- #1

- 3

- 0

_{x})

^{2}, keeping in mind that ψ

_{0}(x) = A

_{0}e

^{−ax2}

where A

_{0}= (2mω0/h)^1/4, and

<x

^{2}> = ∫x

^{2}|ψ|

^{2}dx = h_bar / 2mω

_{0}

<ψ(x)|p

_{x}

^{2}|ψ(x)> = ∫ψ(x)(p

_{op}

^{2})ψ(x) dx

p

_{op}= [h

_{bar}/ i] ([itex]\delta[/itex]/[itex]\delta[/itex]x)

I'm not going to attempt to type out me solving the integral because it would be extremely messy, as I am unfamiliar with syntax on this site.

I take the partial twice of ψ(x), and plug in the value they give for ∫x

^{2}|ψ|

^{2}:

-4h

_{bar}

^{3}a

^{2}/ 2mω

_{0}

This isn't correct, and I'm not too sure what I'm doing wrong. The negative doesn't make sense to me, but that's what the math is giving (the negative comes from squaring i in the momentum operator).

Any help is greatly appreciated.