# How wrong are the classical Maxwell's equations (as compared to QED)?

Now, I don't really mean to say that Maxwell's equations are wrong. I know Maxwell's equations are very accurate when it comes to predicting physical phenomena, but going through high school and now in college, Maxwell's equations are seen as the equations of electricity and magnetism. Now, it's common knowledge among students that, while Newton's laws are generally accurate when applied to everyday experiences, they are also replaced at high velocities by special relativity (and general relativity for very large gravitational fields).

But this is less so the case with Maxwell's equations. I have read that Maxwell's equations are replaced with quantum electrodynamics (which to me has all the effect of mere buzzwords, since I don't know what quantum electrodynamics is) as a more accurate way of describing electromagnetic waves, but what are the limitations of Maxwell's equations?

Or let me phrase this differently. I'm currently an electrical engineering major. I know NASA scientists and engineers can still get away with using Newtonian physics for their calculations, because it's that accurate. I also know, however, that relativity does have to come into play with GPS. So, in what situation as an electrical engineer would Maxwell's equations ever fail me? When (assuming I'm working on such a sufficiently advanced project) would I have to resort to more accurate ways to describe electromagnetic waves?

Maxwellian electrodynamics fails when quantum mechanical phenomena are involved, in the same way that Newtonian mechanics needs to be replaced in that regime by quantum mechanics. Maxwell's equations don't really "fail", as there is still an equivalent version in QM, it's just the mechanics itself that changes.

Let me elaborate on that one for a bit. In Newtonian mechanics, you had a time-dependent position and momentum, $x(t)$ and $p(t)$ for your particle. In quantum mechanics, the dynamical state is transferred to the quantum state $\psi$, whose closest classical analogue is a probability density in phase space in Liouvillian mechanics. There are two different "pictures" in quantum mechanics, which are exactly equivalent.

• In the Schrödinger picture, the dynamical evolution is encoded in the quantum state $\psi$, which evolves in time according to the Schrödinger equation. The position and momentum are replaced by static operators $\hat x$ and $\hat p$ which act on $\psi$; this action can be used to find the expected value, and other statistics, of any measurement of position or momentum.

• In the Heisenberg picture, the quantum state is fixed, and it is the operators of all the dynamical variables, including position and momentum, that evolve in time, via the Heisenberg equation.

In the simplest version of quantum electrodynamics, and in particular when no relativistic phenomena are involved, Maxwell's equations continue to hold: they are exactly the Heisenberg equations on the electric and magnetic fields, which are now operators on the system's state $\psi$. Thus, you're formally still "using" the Maxwell equations, but this is rather misleading as the mechanics around it is completely different. (Also, you tend to work on the Schrödinger picture, but that's beside the point.)

This regime is used to describe experiments that require the field itself to be quantized, such as Hong-Ou-Mandel interferometry or experiments where the field is measurably entangled with matter. There is also a large gray area of experiments which are usefully described with this formalism but do not actually require a quantized EM field, such as the examples mentioned by Anna. (Thus, for example, black-body radiation can be explained equally well with discrete energy levels on the emitters rather than the radiation.)

This regime was, until recently, pretty much confined to optical physics, so it wasn't really something an electrical engineer would need to worry about. That has begun to change with the introduction of circuit QED, which is the study of superconducting circuits which exhibit quantum behaviour. This is an exciting new research field and it's one of our best bets for building a quantum computer (or, depending on who you ask, the model used by the one quantum computer that's already built. ish.), so it's something to look at if you're looking at career options ;).

The really crazy stuff comes in when you push electrodynamics into regimes which are both quantum and relativistic (where "relativistic" means that the frequency $\nu$ of the EM radiation is bigger than $c^2/h$ times the mass of all relevant material particles). Here quantum mechanics also changes, and becomes what's known as quantum field theory, and this introduces a number of different phenomena. In particular, the number of particles may change over time, so you can put a photon in a box and come back to find an electron and a positron (which wouldn't happen in classical EM).

Again, here the problem is not EM itself, but rather the mechanics around it. QFT is built around a concept called the action, which completely determines the dynamics. You can also build classical mechanics around the action, and the action for quantum electrodynamics is formally identical to that of classical electrodynamics.

