I am confused. In my introduction course to particle physics I learned that the parity operator acts on a field like:


Simple enough. Now looking deeper into it I find


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


that to understand how an operator $\hat{O}$ transforms we need to look at.


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.

  • $\begingroup$ 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. $\endgroup$
    – Photon
    Mar 3, 2020 at 15:28
  • $\begingroup$ @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 $P\psi(x) P^{-1}=\gamma^0 \psi(x_P)$. $\endgroup$ Mar 3, 2020 at 16:16
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 3, 2020 at 17:22
  • $\begingroup$ @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. $\endgroup$ Mar 3, 2020 at 18:12

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Mar 3, 2020 at 18:14
  • $\begingroup$ @MarieCurry Happy to help! $\endgroup$
    – Quiver
    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.


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