwise this gauge-fixing will be inconsistent because $\Delta \alpha_2 = 0 \Rightarrow$ $0 = \Delta \partial_t\alpha_2 = - \Delta\phi' \ne 0$ off-shell. In other words, requiring $\phi = 0$, or equivalently $\epsilon^0 = 0$, in order to get rid of unphysical degrees of freedom requires us to be on-shell.
To summarize, we made an off-shell gauge choice $\vec p \cdot \vec \epsilon = 0$, an on-shell gauge choice $\epsilon^0 = 0$ and our equation of motion became $p^2 = 0$. Having exhausted our gauge choices, we find only two physical polarization modes or degrees of freedom.
Now, you understand that merely having an equation of motion does not eat up a degree of freedom. To find the correct number of degrees of freedom, keep on making gauge choices (producing independent constraint equations), some off-shell and some on-shell, until you exhaust your gauge freedom. Then check how many degrees of freedom you are left with. If you notice any unphysical guy showing up, most likely you haven't used up all your gauge freedom and you still have enough flex to shoot this guy dead. Then, count all that you are left with. That's your answer.
Best Answer
In this answer, we summarize the results. The analysis itself can be found in textbooks, see e.g. Refs. 1 & 2.
$\downarrow$ Table 1: Massless spin $j$ field in $D$ spacetime dimensions.
$^1$For massive multiplets, go up 1 spacetime dimension, i.e. change $D\to D+1$ (without changing the number $n$ of spinor components). E.g. the on-shell DOF for massive 4D fields famously has a factor $2j+1$, cf. the row $D=5$ in Table 1.
$^2$ Off-shell DOF = # (components)- # (gauge transformations).
$^3$ On-shell DOF = # (helicity states)= (Classical DOF)/2, where Classical DOF = #(initial conditions).
$n$=# (spinor components). E.g. a Dirac spinor has $n=2^{[D/2]}$ complex components, while a Majorana spinor has $n=2^{[D/2]}$ real components,
$\downarrow$ Table 2: Antisymmetric $p$-form gauge potential in $D$ spacetime dimensions, $p\in\mathbb{N}_0$.
References:
D.Z. Freedman & A. Van Proeyen, SUGRA, 2012.
H. Nastase, Intro to SUGRA, arXiv:1112.3502; chapter 5.