Properties of time reversal operation

Question

During a course of mine time-reversal symmetry was introduced as an anti-linear operator. One property the lecturer pointed out is that \begin{equation} \langle\phi| A\psi\rangle \neq \langle A^\dagger \phi |\psi \rangle \end{equation} but rather we have \begin{equation} \langle\phi| A\psi\rangle = \langle \psi | A^\dagger \phi \rangle = \langle A^\dagger \phi | \psi \rangle^*. \end{equation} I understand that time-reversal involves a complex conjugation, but I don't understand what makes the above equalities true.

The question is: what property of time-reversal itself makes the above hold. Is it only complex conjugation or is there anything else needed for the above to be true?

Answer 1

The equation you listed is the definition of the Hermitian conjugate of an antilinear map. It is true because it is defined that way.

However, I am guessing you are curious about why there must be this extra complex conjugation as compared to the case with linear operators. Suppose we tried to define Hermitian conjugation for antilinear $A$ as we do for linear operators.

$$ \langle\phi| A \psi\rangle = \langle A \phi|\psi\rangle $$

We will find it is not possible to do this in general. To see this, let $\lambda \in \mathbb{C}$ and consider the following manipulations

\begin{align} \begin{aligned} \langle\phi|A\lambda\psi\rangle &= \lambda^* \langle\phi|A\psi\rangle \\ &= \langle\lambda\phi|A\psi\rangle \\ &= \langle A^\dagger \lambda \phi | \psi\rangle \\ &= \langle \lambda^* A^\dagger\phi|\psi\rangle \\ &= \lambda \langle A^\dagger\phi|\psi\rangle \\ &= \lambda \langle \phi|A\psi\rangle \end{aligned} \end{align}

Comparing the first and last line on the right hand side, we see that if $\lambda \neq \lambda^*$, $\langle\phi|A\psi\rangle = 0$. Since this holds for any two states $\phi, \psi$, only $A = 0$ fits the bill! We clearly want other operators besides zero to have Hermitian conjugates. The true definition for antilinear operators that you showed above succeeds in that.

Answer 2

This is a pedantic point, but in general, I would be very careful about using the language $A^\dagger$ when antihermitian operators are involved. The reason I say this is that antihermitian $A$ is not even a linear operator on the Hilbert space, so it's complete nonsense to talk about it as though it were. For the same reason, bra-ket notation is somewhat broken when taking the dual - it is never clear whether the complex conjugation inherent in an antiunitary acts on the bra, the ket or neither.

It is much easier to simply say $A$ with the understanding that if an operator is antiunitary, then

$$\langle A \psi| A \phi\rangle = \langle \psi| \phi \rangle^*$$

and derive all other results you need from the fact that $A^2$ is unitary. See here for a more in-depth discussion.