# Matrix element of momentum

1. Dec 30, 2014

### nigelscott

Can somebody explain how you get:

<pm|p|pn> = (1/2π)∫e-ipmx/h (-ih∂/∂x) eipnx/h dx
.....
= ih∂δ(pm - pn)/∂x

Conceptually, I am having a problem with how the inner product is formed when a derivative is involved (i.e. the ..... steps)

Last edited: Dec 30, 2014
2. Dec 30, 2014

### USeptim

Hi Nigelscott,

The derivative operator just makes a derivative over ∂x over either pn(x) or pm(x). The inner product just makes the multiplication: ∂pn(x)/∂x * pm(x) and this has to be integrated over all the space and multiplied by -i2π/h in order to get <pm|p|pn>.

Best regards.
Sergio

3. Dec 30, 2014

### nigelscott

OK. But the differential gives (ipn/h)eipnx/h ... what happens to the p?

4. Dec 30, 2014

### MisterX

The equation in the initial post is incorrect. It should be $$\langle p_m | x | p_n\rangle = i \hbar \delta^\prime (p_m - p_n)$$

Where
$$\delta^\prime(p) = \frac{d}{dp} \delta(p)$$

5. Dec 30, 2014

### Matterwave

I'm a little confused, since eigenstates are ortho-normal, and $|p\rangle$ is an eigenstate of p, wouldn't the answer simply be $$\langle p_m|p|p_n\rangle = p_n\langle p_m|p_n\rangle = p_n\delta(p_m-p_n)$$?

6. Dec 30, 2014

### MisterX

As for a derivation - first what are the eigenstates of p? We can choose a scheme such that

$$\langle x \mid x' \rangle = \delta(x-x')$$
$$\mid p \rangle = \sqrt{\frac{1}{2\pi}} \int \mathop{dx} e^{i\mathbf{p} \cdot \mathbf{x}/\hbar}\mid x \rangle$$

note that these states are not normalized; $\langle p \mid p' \rangle = \delta(p-p')$, however $\left| \langle p \mid p' \rangle \right|^2$ is not defined under this scheme. Continuing
$$\langle p | x | p'\rangle =$$ $$\frac{1}{2\pi} \int\int \mathop{dxdx'} e^{-i\mathbf{p} \cdot \mathbf{x}/\hbar} x e^{i\mathbf{p'} \cdot \mathbf{x'}/\hbar} \langle x \mid x' \rangle$$
$$\frac{1}{2\pi} \int_{\infty}^\infty \mathop{dx} e^{-i\mathbf{p} \cdot \mathbf{x}/\hbar} x e^{i\mathbf{p'} \cdot \mathbf{x}/\hbar}$$
$$=\frac{1}{2\pi} \int_{\infty}^\infty \mathop{dx} x e^{i(\mathbf{p'} - \mathbf{p}) \cdot \mathbf{x}/\hbar}$$
If you're astute you may already recognize this is going to be related to the derivative of the Dirac delta. For example what we have is a scaled and shifted version of the Fourier transform of $1 x$ which is going to be related to the derivative of the Fourier transform of $1$. Alternately, you could try integrating by parts.

7. Dec 30, 2014

### Matterwave

@MisterX I'm seeing a "p" sandwiched in between the kets in the OP, not an "x", hence my confusion.

8. Dec 30, 2014

### MisterX

The OP was incorrect. Matterwave is correct regarding <p_m |p |p_n > with the "normalization" chosen above.

9. Dec 31, 2014

### nigelscott

Thanks to all. I believe Matterwave has put me on the right track. I was trying to evaluate the integral containing the partial derivative. If I evaluate the integral using the OP p and then substitute the partial derivative at the end everything falls into place.

10. Dec 31, 2014

### MisterX

Except maybe it's not correct. The sign is wrong for one. What does it mean to take the derivative with respect to x of a delta function that clearly involves only p's? Does the derivative apply only to the delta function, or to the expression that might come after it?

$$\langle p_m|p|p_n\rangle = p_n\langle p_m|p_n\rangle = p_n\delta(p_m-p_n)$$
That's it. The pn does not become a derivative. It's a value, and does not mean the momentum operator.

What was the source of this problem?