# Angular momemtum question?

1. Feb 17, 2006

### waht

I'm doing some reading on quantum mechanics and have a quick question about angular momentum.

In the derivation of the eigen values of J (which is either L or S), it starts out like this:

Jz (Pm) = hm (Pm)

Where Pm is Phi sub m, h is h bar and m is the eigen value. (sorry my latex is a little rusty). And then they explain using J+ or J- which I understand.

So my question is why did they include the h bar, where does it come from?

And same thing goes with the derivation of J^2.

2. Feb 17, 2006

### Gokul43201

Staff Emeritus
It comes from :

1. The definition of angular momentum, $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, and

2. The position representation of the momentum operator, $\mathbf{p} = - i\hbar ~\nabla$

The $\hbar$ in 2 above comes from construction. It was chosen, by Dirac if I'm not mistaken, and it works.

Last edited: Feb 17, 2006
3. Feb 17, 2006

Staff Emeritus

My understanding is if the amplitude wave goes through one phase cycle you can imagine that as a point rotating around a circle, and the total action for one rotation would be h, so the average rate of increase of action is h over the angular measure of a complete circle, $$2\pi$$.

4. Feb 17, 2006

### Galileo

If they start the derivation like that, I can only assume that they set out to find the eigenvalues and they name the eigenvalue hm. There's no loss of generality, since at this stage m could be any real number. Only after you find that m must be a multiple of 1/2 you'll see it was a good choice. Same thing for J^2.

5. Feb 17, 2006

### abszero

In most modern texts I've seen it derived by looking at the generators of rotations, and then concluding the commutation relations
$$[L_\imath, L_\jmath] = \imath \epsilon_{\imath \jmath k} \hbar L_k$$ and then realizing that, like how momentum is the generator of space translations, angular momentum is the generator of rotations.

6. Feb 18, 2006

### Perturbation

The h-bar also allows the eigenvalues m to be dimensionless, as h has the same units as angular momentum. As was said above, there's no loss of generality.

7. Feb 20, 2006

### dextercioby

Incidentally [action]=[angular momentum]. Why "h" bar and not simply h...? Well, Dirac invented it since h/2pi always appeared in Schroedinger wave mechanics...

Daniel.