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Conservation of L^2 and Lz in central potential

  1. Sep 20, 2012 #1
    It is a well known fact that in a central potential (spherically symmetric) both Lz and L^2 commutes with H and the expectaitonvalues of these are therefore constants of the motion. On the other hand the proof of this fact seems, in the most cases, to be done in the position basis where it is rather laborous.

    I wondered if someone knew a way to prove this by just working with the operators directly and the fundamental commutation relations?
  2. jcsd
  3. Sep 20, 2012 #2


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    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).
    Last edited: Sep 20, 2012
  4. Sep 20, 2012 #3
    The qualitative way to understand this is to recognize that Lz is the "generator of rotations around the z axis." This means that if we take the operator exp(-i theta Lz / hbar) and apply it to a state, we get the same state but rotated by an angle theta around the z axis. H is clearly invariant under rotations, if the potential is central. So exp(-i theta Lz / hbar) must commute with the Hamiltonian, since the rotation it effects does not change the energy. You can show that this implies that Lz must also commute with H (take theta to be infinitesimal).

    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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