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Expectation values of harmonic oscillator in general state

  1. Nov 19, 2012 #1
    So, this has been bothering me for a while.

    Lets say we have the wavefunction of a harmonic oscillator as a general superposition of energy eigenstates:

    [itex]\Psi = \sum c_{n} \psi _{n} exp(i(E_{n}-E_{m})t/h)[/itex]

    Is it true in this case that <V> =(1/2) <E> .

    I tried calculating this but i get something like

    <V> = < [itex]\Psi |V| \Psi [/itex] > = (1/2)<E> + some other term that does not seem to be zero generally.

    However, it seems to me that <V> =(1/2) <E> should be true even in this case, since
    [itex] <V>_{n} = <\psi_{n} | V | \psi_{n} > = (1/2) <E> [/itex] for every n.
  2. jcsd
  3. Nov 19, 2012 #2


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    The superposition state is no longer stationary. The derivation of the virial theorem (Davydov, Sect 17 from Chapter II) will not be valid.
  4. Nov 19, 2012 #3
    Thank you for the answer.

    I agree with that, but that doesn't necessarily mean that <V>(t) != (1/2) <E>. I still feel the equality should hold.
  5. Nov 19, 2012 #4
    To see that it shouldn't hold, imagine taking the ground state and translating it very far from the minimum of the potential. This doesn't change the kinetic energy, but makes the potential energy very large. Clearly, <V> no longer equals (1/2)<E>.

    This comes from the fact that when you compute <V> in your superposition of energy eigenstates, you will get a bunch of cross terms between the energy eigenstates that spoil the relation <V> = (1/2)<E>
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