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One dimensional harmonic oscillator

  1. Jan 13, 2014 #1
    1. The problem statement, all variables and given/known data
    One dimensional harmonic oscillator is at the beginning in state with wavefunction ##\psi (x,0)=Aexp(-\frac{(x-x_0)^2}{2a^2})exp(\frac{ip_0x}{\hbar })##.
    What is the expected value of full energy?

    2. Relevant equations

    ##<E>=<\psi ^{*}|H|\psi >=\sum \left | C_n \right |^2E_n##

    3. The attempt at a solution

    So I tried to find ##C_n=\int_{-\infty }^{\infty }A^2exp(-\frac{(x-x_0)^2}{a^2})dx##

    But I have no idea on what to do with this integral.. Whatever I do, i keep getting this integral. What can i do here?
     
    Last edited: Jan 13, 2014
  2. jcsd
  3. Jan 13, 2014 #2

    dextercioby

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    Can you check if your wavefunction is normalized to 1 ?
     
  4. Jan 13, 2014 #3
    ##\psi (x,0)=Aexp(-\frac{(x-x_0)^2}{2a^2})exp(\frac{ip_0x}{\hbar })##


    ##A^2\int_{-\infty }^{\infty }\psi ^{*}\psi dx=1## the integral is of course ##1##, therefore ##A^2=1##.

    Yes, it is normalized.
     
  5. Jan 13, 2014 #4

    dextercioby

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    Have you done the integral by yourself?
     
  6. Jan 13, 2014 #5
    Why would I?

    ##\psi ^{*}(x)\psi (x)= \left | \psi (x) \right |^2## which is the probability that we will find a particle at coordinate x. Well, not exactly there, but close to x.

    If I integrate probability from ##-\infty## to ##\infty##, which are all the values for x, the integral will for sure be equal to 1, because the electron has to be somewhere in ##(-\infty, \infty)##
     
  7. Jan 13, 2014 #6

    dextercioby

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    You have an unknown in the wavefunction, namely A. You have to check the wavefunction is properly normalized to unity before applying in probabilistic calculations. By checking this, you get an equation from which you can determine A in terms of x_0 and 'a'.
     
  8. Jan 13, 2014 #7

    CAF123

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    That is only the case if your wave function is normalized. The presence of A in ##\psi(x,0)## suggests that you should actually solve the equation ##\langle \psi | \psi \rangle = 1## for A so as to obtain this normalization constant. It may be the case that such an A is impossible to find but given that the wave function at t=0 describes a physical system, you can be sure it exists.

    If the wave function is normalizable at some arbitrary t, it can be proved it stays normalizable for all t, so you may use your initial wave function in your calculation.
     
  9. Jan 13, 2014 #8
    So..

    ##\int_{-\infty }^{\infty }\left | A \right |^2exp(-\frac{(x-x_0)^2}{a^2})dx=1##

    For ##x-x_0## than ##\int_{-\infty }^{\infty }\left | A \right |^2exp(-\frac{u^2}{a^2})du=1##

    Now with substitution ##t=\frac{u^2}{a^2}## we convert this integral to something we can calculate using Euler's Gamma function, therefore the equation is now ##\left | A \right |^2a\sqrt{\pi }=1##

    That should do it.

    And, I think i just found out how to calculate the integral in my original (first) post... -.- Sorry, for stealing your time.

    However, since i am already here, let me ask you this: So every time I see A in the wavefunction or is there A hint to check if the wavefunction is normalized, this is what I have to do. Right?

    Again, I apologize, i couldn't see that gamma function before...
     
  10. Jan 14, 2014 #9

    CAF123

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    Yes, A is a multiplicative constant to the wave function to make it normalized. So when you want to normalize the wave function, finding the right A is what is meant.
     
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