# Hamiltonian eigenvalues

1. Oct 3, 2006

### JohanL

Find the eigenvalues of the hamiltonian

$$H=a(S_A \cdot S_B+S_B \cdot S_C+S_C \cdot S_D+S_D \cdot S_A)$$

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

I rewrite it as

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

then i define

$$J_1=S_A+S_B+S_C+S_D$$

$$J_2=S_A+S_C$$

$$J_3=S_B+S_D$$

and uses

$$J^2_i |j_1j_2j_3;m_1m_2m_3> = (h^2) j_i(j_i+1)|j_1j_2j_3;m_1m_2m_3>$$

which gives the energies

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

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

Last edited: Oct 3, 2006
2. Oct 3, 2006

### StatMechGuy

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 $$j_2$$ and $$j_3$$ based on $$j_1$$, but the thought process seems right.

3. Oct 3, 2006

### Meir Achuz

Your method is completely correct. Just include the a.
Did it just seem too easy?

4. Oct 3, 2006

### JohanL

Thank you. Yes it seemed too easy

5. Oct 6, 2006

### JohanL

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. Oct 6, 2006

### Meir Achuz

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.