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!

What is the eigenvalue of angular momentum? (Zeeman)

  1. May 15, 2014 #1
    1. The problem statement, all variables and given/known data

    In the calculation of the Zeeman Effect, the most important calculation is

    [tex]\langle L_z + 2S_z \rangle = \langle J_z + S_z\rangle[/tex]

    Suppose we want to find the Zeeman Effect for ##(2p)^2##, meaning ##l=1##.

    In Sakurai's book,

    ZGH5T.png

    My question is, what is ##m##? They say that ##m## is the eigenvalue of ##J_z##, meaning ##J_z = l + m_s = \frac{3}{2}## in this case.


    2. Relevant equations



    3. The attempt at a solution

    Using ##m = \frac{3}{2}##, it gives the wrong Gordon-Clebsch coefficients.

    I have worked out the correct form, which is adding orbital angular momentum ##(l=1)## to its spin ##(\pm\frac{1}{2})##

    [tex]|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1}{3}}|1,1\rangle|-\rangle + \sqrt{\frac{2}{3}}|1,0\rangle|+\rangle[/tex]
     
    Last edited: May 15, 2014
  2. jcsd
  3. May 16, 2014 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Why do you assume ##m_l=l##?


     
  4. May 16, 2014 #3
    my biggest problem is, they say that the eigenvalue of ##J_z## is ##m##, but then ##m = \frac{1}{2}## gives the correct gordon-clebsch coefficients. Does this mean that ##J_z = \frac{1}{2}##? But isn't ##J_z## the TOTAL angular momentum? Shouldn't it be ##J_z = l + s = \frac{3}{2}##?
     
  5. May 16, 2014 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    ##J_z## is the z-component of the total angular momentum. It can take on any value from ##j## to ##-j##. According to the rules about adding angular momenta, ##j## can take on any value from ##l+s## to ##\lvert l-s \rvert##.
     
  6. May 16, 2014 #5
    you seem to be confusing l with ml and s with ms. If l = 1 than ml = -1, 0, 1 are all possible, and if s = 1/2 than ms = +1/2 and -1/2 are possible. m = ml + ms
     
  7. May 16, 2014 #6
    If that's the case, then for ##l=1## and ##m_s = 1##:

    [tex]|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1 - (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|1\rangle|-\rangle + \sqrt{\frac{1 + (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|0\rangle|+\rangle[/tex]

    These coefficients are wrong.
     
  8. May 16, 2014 #7
    For j = 3/2, l = 1, and m = 1/2 I'm getting

    |j,m> = |3/2,1/2> = |1+1/2,1/2> = [(l+m+1/2)/(2l+1)]1/2|m-1/2,1/2> + [(l-m+1/2)/(2l+1)]1/2|m+1/2,-1/2>
    |j,m> = [(1+1/2+1/2)/(2*1+1)]1/2|1/2-1/2,1/2> + [(1-1/2+1/2)/(2*1+1)]1/2|1/2+1/2,-1/2>
    |j,m> = (2/3)1/2|0,1/2> + (1/3)1/2|1,-1/2>
    |3/2,1/2> = (2/3)1/2|1,0>|+> + (1/3)1/2|1,1>|->
     
  9. May 16, 2014 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    For the state ##\left\lvert \frac{3}{2},\frac{1}{2} \right\rangle##, you have ##j=3/2## and ##m=1/2##. In your expression for the coefficients, you're using ##m=3/2##, which is why they're not coming out right.
     
  10. May 17, 2014 #9
    Damn, I should have been more careful. Thanks alot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted