What does sandwiching with an unitary operator and its inverse imply?

Question

I am following the book "An introduction to quantum field theory" by Peskin and Schroeder. In the section 'Discrete symmetries of the Dirac theory', it is written,

$P a^s _p P^{-1} = \eta_a a^s _{-p}$

where $P$ is the parity operator and $a^s _p$ is the annihilation operator of a particle of momentum $p$. And $s=1,2$ depending on up spin or down spin respectively. Why do we need to sandwich $a^s _p$ between $P$ and $P^{-1}$?

Then what does $P a^s _p \ $ imply in the viewpoint of an operator acting on something? I have seen many such equations similar to the first one, but I didn't think about what this sandwiching implies and where it comes from at that time.

Answer 1

This goes back to the paradigm that if a ket transforms as $|\psi\rangle\to \hat{U}|\psi\rangle$ under a unitary symmetry transformation $\hat{U}$, then a bra transforms as $\langle\phi|\to \langle\phi|\hat{U}^{-1}$ and an operator transforms as $\hat{A}\to \hat{U}\hat{A}\hat{U}^{-1}$ to make the sandwich $\langle\phi|\hat{A}|\psi\rangle$ invariant.

Answer 2

A more mundane restatement of the already correct answer by @Qmechanic:
Say, we solve equation $$H|\psi\rangle=E|\psi\rangle$$ We now want to formulate the same equation in a different basis (e.g., the diagonal basis of $H$, but not necessarily.) The wave functions in the two bases are related as $|\psi\rangle=U|\psi'\rangle$, that is we have $$HU|\psi'\rangle=EU|\psi'\rangle,$$ and after multiplying by $U^{-1}$ on the left: $$U^{-1}HU|\psi'\rangle=H'|\psi'\rangle=E|\psi'\rangle\Leftrightarrow H'=U^{-1}HU.$$ If we have reason to believe that the transformation is unitary, then, in addition, $U^{-1}=U^\dagger$.