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Eigenstates of Hamiltonian

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The hamiltonian of a simple anti-ferromagnetic dimer is given by

    H=JS(1)[itex]\bullet[/itex]S(2)-μB(Sz(1)+Sz(2))

    find the eigenvalues and eigenvectors of H.
    2. Relevant equations



    3. The attempt at a solution
    The professor gave the hint that the eigenstates are of S2=(S(1)+S(2))2, S(1)2, S(2)2, and Sz. So I know I should have four eigenvalues. but I still have no Idea how to get this into a form that I recognize as being able to get eigenvalues from. (a matrix, a DiffEQ, etc.)

    Please help
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    vela

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    The hint is to help you to deal with the ##\vec{S}_1\cdot\vec{S}_2## term. Rewrite that term in terms of ##\vec{S}^2##, ##\vec{S}_1^2##, and ##\vec{S}_2^2##.
     
  4. Feb 12, 2013 #3
    When I do that, and apply the spin operators, S2 ket (S,Sz)=s(s+1) ket (s,sz) and Sz ket (S,Sz) = szket (s,sz)(sorry, couldn't find the ket symbol in latex)
    I get
    H = J/2 (s(s+1) - s1(s1+1)-s2(s2+1))-μB(s1z+s2z)

    Is this correct?
     
  5. Feb 12, 2013 #4

    vela

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    Yes. Now you can calculate what H does to simultaneous eigenstates of ##\vec{S}^2##, ##S_z##, ##\vec{S}_1^2##, and ##\vec{S}_2^2##. Recall that these are exactly the states that you got from adding angular momenta.
     
  6. Feb 12, 2013 #5
    I get four answers : J/4 + μB, J/4 -μB, J/4 and - 3J/4. Is this right? These look like the singlet and triplet state energies, but with an added B term.
     
  7. Feb 12, 2013 #6

    vela

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    Yeah, that looks right.
     
  8. Feb 12, 2013 #7
    Thanks so much Vela! you've been a wonderful help.:smile:
     
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