Spin-dependent Hamiltonian of two particles

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 5K views
Jufro
Messages
92
Reaction score
8

Homework Statement


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.


Homework 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)]

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.
 
Physics news on Phys.org
Jufro said:

Homework Statement


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.


Homework 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)]

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

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.
Read about the addition of angular momentum. For this problem, consider ##(\vec{S}_1 + \vec{S}_2)^2##.
 
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.
 
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 :)