- #1

0ddbio

- 31

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The following illustrates both which I pulled from my QM book:

[tex]<x|p_{op}|0>=\int_{-\infty}^{\infty}dp<x|p_{op}|p><p|0>=\int_{-\infty}^{\infty}dp~p<x|p><p|0>[/tex]

[tex]=\int_{-\infty}^{\infty}dp~p\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>[/tex]

[tex]=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp<x|p><p|0>=\frac{\hbar}{i}\frac{d}{dx}<x|0>[/tex]

So, the first thing I am confused about is in the middle expression on the top line.

I know they use the completeness relation [itex]\int_{-\infty}^{\infty}dp|p><p|=\hat{1}[/itex] so I guess they just stick that in the middle and I guess it's ok that they pull the integral to the outside because the other terms don't depend on

*p*.

However,

**what if**it was something like: [itex]<p|p_{op}|0>[/itex] when you stick in that completeness relation, you can't pull the integral to the outside... can you?

The second thing that is bothering me is going from the expression on the right (middle line) to the expression on the left (bottom line). They apparently just change [itex]p\rightarrow\frac{\hbar}{i}\frac{d}{dx}[/itex] which I recognize as the momentum operator... BUT, isn't the

*p*the

**eigenvalue**? It first appears on the expression at the right (top line) from [itex]p_{op}|p>=p|p>[/itex] so the

*p*is the eigenvalue... how on Earth do they justify replacing it with the operator expression?