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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.
    Sergio
     
  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 $$\langle p_m | x | p_n\rangle = i \hbar \delta^\prime (p_m - p_n) $$

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

    Matterwave

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    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

    Matterwave

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    @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?
     
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