1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Spin-spin interaction

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the Hamiltonian
    [tex]H=\alpha L \cdot S[/tex]
    Where L denotes an angular momentum with quantum number l and S a spin with quantum number s.

    Work out
    [tex]L \cdot S |(ls)jj_{z}>[/tex]
    direction. Hint: expand (L+S)2 and go from there.

    2. The attempt at a solution

    I'm highly tempted to start with
    [tex]L \cdot S |(ls)jj_{z}> = \frac{\hbar^{2}}{2}[j(j+1)-l(l+1)-3/4]|(ls)jj_{z}>[/tex]
    except I'm not sure that really buys me much. If I'm on the right track, how do I then handle the ket?
  2. jcsd
  3. Nov 8, 2011 #2
    Ok, so my first mistake was probably in my first crack at the solution; so I should probably have started with
    [tex]L \cdot S |(ls)jj_{z}> = \frac{\hbar^{2}}{2}[j(j+1)-l(l+1)-s(s+1)]|(ls)jj_{z}>[/tex]
    However, my initial self doubts still stand though.
  4. Nov 8, 2011 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's fine so far. Why are you doubting it's right?

    (If you want to derive it, use the hint.)
  5. Nov 9, 2011 #4
    I'm doubting it almost out of habit, lately I've started down the wrong path more often than not.

    So, given that I've actually got this one going correctly, what do I do with the information in the ket?

    And as a more in-depth question; what does the alpha term do to things when I want to calculate the energy (full disclosure: the follow up question actually is to calculate the energy spectrum, where I would tend to want to start from the TISE)?
    Last edited: Nov 9, 2011
  6. Nov 10, 2011 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm not sure what you mean by this.

    It's just a multiplicative constant.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook