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Q.M. harmonic oscillator

  1. Feb 1, 2004 #1
    Here's the problem:

    A one dimensional harmonic oscillator has mass m and frequency w. A time dependent state psi(t) is given at t=0 by:

    psi(0)=1/sqrt(2s)*sum(n=N-s,n=N+s) In>
    where In> are the number eigenstates and N>>s>>1.

    Calculate <x>. Show it varies sinusoidally; find the frequency and amplitude. Compare the amlitude and frequency to the corresponding values of a classical harmonic oscillator.


    Here's how I proceeded:

    <x>=(1/2s) (some constants)*sum(n=N-s,n=N+s)*sum(m=N-s,m=N+s) <n I (a+a') I m> Exp[i(Em-En)t/h]

    (note a' is "a dagger")

    =(1/2s) (some constants)*sum(n=N-s,n=N+s)*sum(m=N-s,m=N+s) {sqrt(m) <n I m-1> + sqrt(m+1) <n I m+1>} Exp[i(En-Em)t/h]

    (note <n I m-1>=delta(n,m-1) and <n I m+1>=delta(n,m+1).

    =(1/2s) (some constants)*sum(n=N-s,n=N+s) {sqrt(m+1) Exp[-iwt] + sqrt(m) Exp[iwt]}

    This is where I get stuck. I don't know if I'm supposed to make some approximation since N>>s>>1, and approximate the term in the {} as sqrt(m) cos (wt), or if I'm just completely wrong from the start. If someone can help, I'd really appriciate it.
     
    Last edited: Feb 1, 2004
  2. jcsd
  3. Feb 3, 2004 #2

    Tom Mattson

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    Gold Member

    If N>>s, then all the numbers in the range of the index of summation [N-s,N+s] are approximately equal to N. That is, it is (approximately) as though you only have a single term.

    That single term is going to be of the form:

    sqrt(N+1)exp(-i&omega;t)+sqrt(N)exp(i&omega;t)

    The thing that is screwing this up from being a sinusoid is the fact that the two terms have different coefficients. Now is the time to invoke N>>1. Do that to approximate as follows:

    sqrt(N)[exp(-i&omega;t)+exp(i&omega;t)]=2sqrt(N)cos(&omega;t)
     
  4. Feb 4, 2004 #3
    I think you're right. Thanks for the help.
     
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