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Homework Help: Sakurai 2.17 - More elegant solution help?

  1. Dec 24, 2008 #1
    Problem

    Show for the one-dimensional simple harmonic oscillator

    [tex]\langle 0 | e^{ikx} | 0 \rangle = \exp{[-k^2 \langle 0 | x^2 | 0 \rangle / 2]} [/tex]

    where [tex]x[/tex] is the position operator (here, [tex]k[/tex] is a number, not an operator, with dimensions 1/length).

    My Solution

    Well, I already know how to do this problem, but my solution isn't as clean. I was searching for a more elegant solution. Here's the outline to my solution:

    1. We know that [tex]e^{ikx}|p'\rangle = |p' + \hbar k\rangle[/tex] (pretty simple to prove).
    2. We can show that
    [tex]\langle 0 |e^{ikx}| 0 \rangle = \int dp' \langle 0|p' \rangle \langle p'-\hbar k | 0 \rangle[/tex] ​
    by putting everything in the momentum basis.
    3. We then just need to find the harmonic oscillator's ground state in the momentum representation, which is just a Fourier transform of the ground state in the position representation.
    4. Substitute into the thing got in #2 and compute away, leading to the equality. (and this works, after expanding everything)

    As we can see, computing the Fourier transform in step 3 and computing everything in step 4 takes some time, and isn't exactly the "winning" solution.:smile: So would someone help me find or post a more elegant solution?
     
  2. jcsd
  3. Dec 24, 2008 #2

    Avodyne

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    I don't know if this is more elegant or not, but ...

    1) Expand in powers of k.
    2) Note that expectation value of odd powers of x is zero, because the ground-state wave function is an even function of x.
    3) Prove that

    [tex]\langle x^{2n}\rangle = (2n{-}1)!!\,\langle x^2\rangle^n.[/tex]

    Useful identify for the proof:

    [tex]\int_{-\infty}^{+\infty}dx\;x^{2n}e^{-cx^2}=\left(-{d\over dc}\right)^{\!n}\int_{-\infty}^{+\infty}dx\;e^{-cx^2}.[/tex]

    4) Resum the series.
     
  4. Dec 24, 2008 #3
    Wow... number three seems to be pretty useful! This was actually my initial first approach... I found a really neat analog to the Catalan numbers, but wasn't able to complete it. I'll go and try out this approach. Thanks a lot for the help :smile:
     
  5. Dec 24, 2008 #4

    Avodyne

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    In quantum field theory #3 is known as Wick's theorem.
     
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