# On the eigenvalues of parity operator

Parity operator $$P : \psi(\mathbf{x}, t)\mapsto\psi(-\mathbf{x}, t)$$ is an example of unitary, self-adjoint involution. As such, its eigenvalues can only be $$\pm 1$$. And yet, citing Wikipedia:

To see that the parity's eigenvalues are phase factors, we assume an eigenstate of the parity operation (this is realized because the intrinsic parity is a property of a particle species) and use the fact that two parity transformations leave the particle in the same state, thus the new wave function can differ by only a phase factor, i.e.: $${\displaystyle P^{2}\psi =e^{i\phi }\psi }$$ thus $${\displaystyle P\psi =\pm e^{i\phi /2}\psi }$$, since these are the only eigenstates satisfying the above equation.

But then all observables in physics should have phase factors in their spectra, for every quantum state is uniquely defined up to a global phase.

Why do we care in this case with $$P$$?

• I don't really understand what the question is here: By asking why we care "in this case", you seem to imply that there are other places where we don't care even if we should by this argument. But can you give an example of where this should matter, too, but we "don't care"? Jul 15, 2022 at 16:44
• What I'm trying to say is... we never consider phase factors as part of the eigenvalue of a self-adjoint operator. So why are we considering them in this case? Jul 15, 2022 at 16:55

It seems to me that we are here considering the parity operator $$P$$ simultaneously an observable (thus selfadjoint) and a symmetry (thus unitary or antiunitary from Wigner's theorem). Antiunitary and selfadjoint is impossible, so that the only possibility is that $$P$$ is both unitary and selfadjoint.

Every unitary operator has spectrum contained in $$\{\lambda \in \mathbb{C} \:|\: |\lambda|=1\}$$. Every selfadjoint operator has real spectrum.

Therefore if $$P$$ is simultaneously unitary and selfadjoint, its eigenvalues must be in the set $$\{\pm 1\}$$ which is the intersection of the sets above.

Barring trivial cases, the set of eigenvalues of $$P$$ must coincide with that whole set $$\{\pm 1\}$$ actually. In the other cases $$P$$ would be a multiple of the identity operator and the parity symmetry is not that trivial.

The fact that a pure state is a unit vector up to phases does not play any role here in my view. (It plays a crucial role in the proof of the Wigner theorem, so we have already used that fact.)

ADDENDUM. A deeper issue related with the definition of pure state in terms of vectors is the following one:

why are we allowed to choose the parity operator selfadjoint?

This fact is physically remarkable because it permits the parity to be interpreted as an observable.

The answer is like this. First of all, the parity operator $$U$$ is unitary or antiunitary as it is a symmetry. The choice depends on specific further requirements on that symmetry. For instance, dealing with non relativistic particles, the dual action of the parity symmetry $$U$$ on position and momentum is the natural one, $$U X U^{-1}= -X\:, \quad U P U^{-1}= -P\:.$$ If $$U$$ were antiunitary the requirements above would be incompatible with the CCR $$[X,P] = -i I\:.$$ The inclusion of the spin does not affect this result, though the analysis becomes more delicate.

From now on, we therefore assume that $$U$$ is unitary.

Evidently, and it is also generally stated in Wigner's theorem, $$U$$ can be re-defined with a multiplicative phase. We want to prove that we may fix the arbitrary phase of $$U$$ in order to make it selfadjoint.

For physical reasons, as $$U$$ represents the parity transformation, the action of $$U^2$$ on pure states has to be the identity. In other words, since states are unit vectors up to phases, $$U^2\psi = c_\psi \psi\:.$$ Let us prove that, actually, $$c_\psi$$ does not depend on $$\psi$$. Consider a Hilbert basis $$\{\psi_j\}_{j\in J}$$ and the further unit vectors $$\phi_j= \frac{1}{\sqrt{2}}(\psi_{j_0} + \psi_j)$$ where $$j_0\in J$$ is fixed and $$J \ni j\neq j_0$$. By hypothesis $$U^2 \phi_j = b_j \phi_j= b_j\frac{1}{\sqrt{2}}(\psi_{j_0} + \psi_j)\:.$$ On the other hand, since $$U^2$$ is linear, $$U^2 \phi_j = \frac{1}{\sqrt{2}}(c_{j_0}\psi_{j_0} + c_j\psi_j)\:.$$ Comparing with the identity above: $$(b_j- c_{j_0})\psi_{j_0}= (c_j-b_j)\psi_j\:.$$ Since $$\psi_{j_0}$$ and $$\psi_{j}$$ are orthogonal, we conclude that: $$b_j = c_j= c_{j_0} =:c \quad \forall j \in J\:.$$ In summary $$U^2 \psi_j = c \psi_j \quad \forall j \in J\:.$$ Decomposing any vector along the basis $$\{\psi_j\}_{j\in J}$$, we conclude that $$U^2 = cI$$ where $$|c|=1$$ as said above. Defining $$P:= c^{-1/2}U\:,$$ since $$c^{-1/2}$$ is still a unit complex number (defined up to a sign), we have that $$P$$ is still unitary, so that $$P^*= P^{-1}$$. But also that $$PP=I\:.$$ The uniqueness property of the inverse operator eventually yields $$P=P^*$$ and this is the wanted result: $$P$$ is simultaneously unitary and selfadjoint.