Angular Momentum and the Role of h-Bar in Quantum Mechanics

[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.

 

[Gokul43201]

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

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.

 

[selfAdjoint]

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.

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.

 

[Galileo]

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.

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.

 

[abszero]

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.

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.

 

[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.

 

[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.