# Momentum in x-basis?

1. Sep 9, 2008

### dod13

<x|p>, something my professor wanted to do in class today.
She then jumped from <x|p> to evaluating <x|p|p> (not sure why).
So now we have <x|pI|p> with the middle p and I being momentum and identity operators respectively.
This is Integral{<x|p|x'><x'|x>}dx'.
I buy all of this, but here's where I don't understand. She claims <x|p|x'>=-i*hbar*DeltaFunction(x-x')*d/dx and that doesn't make much sense to me, though that may be because I'm still pretty unfamiliar with Dirac notation. Where did the delta function pop out from?

Can anybody enlighten me? Thanks.

2. Sep 10, 2008

### h0dgey84bc

The delta functions are the eigenfunctions of the position operator in the coordinate basis.

$$\hat{X} \phi =x_0 \phi$$ and in the coordinate basis $$\hat{X}=x_0$$

i.e. if you measure a particles position, it's wave function collapses to a delta function peaked at that position. Such that $$\int_{x_0}^{x_0+dx} \delta(x-x_0) dx =1$$
i.e. you will find the particle within x->x+dx with probability one.

3. Sep 10, 2008

### Avodyne

Well it's not clear to me what the starting point is supposed to be. Here's how I would approach this. Suppose we have some generic state $|\psi\rangle$. Let $|x\rangle$ be a position eigenstate: $X|x\rangle = x|x\rangle$, where $X$ is the position operator. Writing this in bra from, $\langle x|X=x\langle x|$, and taking the inner product with $|\psi\rangle$, we get

$$\langle x|X|\psi\rangle = x\langle x|\psi\rangle.$$

Then we define the momentum operator $P$ via

$$\langle x|P|\psi\rangle = -i\hbar{\partial\over\partial x}\langle x|\psi\rangle.$$

Now consider a momentum eigenstate $|p\rangle$. Let $|\psi\rangle$ be $|p\rangle$ in the equation above. We get

$$\langle x|P|p\rangle = -i\hbar{\partial\over\partial x}\langle x|p\rangle.$$

But we also have $P|p\rangle=p|p\rangle$, and taking the inner product with the bra $\langle x|$, we get

$$\langle x|P|p\rangle = p\langle x|p\rangle.$$

These two right-hand sides must be equal, which gives us a differential equation for $\langle x|p\rangle$, which I leave it to you to solve.

4. Sep 10, 2008

### Avodyne

Oh, I forgot to answer your question about the delta function. We have the orthonormality condition for position eigenstates,

$$\langle x|x'\rangle = \delta(x-x').$$

Now use the general formula for $\langle x|P|\psi\rangle$ with $|\psi\rangle$ replaced by $|x'\rangle$ to get

$$\langle x|P|x'\rangle = -i\hbar{\partial\over\partial x}\delta(x-x').$$

Note that the derivative acts on the delta-function, and should not be to the right of it, as you have written.

5. Sep 11, 2008

### dod13

Thanks so much for your replies.
Most of that makes a lot of sense. I guess just this bit I take some issue with:

I know that's the momentum operator, but I'm not sure how to deal with that within the Dirac notation framework. How is it that you can just sort of bring the derivative/operator out from between the bra-ket without operating on anything first, and turn the bra-ket into a regular inner product?

6. Sep 11, 2008

### Fredrik

Staff Emeritus
He's not "bringing it out". It was never "in there". The P he's defining acts on kets. The differential operator on the right-hand side acts on wave functions.

You probably don't realize that what you're used to writing as $\psi(x)$ isn't $|\psi\rangle$, it's $\langle x|\psi\rangle$.

7. Sep 16, 2008

### reilly

Note that P|p>= p|p>, where p is a c-number momentum eigenvalue.

Note also the <x|p> = exp(+ipx), apart from normalization constants. So,

<x|P|p> = p exp(+ipx)
, apart from constants.

It's not quite as complicated as some posters indicate. Also, this particular matrix element is often encountered in scattering and particle theory.
Regards,
Reilly Atkinson

8. Sep 16, 2008

### atyy

Try showing that the definition satisfies the commutation relation between X and P operators if <x|psi>=psi(x), or something along those lines.