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I have been reading about quantum magnetism from Assa Auerbach's book, and Equation (2.9) reads

$$J^F \sum_{s s'} c^{\dagger}_{is} c^{\dagger}_{i's'}c_{is'}c_{i's} = -2 J^F\left (\mathbf{S}_i \cdot \mathbf{S}_{i'} + \frac{1}{4} n_i n_{i'} \right)$$

Here $S^{\alpha}_{i}$, $\alpha = x,y,z$ are components of spin one half operators: $$\mathbf{S}_i \equiv \frac{1}{2} \sum_{s s'} c^{\dagger}_{is} \vec{\sigma}_{ss'}c_{is'}$$

Here $\vec{\sigma}$ are the usual Pauli matrices. And $n_i = \sum_{s} c^{\dagger}_{is} c_{is}$

For $s=2$, I can explicitly write the right hand side and obtain the left hand side with some simplification. But that does not seem satisfactory to me. I should be able to algebraically obtain this result at a higher abstraction level.

Please give me ideas on how to start.

Edit:

Given the information by NowIGetToLearnWhatAHeadIs, I do the following $$\mathbf{S}_i \cdot \mathbf{S}_{i'} = \frac{1}{4} \sum_{ss'} c^{\dagger}_{is} c_{is'} \left( \vec{\sigma}_{ss'} \cdot \vec{\sigma}_{ss'} \right) c^{\dagger}_{i's} c_{i's'} \\ = \frac{1}{4} \sum_{ss'} c^{\dagger}_{is} c_{is'} \delta_{s s'}^2 c^{\dagger}_{i's} c_{i's'} \\ = - \frac{1}{4} \sum_{s} c^{\dagger}_{is} c^{\dagger}_{i's} c_{is} c_{i's}$$

Also $$\frac{1}{4} n_i n_{i'} = \frac{1}{4} \sum_{s} c^{\dagger}_{is} c_{is} \sum_{s'} c^{\dagger}_{i's'} c_{i's'} \\ = - \frac{1}{4} \sum_{ss'} c^{\dagger}_{is} c^{\dagger}_{i's'} c_{is} c_{i's'}$$

For $s = s'$, this adds up with the previous expression to give $$- \frac{1}{2} \sum_{s} c^{\dagger}_{is} c^{\dagger}_{i's} c_{is} c_{i's}$$

which describe the diagonal terms for our complete expression. But for $s \neq s'$, we have $$- \frac{1}{4} \sum_{ss'} c^{\dagger}_{is} c^{\dagger}_{i's'} c_{is} c_{i's'}$$

but what I need is $$- \frac{1}{2} \sum_{ss'} c^{\dagger}_{is} c^{\dagger}_{i's'} c_{is'} c_{i's}$$

How do I do this transformation?

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Just check out this completeness relation for the pauli spin matrices:

$$ \vec{\sigma}_{\alpha \beta} \cdot \vec{\sigma}_{\gamma \delta} = 2 \delta_{\alpha \delta}\delta_{\beta \gamma} - \delta_{\alpha \beta}\delta_{\gamma \delta}. $$

You can plug this into the right hand side and everything else should be very easy.

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  • $\begingroup$ See Edit. I am stuck there $\endgroup$ – Cheeku Apr 4 '17 at 11:52
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    $\begingroup$ @Cheeku You are using the same dummy index for two independent sums. If $a=\sum_s a_s$ and $b= \sum_s b_s$, you are doing $ab=\sum_s a_s b_s$, but you should be doing $ab=\sum_{s s'} a_s b_{s'}$ $\endgroup$ – Brian Moths Apr 4 '17 at 13:40
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If you want to derive the right-hand-side of the equation, these are the steps to follow. Below, summation over repeated indices are understood. $$ c^{\dagger}_{is} c^{\dagger}_{j s'}c_{is'}c_{j s} = - c^{\dagger}_{is} c_{is'} c^{\dagger}_{js'} c_{js} = - c^{\dagger}_{i\alpha} c_{i\beta} c^{\dagger}_{j\gamma} c_{j\delta} \delta_{\alpha \delta} \delta_{\beta \gamma} $$ Now using $\boldsymbol{\sigma}_{\alpha \beta} \cdot \boldsymbol{\sigma}_{ \gamma \delta}=2\delta_{\alpha \delta} \delta_{\beta \gamma}-\delta_{\alpha \beta} \delta_{\gamma \delta}$ gives

$$ -\frac{1}{2} \underbrace{c^{\dagger}_{i\alpha} \boldsymbol{\sigma}_{\alpha \beta} c_{i\beta} }_{2 \boldsymbol{S}_i} \cdot \underbrace{c^{\dagger}_{j\gamma} \boldsymbol{\sigma}_{ \gamma \delta}c_{j\delta} }_{2 \boldsymbol{S}_j} -\frac{1}{2} \underbrace{c^{\dagger}_{i\alpha} c_{i\alpha} }_{\hat{n}_i} \cdot \underbrace{c^{\dagger}_{j\delta} c_{j\delta} }_{\hat{n}_j} \\ =-2 \left( \hat{\boldsymbol{S}}_i \cdot \hat{\boldsymbol{S}}_j + \frac{1}{4} \hat{n}_i \hat{n}_j \right) $$

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