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 $$a^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle \implies Pa^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle = \eta_a a^\dagger_{-p,s}Pb^\dagger_{p,s^\prime} = \eta_a\eta_b a^\dagger_{-p,s}b^\dagger_{-p,s^\prime}|0\rangle$$

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.

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.

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 $$a^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle \implies Pa^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle = \eta_a a^\dagger_{-p,s}Pb^\dagger_{p,s^\prime} = \eta_a\eta_b a^\dagger_{-p,s}b^\dagger_{-p,s^\prime}|0\rangle$$

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, i don't think is that important.

$^*$ We are supposing that the vacuum is non degenerate and invariant under parity. So there's like no spontaneous symmetry breaking going on.

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 $$a^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle \implies Pa^\dagger_{p,s}b^\dagger_{p,s^\prime}|0\rangle = \eta_a a^\dagger_{-p,s}Pb^\dagger_{p,s^\prime} = \eta_a\eta_b a^\dagger_{-p,s}b^\dagger_{-p,s^\prime}|0\rangle$$

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.