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Spin-dependent Hamiltonian of two particles

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Two spin-half particles with spins S1 and S2 interact with a spin-dependent Hamiltonian H=λS1*S2 (the multiplication is a dot product and is a positive constant). Find the eigenstates and eigenvalues of H in terms of |m1,m2>, where (hbar)m1 and (hbar)m2 are the z-components of the two spins.


    2. Relevant equations
    Sx |m>=1/2(Sp-Ss) |m>
    Sy |m>=1/2i(Sp+Ss) |m>
    Sz |m>=(hbar)m |m>
    Sp=(hbar)√[s(s+1)-m(m+1)]
    Ss=(hbar)√[s(s+1)-m(m-1)]
    3. The attempt at a solution
    S1*S2=S1xS2x+S1yS2y+S1zS2z

    S1x=S2x=S1y=S2y=0. I said this because the problem only mentioned z-component and most problems only talk about Sz.

    H|m1,m2>=λSz1Sz1|m1,m2>

    H |1/2,1/2> = λ*(hbar)^2 (1/2)(1/2) |1/2,1/2> = λ*(hbar)^2/4 |1/2,1/2>
    H |-1/2,1/2> = λ*(hbar)^2 (-1/2)(1/2) |1/2,1/2> = -λ*(hbar)^2/4 |-1/2,1/2>
    H |1/2,-1/2> = λ*(hbar)^2 (1/2)(-1/2) |1/2,1/2> = -λ*(hbar)^2/4 |1/2,-1/2>
    H |-1/2,-1/2> = λ*(hbar)^2 (-1/2)(-1/2) |-1/2,-1/2> = λ*(hbar)^2/4 |-1/2,-1/2>

    Is this my final answer? Am I close? Or was I completely off.
     
  2. jcsd
  3. Jan 23, 2012 #2

    vela

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    Those are operators. You can't arbitrarily set them equal to 0.

    Read about the addition of angular momentum. For this problem, consider ##(\vec{S}_1 + \vec{S}_2)^2##.
     
  4. Jan 23, 2012 #3
    I get what you are saying about the operators. That was just a bad attempt at trying to simply the problem. I get that (S1+S2)^2 would yield the total spin, but how would that play into the Hamiltonian.
     
  5. Jan 23, 2012 #4

    vela

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    Expand ##(\vec{S}_1 + \vec{S}_2)^2## it out. What do you get?
     
  6. Jan 24, 2012 #5
    Thank you, I had figured it out last night with your hint. The expansion leaves S1^2+S2^2+2S1S2. Then the dot product of S1S2 gave S1xS1y+S2xS2y+S1zS2z. The Z component was simple that was the m(hbar) and the x and y components I wrote in terms of the ladder operators. Thanks again :)
     
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