[Physics] Why a Hermitian linear operator equal to its own complex conjugate

hilbert-spaceoperatorsquantum mechanics

I have been given the following definition for a Hermitian linear operator (HLO):
A is a HLO, if $A=A^{H}$ such that $<Ax|y>=<x|A^{H} y>$.

Using this definition or otherwise, how can it be deduced that the operator is its own complex conjugate?
Am I missing some underlying theory relating Hermitan conjugates and complex conjugates of LO's, when the LO acts on complex vector spaces?

Best Answer

The correct answer is: they're not, necessarily. There are two things that happen when you take the Hermitian conjugate: a transpose, and a complex conjugation. If you break down the Hermitian operators into symmetric and anti-symmetric parts, the symmetric part is real and the anti-symmetric part is imaginary. In equations: $$\begin{align} \mathrm{let:}&\ A^\dagger = A,\ A_s \equiv \frac{A+A^T}{2},\ \mathrm{and}\ A_a \equiv \frac{A - A^T}{2} \Rightarrow \\ A_s^T = A_s & \Rightarrow A_s^\star = A_s \\ A_a^T = -A_a & \Rightarrow A_a^\star = -A_a. \end{align}$$ You can also prioritize real and imaginary parts of $A$ to get: \begin{align} \mathrm{let:}&\ A_r \equiv \frac{A + A^\star}{2},\ \mathrm{and}\ A_i \equiv \frac{A - A^\star}{2} \Rightarrow \\ A_r^\star = A_r & \Rightarrow A_r^T = A_r \\ A_i^\star = -A_i & \Rightarrow A_i^T = -A_i. \end{align}

You can add the $\langle y|$ and $|x\rangle$ to both sides of the above equations, and the conclusion won't be affected - that's just working in terms of the operator's matrix elements.