Exercise 2.25

Suppose \(A\) is an arbitrary operator and \(|v\rangle\) is arbitrary vector, we can compute the inner product of \(|v\rangle\) and \(A^{\dagger}A|v\rangle\) as

(1)\[ \left(|v\rangle, A^{\dagger}A|v\rangle\right) = \langle v|A^{\dagger}A|v\rangle \]

Suppose \(|u\rangle = A|v\rangle\) then eq. (1) becomes \(\langle v|A^{\dagger}A|v\rangle = \langle u|u\rangle\). According to the properties of inner product, we have \(\langle u|u\rangle \geq 0\). That is, for arbitrary operator \(A\) and arbitrary vector \(|u\rangle\), we always have

(2)\[ \langle v|A^{\dagger}A|v\rangle = \langle u|u\rangle \geq 0 \]

Thus, we could conclude that \(A^{\dagger}A\) is positive.