Sakurai 2.17 - More elegant solution help?

[Domnu]

Problem

Show for the one-dimensional simple harmonic oscillator

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

where $$x$$ is the position operator (here, $$k$$ 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 $$e^{ikx}|p'\rangle = |p' + \hbar k\rangle$$ (pretty simple to prove).
2. We can show that

$$\langle 0 |e^{ikx}| 0 \rangle = \int dp' \langle 0|p' \rangle \langle p'-\hbar k | 0 \rangle$$​

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. So would someone help me find or post a more elegant solution?

 

[Avodyne]

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

$$\langle x^{2n}\rangle = (2n{-}1)!\,\langle x^2\rangle^n.$$

Useful identify for the proof:

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

4) Resum the series.

 

[Domnu]

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

 

[Avodyne]

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