Noether's theorem: quantum version

lalbatros

A short question:
Is it right to say that the quantum version of Noether's theorem is simply given by the evolution rule for any observable A:

i hb dA/dt = [H,A]

For example, if A is the angular momentum, the invariance by rotations R = exp(i h L angle) implies [H,A] = 0 and Noether's conclusion follows.

Another way to state my question:
Isn't it right to say that Noether's theorem is "much more popular" in QM than in CM?

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elduderino

To what I have thought about it, yes and yes.

twold

Actually, you've just proved the Noether's theorem also in classical setting. This is so because the easiest way to prove (and also state) the theorem is in Hamiltonian formalism. Namely,

The Hamiltonian (i.e. evolution generator) is invariant w.r.t some symmetry flow if and only if the generator of the symmetry flow is invariant w.r.t the Hamiltonian flow.

The proof of this statement boils down to Lie derivative along the vector fields corresponding to those flows and that is the same thing as the Poisson bracket of the generators (H and A in your case) and the proof is as short as [H, A] = [A, H] iff [H, A] = 0.

Moving to QM, one replaces the flows by exp(-i h t A) and the Poisson brackets by commutators.

A note of caution: when one talks about Noether's theorem in quantum field theory setting, one often means Ward identities. These are identities that correlation functions of quantum fields satisfy whenever the fields have some symmetry.

tom.stoer

The common aspect is that from [H,A]=0 one concludes that A generates a symmetry of the system (it commutes with H and therefore has common eigenvectors with H, that means they can be labelled according to eigenvalues a).

The difference between the Noether theorem and the Hamiltonian setting is that given a symmetry (a transformation law) you can derive A using the Noether theorem, whereas in the above Hamiltonian setting you do not derive A.

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elduderino

The common aspect is that from [H,A]=0 one concludes that A generates a symmetry of the system (it commutes with the H and has common eigenvector with H so they can be labelled according to eigenvalues a).

The difference between the Noether theorem and the Hamiltonian setting is that given a symmetry (a transformation law) you can derive A using the Noether theorem, whereas in the above Hamiltonian setting you do not derive A.
Classically, a change in the phase space variable $$q_i \rightarrow q_i + \epsilon K_i(q)$$ does not cause a first order change in the Lagrangian then it corresponds to a symmetry. The conserved quantity would be $$\sum_i \frac{\partial L}{\partial\dot{q_i}}K_i$$

$$A$$ commuting with the Hamiltonian $$[H,A] = 0$$, corresponds to a symmetry of the time-independent Schrodinger equation. The conserved quantity is A itself.

Considering quantum mechanics to be more fundamental we can explain this the other way as well. If a transformation preserves the Lagrangian then there must exist a function on phase space whose possion bracket with the hamiltonian vanishes. Therefore A is a generating function which can give us the canonical transformation $$K_i(q)$$ we started with.

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tom.stoer

$$A$$ commuting with the Hamiltonian $$[H,A] = 0$$, corresponds to a symmetry of the time-independent Schrodinger equation. The conserved quantity is A itself.

Considering quantum mechanics to be more fundamental we can explain this the other way as well. ... A is a generating function which can give us the canonical transformation ... we started with.
This perspective is not inspired by QM but is essentially the Hamiltonian formalism itself which can be used classically as well. I think the main point for Noether's theorem is that given a symmetry on coordinate space it is rather easy to construct the conserved quantities. But as soon as the coordinate trf. is not known Noether's theorem loses its power. Therefore the canonical formalism which works even w/o any coordinate rep. seems to be more fundamental

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