1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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