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Useful quantum field theories like quantum electrodynamics (QED) suffer from a litany of problems related to the fact that, at least in their usual Lagrangian formulation, interactions between the fields, here a matter field and electromagnetic field, must involve problematic products of operator distributions, which are not mathematically well-defined.

But what if we just admit the Lagrangian is junk and pitch it? How far can we go without it? What I'm thinking is that Maxwell's equations don't involve any product of fields. What stops us from just "quantizing" them directly, by promoting the various terms involved to quantum operators?

What I mean by that is this. A compact relativistic formulation of Maxwell's equations is

$$\partial_\nu \partial^\nu A^\mu = \mu_0 J^\mu$$

where $A^\mu$ is the space-time potential field and $J^\mu$ is the space-time current. What is stopping us from "putting hats" on those like thus:

$$\partial_\nu \partial^\nu \hat{A}^\mu = \mu_0 \hat{J}^\mu$$

that is to say, promoting the fields directly to quantum operators, just like that, and using this equation to do physics?

I suppose one problem is we cannot then account for back-reaction of the EM fields against the charge field, viz. pair-production of electrons and positrons from photons, which is one of the phenomena we would ideally like to be able to account for in the same framework. But as said, I'm not claiming this a complete replacement for regular QED, simply asking "how far can you go?" this way. Can we at least get some low-energy phenomena, i.e. where the energies involved are much less than the rest energy of an electron or positron, out of this?

Is the failure point something else? Is the equation inherently ill-defined? I.e. the partial derivative on the left cannot be had of a distributional field, so we're back in the same boat as with the Lagrangian? Or what? Or is it perfectly consistent, but ends up disagreeing with experimental data (more than "standard" QED) - and if so, why does it end up disagreeing despite that it seems on the surface like a perfectly reasonable quantum model of electromagnetism, following the same recipe you get in your intro QM textbook?

ADD: Now that I see the comments about the gauge, what about if we use the EMF tensor instead? Viz.

$$\partial_\mu \hat{F}^{\mu \nu} = \mu_0 \hat{J}^\nu$$

And then enforce the constraints for space-like position at least,

$$[\hat{E}^i(^{(4)} X), \hat{E}'^i(^{(4)}Y)] = i\hbar \delta({}^{(4)}Y - {}^{(4)} X)$$ $$[\hat{B}^i(^{(4)} X), \hat{B}'^i(^{(4)}Y)] = i\hbar \delta({}^{(4)}Y - {}^{(4)}X)$$

where $\hat{E}'^i$ and $\hat{B}'^i$ are the components of the conjugate electric and magnetic field, and $^{(4)}X$ and $^{(4)}Y$ is how I denote a space-time (viz. four-)point, with these commutator relations chosen on the basis that a free space wave should look like a free linear field? (Note that I'm unsure what to do about the diagonal components viz. $\hat{F}^{\mu \mu}$ because they're all zero in the Lorentz coordinates. Perhaps we can just assume the same, as well as that the state should always suitably render them constant and trivial.)

