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Hamiltonian eigenvalues

  1. Oct 3, 2006 #1
    Find the eigenvalues of the hamiltonian

    [tex]
    H=a(S_A \cdot S_B+S_B \cdot S_C+S_C \cdot S_D+S_D \cdot S_A)
    [/tex]

    where S_A, S_B, S_C, S_D are spin 1/2 objects
    _________________________

    I rewrite it as

    [tex]
    H=(1/2)*a*[(S_A+S_B+S_C+S_D)^2-(S_A+S_C)^2-(S_B+S_D)^2]
    [/tex]

    then i define

    [tex]
    J_1=S_A+S_B+S_C+S_D
    [/tex]

    [tex]
    J_2=S_A+S_C
    [/tex]

    [tex]
    J_3=S_B+S_D
    [/tex]

    and uses

    [tex]
    J^2_i |j_1j_2j_3;m_1m_2m_3> = (h^2) j_i(j_i+1)|j_1j_2j_3;m_1m_2m_3>
    [/tex]

    which gives the energies

    [tex]
    E(j_1,j_2,j_3)=(h^2/2)*a*[j_1(j_1+1)-j_2(j_2+1)-j_3(j_3+1)]
    [/tex]

    Where j_1 is addition of four angular momentum of 1/2 which gives it values of 0 1, 2 and in the same way j_2 and j_3 have values of 0 1.

    Am i doing this the right way? It doesnt feel so :smile:
     
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2
    From the structure of your hamiltonian it almost looks like you could adapt transfer matrix methods, unless your spin things are vectors (I'm not clear on that). I would also say that there are restrictions on [tex]j_2[/tex] and [tex]j_3[/tex] based on [tex]j_1[/tex], but the thought process seems right.
     
  4. Oct 3, 2006 #3

    Meir Achuz

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    Your method is completely correct. Just include the a.
    Did it just seem too easy?
     
  5. Oct 3, 2006 #4
    Thank you. Yes it seemed too easy :smile:
     
  6. Oct 6, 2006 #5
    How about the degeneracy of the energy levels.
    For example E(010)=E(001)=E(111) and then m_1 can take on 9 different values , m_2 and m_3 5 different values. So the degeneracy of this level is 3*9*5*5 ? Is it correct so far?

    But then the j_i in turn are addtions of angular momentums.
    Does this add even more to the degeneracy?
     
  7. Oct 6, 2006 #6

    Meir Achuz

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    I would count the degeneracy of each estate as the product of 2j+1 for each sub j. So I think it is 3+3+27 for your example. You shouldn't just count the m's because they are correlated to give the j estates.
     
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