This regime includes pair creation and annihilation phenomena, and also things like photon-photon scattering, which do seem at odds with classical EM. For example, you can produce two gamma-ray beams and make them intersect, and they will scatter off each other slightly. This is inconsistent with the superposition principle of classical EM, as it breaks linearity, so you could say that the Maxwell equations have failed - but, as I pointed out, it's a bit more subtle than that.

• Is the way classical EM fails in quantum and relativistic the same as the way Newtonian fails in such cases? Mar 11 '15 at 0:48
• @Ooker Essentially yes, though it's hard to compare Newtonian mechanics with full-blown QFT. EM is, mostly, preserved better than other theories in those two transitions. Mar 11 '15 at 11:34

Anna V is wrong when she says Maxwell's equations are inconsistent with black body radiation. In quantum mechanics, even if you ignore radiation there is a charge density which you can calculate (in principle) from Schroedinger's Equation. In a warm body, this charge density is fluctuating by random thermal motion. If you track the time evolution of the charge density, and then apply Maxwell's equations, you will get the correct black body radiation spectrum. You don't have to throw out Maxwell's equations, and you don't need to quantize the energy.

Anna's other examples are similarly false. There is nothing in Maxwell's equations that predicts a continuous spectrum for atoms. You don't use Maxwell's equations to calculate the motion of charge within an atom, you use Schroedinger's equation. From there, you get oscillating charge distributions. If you then use Maxwell's equations to calculate the radiation spectrum, you get exactly the right answer, right down to the linewidths.

Same with lasers. Schroedinger's equation gives you the "population inversions" etc, but the resulting waves resulting from the atomic oscillations are entirely classical: especially the way one resonator, stimulating a nearby resonator, drives it in phase so that the resulting wave is coherent. All comes out of Maxwell's equations. Not the atomic states of course...those come from Schroedinger. But the resulting radiation is all Maxwell.

EDIT: Ruslan has asked (in the comments) how I can calculate the linewidths using just Schroedinger + Maxwell. It's a calculation I've done in my blog. which you can find here. What is critical to understand is that while we like to talk about things like "a hydrogen atom in the pure excited state", what we are actually dealing with normally in an experimental situation is an ensemble of a million hydrogen atoms, some (eg. 1%) of which are in the excited state (100%). In that case I can't do the calculation. But there is no experimental way to distinguish this hypothetical ensemble from an alternate ensemble where all (100%) of the atoms are in a 1% excited state. And then the calculation is obvious.

If you think these states are distinguishable, try writing the density matrix for them.

• I think Anna is discussing the Lamb shift, not the atomic spectra. If you have a way of calculating the lamb shift without invoking QFT, I'm all ears. Mar 10 '15 at 21:22
• Can we see a derivation of the Planck blackbody spectrum without quantization of energy levels? Mar 10 '15 at 23:04
• I think in the first paragraph where I said you don't need to quantize the energy, I meant the e-m field energy. You still get discrete energy levels in the mechanical system when you solve the Schroedinger equation. Those states have stationary charge distributions. But any system in a superposition of those discrete states will have an oscillating charge density, and that oscillating charge emits and aborbs radiation strictly according to classical antenna theory. Mar 11 '15 at 2:14
• But once you rely on Schroedinger's equation you have changed the framework on which you apply Maxwell's equations. The disagreement from the classical calculations using EM uncovered the existence of a deeper framework. The classical interpretations of ME is what fails. You are manipulating the meaning of "failure". Applying classical electromagnetism solutions to the problem of orbiting charges cannot give a line spectrum: the Bohr model was devised. The same with BBR. all these were the impulse for the underlying the quantum nature. The question is from an engineer. @JerrySchirmer Mar 11 '15 at 4:47
• @JerrySchirmer, for a model capable of manifesting Lamb shift that is not quantum field theoretical, see the paper by E. T. Jaynes referenced here: en.wikipedia.org/wiki/… The paper: bayes.wustl.edu/etj/articles/prob.in.qm.pdf Oct 4 '17 at 16:22

It's funny that all the answers so far forgot the simple and elegant following criterion : quantum mechanics appears when $$\dfrac{\hbar\omega}{k_{B}T} > 1$$

with $h=2\pi\hbar\approx6,63.10^{-34}\text{J}\cdot\text{s}$ the Planck constant and $k_{B}\approx1,38.10^{-23}\text{J}\cdot\text{K}^{-1}$ the Boltzmann constant, $\omega$ and $T$ being the (angular) frequency and the temperature, respectively.

