Consider the generally covariant formulation of the relativistic point particle, where the configuration is specified by $x^\mu(\tau)$, and $\tau$ is an arbitrary parameter. In the Hamiltonian picture, the canonical momenta $p_\mu$ are constrained, obeying $$p^2 + m^2 = 0.$$ This is a first-class constraint which corresponds to a gauge symmetry, and there are no second-class constraints. Then we quantize using the usual Poisson brackets, yielding operators $$[\hat{x}^\mu, \hat{p}_\nu] = i \hbar \delta^\mu_\nu.$$ In lecture notes here and here it is claimed that the constraint is imposed as an operator equation on physical states $$(\hat{p}^2 + m^2) |\psi\rangle = 0.$$ This makes sense because it just says the wavefunctions obey the Klein-Gordan equation, but I'm confused as to why this procedure works or how general it is. For example, it certainly doesn't work for QED in Lorenz gauge, because imposing $$\partial_\mu A^\mu | \Psi \rangle = 0$$ is far too stringent. Can somebody explain why first-class constraints can be imposed by the method above? How often does this work, and why doesn't this work for QED? (I imagine there's a huge amount to say here since there are plenty of very powerful quantization methods out there, but I'm hoping there's something relatively elementary that can clear up my confusion.)
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$\begingroup$ $\partial^\mu A_\mu |\Psi\rangle = 0$ corresponds to the Gupta–Bleuler quantization method for QED, hence it does work to some extent, yes. $\endgroup$– SlereahJan 25, 2018 at 21:22
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3$\begingroup$ @Slereah I thought the Gupta-Bleuler condition was $\langle \Psi| \partial^\mu A_\mu |\Psi' \rangle$. $\endgroup$– knzhouJan 25, 2018 at 21:33
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$\begingroup$ Since this is true for every states $\Psi, \Psi'$ and the inner product is positive definite this implies that $\partial^\mu A_\mu |\Psi\rangle = 0$. $\endgroup$– SlereahJan 25, 2018 at 21:45
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1$\begingroup$ In short, a „naive” quantization never works. I urge you to read the only authority on the subject, the book by Marc Henneaux, chapter 13. $\endgroup$– DanielCJan 25, 2018 at 22:06
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1$\begingroup$ @Slereah Gupta-Bleuler reads $(\partial^\mu A_\mu)^+ |\Psi\rangle = 0$, where "$+$" means to take the positive frequency part. $\endgroup$– AccidentalFourierTransformJan 25, 2018 at 22:26
1 Answer
This is basically the Dirac quantization of constrained systems (as opposed to the reduced phase space quantization). Dirac quantization amounts to:
- transform the constraints into operators $C\to\widehat{C}$,
- insist physical states live in the kernel of the constraint operators $\mathcal{H}_{\text{phys}} = \ker(\widehat{C})$.
(For multiple constraints, the physical states live in the intersection of the kernels $\mathcal{H}_{\text{phys}} = \ker(\widehat{C}_{1})\cap\dots\cap\ker(\widehat{C}_{n})$, i.e., physical states must obey all the constraints.)
Reduced phase space quantization first constrains the phase space by satisfying the constraints, then quantizing. When this can be done, it is usually simpler. (Both the Dirac approach and the reduced phase space approach produce equivalent results.)
You don't run into serious problems provided you don't experience any of the usual operator ordering ambiguities in quantizing the constraints...assuming that the constraint analysis has been carried out fully (i.e., you found all the constraints, etc.).
Famously, in general relativity, the Hamiltonian constraint involves a term quadratic in momenta, which cannot be naively quantized without producing a mathematically not-well-defined operator.
For more on quantizing constraints, the only resource that I know of is Henneaux and Teitelboim's Quantization of Gauge Systems.
Question: Why isn't this done in electromagnetism?
Answer: You can do it in electromagnetism, but you need to be careful about the constraints you implement. Brian Hatfield's Quantum Field Theory of Point Particles and Strings discusses the EM situation, carefully counting degrees of freedom killed by various constraints, etc.
