# Hermitian conjugate of an antiunitary transformation

In quantum mechanics, one often considers symmetry transformations which are defined in terms of operators which do not change the norm of states in the Hilbert space. For the Wigner's theorem, this symmetry operators are either unitary operators or antiunitary operators, which generally have the form $A=U K$ where $K$ is complex conjugation and $U$ a unitary operator. Examples of antiunitary operators describe the time-reversal, the particle-hole symmetry, and the chiral symmetry.

Now let us take a generic Hamiltonian $H$. One can define a unitary transformation as $H\rightarrow U H U^\dagger$. How one defines an antiunitary transformation instead? Formally, I would write $$H\rightarrow A H A^\dagger=U K H (U K)^\dagger=U K H K^\dagger U^\dagger=U H^* U^\dagger$$ The questions are:

1) Is this definition of antiunitary transformation correct?

2) What is the hermitian conjugate of an antiunitary operator? Is it correct to write $(U K)^\dagger=K^\dagger U^\dagger$ or is it $(U K)^\dagger=K U^\dagger$ instead?

I apologize for the multiple questions, but I think they are all related to the meaning of the complex conjugation operator $K$ and its hermitian conjugate $K^\dagger$. I already posted this question to the Mathematics StackExchange but I've been completely misunderstood there...

• Someone should explain to me why I got a downvote for this question in only 10 seconds... – sintetico Jul 21 '15 at 21:37
• – Kyle Kanos Jul 22 '15 at 0:11
• Essentially a duplicate of physics.stackexchange.com/q/45227/2451 – Qmechanic Jul 22 '15 at 6:34
• well, almost a duplicate. – sintetico Jul 22 '15 at 11:06

For an antilinear operator, as the antiunitaries and the complex conjugation, the definition of adjoint is changed: $$\langle U^{a*}\psi,\phi\rangle=\overline{\langle \psi,U\phi\rangle}$$ where $a*$ stands for anti-adjoint. It is therefore easy to see that the anti-adjoint of $K$ is $K$ itself (and in general the anti-adjoint of an anti-unitary is anti-unitary and satisfies $U^{a*}U=I$).