Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular momemtum question?

  1. Feb 17, 2006 #1
    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. jcsd
  3. Feb 17, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It comes from :

    1. The definition of angular momentum, [itex]\mathbf{L} = \mathbf{r} \times \mathbf{p} [/itex], and

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

    The [itex]\hbar[/itex] in 2 above comes from construction. It was chosen, by Dirac if I'm not mistaken, and it works.
    Last edited: Feb 17, 2006
  4. Feb 17, 2006 #3


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    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, [tex]2\pi[/tex].
  5. Feb 17, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    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.
  6. Feb 17, 2006 #5
    In most modern texts I've seen it derived by looking at the generators of rotations, and then concluding the commutation relations
    [tex][L_\imath, L_\jmath] = \imath \epsilon_{\imath \jmath k} \hbar L_k[/tex] and then realizing that, like how momentum is the generator of space translations, angular momentum is the generator of rotations.
  7. Feb 18, 2006 #6
    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.
  8. Feb 20, 2006 #7


    User Avatar
    Science Advisor
    Homework Helper

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook