# A question from A. Zee's book

On page 463, it writes in eq. (3) $$4H=\Sigma_\alpha(Q_\alpha Q^\dagger_\alpha+Q^\dagger_\alpha Q_\alpha).\tag 3$$ And then it writes that this is followed by eq.(4) as $$\langle S| H|S\rangle=\frac{1}{2}\Sigma_\alpha\Sigma_{S'}|\langle S'|Q_\alpha|S\rangle|^2\geq 0. \tag 4$$ I just wonder how to get eq.(4) from eq.(3). Specifically, I don't know why we get the square in eq.(4) and what $S'$ actually mean. (I guess $|S'\rangle=Q^\dagger|S\rangle$, but then I can not understand the square and why we have the summation of $S'$). Thanks in advance.

He inserts the identity $I = \sum_{S'}|S'\rangle\langle S'|$. This gives $\langle S|Q_\alpha Q_\alpha^\dagger|S\rangle = \sum_{S'} \langle S|Q_\alpha |S'\rangle \langle S'|Q_\alpha^\dagger|S\rangle =\sum_{S'} |\langle S|Q_\alpha |S'\rangle|^2$.