Timeline for Why can't we "simply" quantize Maxwell's equations without a Lagrangian to create a quantum theory of electrodynamics?
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| Jan 22 at 5:23 | comment | added | The_Sympathizer | @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? | |
| Oct 2, 2023 at 21:09 | comment | added | Andrew | 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. | |
| Oct 2, 2023 at 21:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 29, 2023 at 18:13 | history | edited | The_Sympathizer | CC BY-SA 4.0 |
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| Aug 29, 2023 at 18:00 | answer | added | ACuriousMind♦ | timeline score: 2 | |
| Aug 29, 2023 at 12:53 | comment | added | naturallyInconsistent | @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. | |
| Aug 29, 2023 at 7:58 | answer | added | yaron kedem | timeline score: 1 | |
| Aug 29, 2023 at 6:24 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Aug 29, 2023 at 3:01 | comment | added | The_Sympathizer | @naturallyInconsistent: Digging more though I note that your answer here - 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.) | |
| Aug 29, 2023 at 2:30 | comment | added | The_Sympathizer | 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 :) | |
| Aug 29, 2023 at 2:20 | comment | added | The_Sympathizer | @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? | |
| Aug 29, 2023 at 2:17 | comment | added | naturallyInconsistent | 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. | |
| Aug 29, 2023 at 2:17 | comment | added | The_Sympathizer | , mathematically rigorously. | |
| Aug 29, 2023 at 2:15 | comment | added | The_Sympathizer | 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) | |
| Aug 29, 2023 at 2:15 | comment | added | The_Sympathizer | @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? | |
| Aug 29, 2023 at 1:40 | comment | added | knzhou | 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. | |
| Aug 29, 2023 at 1:38 | comment | added | knzhou | 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. | |
| Aug 29, 2023 at 1:38 | comment | added | The_Sympathizer | @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? | |
| Aug 29, 2023 at 1:36 | comment | added | knzhou | 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. | |
| Aug 29, 2023 at 1:35 | history | edited | The_Sympathizer | CC BY-SA 4.0 |
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| Aug 29, 2023 at 1:29 | history | asked | The_Sympathizer | CC BY-SA 4.0 |