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Wavefunction for particle inside a disk

  1. Jul 14, 2009 #1
    I am trying to solve for the wavefunction for a particle inside a 2 dimensional disk of radius r_0.
    The conditions are:
    [tex]\int^{2\pi}_0 \int^{r_0}_0 |\psi|^2\, r \,dr\, d\theta = 1[/tex]
    Then I tried separations of variables (set psi = R(r)Theta(theta)) to solve the Schrodinger equattion. didnt' really understand how it worked, but tried it anyway.

    I got [tex]R(r)=A_r r^{i\sqrt{2mE_r}\over\hbar}+B_r r^\frac{-i \sqrt{2mE_r}}{\hbar}[/tex] and [tex]\Theta(\theta) = A_\theta \cos \frac{\sqrt{2mE_\theta}}{\hbar} \theta) +B_\theta \sin \frac{\sqrt{2mE_\theta}}{\hbar} \theta) [/tex]
    (substituting [tex]K_i = \frac{\sqrt{2mE_i}}{\hbar}[/tex]) I plugged the wave function to the probability integral. After painstakingly doing that integral I got:
    [tex]\left [ |A_\theta|^2 \left ( \pi+\frac {\sin 4\pi K_\theta \theta}{2K_\theta} \right ) +|B_\theta|^2 \left ( \pi-\frac{\sin 4\pi K_\theta \theta}{2K_\theta}\right ) \right ] |A_r^2+B_r^2| \ln r_0 \left ( \frac{(-1+2 \ln r_0)r_0^2}{2}\right ) \left ( \frac{r^{2iK_r+2}}{2iK_r+2}-\frac{r^{-2iK_r+2}}{2iK_r-2} \right ) = 1[/tex]

    Now i can't solve for the energy eigenvalues because they are stuck in the deep parts of the equation (I set E = E_r + E_theta). Also where is the quantization? I know I screwed up somewhere.
     
    Last edited: Jul 14, 2009
  2. jcsd
  3. Jul 15, 2009 #2
    Hi,
    Just to make sure, what are you using for you Hamiltonian?

    Also, what that condition tells you is that [tex] \Psi(r) [\tex] is zero at r_0 (because the particle has a 100% chance of being between 0 and r_0, so it must be zero outside r_0 and wave functions are assuming to be continuous). Your function though does not appear to hold this to be true. You should wait to integrate and set to zero until you've applied boundary conditions.
     
    Last edited: Jul 15, 2009
  4. Jul 15, 2009 #3
    my hamiltonian was [tex]-\frac{\hbar^2}{2m} \left ( \psi_{rr} +\frac{\psi_r}{r}+\frac{\psi_{\theta\theta}}{r^2}\right ). [/tex] I will have to solve for the energy setting R(r0) = R'(r0) = 0.
    with these conditions I get:
    [tex]A_r r_0^{iK_r}+B_r r_0^{-iK_r}=0=A_r iK_r r_0^{iK_r-1}-B_r iK_r r_0^{-iK_r-1}[/tex]
    Nothing cancels. I can't frickin solve this.
     
    Last edited: Jul 15, 2009
  5. Jul 15, 2009 #4
    Why are you using something of this form?
    [tex]
    A_r {r_0}^{i K_r}
    [/tex]
    Try using a function of this form:
    [tex]
    A e^{i x}
    [/tex]
    (I just woke up, and I'm not generous enough in the mornings to offer hints about what x would be :P). Basically, I mean try something that would be sinusoidal over r.
     
  6. Jul 15, 2009 #5
    the laplacian in polar coordinates is (subscripts are derivatives)[tex] |\nabla|^2 f(r,\theta}) = f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2}[/tex]. (that's how i write the laplacian because it's an inner product of two dels = "absolute value of del squared". less of an abuse of notation than simply writing "del squared") Thus I will pick a function that will multiply r to the denom when it's differentiated, so that all of the terms in the laplacian are put on a common factor. Thus R(r) can't be an exponential. I guessed a function in the form of R(r)=r^c, c constant. I plugged r^c into the equation for R(r) and got my characteristic equation for c. I solved it and got [tex]c=\frac{\pm i\sqrt{2mE_r}}{\hbar}[/tex].
     
    Last edited: Jul 15, 2009
  7. Jul 15, 2009 #6
    Wait a sec, have you tried Bessel functions? It just occurred to me that if you have x=1/f, then the Schroedinger equation looks similar to Bessel's differential equation.
     
  8. Jul 15, 2009 #7
    yeah, i think i'll try bessel functions because my complex power function doesn't work for the equation. the dR/dr term doesn't cancel out.
    don't spoil the answer for me however.
     
    Last edited: Jul 15, 2009
  9. Jul 16, 2009 #8

    Avodyne

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    No, R(r0)=0 only, not also R'(r0)=0. With both the only solution is R(r)=0 for all r. This is like the one-dimensional infinite square well with walls at x=+a and x=-a, where we require the wave function to vanish but not also its derivative.
     
