What is the non-relativistic limit of the quantised electromagnetic field? I’m not a physicist so this question may be naive ... For a real scalar field, quantisation yields the Klein-Gordon equation and the non-relativistic limit of this gives the Schrödinger field. What is the equivalent equation or field starting from the electromagnetic field? I.e. quantising the EM field (E- and B-fields) and dispensing with Lorentz covariance gives what kind of equation or field? I.e. what’s the equivalent of the Schrödinger equation/field for an EM field (rather than a scalar field).
I have an application that observes the structure of the EM field but not Lorentz covariance.
 A: As electromagnetic fields in vacuum always move at speed $c$ there is no nonrelativistic approximation to electromagnetism. In other words photons have zero rest mass and therefore no rest frame.
A: First we have to distinguish between first quantisation and second quantisation. In the first quantisation physical observables like momentum, position and energy become operators, mostly differential operators.
However, in many occasions physicists will consider the Klein-Gordon equation as classical equation for a field $\psi$ that in the second quantisation gets the status of an operator. To reduce the possible confusion: We won't talk about second quantisation here. In the following we will only talk about the "first" quantisation.
What happens in the first quantisation? Generally it consists of considering an energy-momentum relation for a particle as a dispersion-relation of a wave.
So in non-relativistic mechanics the energy momentum-relation of a particle in a potential $U$ is:
$$E = \frac{p^2}{2m} + U$$
We know in particular from the Davisson-Germer experiment that particles can also behave as waves. In order to find the wave equation that belongs to 1-particle motion one considers the energy momentum-relation as dispersion relation of the wave with wave vector $k$:
$$E = \frac{\hbar^2 k^2}{2m} + U$$
be replacing energy by $i\hbar\frac{\partial}{\partial t}$ and the wave vector by $-i\frac{\partial}{\partial x}$ we immediately get the Schroedinger-equation.
$$ i\hbar \frac{\partial}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2 x}  + U$$
The replacement of energy and wave vector by differential operators can be considered as the first-quantisation process. Because now we deal with (differential) operators for the description of the physical quantities.
For the Klein-Gordon-equation is the same. However, one now considers the relativistic energy-momentum-relation (without potential)of a particle:
$$E^2 = (pc)^2 + (mc^2)^2$$
Replacing  energy by $i\hbar\frac{\partial}{\partial t}$ and momentum by $-i\hbar\frac{\partial}{\partial x}$ we get the Klein-Gordon-equation:
$$\frac{1}{c^2}\frac{\partial^2}{\partial^2 t} = \frac{\partial^2}{\partial^2 x} - \frac{m^2 c^2}{\hbar^2} \quad \longrightarrow \frac{1}{c^2}\frac{\partial^2\psi}{\partial^2 t} - \frac{\partial^2\psi}{\partial^2 x} +\frac{m^2 c^2}{\hbar^2}\psi=0  $$
Of course, one lets operate the operators on some wave field or in non-relativistic physics on a wave function $\psi$ that propagates according to the corresponding dispersion relation.
How about the electromagnetic field (EM) ? Actually, in order to make it understandable one only has to put in the right context. Actually, the quantisation process here goes in the other direction, since both precedent cases considered a particle motion which was translated into a wave-motion. However, for the EM-field it is the other way around. One already has a wave-motion (at least this is the way one first gets to know the electromagnetic field) and associates a particle motion to it. This step is formally realised in the second quantisation but because  here we talk about first quantisation it is out of scope. But in order to make the analogy to the two precedent cases perfect we can start from the particle picture of the EM-field, the photons, and get back the corresponding wave equation in the same way as in the two precedent cases. The energy-momentum relation of a photon is:
$$ E=pc \quad\quad \text{or better in this context} \quad\quad E^2 = (pc)^2$$
This is actually a special case of the Klein-Gordon-equation with mass $m=0$.
Therefore by considering the energy-momentum relation as a dispersion relation of a wave followed by the replacement of energy and momentum by the corresponding differential operators ($E = i\hbar \frac{\partial }{\partial t}$ and $p= -i\hbar \frac{\partial}{\partial x}$) we get as wave-equation for the electromagnetic field:
$$\frac{1}{c^2}\frac{\partial^2}{\partial^2 t} = \frac{\partial^2}{\partial^2 x}$$
The is a well-known relation, it actually applies to the for components of the 4-vector potential:
$$\frac{1}{c^2}\frac{\partial^2 A^\mu}{\partial^2 t} = \frac{\partial^2 A^\mu}{\partial^2 x}$$
So we could actually conclude that the wave-function of the photon is the vector potential. Attention, the equation-analogy is one thing, the interpretations of these equations another. Yes one could say to some extent the vector potential is the wave-function of the photon. However, it does not have all the properties of the wave-function that solves the Schroedinger-equation. This is mainly due to the relativistic aspect of the wave-equation of the photon, details are here out of scope.
By the way, the wave-equation of the photon is perfectly Lorentz-covariant.
One might wonder about the right number of degrees of freedom of the photon field, this question is here also out of scope (see gauge freedom).
