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 SchrΓΆdinger 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 SchrΓΆdinger 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.


3 Answers 3


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.


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).


Before you can ask what the quantum theory would look like, you should be asking first what the classical theory would be! Particularly: since field theories are based on the action principle and Lagrangians, what does the Lagrangian and action principle for the classical theory look like? This is essential, because a quantum field theory is formulated in terms of its classical version, and this process requires that there be an action principle and Lagrangian.

Dispensing with Lorentz covariance gives you a field, and a dynamics for it, that is no longer Lorentz covariant. So, you're falling back to a smaller set of symmetries to base covariance on.

The ones that Maxwell used in formulating his field theory were symmetry under spatial translation and time translation; and this is ensured, more or less automatically, by framing the field's dynamics in terms of differential equations. Maxwell's equations are translation-invariant.

He also focused, almost exclusively, on fields whose dynamics were isotropic. What I'm referring to, specifically, are the constitutive laws $𝐃 = Ρ𝐄$ and $𝐁 = μ𝐇$ that link the electric displacement $𝐃$ and magnetic field $𝐇$ to the magnetic induction $𝐁$ and electric field $𝐄$. When isotropic, the two sets of fields are aligned and in fixed ratios ($Ξ΅$ for permittivity, and $ΞΌ$ for permeability) with no directional skewing between $𝐃$ and $𝐄$, or between $𝐁$ and $𝐇$.

The isotropy condition is where issues start arising. Generally speaking, isotropy is a condition that can only hold in one frame of reference. What appears isotropic to a stationary observer may not be so, to a moving observer. More precisely: to an observer moving with a velocity $𝐆$ the isotropy condition (in non-relativistic theory) would take on the form $𝐃 = Ξ΅(𝐄 + 𝐆×𝐁)$ and $𝐁 = ΞΌ(𝐇 - 𝐆×𝐃)$. Now you may have $𝐃$ versus $𝐄$ skewing and $𝐁$ versus $𝐇$ skewing.

The basic reason is that $𝐁$ and $𝐃$ have the same appearance for moving observers as they do for stationary observers (in non-relativistic theory), while $𝐄$ and $𝐇$ do not. Maxwell actually fudged this a little by sticking the velocity-dependent term $𝐆×𝐁$ into his definition of $𝐄$. So his $𝐄$ is actually our $𝐄 + 𝐆×𝐁$; but failed to do a similar thing with $𝐇$ and apparently failed to note that it transforms to $𝐇 - 𝐆×𝐃$ for an observer moving with velocity $𝐆$.

As such, using the isotropy condition ties you down to a specific frame of reference. As a consequence, a distinction was drawn - before the 20th century and up into the early part of it - between the "stationary" versus "moving" form of Maxwell's dynamics. The "stationary" term means: stationary with respect to the frame of isotropy. The "moving" term, here, was the same meaning and sense of the word in the title of Einstein's 1905 paper "On the electrodynamics of moving bodies". That's what the word was actually referring to.

So, your Lagrangian and action principle ... and your field dynamics ... will now only have covariance with respect to spatial and time translations and with respect to rotations; i.e. reorientation of spatial directions; but not with respect to "boosts". A "boost" is the name given for the transformation from one frame of reference to another moving with respect to it.

That means you have to treat even the vacuum of outer space as something that has a reference frame attached to it. For non-relativistic theory, that forces you into a conundrum. The vacuum has to be treated as a medium of some kind that contains a reference frame and attributes of a medium: the vacuum permittivity $Ξ΅_0$ and permeability $ΞΌ_0$, along with a natural speed $1/\sqrt{Ξ΅_0ΞΌ_0}$ for electromagnetic waves traveling through it, with the speed taken relative to the distinguished frame of reference.

In relativity, you actually still have a similar issue. In general, the isotropy condition does not remain so to a moving observer, and the isotropy condition would have the appearance $𝐃 + (1/c)^2 𝐆×𝐇 = Ξ΅(𝐄 + 𝐆×𝐁)$ and $𝐁 - (1/c)^2 𝐆×𝐄 = ΞΌ(𝐇 - 𝐆×𝐃)$. That's, today, known as the Maxwell-Minkowski relations. The earlier-cited relations are the non-relativistic versions of them ... and were equivalent to what Lorentz posed, when he wrote down his equations for a moving media. No likeness of any of the $(1/c)^2$ terms was present in his equations.

