I have a system of $4$ spin $\frac{1}{2}$ particles, whose Hamiltonian looks like this:
$$ H = \alpha (s_1 \cdot s_2 + s_1 \cdot s_4 +s_2 \cdot s_3 +s_3 \cdot s_4 )$$
In order to find its eigenvalues and respective degeneracies I tried the typical method:
$$ \hat{\overrightarrow{S}} = s_1+ s_2+s_3+s_4 $$
$$ \hat{S}^2 = (s_1+ s_2+s_3+s_4) \cdot (s_1+ s_2+s_3+s_4)$$
which leads me to:
$$ \frac{1}{2} \left( \hat{S}^2 – \sum_{i=1}^4 |s_i|^2\right) = s_1 \cdot s_2 + s_1 \cdot s_4 +s_2 \cdot s_3 +s_3 \cdot s_4 + s_1 \cdot s_3 +s_2 \cdot s_4 $$
There are two extra terms $s_1 \cdot s_3$ and $s_2 \cdot s_4$ that will still appear in the Hamiltonian. How do I handle these in order to get the possible eigenvalues and degeneracies of this particular system?
Best Answer
You are composing four doublets, hence you are inspecting a $2^4\times 2^4= 16\times 16$ matrix. This reduces to the familiar spin 2, spin 1, and spin 0 blocks, specifically a quintet, three triplets, and two singlets, 16 states in all, $$ 1/2\otimes 1/2\otimes 1/2\otimes 1/2= 2\oplus (3) ~~1\oplus (2) ~~0. $$
But your hamiltonian $\alpha(𝑠_1+𝑠_3)⋅(𝑠_2+𝑠_4)$ is special: it does not care how spins 1 & 3 compose together symmetrically or anti symmetrically, in a triplet or a singlet, and likewise for 2 & 4. These two options are degenerate, respectively, in both cases, so you are really composing just $$ \vec S_a = {\vec s_1+\vec s_3} ~~~\hbox{with} \\ \vec S_b = {\vec s_2+\vec s_4}, $$ where the spins of each $S_a$ and $S_b$ are 0 and 1; so, four combinations in all for $\vec S= \vec S_a+\vec S_b$, $$ H= {\alpha\over 2} (S^2-S_a^2-S_b^2). $$
Recalling that the Casimirs for spin 0; 1; and 2 are 0; 2; and 6, respectively: you get the singlet-singlet combination (singlet) to have 0 energy eigenvalue; the two singlet-triplet combinations (triplets) to have 0 energy eigenvalue, as well.
The triplet-triplet combination reduces to the quintet, with eigenvalue α; a triplet with eigenvalue -α; and a singlet with eigenvalue -2α.
Not all products of the reduction with a common spin are treated identically by the (a)symmetric hamiltonian.