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Quantum mechanics: free partical in spherical coordinates

  1. Jun 21, 2008 #1
    1. The problem statement, all variables and given/known data
    My wavefunction is [tex]\psi (r, \theta, \phi )=N cos(\theta) e^{-(r/R_0)^2}[/tex].

    I need to calculate [tex]<p_r>[/tex] and [tex]\Delta p_r[/tex] where [tex]p_r[/tex] is the radial momentum.

    2. Relevant equations
    I think i know [tex]p_r=\frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right) [/tex].

    3. The attempt at a solution

    When I try to calculate the observation value I got infinity (the integral does not seem to converge):

    [tex]<\psi | p_r | \psi > = \int_0^{2 \pi}d\phi \int_0^\pi d\theta \int_0^{\infty}dr [\psi^\star p_r \psi] [/tex]

    Are the limits for the integral correct? What am I doing wrong? :(

  2. jcsd
  3. Jun 21, 2008 #2


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    [tex] ( \psi |p_r|\psi) = \frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right) N^2 cos^2( \theta) e^{-2(r/R_0)^2} [/tex]...agreed?


    [tex] < \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr [/tex]

    [tex] = \frac{ N^2 \hbar \pi^2}{i} \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr [/tex]

    Since the theta int. goes to pi/2 and the phi int. goes to 2pi

    Is this the integral you ended up computing?

  4. Jun 21, 2008 #3
    Hi Josh,
    Thanks for answering.

    I ended up with this:
    [tex]< \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left( \frac{d}{dr}+ \frac{1}{r} \right) \right e^{-2(r/R_0)^2} dr [/tex]

    My problem is with [tex]\int_0^\infty \frac{1}{r} e^{-2(r/R_0)^2} dr[/tex] which does not converge.
  5. Jun 21, 2008 #4


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    Ok, we're on the same page then with the equations.

    The r integral will separate into two integrals after you distribute the variables in parenthesis.... that is, one term is d/dr of the e-fcn, and the other is 1/r times the e-fcn.

    Because of this, its ok for the 1/r term to blow up as long as the other one does not (I haven't computed either one of them, so let me know if there is an additional problem..)

    Easiest way to integrate these, in case you didnt already know, is with the wolfram online integrator:

  6. Jun 21, 2008 #5
    Are you sure? Can I just disregard the inf in the equation?!?

    thanks again for answering.
  7. Jun 21, 2008 #6


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    Yes, if a term goes to infinity it is called 'blowing up'

    This is a common situation when evaluating boundary criterion in, for example, the finite square well, where the psi contains e^x and e^-x for x>0, and since as x-> infty e^x blows up, psi reduces to just the e^-x term.
  8. Jun 21, 2008 #7
    the integral you have is incorrect, i believe.
    [tex] dV = r^{2} sin\theta d\theta d\phi dr [/tex]

    Edit: in your equation you have:
    [tex] dV = dr d\phi d\theta [/tex]
  9. Jun 21, 2008 #8
    the r^2*sin(t) is the |Jacobian| isn't it? I thought it should only be added when making a transformation between Cartesian and spherical coordinates, i.e [tex]\int\int\int dxdydz = \int\int\int r^2 sin(\theta) d\phi d\theta dr[/tex]...

    although the r^2 sin(t) will solve the integral convergence problem, i don't understand yet if it is necessary.
    Last edited: Jun 21, 2008
  10. Jun 21, 2008 #9
    Well, the volume element I suggested to you is just the general form of a volume integral in spherical coordinates. Check it on a sphere to see what you get. The transformation you give is correct. You are essentially doing a transformation from cartesian to polar spherical here (if you want to look at it like you started in cartesian). Regardless of which, you must use dV = r^2 sin(theta) dphi dtheta dr if you're integrating over phi, theta, and r, in spherical coordinates. You should draw out the element, and you will see why you must do this. dV = dr dtheta dphi doesn't make sense as you will see if you try to draw out the element.
  11. Jun 21, 2008 #10
    \int\int\int_{D} \Psi(x,y,z)^{*}p\Psi(x,y,z) dxdydz = \int\int\int_{D} \Psi(r,\phi,\theta)^{*}p\Psi(r,\phi,\theta) r^2 sin(\theta) d\phi d\theta dr[/tex]
  12. Jun 21, 2008 #11
    kreil and EngageEngage: thanks a lot!

    I now get that the observation value for the radial momentum is 0. does it have a physical meaning? is there any way of anticipating that result by looking at the wavefunction? and should I even try to search physical meanings in quantum mechanics questions?

    thanks again.
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