# Angular Momentum squared operator (L^2) eigenvalues?

1. Dec 3, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
Find the eigenvalues of the angular-momentum-squared operator (L2) for hydrogen 2s and 2px orbitals...

2. Relevant equations
Ψ2s = A (2-r/a0)e-r/(2a0)
Ψ2px = B (r/a0)e-r/(2a0)
3. The attempt at a solution
If I am not wrong, is the use of L2 in eigenfunction L2Ψ = ħ2 l(l+1) Ψ?

...so I wonder if eigenvalues are l(l+1) from this part, ħ2 l(l+1) Ψ, as in:

for Ψ2s, because for s orbital, l = 0--> l(l+1) = 0(0+1) = 0
for Ψ2px, because for p orbital, l = 1--> l(l+1) = 1(1+1) = 2 ?

2. Dec 4, 2014

### Staff: Mentor

That's only the radial part. You are missing the angular part, which is essential ;)

How is an eigenvalue defined?

3. Dec 4, 2014

### terp.asessed

OMG, I was like, why does the equation look funny----the correct one for Ψ2px would be B (r/a0)e-r/(2a0)sinθcosφ, right?

Also, as how an eigenvalue is defined....well, from one previous eigenfunction example I solved, the one involving Hamiltonian operator, HΨ = EΨ, E is the eigenvalue?
So, is the eigenvalue for Ψ2px2?

4. Dec 4, 2014

### Staff: Mentor

Right.

Yes. For an operator $\hat{A}$, $\hat{A} \psi = a \psi$ where $a$ is a scalar means that $\psi$ is an eigenfunction of $\hat{A}$ and $a$ its corresponding eigenvalue.

Yes.

I don't know what the level of the question is, so I don't know how simple the answer can be. If this is an advanced problem, you probably are expected to actually derive the eigenvalue.

5. Dec 4, 2014

### terp.asessed

Thank you! I am quite surprised myself at how simple the answer turns out to be.....but still thank you.