2楼: Originally posted by simonrac at 2016-01-20 13:14:28
1. Operator N_i = a_{i}^{\dagger} a_i 本身就是厄米算符； 很明显

N_i^{\dagger} = (a_i^{\dagger} a_i)^{\dagger} = a_i^{\dagger} a_i = N_i

但是这里它也是projection operator (2.6.11) ; self-adjoint ...

3楼: Originally posted by Sarah_Bless at 2016-01-20 22:12:21
谢谢您，但是我还是不明白那小于等于号是怎么来的，您可以解释一下吗？...

4楼: Originally posted by simonrac at 2016-01-20 23:16:27
用柯西不等式证明就好了，一般代数书上都有的：

\langle \Psi | P^2 | \Psi \rangle = \langle P\Psi , P \Psi \rangle = \| P \Psi \|^2   

= \langle P \Psi , \Psi \rangle \leq \|P \Psi \| \dot \| \Ps ...