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Coherent States of the Quantum Harmonic Oscillator

  1. Nov 19, 2008 #1
    Does anyone know why a harmonic potential gives rise to coherent states? In other words, what is special about a quadratic potential that causes the shifted ground state to oscillate like a classical particle without dispersing so as to saturate the uncertainty principle? Any help or insight would be much appreciated. Thanks

  2. jcsd
  3. Nov 19, 2008 #2


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    It comes from the underlying dynamical symmetry algebra for this system, which in this
    case is the Heisenberg algebra. One can generalize the whole concept of coherent states
    to other systems (with different dynamical symmetry groups).
  4. Nov 19, 2008 #3


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    One requirement is that the energy eigenvalues be spaced by integer multiples of some energy [itex]\hbar\omega[/itex] so that the quantum state is exactly periodic with period [itex]2\pi/\omega[/itex]. This is necessary but not sufficient, since the wave functions could still spread out at intermediate times. (This is what happens for a one-dimensional particle in a box.)

    strangerep, I thought the Heisenberg algebra was [x,p]=c; how does that help us see that the harmonic oscillator has coherent states?
  5. Nov 21, 2008 #4
    yea, im a little confused as to how the canonical commutation relation allows a quadratic potential to have coherent states?
  6. Nov 24, 2008 #5
    ^ You can derive the existence of number states from the canonical commutation relations and the quadratic form of the Hamiltonian alone. Once you have the number states, it is a trivial matter to find the coherent states in terms of them.
  7. Nov 28, 2008 #6


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    Sorry for taking so long to reply. I'm severely pressed for time right now.

    The short (probably inadequate) answer is that it's much easier to see in
    Fock representation. From the Heisenberg algebra (expressed in terms of
    the usual annihilation and creation operators) one can look for
    eigenstates of the annihilation operator, ie: [itex]a |v\rangle = v
    |v\rangle[/itex], and it then takes only a few lines to show that such
    states have the required minimum uncertainty product.

    With a little more work, one can show that such states can be
    expressed as

    |v\rangle = e^{va^\dagger - v^* a} |0> ~~(1)

    See, eg, Mandel & Wolf, "Optical Coherence & Quantum Optics" ch11,
    for more details.

    Enlarging the basic Heisenberg algebra to include the Hamiltonian (or
    in this case the number operator) [itex]H \propto a^\dagger a[/itex]),
    we get the "oscillator algebra" and the important feature here is that
    [itex][H,a] \propto a[/itex] and [itex][H,a^\dagger] \propto
    a^\dagger[/itex]. This means that the action of H on a coherent state
    [itex]|v\rangle[/itex] always yield some other coherent state
    [itex]|w\rangle[/itex] (which follows from (1) above).

    Standing back a bit, the key point of all this is that we have a
    basic algebra (Heisenberg) which the Hamiltonian preserves, meaning
    that the adjoint action of the Hamiltonian on this algebra (both
    considered as elements of a larger enveloping algebra) yields
    another element of that same basic algebra. In the present case,
    this means that although coherent states are not eigenstates of
    the Hamiltonian, we can say that coherent states evolve into (other)
    coherent states
    and we already know that all coherent states have
    minimum uncertainty product.

    There is a generalization of all this: given a dynamical algebra "A",
    we find the largest commuting subalgebra "C" containing the
    Hamiltonian, and then construct states by cyclically acting on a vacuum
    state with the (exponentiated) remaining generators in A/C. Details can
    be found in the review paper:

    Zhang, Feng, Gilmore:
    "Coherent states: Theory and some applications"
    Rev Mod Phys 62:867-927, 1990.

    Meanwhile, let me try to answer the OP's original question a bit better, i.e.,
    It's because there are certain states constructable from the Heisenberg
    algebra alone, all of which have minimum uncertainty product. We call
    these "coherent states". Then, when we consider the Hamiltonian
    together with the Heisenberg algebra, we find the latter is preserved
    by the action of the Hamiltonian and that coherent states evolve into
    other coherent states. So it's not so much the "quadratic potential"
    that's at work here, but more how the total Hamiltonian preserves
    the Heisenberg algebra.
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