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When does a wavefunction inherit the symmetries of the hamiltonian?

  1. Oct 13, 2012 #1
    As the title suggests, I am interested in symmetries of QM systems.

    Assume we have a stationary nonrelativistic quantum mechanical system [itex]H\psi = E\psi[/itex] where we have a unique ground state.

    I am interested in the conditions under which the stationary states of the system inherit the symmetries of the hamiltonian.

    I am aware of some examples. If the hamiltionian is symmetric in one axis, the wave function will be symmetric or antisymmetric. In particular, the ground state will be symmetric. Similarly, if H is invariant under interchange of two coordinates, so will the ground state be (for example in the hydrogen atom).

    In general, if the Hamiltonian is invariant under the action of a group, will the ground state also be? Can anyone supply a proof?

    Also, even if there is not a unique ground state, will the action of such a group on one ground state always produce another ground state (such that the set of all ground states is invariant under the group action)?

    Thank you for your time.
  2. jcsd
  3. Oct 13, 2012 #2
    In general the whole eigensubspace will be invariant under any transformation that leaves the hamiltonian invariant. For a one dimensional ground state that means you have invariance up to a phase factor.
  4. Oct 13, 2012 #3
    I think the general statement is something like: if the Hamiltonian has some symmetry group, then the energy eigenstates can be organized into multiplets that form representations of that symmetry group, with the members of each multiplet having the same energy. The hydrogen atom Hamiltonian is invariant under the rotation group, so the energy eigenstates can be organized into degenerate multiplets that form representations of the rotation group (labeled by the quantum number L).

    The intuitive content of this is: if the Hamiltonian is invariant under rotations, then we can take any energy eigenstate and rotate it to get another eigenstate with the same energy. So we expect the energy eigenstates to come in degenerate sets consisting of a bunch rotated versions of the same wave function.

    The ground state need not be invariant under the symmetry group. [That is, there can be many degenerate ground states, all related to each other by the symmetry]. When this happens we get spontaneous symmetry breaking.
  5. Oct 13, 2012 #4


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    Of course one can prepare quantum states which do solve the Schrödinger equation

    [tex](H-E)|\psi\rangle = 0[/tex]

    but which do not have the corresponding symmetry of the Hamiltonian H. Think about the hydrogen atom. You have multiplets |nlm>. You can prepare a state

    [tex]|\psi\rangle = \sum_{lm} \psi_{nlm}|nlm\rangle[/tex]

    (no summation over n)

    These states are energy eigenstates but not necessary eigenstates of L2 or Lz

    So the symmetry of a Hamiltonian is not represented by a single state but by the multiplets.
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