Why does the classical Noether charge become the quantum symmetry generator?

It is often said that the classical charge $Q$ becomes the quantum generator $X$ after quantization. Indeed this is certainly the case for simple examples of energy and momentum. But why should this be the case mathematically?

For clarity let's assume we are doing canonical quantization, so that Poisson brackets become commutators. I assume that the reason has something to do with the relationship between classical Hamiltonian mechanics and Schrodinger's equation. Perhaps there's a simple formulation of Noether's theorem in the classical Hamiltonian setting which makes the quantum analogy precisely clear?

Any hints or references would be much appreciated!

Mathematical Background

In classical mechanics a continuous transformation of the Lagrangian which leaves the action invariant is called a symmetry. It yields a conserved charge $Q$ according to Noether's theorem. $Q$ remains unchanged throughout the motion of the system.

In quantum mechanics a continuous transformation is effected through a representation of a Lie group $G$ on a Hilbert space of states. We insist that this representation is unitary or antiunitary so that probabilities are conserved.

A continuous transformation which preserves solutions of Schrodinger's equation is called a symmetry. It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator.

We can equivalently view a continuous transformation as the conjugation action of a unitary operator on the space of Hermitian observables of the theory

$$A \mapsto UAU^\dagger = g.A$$

where $g \in G$. This immediately yields a representation of the Lie algebra on the space of observables

$$A \mapsto [X,A] = \delta A$$

$$\textrm{where}\ \ X \in \mathfrak{g}\ \ \textrm{and} \ \ e^{iX} = U \ \ \textrm{and} \ \ e^{i\delta A} = g.A$$

$X$ is typically called a generator. Clearly if $U$ describes a symmetry then $X$ will be a conserved quantity in the time evolution of the quantum system.

Edit

I've had a thought that maybe it's related to the 'Hamiltonian vector fields' for functions on a symplectic manifold. Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone?

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Comments: The first part of the question (v2) is related to this Phys.SE question. Comment to the second part of the question (v2): Note that a symmetry of the Euler-Lagrange eqs. is not necessarily a symmetry of the action, so that Noether's theorem (in its original action formulation) does not apply to such situation. See also this and this Phys.SE questions. –  Qmechanic Aug 21 '13 at 13:12
@Qmechanic - many thanks for your comments. The related question doesn't really answer mine. I'm not just asking about the path integral formalism in field theory. I'm interested in the canonical approach, principally in QM (although also in QFT). –  Edward Hughes Aug 21 '13 at 13:15
@EdwardHughes As a side note, recall that in converting classical observables to quantum quantities, there are all sorts of issues that make the conversion non-systematic. For example, there are operator ordering ambiguities. Therefore, there isn't a completely unambiguous procedure for converting a classical charge into a quantum symmetry generator without extra physical input. –  joshphysics Aug 21 '13 at 16:05
@joshphysics: I am indeed aware of that. I'm just trying to pin down exactly where the physical input comes in! –  Edward Hughes Aug 21 '13 at 16:18
A couple of related questions I found: physics.stackexchange.com/q/14481 and physics.stackexchange.com/q/37711. Unfortunately neither of them address the Noether charges. –  Edward Hughes Aug 22 '13 at 9:25

The canonical quantization after Dirac should fulfill the following axioms:

• Q1: The map $f \to \hat f$ that assigns a operator to every function on the phase space is linear and the constant 1-functions get mapped to the 1-operator

• Q2: The Poisson bracket maps to the commutator decorated with $\hbar$

• Q3: A complete system of functions in involution maps to a complete system of commutative operators.

It is the last condition which ensures that $G$ is a symmetry on the quantum side (the assignment $f \to \hat f$ needs to be a irreducible representation of the symmetry generators). But the No-Go theorems of Groenwald und Van Hove shows that a quantization for all observables with Q1-Q3 is not possible. The two mayor solutions are: Weaken Q2 and only require that it holds only up to first order of $\hbar$ - this leads to deformation quantization. On the other hand, geometric quantization modifes Q3 in the sense that it should hold only for some reasonable subalgebra of functions (eg which contains momentum ect.).

