Verify that $\mathbf{E}$ and $\mathbf{B}$ in (6.55) are unchanged by the gauge transformation
$$
A_\mu \rightarrow A_\mu^{\prime}=A_\mu+\partial_\mu \chi,
$$
where $\chi$ can be any function of $x$. Use this freedom to write Maxwell's equations in the form
$$
\square^2 A^\mu=j^\mu \quad \text { with } \partial_\mu A^\mu=0 .
$$
The requirement $\partial_\mu A^\mu=0$ is known as the Lorentz condition. However, even after imposing this, there is still some residual freedom in the choice of the potential $A^\mu$. We can still make another gauge transformation,
$$
A_\mu \rightarrow A_\mu^{\prime}=A_\mu+\partial_\mu \Lambda
$$
where $\Lambda$ is any function that satisfies
$$
\square^2 \Lambda=0 .
$$
This last equation ensures that the Lorentz condition is still satisfied.
Turning now from classical to quantum mechanics, we see that the wavefunction $A^\mu$ for a free photon satisfies the equation
$$
\square^2 A^\mu=0,
$$
which has solutions
$$
A^\mu=\varepsilon^\mu(\mathbf{q}) e^{-i q \cdot x} .
$$
The four-vector $\varepsilon^\mu$ is called the polarization vector of the photon. On substituting into the equation, we find that $q$, the four-momentum of the photon, satisfies
$$
q^2=0, \quad \text { that is, } m_\gamma=0 .
$$
The polarization vector has four components and yet it describes a spin-1 particle. How does this come about? First, the Lorentz condition, $\partial_\mu A^\mu=0$, gives
$$
q_\mu \varepsilon^\mu=0,
$$
and this reduces the number of independent components of $\varepsilon^\mu$ to three. Moreover, we have to explore the consequences of the additional gauge freedom (6.60). Choose a gauge parameter
$$
\Lambda=i a e^{-i q \cdot x}
$$
with $a$ constant so that (6.61) is satisfied. Substituting this, together with (6.63), into (6.60) shows that the physics is unchanged by the replacement
$$
\varepsilon_\mu \rightarrow \varepsilon_\mu^{\prime}=\varepsilon_\mu+a q_\mu .
$$
In other words, two polarization vectors $\left(\varepsilon_\mu, \varepsilon_\mu^{\prime}\right)$ which differ by a multiple of $q_\mu$ describe the same photon. We may use this freedom to ensure that the time component of $\varepsilon^k$ vanishes, $\varepsilon^0 \equiv 0$; and then the Lorentz condition (6.65) reduces to
$$
\boldsymbol{\varepsilon} \cdot \mathbf{q}=0 .
$$
This (noncovariant) choice of gauge is known as the Coulomb gauge.
From (6.67), we see that there are only two independent polarization vectors and that they are both transverse to the three-momentum of the photon. For example, for a photon traveling along the $z$ axis, we may take
$$
\boldsymbol{\varepsilon}_1=(1,0,0), \quad \varepsilon_2=(0,1,0) .
$$
A free photon is thus described by its momentum $q$ and a polarization vector $\varepsilon_i$. Since $\varepsilon_i$ transforms as a vector, we anticipate that it is associated with a particle of spin 1.