- #1
pellman
- 683
- 6
- TL;DR
- What is the significance of the hermitian conjugate and the unitary condition for anti-linear operators? (the context here is that of the time reversal operator in QM but there is no physics in the question really.)
The definition of the hermitian conjugate of an anti-linear operator B in physics QM notation is
\langle \phi | (B^{\dagger} | \psi \rangle ) = \langle \psi | (B | \phi \rangle )
where the operators act to the right, since for anti-linear operators ( \langle \psi |B) | \phi \rangle \neq \langle \psi | (B | \phi \rangle ). Contrast with the definition for linear operators
\langle \phi | A^{\dagger} | \psi \rangle = \langle \psi | A | \phi \rangle^{*}
For linear operators the hermitian conjugate frequently shows up because \langle \psi | A^{\dagger} is the bra corresponding to A | \psi \rangle, and in \langle \psi | A^{\dagger} we can treat A^{\dagger} as an operator acting to the right. Thus, the significance of the unitary condition A^{\dagger} A = 1: the inner product is preserved under transformations by A.
But for anti-linear operators the bra corresponding to B | \phi \rangle is not \langle \psi | B^{\dagger} . That is, the inner product of B | \psi \rangle and B | \phi \rangle is not \langle \psi | B^{\dagger} B | \phi \rangle . ( or is it?)
So what is the significance of the unitary condition B^{\dagger} B= 1 ?
Using mathematician notation here can clarify, so instead we write \langle B \psi , B \phi \rangle , and it is clearer what is acting on what. But I still don't see the significance of the unitary condition. Where does B^{\dagger} B show up if not in \langle B \psi , B \phi \rangle?
\langle \phi | (B^{\dagger} | \psi \rangle ) = \langle \psi | (B | \phi \rangle )
where the operators act to the right, since for anti-linear operators ( \langle \psi |B) | \phi \rangle \neq \langle \psi | (B | \phi \rangle ). Contrast with the definition for linear operators
\langle \phi | A^{\dagger} | \psi \rangle = \langle \psi | A | \phi \rangle^{*}
For linear operators the hermitian conjugate frequently shows up because \langle \psi | A^{\dagger} is the bra corresponding to A | \psi \rangle, and in \langle \psi | A^{\dagger} we can treat A^{\dagger} as an operator acting to the right. Thus, the significance of the unitary condition A^{\dagger} A = 1: the inner product is preserved under transformations by A.
But for anti-linear operators the bra corresponding to B | \phi \rangle is not \langle \psi | B^{\dagger} . That is, the inner product of B | \psi \rangle and B | \phi \rangle is not \langle \psi | B^{\dagger} B | \phi \rangle . ( or is it?)
So what is the significance of the unitary condition B^{\dagger} B= 1 ?
Using mathematician notation here can clarify, so instead we write \langle B \psi , B \phi \rangle , and it is clearer what is acting on what. But I still don't see the significance of the unitary condition. Where does B^{\dagger} B show up if not in \langle B \psi , B \phi \rangle?