The_Sympathizer
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    That isn't really a full quantum theory -- this is like proposing to quantize the harmonic oscillator by writing $m \partial_t^2 x = - k x$ as $m \partial_t^2 \hat{x} = - k \hat{x}$. That doesn't actually fix $\hat{x}(t)$ unless you bring in the canonical momentum $\hat{p}$ as well. But when you try to do that with E&M you run into the same problems as in the standard QFT textbooks, which arise because $A_\mu$ is a gauge field. – knzhou Aug 29 '23 at 01:36
  • @knzhou : That's interesting. Can you explain more? I.e. don't we have a natural conjugate that we could assume from prior theory, like just the usual wave conjugate, knowing that in free space we should have linear EM waves? – The_Sympathizer Aug 29 '23 at 01:38
  • That said, you can develop your idea a bit further and you'll get a quantized EM field coupled to a classical source -- which is a perfectly fine theory. But it won't give you any of the signature effects of QED which made it famous, like the $g-2$ correction. – knzhou Aug 29 '23 at 01:38
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    Sure, to elaborate, if you go ahead and compute the conjugate momentum in the usual way, you'll find that the one for $A^0$ just vanishes. You can get rid of this problem by gauge fixing $A^0 = 0$, but then you lose Lorentz invariance. – knzhou Aug 29 '23 at 01:40
  • @knzhou: Ah yes, I see the problem now. Even looking at a classical EM wave, it is impossible to get $A^0$, which is the voltage component, unique, because the E-field is not conservative (the wave is carrying energy, after all), so there is no well-defined notion of "voltage" for a free-space EM wave. So basically the problem with quantizing EM is that the gauge redundancy, which is irrelevant in the classical realm, becomes unmasked in the quantum realm? – The_Sympathizer Aug 29 '23 at 02:15
  • Or to put it even more generally, quantization is simply the "Wrong way" to build a quantum EM theory, because quantum mechanics is richer than classical; we need information that the classical theory does not provide, period, and it's rather more luck that it worked as far as it did with simple non-relativistic systems. And there just isn't enough of that information we have to build a theory that can, say, account for a hydrogen atom in full quantum-relativistic glory simultaneously (i.e. assuming a scale of precision much below an electron Compton wavelength and fully Lorentz invariant) – The_Sympathizer Aug 29 '23 at 02:15
  • , mathematically rigorously. – The_Sympathizer Aug 29 '23 at 02:17
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    There are many misconceptions on this question. For example, Maxwell's equations are usually derived from assuming the Lagrangian for it, and we use the same Lagrangian for QED, just swapping the fields for operators. knzhou correctly points out that the derivatives will need reinterpretation. I want to point out that "problematic products of operator distributions" is a solved problem. Under causal perturbation theory, careful operator product splitting makes all the UV divergence go away, so all terms are well-defined. – naturallyInconsistent Aug 29 '23 at 02:17
  • @naturallyInconsistent : This could also serve as a good answer to the posed question. But is that "causal perturbation theory" a well-accepted theory? Does it provably (mathematically) match all experimental data for QED up to the limits of error? ADD: I see it actually goes back a long ways. Why then is there still debate as to the "rigorousness" of quantum field theories, if this has been there for a long time and it works as good as you say it does? – The_Sympathizer Aug 29 '23 at 02:20
  • FWIW I'd suggest both of yous expand these comments into answers to the question so I can accept one, as is customary for the site. thanks :) – The_Sympathizer Aug 29 '23 at 02:30
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    @naturallyInconsistent: Digging more though I note that your answer here - https://physics.stackexchange.com/q/768988/ - seems to present a rather less rosy picture of the situation than this comment would seem to imply. What's going on? (Note also that being able to solve the equations, by the way, I'd differentiate from the idea of them being well-defined in principle. A 3-body problem is "not easy to solve", but it is well-defined in principle. And it's the idea of WDIP that these lines of questioning are intended to explore.) – The_Sympathizer Aug 29 '23 at 03:01
  • @The_Sympathizer note that I specifically wrote that the perturbation series is naïvely treated, then it would be infinite. Now, I didn't have enough space earlier to write more, but that the UV divergence can be tamed for every term in the perturbation series, but the series can still sum to be divergent, which it has to, if the usual analysis is to work. However, it is not longer obvious that the divergence is due to some failure, or merely to mathematical silliness. For example, it is a known issue that a Taylor expansion of a convergent series of fractional power will diverge. – naturallyInconsistent Aug 29 '23 at 12:53
  • Even if your approach worked, it wouldn't really handle the key difficulties, even for QED. For an interacting theory you would need to couple your photon to some charged matter (eg, a scalar such as the $\pi^+$ particle or a spin-1/2 fermion such as the electron), then the equations of motion would be non-linear. We know to quantize a "free" theory (linear eoms / quadratic lagrangian / quadratic hamiltonian) exactly; the problem comes from including interactions. – Andrew Oct 02 '23 at 21:09
  • @knzhou: I came back to this again and am wondering. Why is it a problem if $A^0$ violates the Lorentz invariance if the gauge field is not observable, only the derived fields $E^i$ and $B^i$? Shouldn't the observable part of the theory still be fully Lorentz-invariant? Why would the gauge cause problems that cause the full theory to fail to give a well-defined state space and time evolution? Where exactly does it go wrong, i.e. at what step? – The_Sympathizer Jan 22 '24 at 05:23

2 Answers2

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What is stopping us from "putting hats" on those like thus: $$\partial_\nu \partial^\nu \hat{A}^\mu = \mu_0 \hat{J}^\mu$$ that is to say, promoting the fields directly to quantum operators, just like that, and using this equation to do physics?

That's not how quantum physics works. You've just written down an equation for some operator $\hat{A}^\mu$, but you can't really do any physics with this yet! Where do the states come from? How would you even start computing a scattering amplitude from just this?

Remember, on an abstract level, we get states as representations of canonical (anti-)commutation relations, which in turn are closely related to Lagrangians/Hamiltonians. You can't even get to what a wavefunction has to look like if you try to start normal QM from some equation like $\ddot{\hat{x}} = 0$. I think you can only claim that you can "do physics" with this if you can answer how we get to "states are wavefunctions $\psi(x)$" from that and show e.g. how an initial Gaussian $\psi(x)$ spreads with time.

Fundamentally, what we want from a QED-like theory - or any quantum theory - is to give us a notion of states, some of which we want to interpret at photons and (anti-)electrons, and a rule for how these states evolve in time and interact with each other. This is what quantum physics usually realizes via the time-evolution operator, which is formally something like $U(t) = \mathrm{e}^{\mathrm{i}\int H(t)\mathrm{d}t}$.

Phrased differently, a fundamental notion of quantum physics is time translation acting on arbitrary physical states, and the generator of time translation is the Hamiltonian. The "equation of motion" for the operators alone is pretty worthless since the solutions don't tell you anything about the states.

While the Heisenberg and the Schrödinger picture are equivalent, the Heisenberg picture needs a general equation that tells you the evolution equation for all operators - via the commutator with the Hamiltonian - not just the equation of motion for one particular operator.

ACuriousMind
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  • Thanks. Yeah, I did not include all that detail because I wasn't sure it'd be appropriate for the site but also because I wanted to just indicate a course or direction of approach to ask about, not a complete theory! I did add a bit more though just now, hopefully trying to clarify the approach I imagined for the EM fields at least. – The_Sympathizer Aug 29 '23 at 18:42
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What is really missing in your description is the Lorentz force. Note that by writing a Lagrangian you get both Maxwell's equation and the Lorentz force. So even without considering any quantum effects, such as pair productions, there is important physics that is not described by this equation.

One option is consider the currents as static, i.e. neglect the Lorentz force, and just quantize the electric and magnetic fields. This is a widespread procedure that leads to useful results.

Another option is to try writing the Lorentz force using the quantum operators you proposed. I do not see any fundamental reason why this could not work. However, it will be quite messy and one will have to work hard to get things that comes free with the Lagrangian formulation, such as Lorenz covariance and conservation laws.

yaron kedem
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