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.


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?


2 Answers 2


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.

  • $\begingroup$ See Edit. I am stuck there $\endgroup$
    – Cheeku
    Apr 4, 2017 at 11:52
  • 2
    $\begingroup$ @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 $ab=\sum_{s s'} a_s b_{s'}$ $\endgroup$ Apr 4, 2017 at 13:40

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.