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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 positionoperator(here, [tex]k[/tex] is anumber, 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. So would someone help me find or post a more elegant solution?

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# Homework Help: Sakurai 2.17 - More elegant solution help?

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