Parity operator in quantum field theory In the quantum thory of a Dirac field we require that parity is implemented by an operator $P$ such that:
$P^{-1}=P^\dagger$
$P^{-1}=P$
$P a_{r}(\vec k) P^{-1}=a_{r}(-\vec{k})$
$P b_{r}(\vec k) P^{-1}=-b_{r}(-\vec{k})$
where $a_{r}(\vec k)$ and $b_{r}(\vec k)$ are distruction operator of particles and antiparticles respectively, thus we get the nice result:
$P \Psi (t, \vec x) P^{-1} = \gamma^{0} \Psi (t, -\vec x) $
But shouldn't we prove that the 4 conditions above are consistent? If so, I haven't found in textbooks such a proof or an explicit expression for the $P$ operator satisfying them. Any idea or reference? 
Thanks
 A: You've given two formulas for the action of $P$ here, one in terms of annihilation operators and one in terms of the field operator.
I'd say these are already fairly explicit representations for $P$.  But if you want to be more explicit, you can re-write them as the action of $P$ on states instead of operators.  For the particle momentum eigenstates $|\vec{k},r\rangle_a = a_r^\dagger\big(\vec{k}\big)\space|0\rangle$:
$ P |\vec{k},r\rangle_a = |-\vec{k},r\rangle_a$
$\langle\vec{k},r|_a P^{\dagger} = \langle -\vec{k},r|_a$
And for antiparticle momentum eigenstates $|\vec{k},r\rangle_b = b_r^\dagger\big(\vec{k}\big)\space|0\rangle$:
$ P |\vec{k},r\rangle_b = -|-\vec{k},r\rangle_b$
$\langle\vec{k},r|_b P^{\dagger} = -\langle -\vec{k},r|_b$
This is as about as explicit as you can get in terms of representing P as an operator acting on the Hilbert space--it's a reflection in the space of states, about the point $\vec{k}=0$.
The first condition $P^{-1} = P^\dagger$ says that $P$ is unitary, in other words it is a 1-to-1 mapping between the states.  And the second one is equivalent to $P^2 = I$.  Both of those follow immediately from the fact that $P$ is a reflection, where you get back the same thing when a state is reflected twice.
In terms of the fields, $\Psi(t, \vec{x})\space |0\rangle$ represents a local excitation of a Dirac spinor field at a particular point in space, for example an electron in a position eigenstate.
In the Weyl representation, the two upper components of a Dirac spinor transform under the Lorentz group as a left-handed Weyl spinor, and the two lower components transform as a right-handed Weyl spinor.  If we just did a straightforward reflection in space by replacing $\vec{x}$ with $-\vec{x}$, we'd be left with a right-handed spinor in the upper two components and a left-handed spinor in the lower two.  So if we want to maintain the standard convention of having the top half of a Dirac spinor represent the left-handed part, so we can still use same equations for Lorentz transforming it, we have to fix that by swapping the upper and lower spinors... which is exactly what multiplying by $\gamma^0$ accomplishes.