The reason Relativity gets away with having boost-symmetry is that these equations are equivalent to the stationary form $𝐃 = Ρ𝐄$ and $𝐁 = μ𝐇$, at least for observers moving with speed $|𝐆| < V$, if $Ξ΅$ and $ΞΌ$ take on values $Ξ΅ = Ξ΅_0$ and $ΞΌ = ΞΌ_0$ whose corresponding wave speed $V = 1/\sqrt{Ξ΅_0ΞΌ_0}$ is equal to light speed $c$. Just rewrite $(1/c)^2 = Ξ΅_0ΞΌ_0$ and substitute $$𝐃 + Ξ΅_0ΞΌ_0 𝐆×𝐇 = Ξ΅_0(𝐄 + 𝐆×𝐁), \hspace 1em 𝐁 - Ξ΅_0ΞΌ_0 𝐆×𝐄 = ΞΌ_0(𝐇 - 𝐆×𝐃)\\ 𝐃 - Ξ΅_0𝐄 = Ξ΅_0 𝐆×\left(𝐁 - ΞΌ_0𝐇\right), \hspace 1em 𝐁 - ΞΌ_0𝐇 = -ΞΌ_0𝐆×\left(𝐃 - Ξ΅_0𝐄\right) $$ Equations of the form $𝐚 = Ξ΅_0 𝐆×𝐛$ and $𝐛 = -ΞΌ_0 π†Γ—πš$ can only have the solution $𝐚 = 𝟎$ and $𝐛 = 𝟎$, if $|𝐆| < V$, because they imply $|𝐚| ≀ Ξ΅_0 |𝐆||𝐛|$ and $|𝐛| ≀ ΞΌ_0|𝐆||𝐚|$ or that $|𝐚||𝐛| ≀ Ξ΅_0ΞΌ_0|𝐆|^2|𝐚||𝐛|$, which is not possible if $Ξ΅_0ΞΌ_0|𝐆|^2 < 1$, unless $|𝐚||𝐛| = 0$. Therefore, applied to the situation above, this means $𝐃 = Ξ΅_0𝐄$ and $𝐁 = ΞΌ_0𝐇$.

So, in Relativity, you can get away with declaring that the dynamics are boost-invariant, in a vacuum, without requiring any specific frame of reference, provided you have $V = c$, i.e. $Ξ΅_0ΞΌ_0 = (1/c)^2$, and you recover the full set of Lorentz symmetries. In Relativity, $c$ is an absolute speed, not relative. (And, yes, Einstein noted the irony of the exonym "Relativity", which is why he originally rejected that name.)

However, "boost", here, is referring to the Lorentz transforms, not to the Galilean transforms.

For Galilean boosts, the only way you can get boost-invariance is for media in which $Ρμ = 0$, which completely destroys the viability of the formulation of any sensible dynamics. So, you're stuck with a field dynamics that breaks boost-symmetry and calls out a single frame of reference.

A Lagrangian formulation for the dynamics would then require a Lagrangian function $𝔏(𝐁,𝐄,β‹―)$, that is a function of the $𝐁$, $𝐄$ fields, as well as other fields. As is already the case, the fields are to be udnerstood as as expressions taken in terms of their corresponding potentials $𝐁 = βˆ‡Γ—π€$, $𝐄 = -βˆ‡Ο† - βˆ‚π€/βˆ‚t$, where $Ο†$ is the electric potential and $𝐀$ the magnetic potential.

The other fields are defined in terms of these by: $$𝐃 = \frac{βˆ‚π”}{βˆ‚π„}, \hspace 1em 𝐇 = -\frac{βˆ‚π”}{βˆ‚π},$$ and application of the principle of least action, with this Lagrangian will then produce the source-free Maxwell equations $$βˆ‡Β·πƒ = ρ = 0, \hspace 1em βˆ‡Γ—π‡ - \frac{βˆ‚πƒ}{βˆ‚t} = 𝐉 = 𝟎$$ as a result. If taking into account the other fields that are present, they may involve the potentials $Ο†$ and $𝐀$ in them, so that the Lagrangian also acquires an implied dependency on the potentials. In that case, one would also have expressions: $$ρ = -\frac{βˆ‚π”}{βˆ‚Ο†}, \hspace 1em 𝐉 = \frac{βˆ‚π”}{βˆ‚π€},$$ for the charge density $ρ$ and current density $𝐉$.