The criterion comes into play as follows : the Planck constant is the characteristic energy (to frequency) scale of the quantum world, and the Boltzmann constant is the characteristic energy (to temperature) scale of the statistical world.

In electrical circuits, most of the properties one is interested in invoke a flock of electrons flowing from one point to an other. Actually, the number of electrons flowing is pretty huge, say $\left(10^{10}-10^{30}\right)$ electrons (large window). The branch of physics discussing many-body problems is the statistical physics. In contrary, quantum mechanics (especially quantum optics) was constructed with one-body problems in head (one electron orbiting around a nucleus, one particle tunnelling, ...). This is still true (but tend to be less prominent) in high-energy physics. The branch of physics dealing with quantum property of the bunch of electrons flowing in matter is called condensed matter physics. In fact, one can not understand most of the materials properties without quantum mechanics (band theory, optical response of materials, ...) So every electrical engineer use quantum mechanics, the point is that they use statistical averaging of these quantum properties, which behave quite well as in the classical regime. Say differently, it might well be quantum effects introduced in the calculation of the capacitance, inductance and resistance of materials, but once you know these quantities, you can simply use Kirchhoff's laws.

Thats the most important secret of physics : a theory is always effective. Kirchhof's laws are correct at high-temperatures (but not too high, say at room temperature) and for quasi-static phenomenon (quite low frequencies). That's how hey are derived from Maxwell's theory.

A property of matter you can not explain without quantum mechanics, but is well known in microwave regime is the magnetism. Yet it exists a circuit theory for these materials called magnetic circuit theory, which really looks like the Maxwell's equations. But to calculate the macroscopic quantities the materials exhibit you need quantum mechanics ... or to measure and tabulate them !

As an historical perspective, when people tried to apply the same rules to the one-body problems and to the many-body problems (and also the generalisation to relativistic problems), they start constructing the quantum-field-theory. Most of the answers given so far discuss essentially the quantum-field-theory in vacuum, when in fact Maxwell's equations are always valid, provided you define the electric and magnetic fields as average over the quantum properties. Indeed, sometimes you have to quantise the exchange of excitations between isolated object and the fields (Anna gave some examples of this quantisation in her answer). The Maxwell's equations are recovered when you sum-up many one-particle configuration (to say it quickly). Also sometimes this quantisation requires to generalise the symmetry of the fields, there are then generalisation of the electric and magnetic fields, called non-Abelian gauge-fields in that context. But we are then far from the description of electrons in matter, since these theories are required to discuss elementary constituents of the nucleus, some energy scales electrical engineering is not concerned with, at all ! (I cross my finger since we should never bet on technological future, say at least for the next few decades ... )

When you reduce the number of particles involved in circuits, when you decrease temperature, and when you increase the frequency, the systems start to behave differently than in the Kirchhof's laws. In a way, all the quantum properties which were hidden in the statistical averaging discussed above start to become more and more prominent.

The discussion of electronic properties at low temperatures (a few Kelvin), low dimensional scales (nanometer / micrometer scale) and high frequencies (megaHertz / gigaHertz ; for opticians it's not that high, but for electrical engineers it is) is the central topics of what is called mesoscopic physics, an intermediary physics between classical and quantum worlds. The circuit QED mentioned by Emilio Pisanty in his answer on this page is just a sub-topic of mesoscopic physics. It might well be the "most-quantum" one, since the electronic properties of superconductors can not be described by classical physics, though you can correct the Maxwell's equations to discuss the electromagnetism of superconductors. This set of modified laws are called London's equations, the effective theory for superconducting circuits, which can not explain the Josephson effect, though. Exploring a bit these laws, you will see that they fail to be gauge-invariant ... something you can not explain without the big apparatus of quantum-field-theory. It took 10-20 years to explain the London's theory from microscopic calculation, and again 10 years to relate the failure to gauge-invariance to the Anderson-Higgs-mechanism. In the mean-time, electrical-engineers were applying the London's equations and were tabulating the superconductors with great accuracy !

