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Expectation value of Energy Quantum

  1. Sep 27, 2006 #1
    I'm still really confused on how to go about calculating this for non eigenstates. I'm trying to do the problem below, and am wondering how to go about it.

    [itex] \Psi (x,0) = A (1-2 \sqrt {\frac{m \omega}{\hbar}} x)^2 e^ {-\frac{m \omega x^2}{2 \hbar}}[/itex]

    So I can't calculate the expectation value using Psi (x,) because this is not a stationary state, therefore i have to get Psi (x,t), correct? To do this i need to first normalize the equation and get A, and then plug Psi(x,0) into the equation:

    [itex] Cn = \int(\psi_n(x)*f(x) dx)) [/itex]

    [itex] \psi_n(x) [/itex] is pretty complex for a harmonic oscillator, being
    [itex] \frac{1}{sqrt(n-factorial)} (A_+)^n \psi_0(x)[/itex]

    having solved for Cn...assuming i can..which i seriously doubt that i can..i could then go about calculating the energy for Psi(x,t). However, i can't honestly believe this is what i'm supposed to do...it sounds too complex, so i'm guessing either i can for some reason just use the Hamiltonian operator on Psi(x,0), or there is some simpler way to get the expectation energy
    Last edited: Sep 27, 2006
  2. jcsd
  3. Sep 27, 2006 #2
    Second question..i've noticed in the book it says i can write Psi(x,0) as a linear combination of stationary states. So i'm trying to tell what linear combination this function is. However, i was wondering if anyone had the 2nd stationary state handy. I calculated it but am unsure if it is right..it doesnt' appear to me to be related to Psi(x,0), but then my mind is fried from so many hours of physics in a row...anyways, i'm off to sleep for 3 hrs before getting up to study japanese. I'll probably check back here sometime tomorrow or ask a friend, and then turn in my homework a little late.
  4. Sep 27, 2006 #3
    For your problem, the method with the integral seems rather "oversized" - try calculating the first three eigenstates of the harmonic oszillator yourself with the [itex] a_+ [/itex]-Operator (or looking them up in some textbook), then you can see what the coefficients of your [itex] \Psi(x,0) [/itex] must be and determine A. (Remember normalization).
  5. Sep 27, 2006 #4
    so ya basically a friend showed me what they did but i still don't really understand how they got Cn for this equation without integrating. I think they set the constants equal to what was already in Psi(x,0), but it was really hard to place all the factors they got
  6. Sep 27, 2006 #5
    [itex] \Psi (x,0) = A (1-2 \sqrt {\frac{m \omega}{\hbar}} x)^2 e^ {-\frac{m \omega x^2}{2 \hbar}} = A (1 - 4 \sqrt {\frac{m \omega}{\hbar}} x + 4 \frac{m \omega}{\hbar} x^2)e^ {-\frac{m \omega x^2}{2 \hbar}} [/itex]

    The first three eigenstates are all polynomials up to 2nd degree times your e-function. Knowing this, you can determine the c_0, c_1, c_2 just by comparison.

    Basically yes, but the important thing here is not the time dependence, but getting the c_n's so you can write your wavefunction as a sum of eigenfunctions, for which you already know that the energy-expectation-value is just their eigenvalue [itex]E_n[/itex].
  7. Sep 27, 2006 #6
    Whoops, didn't see your 2nd posting before I replied. Second eigentstate up to a constant:

    [itex] \phi_2 = (4 \frac{m \omega x^2}{ \hbar} } - 2) e^{-\frac{m \omega x^2}{2 \hbar}}[/itex]
    Last edited: Sep 27, 2006
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