Exercise 2.24

According to the Hint, we can show at first that an arbitrary operator \(A\) can be written as \(A = B+iC\) where \(B\) and \(C\) is Hermitian. We can construct

(1)\[ B = \frac{A + A^{\dagger}}{2}, C = \frac{A - A^{\dagger}}{2i} = \frac{iA^{\dagger} - iA}{2} \]

and with above construction, we can recover \(A\) by

(2)\[ B + iC = \frac{A + A^{\dagger}}{2} +i\frac{A - A^{\dagger}}{2i} = \frac{A + A^{\dagger}}{2} +\frac{A - A^{\dagger}}{2} = A \]

It is easy to verify that \(B\) and \(C\) are Hermitian, since

(3)\[ B^{\dagger} = \frac{A^{\dagger}+A}{2} = B, C^{\dagger} = \frac{-iA +i A^{\dagger}}{2} = C \]

where we use

(4)\[ \left(\sum_{i} a_i A_i\right)^{\dagger} = \sum_{i} a^{*}_i A^{\dagger}_i \]

Then for arbitrary vector \(|v\rangle\), using \(A = B + iC\) we have

(5)\[ \langle v|A|v\rangle = \langle v|B|v\rangle + i\langle v|C|v\rangle \]

Note that \(B\) and \(C\) are Hermitian, so we can use spectral decomposition and compute

(6)\[\begin{split} \begin{align} \langle v|A|v\rangle =& \langle v|B|v\rangle + i\langle v|C|v\rangle \\ =& \left\langle v\left|\left(\sum_{i}\lambda_i |b_i\rangle\langle b_i|\right)\right|v\right\rangle + i\left\langle v\left|\left(\sum_{x}\mu_x |c_x\rangle\langle c_x|\right)\right|v\right\rangle \\ =& \left(\sum_j v^*_{bj}\langle b_j|\right)\left|\left(\sum_{i}\lambda_i |b_i\rangle\langle b_i|\right)\right|\left(\sum_k v_{bk}|b_k\rangle\right) \\ &+ i\left(\sum_y v^*_{cy}\langle c_y|\right)\left|\left(\sum_{x}\mu_x |c_x\rangle\langle c_x|\right)\right|\left(\sum_z v_{cz}|c_z\rangle\right) \\ =& \sum_{i,j,k}v^*_{bj}v_{bk}\lambda_i \langle b_j|b_i\rangle\langle b_i|b_k\rangle + i\sum_{x,y,z}v^*_{cy}v_{cz}\mu_x \langle c_y|c_x\rangle\langle c_x|c_z\rangle \\ =& \sum_{i}|v_{bi}|^2\lambda_i + i\sum_{x}|v_{cx}|^2\mu_x \\ \end{align} \end{split}\]

Note that generally \(B\) and \(C\) do not have common eigenvector, but their own eigenvectors are orthogonal to each other, and we adopt two orthonormal basis \(\{|b_i\rangle\}\) and \(\{|c_i\rangle\}\) to decompose \(|v\rangle\) in the above calculation.

For positive operator \(A\), we should have \(\langle v|A|v\rangle >0\) is a real number. So from eq. (6) we have

(7)\[ \langle v|A|v\rangle = \sum_{i}|v_{bi}|^2\lambda_i + i\sum_{x}|v_{cx}|^2\mu_x >0 \iff \lambda_i \geq 0, \mu_x = 0 \]

Therefore, for positive operator \(A\) we have \(A = B + iC\) where \(B\) is Hermitian and \(C = 0\), which indicates that \(A\) is Hermitian.