# 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.

• As far as I remember it is not about being an operator or not but rather about being a scalar, spinor or vector. Scalars don't transform at all (only the argument changes from $\vec x$ to $-\vec x$), vectors transform with one transformation matrix, spinors transform in a weird way, in this case involving $\gamma_0$. But it's long ago I dealt with QFT, so I'm not too sure about it - thus posting as a comment. Mar 3, 2020 at 15:28
• @CosmasZachos I am positive. To be precise I got that particular formula from "Dynamics of the Standard Model" by Donoghue, Golowich, Holstein. They define $x_P=(x^0,-\mathbf{x)}$ and $P\psi(x) P^{-1}=\gamma^0 \psi(x_P)$. Mar 3, 2020 at 16:16
• so you are positive both first and second formula are in the same text? Your second one is the standard parity transform of a spinor field operator. Your first one is the parity transform of a scalar wavefunction. Mar 3, 2020 at 17:22
• @CosmasZachos Iam sorry, I misunderstood you. They are not from the same text. Your comment together with the other answers helped. The use of the same variables in both my sources confused me. Mar 3, 2020 at 18:12

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.

• The little bits of knowledge I had before this confused me. Thanks for clearing it up. Regarding my second equation: the question I linked in my original question deals with exactly that convention. Mar 3, 2020 at 18:14
• @MarieCurry Happy to help! Mar 3, 2020 at 18:15

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.