# Dirac Notation and completeness relation

I am confused about two minor things right now.
The following illustrates both which I pulled from my QM book:

$$<x|p_{op}|0>=\int_{-\infty}^{\infty}dp<x|p_{op}|p><p|0>=\int_{-\infty}^{\infty}dp~p<x|p><p|0>$$
$$=\int_{-\infty}^{\infty}dp~p\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>$$
$$=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp<x|p><p|0>=\frac{\hbar}{i}\frac{d}{dx}<x|0>$$

So, the first thing I am confused about is in the middle expression on the top line.
I know they use the completeness relation $\int_{-\infty}^{\infty}dp|p><p|=\hat{1}$ 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: $<p|p_{op}|0>$ 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 $p\rightarrow\frac{\hbar}{i}\frac{d}{dx}$ 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 $p_{op}|p>=p|p>$ so the p is the eigenvalue... how on earth do they justify replacing it with the operator expression?

## Answers and Replies

dextercioby
Science Advisor
Homework Helper
[...]
However, what if it was something like: $<p|p_{op}|0>$ when you stick in that completeness relation, you can't pull the integral to the outside... can you?[...]

Well, you can't, because $\hat{p}\langle p| = p \langle p |$ and the integral is wrt p.

0ddbio said:
[...] 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 $p\rightarrow\frac{\hbar}{i}\frac{d}{dx}$ 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 $p_{op}|p>=p|p>$ so the p is the eigenvalue... how on earth do they justify replacing it with the operator expression?

There's no operator expression, but a bracket $\langle x|p\rangle$ which is picked up from the complex exponential and the numerical factor in front of it.

Thanks for the response.

There's no operator expression, but a bracket $\langle x|p\rangle$ which is picked up from the complex exponential and the numerical factor in front of it.
I am so sorry, I mentioned the wrong portion of it. I meant to say that about going from the first expression on the middle line to the second. There was no $\langle x|p\rangle$

I'll just rewrite it here for your convenience. The part I was referring to was this:
$$\int_{-\infty}^{\infty}dp~p\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\langle p|0\rangle=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\langle p|0\rangle$$

and all they did was change the p into it's operator.. but isn't p there just the eigenvalue? So how are they doing that?

thanks.

Bill_K
Science Advisor
All they did was use the algebraic fact that p exp(ipx/h) = h/i d/dx exp(ipx/h). No operators involved in that step.

All they did was use the algebraic fact that p exp(ipx/h) = h/i d/dx exp(ipx/h). No operators involved in that step.

ooh, yes of course!
thank you both very much.

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
However, what if it was something like: $<p|p_{op}|0>$ when you stick in that completeness relation, you can't pull the integral to the outside... can you?
Then the p in $\int_{-\infty}^{\infty}dp|p><p|=\hat{1}$ is a dummy variable, and exists only within the expression on the left hand side. In particular, it is not the p in $<p|p_{op}|0>$. Therefore, you can still pull the integral outside.

However, you really should use a different dummy variable so that you don't forget which p's are which.