Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix element of momentum

  1. Dec 30, 2014 #1
    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. jcsd
  3. Dec 30, 2014 #2
    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.
  4. Dec 30, 2014 #3
    OK. But the differential gives (ipn/h)eipnx/h ... what happens to the p?
  5. Dec 30, 2014 #4
    The equation in the initial post is incorrect. It should be $$\langle p_m | x | p_n\rangle = i \hbar \delta^\prime (p_m - p_n) $$

    $$ \delta^\prime(p) = \frac{d}{dp} \delta(p)$$
  6. Dec 30, 2014 #5


    User Avatar
    Science Advisor
    Gold Member

    I'm a little confused, since eigenstates are ortho-normal, and ##|p\rangle## is an eigenstate of p, wouldn't the answer simply be $$\langle p_m|p|p_n\rangle = p_n\langle p_m|p_n\rangle = p_n\delta(p_m-p_n)$$?
  7. Dec 30, 2014 #6
    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.
  8. Dec 30, 2014 #7


    User Avatar
    Science Advisor
    Gold Member

    @MisterX I'm seeing a "p" sandwiched in between the kets in the OP, not an "x", hence my confusion.
  9. Dec 30, 2014 #8
    The OP was incorrect. Matterwave is correct regarding <p_m |p |p_n > with the "normalization" chosen above.
  10. Dec 31, 2014 #9
    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.
  11. Dec 31, 2014 #10
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook