# Gauge theory and eliminating unphysical degrees of freedom

In free space we can express Maxwell's equations as \begin{align} \varepsilon^{abcd}\partial_bF_{cd}=0 ~~\text{ and }~~ \partial_aF^{ab}=0 \tag{1} \end{align} where $F^{ab}=-F^{ba}$. The most general solution to the above equations is \begin{align} F_{ab}=\partial_aA_b-\partial_bA_a. \tag{2} \end{align} Since partial derivatives commute it is evident that the field strength $F^{ab}$ is invariant under a gauge transformation of the form $\delta A_a=\partial_a\phi$ where $\phi=\phi(x)$ is an arbitrary scalar function on space-time.

Due to its anti-symmetry, the field $F^{ab}(x)$ has only six independent components. The first relation in (1) imposes four relations between these six components, leaving two independent degrees of freedom (the two helicity states of a photon). The second relation is used usually used to determine the components of $F^{ab}$ in terms of the current and charges densities. So we have two degrees of freedom for the observable field strength $F^{ab}$. However, equation (2) states that $F^{ab}$ has four independent degrees of freedom, the components of the field $A_a(x)$. We must conclude that there are redundant degrees of freedom contained within the 'gauge field' $A_a$.

To eliminate one of these unphysical degrees of freedom we choose a 'gauge fixing condition' where we impose a relation among the components of $A_a$. For example we might require that it be divergenceless, $\partial_aA^a=0$, this is known as the Lorenz gauge. If we impose this gauge then there is still some residual gauge freedom in that $A_a$ is now defined up to some arbitrary function on space-time which satisfies $\square\phi(x)=0$ (since such a redefinition respects the Lorenz gauge and leaves $F^{ab}$ invariant).

My questions are:

• What kind of gauge fixing conditions can we impose? Can they be completely arbitrary or must they satisfy some requirement? For example, could I (for no good reason) impose the condition $A_aA^a=0$?
• In this example, choosing a gauge condition eliminates only one of the two unphysical degrees of freedom in the gauge field $A_a$. How do we eliminate the second? I would think that it has something to do with the fact that it is defined up to a term $\partial_a\phi$, but I don't see directly how we can use this to kill another degree of freedom.
• In general, if you are constructing a gauge theory and you do not know in advance the dynamical equations of motion a certain field must obey (in this example equations (1)), how do you know how many physical degrees of freedom the theory must possess so that you can impose the neccesary gauge conditions to eliminate the unphysical degrees of freedom? Is it determined by the type of particle you are trying to describe and the spin degrees of freedom it possess?

You can fix the gauge doing the transformation $A^\mu \rightarrow {A^\prime}^\mu = A^\mu + \partial^{\,\mu} f$ and choosing an appropriate function $f$. You cannot fix it by impositing whatever condition on ${A^\prime}^\mu$ you want. You have to guarantee there is a function $f$ which can be used so as your chosen condition on ${A^\prime}^\mu$ can be achieved. An example is the Lorenz gauge condition $\partial_\mu {A^\prime}^\mu \equiv 0$, which is achieved from the transformation above if we use any function $f$ that satisfies $\square\,f=-\partial_\mu A^\mu$. This is so because

${A^\prime}^\mu = A^\mu + \partial^{\,\mu} f \Longrightarrow \partial_\mu{A^\prime}^\mu \equiv 0 = \partial_\mu A^\mu + \square\,f \Longrightarrow \square\,f = -\partial_\mu A^\mu.$

The 4-potential $A^\mu$ has four degrees of freedom (d.o.f.) but two of these are unphysical and can be eliminated exploiting electromagnetism's invariance under gauge transformations $A^\mu \rightarrow {A^\prime}^\mu = A^\mu + \partial^{\,\mu} f$. For instance, we can take $\partial_\mu {A^\prime}^\mu \equiv 0$ as long as $f$ satisfies $\square\,f=-\partial_\mu A^\mu$, as seen in the previous answer. Condition $\partial_\mu {A^\prime}^\mu \equiv 0$ gives you one equation relating $A_0, A_1, A_2$ and $A_3$, therefore it removes one out of four d.o.f.

The last unphysical d.o.f. is removed because there is still some gauge freedom to be explored. To ensure $\partial_\mu {A^\prime}^\mu \equiv 0$ all we have to do is use a function $f$ satisfying $\square\,f = -\partial_\mu A^\mu$. It is clear any function $f_0$ satisfying $\square\,f_0\equiv 0$ can be added to $f$ while keeping $\partial_\mu {A^\prime}^\mu \equiv 0$ intact. This last freedom removes one more d.o.f. as we can take ${A^{\prime\prime}}^\mu = {A^\prime}^\mu + \partial^{\,\mu} f_0$ and choose an appropriate solution of $\square\,f_0\equiv 0$.

The last paragraph may be best understoond with an example. In free space, using the Lorenz gauge $\partial_\mu {A^\prime}^\mu \equiv 0$, the 4-potential satisfies $\square\, {A^\prime}^\mu = 0$. Consider the solution for a plane wave propagating along the $z$ axis: ${A^\prime}^\mu = \mathcal{A}\, \varepsilon^\mu \cos(kz-\omega t)$, where $\mathcal{A}$ is the wave's amplitude, $\omega=|k|$ (I'm using natural units where $c\equiv 1$), and $\varepsilon^\mu$ is the polarization vector. You can check gauge condition $\partial_\mu {A^\prime}^\mu \equiv 0$ is satisfied for the polarization vector $\varepsilon^\mu$ chosen as $\varepsilon^\mu_{(x)} = (0,1,0,0)$ or $\varepsilon^\mu_{(y)} = (0,0,1,0)$, or some combination of these. These correspond to the two physical d.o.f. associated with horizontal and vertical polarizations.

For this example, the two unphysical d.o.f. are related to polarization vectors $\varepsilon^\mu_{(t)} = (1,0,0,0)$ and $\varepsilon^\mu_{(z)} = (0,0,0,1)$, corresponding to "temporal" and longitudinal polarizations, respectively. Although ${A^\prime}^\mu$ with $\varepsilon^\mu = \varepsilon^\mu_{(t)}$ or $\varepsilon^\mu = \varepsilon^\mu_{(z)}$ satisfies the equation of motion $\square\,{A^\prime}^\mu=0$, it won't satisfy the Lorenz gauge $\partial_\mu {A^\prime}^\mu \equiv 0$. Nevertheless, if the polarization vector is taken as the combination $\varepsilon^\mu = c_1 \varepsilon^\mu_{(t)} + c_2 \varepsilon^\mu_{(z)}$, for constants $c_1$ and $c_2$, Lorenz gauge is satisfied once we impose $c_1 = c_2$. Because of that, we are left with only one unphysical d.o.f. Finally, choosing $f_0$ (from previous paragraphs) as $\mathcal{A}\,\frac{c_1}{\omega}\sin(kz-\omega t)$, you can check the solution ${A^{\prime\prime}}^\mu = \mathcal{A}\,c_1(\varepsilon^\mu_{(t)}+\varepsilon^\mu_{(z)})\cos(kz-\omega t) + \partial^{\,\mu} f_o$ identically vanishes, meaning the last unphysical d.o.f. doesn't propagate indeed.