Spherical symmetri of eigenfunctions?

[Azael]

I have been given these 4 eigenfunctions of the hydrogen atoms first 2 n-shells.

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

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

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

\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}

Where a_0 is the Bohr radius.

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

Now what I don't know is how do I show spherical symmetri(not even generaly and not just in this particular case). Any hints?

 

[Azael]

do I basicly just show that the superposition has the same value in a arbitrary r and -r? and the same for \phi and \theta

 

[Azael]

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?

 

[Azael]

the first one
|\psi_{100}|^2 = \frac{1}{\pi a_0^3}e^{-\frac{2r}{a_0}}
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

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

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

third one

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

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 sin\theta e^{\pm i\phi} should I read that as sin(\theta e^{\pm i\phi}) or sin(\theta) e^{\pm i\phi} ??