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Time until spontaneous emission in a QHO

  1. Jul 9, 2013 #1
    I have a quantum harmonic oscillator at energy level [itex]n_1[/itex], and I want to know how much time will pass before it spontaneously drops to energy level [itex]n_2[/itex]. I know about the position operator, momentum operator, energy operator, and angular momentum operators. Is there a "time until a system changes from one state to another" operator?
  2. jcsd
  3. Jul 10, 2013 #2


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    The energy eigenstates are stationary. This means if you consider only the HO, you won't get transitions between states at all. Spontaneous emission happens because of the presence of another physical system, the electromagnetic field.

    The decay time is given by the strength of the coupling between the initial state and the state of the electromagnetic field. You can include it phenomenologically in your Hamiltonian by introducing an imaginary part (like this for example) but you can't calculate it without taking into account the electromagnetic field somehow.
  4. Jul 10, 2013 #3
    This is a crucial question in QM and marks a point where QM fails and one needs to move toward QFT. As kith has explained, in ordinary QM, the energy eigenstates are stationary, so for example, if we had a hydrogen atom in empty space (no other matter or sources to the EM field), and it is in the first excited state, then according to elementary QM, the H atom will remain in the excited state forever. However, this prediction is WRONG because any experiment will show that the atom actually does decay in a rather short time. The same would apply to a harmonic oscillator, but finding a HO to do an experiment on is a bit harder than finding a hydrogen atom.

    To actually get a reasonable estimation for the decay time of an atom in vacuum, one needs to incorporate the "vacuum fluctuations" which act as a perturbation to the Hydrogen hamiltonian. But this is a QFT topic. In the language of ordinary QM, perturbations that change rapidly generally can cause the eigenstates to wiggle around so that what was initially an eigenstate could have overlaps with different eigenstates a moment later, allowing for transitions. The rate depends on the specifics of the perturbation.

    If you do not want to concern yourself with "vacuum fluctuations" and would rather stay in the realm of elementary QM, you can easily calculate the transition rates (either decay or absorption) between energy eigenstates in the presence of a classical perturbation such as an electromagnetic wave. The equation that gives the transition rates is called Fermi's Golden Rule. It applies equally well to an idealized harmonic oscillator, a real hydrogen atom, and pretty much anything with energy eigenstates. This rule is actually very useful in experiment because in most laboratory settings, you can't filter out all the external EM waves, so the perturbation the EM field causes is much more significant than that of the vacuum fluctuations, and in that case Fermi's Golden Rule gives a good approximation.

    This looks like a pretty nice discussion of Fermi's Golden Rule: http://www.google.com/url?sa=t&rct=...L5xtbjkBm6q1V6BVG_fb1EQ&bvm=bv.48705608,d.cGE
    Beware with this equation though--it often gets thrown at students without much explanation, and it often gets misapplied. This is because it is actually a rather hard equation to apply correctly, due to the [itex]\rho[/itex] or [itex]\delta(E-E_0)[/itex] term that appears. Usually, both sides of the equation need to appear under some sort of integral for it to make sense.
    Last edited: Jul 10, 2013
  5. Jul 10, 2013 #4
    Thank you both, very helpful!
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