  10. Aug 5, 2009 #9
    but which bessel function should i use? according to my knowledge there are several types of bessel functions
     
  11. Aug 5, 2009 #10

    Avodyne

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    First you have to separate variables properly to get two equations, one for R(r) alone and one for Theta(theta) alone. Then you should find that your radial equation has two solutions (that is, two different Bessel functions solve it, before imposing boundary conditions), but that only one of these can be normalized; the other blows up at r=0. Finally, imposing R(r0)=0 will give you a condition that determines the energy eigenvalues in terms of the zeroes of the Bessel function.
     
  12. Aug 6, 2009 #11
    since i assumed [tex]\psi(r,\theta) = R(r) \Theta(\theta)[/tex]
    for the radial part of the wave equation i got r^2 R'' + r R' - (2mE_r / hbar^2)R = 0. this is not a bessel equation, it's an euler equation. so the general solution is not a bessle function. should be
    R(r) = A r^(-1 + sqrt(1 - (2mE_r / hbar^2))) + B r^(-1 - sqrt(1 - (2mE_r / hbar^2))).

    now, i don't want R to bolw up at 0, so sqrt(1 - (2mE_r / hbar^2)) can't be ±1.
     
    Last edited: Aug 6, 2009
  13. Aug 6, 2009 #12

    Avodyne

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    Science Advisor

    Your radial wave equation is wrong in two respects. First, the energy term should have a factor of r^2 (and what you call E_r is just the energy E). Second, you are missing the angular-momentum term that comes from introducing a separation constant when separating variables. Your equation (after fixing the first mistake) assumes Theta(theta)=constant.
     
    Last edited: Aug 6, 2009
  14. Aug 6, 2009 #13
    i did not assume the angular component was constant. infact you can see i tried to solve for it:
    [tex]\Theta(\theta) = A_\theta \cos \frac{\sqrt{2mE_\theta}}{\hbar} \theta +B_\theta \sin \frac{\sqrt{2mE_\theta}}{\hbar} \theta [/tex]

    oh really? where does the r^2 come from?
     
  15. Aug 6, 2009 #14

    Avodyne

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    Science Advisor

    Inside the disk, the time-independent Schrodinger equation is

    [tex]-{\hbar^2\over 2m}\nabla^2\psi(r,\theta) = E\psi(r,\theta).[/tex]

    Define [itex]k^2 \equiv 2mE/\hbar^2[/itex]. Then the Schrodinger equation becomes

    [tex](\nabla^2+k^2)\psi(r,\theta) = 0.[/tex]

    In polar coordinates,

    [tex]\nabla^2 = {\partial^2\over\partial r^2} + {1\over r}{\partial\over\partial r} + {1\over r^2}{\partial^2\over\partial\theta^2}.[/tex]

    So if we assume [itex]\psi(r,\theta) = R(r)\Theta(\theta)[/itex], then we have

    [tex]0 = (\nabla^2+k^2)\psi(r,\theta)[/tex]

    [tex]{} = \left[{\partial^2\over\partial r^2} + {1\over r}{\partial\over\partial r} + {1\over r^2}{\partial^2\over\partial\theta^2}+k^2\right]\!R(r)\Theta(\theta)[/tex]

    [tex]{}=\Theta(\theta)\!\left[{\partial^2\over\partial r^2} + {1\over r}{\partial\over\partial r}\right]\!R(r) + R(r)\!\left[{1\over r^2}{\partial^2\over\partial\theta^2}\right]\!\Theta(\theta) + k^2 R(r)\Theta(\theta)[/tex]

    [tex]{}=\Theta(\theta)[R''(r)+r^{-1}R'(r)+k^2 R(r)] + r^{-2}R(r)\Theta''(\theta).[/tex]

    Now divide both sides by [itex]r^{-2}R(r)\Theta(\theta)[/itex]; we get

    [tex]0={r^2[R''(r)+r^{-1}R'(r)+k^2 R(r)]\over R(r)} + {\Theta''(\theta)\over\Theta(\theta).[/tex]

    So we have two terms that sum to zero. One is a function of [itex]r[/itex] only, the other is a function of [itex]\theta[/itex] only. If we vary [itex]r[/itex] while holding [itex]\theta[/itex] fixed, the first term can vary, but the second cannot. But since they sum to zero, the first term cannot vary either; it must equal a constant. Call the constant [itex]\ell^2[/itex]. Then we have

    [tex]{r^2[R''(r)+r^{-1}R'(r)+k^2 R(r)]\over R(r)} = \ell^2[/tex]

    and

    [tex]{\Theta''(\theta)\over\Theta(\theta)}=-\ell^2.[/tex]

    The two linearly independent solutions of the second equation are [itex]\Theta(\theta)=\exp(+i\ell\theta)[/itex] and [itex]\Theta(\theta)=\exp(-i\ell\theta)[/itex]. We can treat both at once by taking [itex]\Theta(\theta)=\exp(+i\ell\theta)[/itex] but letting [itex]\ell[/itex] be positive or negative. The angle [itex]\theta[/itex] is periodic with period [itex]2\pi[/itex], and [itex]\Theta(\theta)[/itex] must be single valued, so [itex]\ell[/itex] must be an integer (positive, negative, or zero).

    The two linearly independent solutions of the first equation are then the Bessel functions [itex]J_\ell(kr)[/itex] and [itex]Y_\ell(kr)[/itex].
     
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