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Homework Help: Expectation value of raising/lowering operators

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data

    This has been driving me CRAZY:

    Show that [tex] \langle a(t)\rangle = e^{-i\omega t} \langle a(0) \rangle [/tex]

    and

    [tex] \langle a^{\dagger}(t)\rangle = e^{i\omega t} \langle a^{\dagger}(0) \rangle [/tex]

    2. Relevant equations

    Raising/lowering eigenvalue equations:

    [tex] a |n \rangle = \sqrt{n} |n-1 \rangle [/tex]
    [tex] a^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle [/tex]

    Time development of stationary states: psi(x)*exp(-i*En*t/hbar)=psi(x)*exp(-i*w_n*t)

    3. The attempt at a solution

    Suppose we've got the system in some state [tex] \psi(0) [/tex].

    Then expanding into the [tex] | n(0)\rangle [/tex] basis (looking just at a, not a dagger here) we have

    [tex] \langle a(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle = \langle \sum_k c_k n_k | a | \sum_k c_k n_k \rangle = \sum_k \sum_l c_k^* c_l \langle n_k | a | n_l \rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \langle n_k | a | n_{l-1}\rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \delta_{k, l-1} [/tex]

    so for non-trivial <a(0)>, [tex] \langle a(0) \rangle = \sum_k |c_k|^2 \sqrt{n_k} [/tex]

    But now

    [tex] \langle a(t) \rangle = \langle \psi(0) e^{-i \omega t} | a | \psi(0) e^{-i \omega t} \rangle = \langle \psi(0)| e^{i \omega t} a e^{-i \omega t} | \psi(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle \neq e^{-i \omega t} \langle a(0) \rangle [/tex]

    because as far as I know, a does not act on exp(i*w*t) and the two exponential terms cancel out!! I get the same sort of problem with a dagger. So... what's the deal?? This is really bugging me, I'd love to know how to do this problem...
     
  2. jcsd
  3. Dec 15, 2009 #2

    diazona

    User Avatar
    Homework Helper

    First of all, thankyouthankyouthankyou for making a decent attempt at the problem o:)

    Here's the catch: when you did that calculation, you implicitly assumed that [itex]|\psi(0)\rangle[/itex] was an energy eigenstate with a particular eigenvalue [itex]E = \hbar \omega[/itex]. It's not, in general, an eigenstate, and so it doesn't necessarily evolve according to a single exponential factor [itex]e^{-i\omega t}[/itex]. You can't write
    [tex]|\psi(t)\rangle = |\psi(0) e^{-i\omega t}\rangle[/tex]
    unless you know that the state is an energy eigenstate.

    Before you try to generalize that calculation, take another look at this:
    Note that you can calculate [itex]\langle a(t)\rangle[/itex] the same way, just make the coefficients [itex]c_k[/itex] functions of time. Specifically,
    [tex]|\psi(t)\rangle = \sum_k c_k(t) |n_k\rangle = \sum_k c_k e^{-i\omega_k t} |n_k\rangle[/tex]
    Since this is a harmonic oscillator, you know what [itex]\omega_k[/itex] is in terms of [itex]k[/itex].
    Be careful there: you started out with a delta function that requires [itex]k = l-1[/itex]. So you shouldn't be winding up with [itex]|c_k|^2[/itex], since that results from the term where [itex]k = l[/itex].
     
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