The hermitian conjugate of anti-linear operator

Some quantum mechanics books tell us that the definitions of hermitian are

If $$\langle\psi|A\phi\rangle=\langle B\psi|\phi\rangle$$ for linear operators, then $$B=A^\dagger$$

If $$\langle\psi|C\phi\rangle=\langle D\psi|\phi\rangle^*$$ for anti-linear operators, then $$D=C^\dagger$$

1. Why are definitions different between linear operators and anti-linear operators ?
2. Are they equivalent to the definition $$Q^\dagger=(Q^T)^*$$ in linear algebra, where $$Q^T$$ is the transpose of $$Q$$ ?
• – Qmechanic Dec 5 '18 at 3:33

Why are defintions different between linear operators and anti-linear operators?

Let $$\lambda$$ be any complex scalar. Then $$\langle \psi|C(\lambda \phi)\rangle = \langle\psi|\overline \lambda C\phi\rangle = \overline \lambda\langle\psi|C\phi\rangle$$

by the definition of anti-linearity. If we use the definition we normally use for linear operators, then we would find $$\langle \psi | C(\lambda \phi)\rangle = \langle D\psi|\lambda \phi\rangle$$ but also that $$\overline \lambda \langle\psi|C\phi\rangle = \overline \lambda\langle D \psi|\phi\rangle = \langle D\psi|\overline \lambda \phi\rangle$$

which is inconsistent. Using the slightly modified definition fixes this problem.

Are they equivalent to the definition $$Q^\dagger=(Q^T)^∗$$ in linear algebra, where $$Q^T$$ is the transpose of $$Q$$?

The first one is, yes.

If you imagine two arbitrary column vectors $$\mathbf x$$ and $$\mathbf y$$ and matrices $$A$$ and $$B$$, then the first definition becomes $$\mathbf x^\dagger A \mathbf y = (B\mathbf x)^\dagger \mathbf y \implies B=A^\dagger$$

where the dagger denotes conjugate transposition. But this should be clear, since $$(B\mathbf x)^\dagger = \mathbf x^\dagger B^\dagger$$ so $$\mathbf x^\dagger A \mathbf y= \mathbf x^\dagger B^\dagger \mathbf y$$ If that equality holds for all $$\mathbf x$$ and $$\mathbf y$$, then $$A=B^\dagger \iff A^\dagger = B$$.

The case of anti-linear operators is a bit more subtle because an anti-linear operator cannot be written as a matrix all by itself; it has to be a matrix plus a complex conjugation, and therefore cannot be realized if we restrict ourselves to standard matrix algebra.