This answer got far longer than what I thought when starting writing, certainly because the subject is fascinating :-)

I'd like to give you an other perspectives about Maxwell's equations and effective theory. A few decades after they appeared, people thought the Maxwell's equations were unifying all the radiations possible ever. In fact, they unified all the radiations known at that time, which were already a big step, isn't it ? So its the interpretation of universality naively associated to the Maxwell's equations which failed, not the Maxwell's equations themselves, since they were discussed in a clear historical and experimental context. Maxwell's equations have to be corrected (or radically changed) when going to high frequencies (which means : at high-energies), or when going to low temperatures (which means : at low-energy). One more subtle point (I hope I explain it carefully above) is : Maxwell's equations also fail to describe individual object (say, individual photons). But clearly they were never designed for this... and in fact that's all the mechanical description which has to be changed to go to this limit. Emilio discussed this in details.

An important point left over in the discussion above is : of course the big merit of Maxwell was to unify electricity and magnetism, not the the radiations as I said above. In fact radiation comes naturally from the electromagnetic theory. As a final remark, you could also say that Maxwell's equations define electromagnetism. That's the common point-of-view, and that's why people are talking about quantum-electromagnetism when they use the quantum version of the Maxwell's equations, ...

Classical electromagnetic waves emerge from the underlying quantum electrodynamic description in a smooth and consistent manner. The quantum framework means that the classical waves are built out of photons, and the only time one has to worry about more detail than what Maxwell's equations provide is at the level of particle physics and wherever quantization has been found to eliminate discrepancies of classical predictions with data.

1. the atomic spectra that display lines unexpected from the smooth solutions of ME

2. the black body radiation that needs quantization of energy

3. lasing and similar phenomena dependent on atomic physics

4. situations where photons (particles) are energetic enough for creation and annihilation of particles,have measurable photon photon interactions, etc which cannot be described with solutions of ME.

Classical solutions of Maxwell's equations are fine for most situations .

Edit: Looking at the discussion in the answer by Marty Green, where he insists that Maxwells equations do not fail by using the Schrodinger atom and hand waving the emission and absorption on the lines, I think that in the complete answer by Emilio Pisanty above this is cleared:

In the simplest version of quantum electrodynamics, and in particular when no relativistic phenomena are involved, Maxwell's equations continue to hold: they are exactly the Heisenberg equations on the electric and magnetic fields, which are now operators on the system's state ψ. Thus, you're formally still "using" the Maxwell equations, but this is rather misleading as the mechanics around it is completely different.

The form is the same but the way it is applied in the quantum regime is drastically different, as different as differential operators are (quantum) from real variables (classical).

The quote elucidates what I handwaved as "Classical electromagnetic waves emerge from the underlying quantum electrodynamic description in a smooth and consistent manner."

EDIT : As this was declared the duplicate of a recent question I want to add two links. The Maxwell wavefunction of the photon is given here., Lubos Motl has this blog entry on How classical fields, particles emerge from quantum theory

The problem is with one charge in empty space, even it is run with acceleration, it still cannot sent out the wave. It cannot sent out the radiation. For an emitter to send radiation wave out, it need an absorber to receive the wave. This is what the Wheeler-Feynman's absorber theory tell us.

If you assume one charge can always send radiation wave out even without an absorber, you make mistakes. According to the Maxwell's theory, the source current always sends wave out, that is not true.

According to "mutual energy principle" and "self-energy principle", only when the retarded wave can find a matched advanced wave, a photon can be sent out. That means only the retarded wave and advance wave synchronized, the radiation can be produced.

Hence there must two group of Maxwell equations worked together, one is for the retarded wave, another is for the advanced wave.

Without a matched advanced wave, the retarded wave perhaps is also sent out but it returns with a time-reversal process.

Conclusion: Maxwell equations are only true with some probability, depending whether or not a male wave can marry with a female wave. That is the reason in QET, the wave is probability wave.

See my publication: http://www.openscienceonline.com/journal/archive2?journalId=726&paperId=4042