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Many thanks for your answer. Do you think you could expand on your logic about the third condition, and/or point me to some references on the subject? I'd then be happy to accept your answer! Cheers. –  Edward Hughes Aug 21 '13 at 23:44
A complete system of involutive functions is a collection of function on the phase space such that every Poisson bracket between them vanishes (ie you can measure them independently) and they completly determine the state of the system. Such systems are also called integrable. A complete system of commutative operators is the same on the quantum side (simultaneous measurable and only other operator which comutes with them all is the 1). Literature on the subject: Abraham/Marsden contains a discussion of quantization, also all references to geometric quantization are appropriate (eg Woodhouse) –  Tobias Diez Aug 22 '13 at 10:53
They main idea behind Q3 is that you need an irreducible representation of the symmetry generators because otherwise your classical symmetry gets mapped to a possible smaller symmetry-class on the quantum level. What you should keep in mind here (the 'take-away-message'): The correspondence of the Poisson bracket and the commutator is not enough to ensure that classical symmetries are represented (in a well defined way) as quantum symmetries. –  Tobias Diez Aug 22 '13 at 10:58
Okay - but why should $Q$ precisely map to the correct generator of the quantum symmetry, given that it is typically a function of position and momentum...? I understand why it gives a symmetry, just not why it gives the correct symmetry we expect. Is this to do with Hamiltonian vector fields, as I mentioned in my edit. I'll certainly accept your excellent answer if you can clarify this! Many thanks for all your help. –  Edward Hughes Aug 22 '13 at 11:14
It gives the correct symmetry by definition. First and foremost you do not know what symmetry corresponds to a given classical one and you define the quantum symmetry to be what comes out of your quantization prescription. –  Tobias Diez Aug 22 '13 at 19:20

Consider a quantum field formalism, where fields are operators. For instance, consider, for simplicity, a charged scalar field , with action $S = \int d^4x \partial_\mu \phi \partial^\mu \phi^*$, with a field :

$$\phi(x) = \int ~ \frac{d^3k}{2E_k} ~(a(p)e^{-ip.x} + b^+(p)e^{ip.x} )\tag{1}$$

$$\phi^*(x) = \int ~ \frac{d^3k}{2E_k} ~(b(p)e^{-ip.x} + a^+(p)e^{ip.x} )\tag{2}$$

where $a^+(p), a(p)$ are creation/anihilation operators for the particle, and $b^+(p), b(p)$ creation/anihilation operator for the anti-particle.

If we have a infinitesimal transformation, $\delta \phi(x)$ , leaving unchanged the action $S$, the conserved current is $j^\mu(x) = [\frac{\partial L}{\partial(\partial_\mu \phi)}~\delta \phi(x) + \frac{\partial L}{\partial(\partial_\mu \phi^*)}~\delta \phi^*(x)]$ (skipping the infinitesimal parameters), and the conserved generalized charge is $Q =~:\int d^3x~ j^0(x): ~= ~:\int d^3x ~[\Pi(x)~ \delta\phi(x) +\Pi^*(x)~ \delta\phi^*(x)]:$ - where the $\Pi(x),\Pi^*(x)$ are the conjugated momenta of $\phi(x),\phi^*(x)$, and the sign $:$ is for normal ordered product (putting anihilation operators at right).

We see, of course, that $Q$ is an operator.

An example : the standard (electric) charge is the conserved quantity which corresponds to the (global) transformation $\phi(x) \rightarrow e^{-i\Lambda}\phi(x)$, here the infinitesimal transformation is $\delta \phi(x) = -i\epsilon ~\phi(x)~$, so we have :

$$Q = -i:\int ~ d^3x ~(\Pi(x) ~\phi(x) - \Pi^*(x) ~\phi^*(x) ): \tag{3}$$

Or, equivalently :

$$Q = \int ~ \frac{d^3k}{2E_k} ~(a^+(p)a(p) - b^+(p)b(p) )\tag{4}$$

And we have : $$[Q,\phi(x)]=-\phi(x) \tag{5}$$ (This can be checked from the fundamental commutation relations between operators, like $[a(k),a^+(k')]=\delta^3(k-k')$)

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Thanks for your answer. Unfortunately I'm not interested in special cases. I want to know why in general $Q$ gives precisely the quantum generator. I'm aware that it works for specific examples, but I expect it's a general principle. –  Edward Hughes Aug 21 '13 at 12:01
@EdwardHughes : But this is just the general expression cited in the answer :$Q =~:\int d^3x ~[\Pi(x)~ \delta\phi(x) +\Pi^*(x)~ \delta\phi^*(x)]:$, so you may apply it with any variation $\delta\phi(x)$, and you have $[Q,\phi(x)] = -i \delta\phi(x)$ –  Trimok Aug 21 '13 at 14:14
@EdwardHughes: Weinberg proves this in chap7 of his QFT textbook, for a large class of field theories, and for a reasonably large class of symmetries(space-time symmetries&internal linear symmetries), the procedure is not very much different from what Trimok showed in his example. –  Jia Yiyang Aug 21 '13 at 14:56
@JiaYiyang: but what about the case for quantum mechanics? And how about if the internal symmetries aren't linear? I'm looking for a general mathematical reason, rather than just an algebraic derivation. –  Edward Hughes Aug 21 '13 at 15:05