# Homework Help: Dirac Notation - Position and Momentum

1. Sep 29, 2011

### atomicpedals

1. The problem statement, all variables and given/known data

Show that $\left\langle$x|p|x'$\right\rangle$ = $\hbar$/i $\partial$/$\partial$x $\delta$(x-x')

2. The attempt at a solution

$\left\langle$x|p|x'$\right\rangle$ = i$\hbar$ $\delta$(x-x')/(x-x') = i$\hbar$ $\partial$/$\partial$x' $\delta$(x-x') = $\hbar$/i $\partial$/$\partial$x $\delta$(x-x')

For the sake of formality I think I need an integral after my first equals sign which I think would be:

$\int$$\delta$(x-x') p $\delta$(x-x') dx'= p$\int$$\delta$(x-x') p $\delta$(x-x') dx'

However I'm not sure if a) it's needed, or b) if I set it up correctly. Any help would be appreciated!

2. Sep 29, 2011

### vela

Staff Emeritus
Suggestion: If you're going to use LaTeX, use it for the entire expression instead of individual symbols. Fixed your post up:
I actually don't follow what you're doing here at all. Try inserting a complete set:
$$\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp$$

3. Sep 29, 2011

### atomicpedals

So, I was right... in that I was wrong! :)

I'll go back and work with the complete set.

4. Oct 3, 2011

### atomicpedals

I'm going to call out my own ignorance (probably related to having spent too much time with a QM book causing less clarity instead of more):

$$\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp$$

I don't see how this is the complete set. (If that question made sense, I'm on a good start.)

Shouldn't we have:

$$\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle dp \langle p |x' \rangle \$$?

Last edited: Oct 3, 2011
5. Oct 3, 2011

### atomicpedals

Wait, does this work?

$$\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp$$

$$\langle x| \hat{p} |x' \rangle = \hbar /i \int \delta(x-x') \langle p |x' \rangle \,dp$$

$$\langle x| \hat{p} |x' \rangle = ( \hbar /i ) ( d/dx ) \delta(x-x')$$

6. Oct 3, 2011

### vela

Staff Emeritus
What are $\langle p \vert x' \rangle$ and $\langle x \lvert \hat{p} \vert p \rangle$ equal to?

7. Oct 3, 2011

### diazona

Just a side note:
They're the same thing. $\mathrm{d}p$ is a pure number and thus commutes with everything, in particular $\langle p|x'\rangle$.

8. Oct 4, 2011

### atomicpedals

The easier question to answer is what is $\langle p \vert x' \rangle$ equal to

$\langle p \vert x' \rangle = 1/ \sqrt{2 \pi \hbar} exp( (i/ \hbar) p x')$

The more important question is then what does $\langle x \lvert \hat{p} \vert p \rangle$ equal, because I don't think I know anymore! Except to say that it has matrix elements (and I'm not sure what those are).

9. Oct 4, 2011

### vela

Staff Emeritus
Surely you must know what $\hat{p}\vert p\rangle$ is equal to.

10. Oct 4, 2011

### atomicpedals

Well my first gut reply is that

$\hat{p}\vert p\rangle = p\vert p\rangle$

11. Oct 4, 2011

### vela

Staff Emeritus
Right, so $\langle x | \hat{p} | p \rangle = \langle x | p | p \rangle = p\langle x | p \rangle$.

12. Oct 4, 2011

### atomicpedals

Ah right! Ok, so what's the mathematical reasoning that gets me from $\langle x| \hat{p} |x' \rangle$ to $\int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp$ ? I'm really trying to understand the basic underpinnings of what's going on.

13. Oct 5, 2011

### atomicpedals

I'm considerably more comfortable with the original problem. Now I've got a follow up (thanks to yesterday's class). We have

$\langle x | \hat{p} | p \rangle = \langle x | p | p \rangle = p\langle x | p \rangle$

If $p = i / (\hbar) d/dx$ and $\langle x | p \rangle = 1 / ( \sqrt{2 \pi \hbar} ) exp[ (i / \hbar) p x) ]$ How do I go about determining matrix elements? Is it as simple as the product of p and $\langle x | p \rangle$ and then taking x = 0,1,2,...?

14. Oct 5, 2011

### vela

Staff Emeritus
No. I don't understand what you're trying to do.

15. Oct 5, 2011

### atomicpedals

I would like to determine matrix elements for $\langle x | \hat{p} | p \rangle$.

16. Oct 5, 2011

### vela

Staff Emeritus
I think you're confusing the operator $\hat{p}$ with the eigenvalue $p$ of the momentum eigenstate $\vert p \rangle$.