I define the photon wave function in a covariant formulation which has four polarisation states, two of which are not observable. Some authors use only transverse states, but the other two states would appear on Lorentz transformation, and they appear to be necessary to derive the classical correspondence correctly.
For momentum $p=(P^0,\mathbf p)$, define a longitudinal unit 3-vector,
$$\mathbf w(\mathbf p,3) = \frac {\mathbf p} {|\mathbf p |} $$
and orthogonal transverse unit 3-vectors $\mathbf w(\mathbf p,1),\mathbf w(\mathbf p,2)$ such that for $r,s= 1,2,3$
$$ \mathbf w(\mathbf p,r) \cdot \mathbf w(\mathbf p,s) = \delta_{rs} $$
Define normalised spin vectors,
$$ \mathbf w(\mathbf p,0) = (1,\mathbf 0)$$
$$ \mathbf w(\mathbf p,r) = (0,\mathbf w(\mathbf p,r))$$
For momentum $p$ a photon plane wave state is given by the wave function,
$$\langle x|p,r\rangle = \lambda(|\mathbf p |,r) w(\mathbf p,r) e^{-ix \cdot p} $$
where $p^2 = 0$ and $ \lambda $ is determined by relativistic considerations
$$ \lambda(|\mathbf p|, r) = \frac 1{(2\pi)^{3/2}} \frac 1{\sqrt{2p^0}} $$
You can then express a photon wave function as an integral
$$f^a(x) = \frac 1{(2\pi)^{3/2}} = \sum\limits_{r=0}^3 \int \frac{d^3\mathbf p}{\sqrt{2p^0}} \mathbf w(\mathbf p,r) e^{-ix \cdot p} \langle \mathbf p, r|f\rangle$$
I took this from lecture notes at Cambridge, and I am not sure which books do things much the same way (there are some normalisation choices, as well as choice of gauge). I have given more detail in A Construction of Full QED Using Finite Dimensional Hilbert Space and in The Mathematics of Gravity and Quanta