- #1

- 560

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I wondered if someone knew a way to prove this by just working with the operators directly and the fundamental commutation relations?

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- Thread starter center o bass
- Start date

- #1

- 560

- 2

I wondered if someone knew a way to prove this by just working with the operators directly and the fundamental commutation relations?

- #2

- 13,058

- 606

This stems directly from the quantum Noether theorem and the fact that spherical symmetry means that both at classical and at quantum level SO(3) is a symmetry group, hence the Casimir of its Lie algebra commutes with the Hamiltonian.

A proof of the quantum Noether theorem should be in the set of books by Greiner et al., if I'm not mistaking in the Symmetries one (Quantum Mechanics - Symmetries).

A proof of the quantum Noether theorem should be in the set of books by Greiner et al., if I'm not mistaking in the Symmetries one (Quantum Mechanics - Symmetries).

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- #3

- 1,006

- 105

The same argument establishes that Ly and Lz commute with H, therefore so does L^2 = Lx^2 + Ly^2 + Lz^2.

The general idea is: if an operator A generates a symmetry of the Hamiltonian, then A commutes with the Hamiltonian. The other important "generators" to know about are the momentum operator, which generates spatial translations, and the Hamiltonian itself, which generates time translations (i.e. time evolution).

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