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Units issue when normalizing momentum space wave function

  1. Aug 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Check that a given momentum space wave function is normalized. I've done the integral, but the result is not dimensionless. Here is the wave function:

    [tex]\overline{\phi} = \frac{1}{\pi} ( \frac{2 a_{0}}{\bar{h}})^{3/2} \frac{1}{(1+(a_{0} p / \bar{h})^2)^2}[/tex]

    The units of this function are [tex] p^{-3/2} [/tex]. This implies that the normalization should be done over all of momentum space, but I'm getting tripped up because it has been assumed that p = pz, so I don't know how to integrate over eveything. I tried:

    [tex] \int^{-\infty}_{\infty} \left|\overline{\phi}\right|^{2} dp = 1[/tex]

    and I got

    [tex] \frac{5a_{0}^2}{2\bar{h}^2\pi} [/tex]

    but that is not dimensionless, nor is it equal to 1. I get the feeling I need to use some kind of delta function, since the wave function vanishes everywhere except along the p_z axis. I can't seem to wrap my head around this, can anyone help? Thanks.
     
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  3. Aug 4, 2008 #2

    nrqed

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    Are you sure it's not a three-dimensional wavefunction?
    Check if


    [tex] \int^{-\infty}_{\infty} \left|\overline{\phi}\right|^{2} d^3p = 1[/tex]

    Or sometimes people define the measure in momentum space to have a factor of [tex]1/(2 \pi)^3 [/tex] or [tex] 1/(2 \pi)^{3/2} [/tex] so it's possible that you need one of those factors
     
  4. Aug 4, 2008 #3

    Dick

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    It must be three dimensional if the wavefunctions dimensions are p^(-3/2). That gives phi^2 dimension of p^(-3) and d^3p has dimensions of p^3. So the integral is dimensionless. And I get that it is 1 without any extra factors. Hint: use spherical coordinates.
     
    Last edited: Aug 4, 2008
  5. Aug 4, 2008 #4
    Are you sure I should be using spherical coordinates? p vanishes everywhere except the p_z axis, so the symmetry is around the p_z axis, which implies cylindrical coordinates doesn't it? I'll try integrating spherically taking the p to mean p_r, but if that's the case, I am probably misinterpreting p = pz.
     
  6. Aug 4, 2008 #5
    I tried this integral:

    [tex] \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty} \left|\phi(p_r)\right|^{2} p_{r}sin(p_{\theta})dp_{\phi}dp_{\theta}dp_{r}[/tex]

    But my result was

    [tex] - \frac{16}{3\pi}\frac{a_0}{\bar{h}}[/tex]

    When you look at the integrand, it's obvious that there's a unit issue. The extra factor of p_r due to the spherical volume element means initially the units are p^1. But when you integrate the angular components, dp_phi and dp_theta disappear so you finally end up with units of p^-1, as in my result above. I really am having trouble understanding these units.
     
  7. Aug 4, 2008 #6

    nrqed

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    The volume element in spherical coordinates is
    [tex] r^2 \sin \theta ~d\theta d \phi dr [/tex]
     
  8. Aug 4, 2008 #7
    Okay I realized I missed a factor of p_r for the spherical volume element. I computed the integral and got 1. So now I need to understand why I'm integrating radially over all space. I thought there was only momentum in the z direction, and doesn't that mean in phase space the wave function is zero unless p_x = p_y = 0?
     
  9. Aug 4, 2008 #8

    Dick

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    They didn't say p_z in the problem and a0 doesn't look like a vector in the prefactor. So p in the wavefunction probably means |p|.
     
    Last edited: Aug 4, 2008
  10. Aug 4, 2008 #9

    nrqed

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    You have to square the p in your wavefunction. How do you interpret squaring the vector? It's simply taking the dot product with itself so after squaring you get p^2 where now p is themagnitude of the vector.




    You may align your z axis with the vector p but in the end you are integrating over all directions anyway.
     
  11. Aug 4, 2008 #10
    This was the bit that was tripping me up. I now see that we have spherical symmetry, since if the position wave function is spherically symmetric, the problem is invariant under rotations of the coordinate system, so the momentum space wave function cannot have an angular dependence.

    What I still don't understand is why we "set" the polar axis along the direction of p, or "choose a convenient direction for p" by setting it p = p z. What does it mean when we do this? Clearly we're not saying the momentum is always in a single direction.

    I think the answer lies in examining the formula for the momentum space wave function:

    [tex] \bar{\phi}(\textbf{p}) = \frac{1}{(2\pi\bar{h})^{3/2}}\int d^{3}re^{-i\textbf{p}.\textbf{r}/\bar{h}}\phi(\textbf{r})[/tex]

    To find the momentum probability amplitude for some momentum (p_x,p_y,p_z), we integrate over all space with that integrand. But first, we transform into a new coordinate system in which both p_x' and p_y' are zero. Then we integrate over all space in the new coordinate system, which is easy since phi(r)=phi(r), ie the position wave function didn't change when we moved to the new coordinate system.

    This makes sense to me. Does this imply that if phi(r) had an angular dependence, we might not be able to proceed as above?
     
  12. Aug 5, 2008 #11

    Dick

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    I think the question is just what 'p' means in the wavefunction. I think it means sqrt(p.p). If it means p_z then it's a perfectly fine wavefunction, but it's not normalized. If you say, p = p z, then you are saying the momentum is always in the z direction. I just don't think that's what the problem means.
     
  13. Aug 5, 2008 #12
    Well considering that this is the momentum space wave function for the ground state of hydrogen, 'p' must mean sqrt(p.p) due to symmetry. Thanks for the help, this has given me a much improved understanding of momentum space and fourier transforms now.
     
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