1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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

    Gokul43201

    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

    selfAdjoint

    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

    Galileo

    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

    dextercioby

    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...


    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Angular momemtum question?
Loading...