Quadruplet interaction¶
In Spirit, the quadruplet interaction term is written as
First, notice, that \(i \ne j\) and \(k \ne l\), otherwise the Hamiltonian is constant.
Second, a special case of this Hamiltonian have been discussed in one of the previous pages. This page includes it as a particular case:
\(i = k\) and \(j = l\) (or equivalently \(i = l\) and \(j = k\))
This is a case of Biquadratic exchange.
In general the Hamiltonian maps to the \(\mathcal{H}_4\)
\[\begin{split}\mathcal{H}_4 = C_4 \sum_{\substack{\mu_1, \mu_2, \mu_3, \mu_4, \\ \alpha_1, \alpha_2, \alpha_3, \alpha_4, \\ i_1, i_2, i_3, i_4}} V_{\mu_1, \mu_2, \mu_3, \mu_4; \alpha_1, \alpha_2, \alpha_3, \alpha_4}^{i_1, i_2, i_3, i_4} \cdot X_{\mu_1; \alpha_1}^{i_1} \cdot X_{\mu_2; \alpha_2}^{i_2} \cdot X_{\mu_3; \alpha_3}^{i_3} \cdot X_{\mu_4; \alpha_4}^{i_4}\end{split}\]
By expanding the dot product, one gets
where the tensor \(K_{ijkl}^{i_1, i_2, i_3, i_4}\) is defined as
and all other components are zero. Then, one renames the indices as \(i \rightarrow (\mu_1, \alpha_1)\), \(j \rightarrow (\mu_2, \alpha_2)\), \(k \rightarrow (\mu_3, \alpha_3)\), \(l \rightarrow (\mu_4, \alpha_4)\) and correspondence becomes clear