However, a quantum field theory is normally only formulated for the source-free field. The one which Quantum Electrodynamics (or QED) is based on is the one that produces the above-mentioned laws, $𝐃 = Ξ΅_0𝐄$ and $𝐁 = ΞΌ_0𝐇$, with zero sources: $𝐉 = 𝟎$ and $ρ = 0$, and that's the Maxwell-Lorentz Lagrangian: $𝔏 = Β½(Ξ΅_0 |𝐄|^2 - |𝐁|^2/ΞΌ_0)$.

It has the full set of symmetries, including boost-symmetry, while ours can't. Instead, the best we can do is include isotropy as a condition. That means there should be no directional-dependence on any of the fields, thereby forcing the Lagrangian to be a function of only the scalar combinations of the fields. There are three of them, which we'll label: $$β„‘ = \frac{|𝐄|^2}2, \hspace 1em 𝔍 = 𝐄·𝐁, \hspace 1em π”Ž = \frac{|𝐁|^2}2.$$ The simplest case, and one which produces linear relations between the fields with constant coefficients, is: $$𝔏 = Ξ΅ β„‘ + ΞΈ 𝔍 - \frac{1}{ΞΌ} π”Ž,$$ with a fudge used in the last term to ensure backwards-compatibility. The corresponding relations derived from this would be: $$𝐃 = Ξ΅ 𝐄 + ΞΈ 𝐁, \hspace 1em 𝐇 = \frac{𝐁}{ΞΌ} - ΞΈ 𝐄, \hspace 1em 𝐉 = 𝟎, \hspace 1em ρ = 0.$$ In the formulation, it is assumed that $Ξ΅ > 0$, $ΞΌ > 0$, and that the wave speed $V = 1/\sqrt{Ρμ}$ is just some speed, and that's it.

The $ΞΈ$ term will actually be superfluous, if $ΞΈ$ is a constant, since you could just redefine $𝐃$ as $𝐃 - ΞΈ 𝐁$ and $𝐇$ as $𝐇 + ΞΈ 𝐄$ and pretend that those were the actual $𝐃$ and $𝐇$ all along. It won't affect any of Maxwell's equations; since either alternative satisfies the very same equations. The coefficient $ΞΈ$ would only be relevant if it were a function of position and time. In fact, that's what the axion field is; only the Lagrangian for axion dynamics has additional terms for $ΞΈ$ in it.

So, here's the thing: this Lagrangian density is not boost-invariant. These expressions are tied specifically to the isotropy frame and are only the "stationary form". To make this applicable to all frames requires explicitly including the velocity vector $𝐆$. That means making the replacements for the various fields, from stationary observers to moving observers, $𝐄 β†’ 𝐄 + 𝐆×𝐁$ so that the actual Lagrangian density would read as follows: $$\begin{align} 𝔏 &= \frac{Ξ΅ {|𝐄 + 𝐆×𝐁|}^2}{2} + ΞΈ (𝐄 + 𝐆×𝐁)·𝐁 - \frac{|𝐁|^2}{2ΞΌ} \\ &= \frac{Ξ΅ {|𝐄 + 𝐆×𝐁|}^2}{2} + ΞΈ 𝐄·𝐁 - \frac{|𝐁|^2}{2ΞΌ}, \end{align}$$ or if the $ΞΈ$ term is removed: $$𝔏 = \frac{Ξ΅ {|𝐄 + 𝐆×𝐁|}^2}{2} - \frac{|𝐁|^2}{2ΞΌ}.$$

You could probably make a quantum field theory off of that. Its waves will be traveling at the speed $V = 1/\sqrt{Ρμ}$, relative to the frame in which $𝐆 = 𝟎$. In other frames, light waves will propagate from a point outward in a sphere whose center is moving with speed $𝐆$.

None of this is actually tied specifically to non-relativistic theory, because you could also do the same thing for the relativistic theory of moving media for the previously-mentioned Maxwell-Minkowski relations. The Lagrangian will have an explicitly $𝐆$-dependent form similar to that above, but more involved, because of the extra $(1/c)^2$ terms. This would give rise to an effective quantum theory for the electromagnetic field in the medium.

Oh, and by the way: field theories with reference velocities like this ... that's an active area of research.

Einstein-Aether Theory

Even luminaries like Jacobson have written papers on it. The only difference is, their velocity field is not fixed but is included in Lagrangian, too, and in the dynamics and affected by the dynamics.


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