Confusion on parity transformation I am confused. In my introduction course to particle physics I learned that the parity operator acts on a field like:
$\hat{P}\psi(\vec{x},t)=\psi(-\vec{x},t)$.
Simple enough. Now looking deeper into it I find
$\hat{P}\psi(\vec{x},t)\hat{P}^{-1}=\gamma^0\psi(-\vec{x},t)$
which reminds me of a unitary transformation. But I can't figure out how both these expressions relate to each other. Why are we adding the $\hat{P}^{-1}$?
While I tried to figure it out on my own I found this question, where they conclude from
$\hat{P}(\hat{O}|v\rangle)=(\hat{P}\hat{O}\hat{P}^{-1})(\hat{P}|v\rangle)$
that to understand how an operator $\hat{O}$ transforms we need to look at.
$\hat{O}\mapsto\hat{P}\hat{O}\hat{P}^{-1}$.
This seems good to me, but looking back at the beginning, $\psi(\vec{x},t)$ is not an operator. But it transforms as one(?).
As you can probably tell there is confusion on my side.
Maybe someone can help me to understand the parity transformation.
 A: Parity, like every operator, acts on a state only by
$$ P|p,s,a\rangle = \eta_a|-p,s,a\rangle $$ where $|p,s,a\rangle$ defines the state of a given particle $a$ with momentum $p$ and spin $s$. In quantum field theory, states are given by acting on the vacuum $|0\rangle$ by a suitable creation operator.
By this you can easily see that, for a parity operation we first impose an operator condition on the creation and annihilation operators (remember that for parity $PP^\dagger = 1 \implies P^\dagger = P$ since by Wigner theorem the symmetry can be implemented by a unitary operator)
$$Pa^\dagger_{p,s}P = \eta_a a^\dagger_{-p, s} \qquad  Pb^\dagger_{p,s}P = \eta_b b^\dagger_{-p, s} \\
Pa_{p,s}P = \eta_a a_{-p, s} \qquad  Pb_{p,s}P = \eta_b b_{-p, s}$$
since this is the only way to impose the first definition on a generic multiparticle state. In fact, if we take a two particle state $$a^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle \implies Pa^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle = \eta_a a^\dagger_{-p,s}Pb^\dagger_{p,s^\prime}PP|0\rangle = \eta_a\eta_b a^\dagger_{-p,s}b^\dagger_{-p,s^\prime}|0\rangle$$
you see that you'll get the desired result.$^*$
At this stage you see that the field
$$\Psi(x) = \int\frac{d^3 p}{(2\pi)^3\sqrt{E_p}} \sum_s\left(a_{p,s}u^{s}(p)e^{-ipx}+b^\dagger_{p,s}v^s(p)e^{ipx}\right)$$
has to transform, under parity, like $$P\Psi(x)P$$ which, by doing the calculation, you can easily see that $$ P\Psi(x)P = \gamma^0\Psi(x) $$
In fact, i'm led to believe that the second equation you gave is not true. But it might be a matter of convention. For the moment is not so important.
The error you're doing is by considering a state to be described by a wavefunction like in non relativistic quantum mechanics. In QFT a state is given only by acting on the vacuum with creation and annihilation operators. You could even act on the vacuum with the field operator to get once again a particle but this time in a given position and no specific momenta.
$^*$ We are supposing that the vacuum is non degenerate and invariant under parity. So there's like no spontaneous symmetry breaking going on.
A: The parity operators in your first and second equations are not the same thing. 
In the first equation, you have a scalar representation (1D space). Here the parity operator just act on the scalar object. So $P\psi \rightarrow \psi$. Since $\psi$ is space-dependent and parity reverses space components, so you get $\psi(-\mathbf{x},t)$ in the end.
In the left hand side of the second equation, the parity operator is in the Hilbert space representation. It acts on Hilbert space operators. Remember that $\psi(x)$ is itself an operator in the physical Hilbert space. You can then expand $\psi(x)$ in terms of creation and annihilation operators to evaluate the left hand side. The result is, of course, the right hand side. In this case, $\gamma^0$ is the parity operator in the spinor representation. 
