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

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.

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.
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, $$U^{-1}=U^\dagger$$.