# When does 'naively' quantizing classical constraints work?

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.)

• $\partial^\mu A_\mu |\Psi\rangle = 0$ corresponds to the Gupta–Bleuler quantization method for QED, hence it does work to some extent, yes. Commented Jan 25, 2018 at 21:22
• @Slereah I thought the Gupta-Bleuler condition was $\langle \Psi| \partial^\mu A_\mu |\Psi' \rangle$. Commented Jan 25, 2018 at 21:33
• Since this is true for every states $\Psi, \Psi'$ and the inner product is positive definite this implies that $\partial^\mu A_\mu |\Psi\rangle = 0$. Commented Jan 25, 2018 at 21:45
• 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. Commented Jan 25, 2018 at 22:06
• @Slereah Gupta-Bleuler reads $(\partial^\mu A_\mu)^+ |\Psi\rangle = 0$, where "$+$" means to take the positive frequency part. Commented Jan 25, 2018 at 22:26

This is basically the Dirac quantization of constrained systems (as opposed to the reduced phase space quantization). Dirac quantization amounts to:

1. transform the constraints into operators $C\to\widehat{C}$,
2. 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.

• “Reduced PS quantization” refers to quantizing only gauge invariant functions. It does not mean “quantizing after constraining” which is also the case for Dirac quantization. Commented Apr 26 at 6:15
• @zixuanfeng Yeah, gauge-invariant functions live on the constraint surface, and "quantizing after constraining" refers to those functions which live on the constraint surface. You seem confused about Dirac quantization, which does not quantize after constraining (that's literally reduced phase space quantization), which is why the constraints are quantized as operators, and you find their kernels. Commented Apr 26 at 14:18
• As far as I knew, Dirac quantization means “quantize functions on constraint surface $\Sigma$ no matter whether they are gauge invariant” and then “impose gauge invariance on physical states $G_a \vec{\psi}=0$”. Maybe I’m wrong thanks to your noticing. Do you mean that Dirac quantization quantize “function on big phase space P” and then impose “$G_a \vec{\psi}=0$ “ will do two steps “identify two functions that coincide on constraint surface $\Sigma$” and “imposing gauge invariance $[F, \gamma_a]=0$” at once? Commented Apr 28 at 12:23
• @zixuanfeng This is covered in Henneaux and Teitelboim's Quantization of Gauge Systems in greater detail. The main steps are: (1) quantize the full system (I guess this is your "big phase space P"), then (2) impose the constraints as quantum operators on the Hilbert space. Modulo technical difficulties with operator ordering ambiguities (a thorn in quantization in general), this intuitively corresponds to working on the "quantized" constraint surface in the Hilbert space. This is the Dirac approach. Commented Apr 28 at 14:46
• thanks! I’ve started reading Henneaux’s book these days and it’s an excellent book. (And the notations I use is the same as in this book.) The only point confusing me (and maybe I’ve misunderstood) is that: Dirac quantize $C^\infty(P)$, instead of $C^\infty(\Sigma)$, and then impose $G_a\vec{\psi}=0$, right? Commented Apr 28 at 17:56