Can the Orbital Angular Momentum Quantum Number Only Be Integer Values?

[king vitamin]

I'm being asked on a homework to show that the m orbital angular momentum quantum number can only take integer values. Using ladder operators I know how to prove that m is restricted to half-integers, but I'm having trouble with a further restriction. I'm quite certain the problem does not want me to generally solve Laplace's Equation for spherical harmonics.

The method which I've seen but has trouble convincing me is the following:

$$-i \hbar \frac{\partial \Phi}{\partial \phi^2} \ = \ m \ \hbar \ \Phi$$
$$\Phi \ = \ e^{im \phi}$$

Now this requirement:
$$\Phi(\phi + 2 \pi) \ = \ \Phi(\phi)$$

Producing the desired quantization. This feels a little artificial though - on one hand it seems obvious that the wavefunction should be single-valued at a point, the addition of a phase factor wouldn't change any predictions, would it? Any calculations of an observable for a point would be the same under a full rotation, unless I'm missing something. Is there a way to either make this more concrete or a better way to show this quantization?

 

[thebigstar25]

im taking the same course now, and we are using "Stephen Gasiorowicz" book .. I am not sure what is your question, would you please clarify it maybe I can be helpful..

 

[king vitamin]

How do you prove m must be an integer?

 

[thebigstar25]

well, I am not sure from what I am going to say, but as far as I think, since the solution should be periodic, only integer values of m would fulfil the requirement of the solution..

 

[king vitamin]

yes, but non-integer valued m values would only multiply the wavefunction by a phase factor of magnitude 1, which, as far as I know, has no effect on any predictions that can be made with quantum theory. unless there's something I'm not thinking of?

 

[thebigstar25]

$$-i \hbar \frac{\partial \Phi}{\partial \phi^2} \ = \ m \ \hbar \ \Phi$$
$$\Phi \ = \ e^{im \phi}$$

Now this requirement:
$$\Phi(\phi + 2 \pi) \ = \ \Phi(\phi)$$

I think at this point to have a complete picture of what m should be, you have to consider

$$\Phi(\phi + 2 \pi) \ = \ \Phi(\phi)$$

having integer m values will result in having (n2pi) .. the thing which would complete the requirement of the solution.. (that is atleast what I think, I am not sure about the other members) ..

 

[king vitamin]

I think at this point to have a complete picture of what m should be, you have to consider
$$\Phi(\phi + 2 \pi) \ = \ \Phi(\phi)$$

Why can we assume this?

 

[thebigstar25]

because we are considering the phi dependence here..

 

[king vitamin]

What I've been trying to say since post 1 is that, even though a periodic phi is required for the function to be single valued at a point, if m were not an integer, it wouldn't actually give different answers since the mod of the wave function would not be effected.

That is, if you multiply any given wave function by $$e^{i \phi}$$ where phi is any real number, the wave function would give the same results as the original.

I understand that if we need the wave function to be single valued, we need m to be an integer, but is there such a constraint? It seems to me that any well-defined state has an infinite number of possible values which differ by only a phase factor.

 

[nickjer]

You are basically saying that you can multiply the wavefunction by any phase factor, and the probability of the wavefunction will be unaffected by this factor. That isn't entirely true for all cases. One case being a superposition of two wavefunctions with different phase factors out front.

But to make a long story short, your wavefunction must satisfy the Hamiltonian you are given. Since there are no discontinuities in the Hamiltonian (the potential is usually a smooth continuous function). Then your wavefunction should also be continuous. If your wavefunction was discontinuous at the 0 and 2*Pi, then you would be unable to satisfy the Hamiltonian you were given.

An example where you would see a discontinuity is if you had a dirac-delta potential. In this case you are required to have a discontinuous first derivative to satisfy the Hamiltonian.