# Electron orbital angular momentum

1. Jan 6, 2008

### ijustlost

Why can't an electron in a coulomb field have an orbital angular momentum quantum number higher than it's principal quantum number (ie there is no 1p state etc.)

I think is probably something I learnt at one point, but I've forgotten and can't seem to find anything about it anywhere. I know that it's a consequence of solving the radial Schrodinger equation, but I'd like to figure out why this should be, at least in vague terms.

2. Jan 6, 2008

### olgranpappy

"vague terms", huh? I'll give it my best shot:

the ground state must have the same symmetry as the potential, thus the ground state must be spherically symmetric, thus $\ell=0$.

3. Jan 6, 2008

### ijustlost

Cheers, that makes sense. But why then may the first excited state have l=0,1 but not higher, the 2nd excited state have l=0,1,2 etc. I'm guessing the higher the energy the more the state may 'break' the symmetry of the potential.

And also in a nuclear potential there isn't any such restriction (ie there are 1p, 1d, 1f states etc), but the nuclear potential is spherically symmetric. Why is this?

4. Jan 6, 2008

### olgranpappy

I don't know if this is the best line of reasoning to continue--I was just giving you a cute, vague reason why the ground state should have l=0. I think that the only real way to understand the behavior of the quantum numbers really is through the analysis of the solutions to the differential equation to which they are solutions. I.e., in order for the solutions to be normalizable (physical), the quantum number l must be leq to the quantum number n. Remember, when you solve for the bound states you try a power series ansatz and get a condition on n and l in order to make the series finite...

Is it? I think the nuclear potential is a pleasant fiction

The nuclear problem is a many body problem with unknown "potentials". It's quite a bit different than the hydrogen atom... It's funny that a shell model works at all...

5. Jan 7, 2008

### Marco_84

Well if you solve schrodinger equation factorizing the solution (as usual for a PDE) you can find the solution using Fouier tools (series, orthonormal sets...) whence you have found the coefficients you see how they are related... there are many books that make this calculation... i did it once it took me two days to get every minus or plus sign....

So... good work... take the tables of the laplacian in spherical coordinates and... blah bla bla...

ciao

6. Jan 7, 2008

### Staff: Mentor

In the pre-QM Bohr-Sommerfeld model, the electron has an elliptical orbit, in general. Orbits with different ellipticities have different amounts of orbital angular momentum, for the same energy. Zero angular momentum corresponds to an "orbit" that is actually a straight line passing through the nucleus. The maximum angular momentum for a classical orbit in general, is for a circular orbit.

I'm in a hurry right now and don't have time to work this out, but I wouldn't be surprised to find that the maximum value of the orbital quantum number corresponds to a circular orbit in the Bohr-Sommerfeld model, or at least nearly a circular orbit, in the limit of large values of n (energy level). It clearly wouldn't work for the ground state, in which the orbital angular momentum is zero.

And of course we know the Bohr-Sommerfeld model is ultimately wrong, so it's dangerous to make analogies using it...

7. Jan 7, 2008

### reilly

The issue is much the same as throwing out the harmonic oscillator solutions that go as exp(+r^^2). Change the sign, and we get the normal solutions of Hermite Functions.

ijustlost -That is, this is a very standard problem in differential equations. To see what's going on, use the standard approach of factoring out the behavior at 0 and infinity, and then go for a power series solution. It will all become clear.

jtbell -- Interesting approach. Very likely will work OK for large n, the principle quantum number.

Regards,
Reilly Atkinson

8. Jan 13, 2008

### ijustlost

reilly - I understand how it comes from solving the equations, I was just wondering if there was a nice 'classical' picture, but I guess there isn't really one that isn't terribly inaccurate.

jtbell - yeah makes sense