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

Homework Help: Spherical symmetri of eigenfunctions?

  1. Sep 10, 2006 #1
    I have been given these 4 eigenfunctions of the hydrogen atoms first 2 n-shells.

    [tex] \psi_{100}(r, \theta, \phi )=\frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0} [/tex]

    [tex] \psi_{200}(r, \theta, \phi )=\frac{1}{\sqrt{8\pi a^3_0}}(1-\frac{r}{2a_0})e^{-r/2a_0} [/tex]

    [tex] \psi_{210}(r, \theta, \phi )=\frac{1}{4\sqrt{2\pi a^3_0}}(\frac{r}{a_0})e^{-r/2a_0} cos\theta[/tex]

    [tex] \psi_{21\pm 1}(r, \theta, \phi )=\pm\frac{1}{8\sqrt{\pi a^3_0}}(\frac{r}{a_0})e^{-\frac{r}{2a_0}} sin\theta e^{\pm i\phi}[/tex]

    Where [tex] a_0 [/tex] is the Bohr radius.

    I am suposed to show that the superposition
    [tex] |\psi_{nlm}(r, \theta, \phi )|^2 [/tex] is sphericaly symmetric within each shell.

    Now what I dont know is how do I show spherical symmetri(not even generaly and not just in this particular case). Any hints?:confused:
  2. jcsd
  3. Sep 10, 2006 #2
    do I basicly just show that the superposition has the same value in a arbitrary r and -r? and the same for [tex] \phi [/tex] and [tex] \theta [/tex]
  4. Sep 11, 2006 #3
    I guess I should rephrase.

    How do I show that any function is sphericaly symetric? Right now my brain has frozen even though I know it must be ridicilously simple.

    Ignore the -r brainfart in the previous post btw.

    What I mean is that should I show the function is the same if I hold r, theta constant and replace phi with -phi and then if I hold r and phi constant and replace theta with -theta?
    Last edited: Sep 11, 2006
  5. Sep 11, 2006 #4
    the first one
    [tex] |\psi_{100}|^2 = \frac{1}{\pi a_0^3}e^{-\frac{2r}{a_0}} [/tex]
    this is obviously sphericaly symmetric since it only depend on the radius. Its a sphere pure and simple. But is that proof enough?

    The second one

    [tex] |\psi_{200}|^2 = \frac{1}{8\pi a_0^3}[1-\frac{r}{2a_0}]^2 e^{-\frac{r}{a_0}} [/tex]

    same as above? only radialy dependant=the function gest weaker in the radial direction, obviously spericaly symetric.

    third one

    [tex] |\psi_{210}|^2=\frac{r}{32\pi a^3_0}e^{-\frac{r}{a_0}} (cos\theta)^2[/tex]

    Now this one. If I hold r constant and replace theta with -theta I get the same answere. So does that mean sphericaly symetric? Do I just have to say that to prove it?

    In the fourth one at the end

    The [tex] sin\theta e^{\pm i\phi} [/tex] should I read that as [tex] sin(\theta e^{\pm i\phi}) [/tex] or [tex] sin(\theta) e^{\pm i\phi} [/tex] ??
    Last edited: Sep 11, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook