Dirac Notation - Position and Momentum

[atomicpedals]

Homework Statement

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

2. The attempt at a solution

\left\langlex|p|x'\right\rangle = i\hbar \delta(x-x')/(x-x') = i\hbar \partial/\partialx' \delta(x-x') = \hbar/i \partial/\partialx \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!

 

[vela]

Suggestion: If you're going to use LaTeX, use it for the entire expression instead of individual symbols. Fixed your post up:

Homework Statement

Show that \langle x|\hat{p}|x'\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x}\delta(x-x')

2. The attempt at a solution
\langle x| \hat{p} |x&#039; \rangle<br /> = i \hbar \,\delta(x-x&#039;)/(x-x&#039;) <br /> = i\hbar \frac{\partial}{\partial x&#039;} \delta(x-x&#039;)<br /> = \frac{\hbar}{i} \frac{\partial}{\partial x} \delta(x-x&#039;)
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&#039;) \hat{p} \delta(x-x&#039;)\,dx&#039;= p\int\delta(x-x&#039;) \hat{p} \delta(x-x&#039;)\, dx&#039;
However I'm not sure if a) it's needed, or b) if I set it up correctly. Any help would be appreciated!

I actually don't follow what you're doing here at all. Try inserting a complete set:
\langle x| \hat{p} |x&#039; \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x&#039; \rangle \,dp

 

[atomicpedals]

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

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

 

[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&#039; \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x&#039; \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&#039; \rangle = \int \langle x | \hat{p} | p\rangle dp \langle p |x&#039; \rangle \?

 

[atomicpedals]

Wait, does this work?

\langle x| \hat{p} |x&#039; \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x&#039; \rangle \,dp

\langle x| \hat{p} |x&#039; \rangle = \hbar /i \int \delta(x-x&#039;) \langle p |x&#039; \rangle \,dp

\langle x| \hat{p} |x&#039; \rangle = ( \hbar /i ) ( d/dx ) \delta(x-x&#039;)

 

[vela]

What are \langle p \vert x&#039; \rangle and \langle x \lvert \hat{p} \vert p \rangle equal to?

 

[diazona]

Just a side note:

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&#039; \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x&#039; \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&#039; \rangle = \int \langle x | \hat{p} | p\rangle dp \langle p |x&#039; \rangle?

They're the same thing. \mathrm{d}p is a pure number and thus commutes with everything, in particular \langle p|x&#039;\rangle.

 

[atomicpedals]

The easier question to answer is what is \langle p \vert x&#039; \rangle equal to

\langle p \vert x&#039; \rangle = 1/ \sqrt{2 \pi \hbar} exp( (i/ \hbar) p x&#039;)

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).

 

[vela]

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

 

[atomicpedals]

Well my first gut reply is that

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

 

[vela]

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

 

[atomicpedals]

Ah right! Ok, so what's the mathematical reasoning that gets me from \langle x| \hat{p} |x&#039; \rangle to \int \langle x | \hat{p} | p\rangle\langle p |x&#039; \rangle \,dp ? I'm really trying to understand the basic underpinnings of what's going on.

 

[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,...?

 

[vela]

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

 

[atomicpedals]

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

 

[vela]

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