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Quantum 2

  1. Apr 9, 2009 #1
    The spin-orbit interaction in Hydrogen adds an extra term [itex]\alpha \mathbf{L} \cdot \mathbf{S}[/itex] to the Hamiltonian of the system. If the electron is in an energy eigenstat show that it cannot also be in an eigenstate of either [itex]L_z[/itex] or [itex]S_z[/itex].

    I have that the modified Hamiltonian is given as [itex]\hat{H}_{S-O}=f(r) \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}[/itex]. i have in my notes that [itex]f(r)=\frac{1}{2M^2c^2r} \frac{dV(r)}{dr}[/itex]. this is a past exam question so im guessing its probably asking a bit much to memorise exactly what f(r) is so i reckon it should be manageable using just [itex]\hat{H}_{S-O}=f(r) \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}[/itex].

    but have no idea how to proceed...

    in my notes they somehow substitute [itex]\mathbf{\hat{L}} \cdot \mathbf{\hat{S}}=[\hat{J}^2-\hat{L}^2-\hat{S}^2][/itex].
     
  2. jcsd
  3. Apr 9, 2009 #2
    I don't really know what you're talking about in the middle section but to show that "If the electron is in an energy eigenstat show that it cannot also be in an eigenstate of either [itex]L_z[/itex] or [itex]S_z[/itex]." You just have to show that the commutator of H with Lz and Sz is non-zero (i.e. they are not simultaneous observables). If the original hamiltonian commutes with Sz then all you have to show is that [Hso,Sz] and [Hso,Lz] don't equal zero. As for your relation at the bottom that just comes from:

    [itex]J^2=(L+S)^2=L^2+S^2+2L \cdot S[/itex]
    [itex]L \cdot S =\frac{J^2-L^2-S^2}{2}[/itex]
     
  4. Apr 10, 2009 #3
    so we assume that [itex][\hat{H}.\hat{S_z}]=0[/itex]

    now if this "perturbation" changes [itex]\hat{H} \rightarrow \hat{H}'[/itex] then

    [itex][\hat{H}',\hat{S_z}]=\hat{H}' \hat{S_z} - \hat{S_z} \hat{H}'[/itex]
    [itex]=\left(\hat{H}+\hat{L} \cdot \hat{S} \right) \hat {S_z} - \hat{S_z} \left(\hat{H}+\hat{L} \cdot \hat{S} \right)= \left( \hat{L} \cdot \hat{S} \right) \hat{S_z} - \hat{S_z} \left(\hat{L} \cdot \hat{S} \right)[/itex]
    i don't understand why this would be non-zero?
     
  5. Apr 10, 2009 #4
    Well, what is [tex]\vec{L}[/tex] [tex]\vec{S}[/tex]?
    Try to expand it into J, L, S.
    And you know, if Sz commute with each one of them, then it commutes with [tex]\vec{L}[/tex] [tex]\vec{S}[/tex].
     
  6. Apr 10, 2009 #5
    so i can expand L.S =1/2(J^2-L^2-S^2)

    so S_z will commute with S^2

    but how do i know about J^2 and L^2?
     
  7. Apr 10, 2009 #6
    my bad, I just run back and read my note, and there is an even more simple way to do it
    What is [tex]\vec{L}[/tex]dot[tex]\vec{S}[/tex], by definition.
    It is actually [tex]L_x[/tex][tex]S_x[/tex]+[tex]L_y[/tex][tex]S_y[/tex]+[tex]L_z[/tex][tex]S_z[/tex]
    Now, could you tell why they don't commute?
     
  8. Apr 10, 2009 #7
    is it because

    [itex][\hat{S_i},\hat{S_j}]=i \hbar \epsilon_{ijk} \hat{S_k} \neq 0[/itex]
     
  9. Apr 10, 2009 #8
    Bingo. And later on, or right now, you'll find out that J is in fact the one that commutes with L dot S. And although individual L_z, S_z don't commute with L dot S, their linear combination, J_z does.
     
  10. Apr 10, 2009 #9
    Just one more note. Although this is the essential reason, you probably want to do the evaluation because it is a nice practice.
     
  11. Apr 10, 2009 #10
    ok. can we just quickly recap the theory behind this.

    we want to show [itex][\hat{H}',\hat{S_z}] \neq 0 , [\hat{H},\hat{L_z}] \neq 0[/itex] as then we know that they aren't simultaneous observables and hence if we are in an energy eigenstate, we cann;t also be in an eigenstate of either [itex]\hat{S_z}[/itex] or [itex] \hat{L_z}[/itex] by the Compatibility Theorem. Is this true?
     
  12. Apr 10, 2009 #11
    Yeap, you got it. That thm is really powerful. You'll probably encounter it through our your class (although now I am wondering whether it is if and only if. It should anyway).
     
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