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Algebraic (ladder) solutions in QM

  1. Oct 15, 2009 #1
    When solving the quantum harmonic oscillator often ladder operators (that send energy eigenstates to higher or lower energy eigenstates) are introduced that allow one to algebraically solve the system. Similarly (but with much more difficulty) such operators can be introduced to solve the non-relativistic Hydrogen atom.

    I was wondering for what systems can these ladder-type operators be found, and for such systems is there a canonical method of finding them? (So for example do such operators exist for the infinite square well or the Dirac Hydrogen atom?)
  2. jcsd
  3. Oct 16, 2009 #2


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    Gold Member

    Starting with the SHO Hamiltonian,


    If you have position and momentum operators with,

    [tex][x,p]=ih [/tex]

    then it is possible to define operators

    a=\sqrt{\frac{1}{2\hbar\omega}}\left( \sqrt{\frac{1}{m}}p-i\sqrt{g}x \right)

    a^*=\sqrt{\frac{1}{2\hbar\omega}}\left( \sqrt{\frac{1}{m}}p+i\sqrt{g}x \right)

    so that

    [tex][a,a^*]=1 [/tex]

    and the Hamiltonian takes a simple form

    I'm not sure how general this prescription is.
    Last edited: Oct 16, 2009
  4. Oct 16, 2009 #3


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    I believe the use of ladder operators is possible in a surprisingly large
    number of cases...

    Start with the dynamical algebra -- the Heisenberg/oscillator
    algebra for the SHO case here, SO(4,2) for the non-rel H atom, etc.

    The prescription relies on having some kind of ground state which
    is invariant under the Hamiltonian. Then find which generators
    commute with the Hamiltonian, these are sometimes called the
    "symmetry" generators. Then look at the remaining generators
    which don't commute with the Hamiltonian.

    In the SHO case, the only generator that commutes with the
    Hamiltonian is the central element [itex]1[/itex]. The other generators
    [itex]a, a^*[/itex] satisfy relations like [itex][a,H] \propto -a[/itex] and [itex][a^*,H] \propto a^*[/itex].
    I.e., the action of the Hamiltonian doesn't mix up the two generators [itex]a, a^*[/itex].
    That's the crucial bit for getting the ladder behaviour wrt the eigenvalues of the Hamiltonian. (Exercise.)

    A similar thing happens when calculating the unitary irreducible
    representations of SO(3), i.e., the angular momentum spectrum.
    One takes [itex]J^2, J_z[/itex] as a maximal set of mutually commuting generators,
    and then notes that the combinations [itex]J_+ := J_x + iJ_y[/itex] and [itex]J_- := J_x - iJ_y[/itex]
    satisfy commutation relations like
    [J_z, J_+] ~=~ J_+ ~~~;~~~~~~
    [J_z, J_-] ~=~ -J_- ~~~;~~~~~~
    [J_+, J_-] ~=~ 2J_z

    These are what makes [itex]J_+, J_-[/itex] useful as operators for
    raising and lowering the usual "m" eigenvalue of [itex]J_z[/itex].

    Similar magic in many other dynamical groups makes possible a theory
    of generalized coherent states. The cases mentioned above underpin
    ordinary (Glauber) coherent states and spin-coherent states respectively.
    But coherent states are known for heaps of other groups.

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