Exchange striction¶
Warning
This is wrong, the mapping can not be performed in the formalism of the attempt #1. See attempt #3, there this problem is resolved.
Exchange striction involves two sites and two entities. Therefore, it can be expressed via the term \(\mathcal{H}_{2, 2}\).
\(k = 2\)
\(l = 2\)
\(m_{2,2} = 2\)
The summary of the mapping for each code is given in the table below.
Code |
|
|---|---|
\(\mathcal{H}\) |
\[\begin{split}\hat{\mathcal{H}}
=
-
\dfrac{1}{2}
\sum_{\substack{nn^{\prime},\alpha\beta\alpha^{\prime}\gamma=1,2,3\\\beta^{\prime}=1-6}}
\Biggl(
\dfrac{\partial \mathcal{J}_{\alpha\beta}}{\partial \epsilon_{\beta^{\prime}}}
+
\dfrac{\partial \mathcal{J}_{\alpha\beta}}{\partial R_{nn^{\prime}}^{\alpha^{\prime}}}
\dfrac{\partial \epsilon_{\alpha^{\prime}\gamma} R_{nn^{\prime}}^{\gamma}}{\partial \epsilon_{\beta^{\prime}}}
\Biggr)
\epsilon_{\beta^{\prime}}
\hat{\mathcal{I}}_{\alpha}^n
\hat{\mathcal{I}}_{\beta}^{n^{\prime}}\end{split}\]
|
Indices renaming |
\(n \rightarrow (\mu_1, \alpha_1)\), \(n^{\prime} \rightarrow (\mu_2, \alpha_2)\), \(\alpha \rightarrow i_1\), \(\beta \rightarrow i_2\) |
\(C_{2, 2}\) |
\(-\dfrac{1}{2}\) |
\(V_{\mu_1, \mu_2; \alpha_1, \alpha_2}^{i_1, i_2}\) |
\[\begin{split}\sum_{\alpha^{\prime}\gamma=1,2,3\\\beta^{\prime}=1-6}
\Biggl(
\dfrac{\partial \mathcal{J}_{i_1,i_2}}{\partial \epsilon_{\beta^{\prime}}}
+
\dfrac{\partial \mathcal{J}_{i_1,i_2}}{\partial R_{\mu_1,\mu_2;\alpha_1,\alpha_2}^{\alpha^{\prime}}}
\dfrac{\partial \epsilon_{\alpha^{\prime}\gamma} R_{\mu_1,\mu_2;\alpha_1,\alpha_2}^{\gamma}}{\partial \epsilon_{\beta^{\prime}}}
\Biggr)
\epsilon_{\beta^{\prime}}\end{split}\]
|
\(X_{\mu_1; \alpha_1}^{i_1}\) |
\(\hat{\mathcal{I}}_{i_1}^{\mu_1,\alpha_1}\) |
\(X_{\mu_2; \alpha_2}^{i_2}\) |
\(\hat{\mathcal{I}}_{i_2}^{\mu_2,\alpha_2}\) |