Hamiltonian eigenvalues

  • Thread starter JohanL
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  • #1
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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:

Answers and Replies

  • #2
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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.
 
  • #3
Meir Achuz
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Your method is completely correct. Just include the a.
Did it just seem too easy?
 
  • #4
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Meir Achuz said:
Your method is completely correct. Just include the a.
Did it just seem too easy?

Thank you. Yes it seemed too easy :smile:
 
  • #5
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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?
 
  • #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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