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Fermion wavefunctions are antisymmetric under the interchange of two particles. Spatial inversion flips the spatial coordinate, but does not interchange particles. In other words, let's say we have a two particle wave function, $\psi(x_1, x_2)$ (where $x_1$ is the position of particle 1, and $x_2$ is the position of particle 2). Being odd under parity ...


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well, fermions' "spatial wave function" can also be antisymmetric. I think it's the whole wave function(spin+spatial) that matters.


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I find things are clearer using the dotted and undotted spinor notation. The L-spinors $\chi_{L}$ are dotted vectors $\chi^{\dot{A}}$ and the R-spinors $\xi_{R}$ are undotted vectors $\xi^{A}$ with index $A=1,2$. The parity operator has to be a tensor $P^{\dot{A}}_{B}$ and another tensor $P^{A}_{\dot{B}}$ in order to change the way each type of spinor ...


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You are looking for a unitary representation of partity on spinors. That it should be unitary can be seen from the fact, that partity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation. As demonstrated parity ...


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The pion is a pseudoscalar particle, which behaves like a scalar, except that it changes sign under a parity inversion while a true scalar does not. For details, see this post by @LuboŇ° Motl and links there: What is a Pseudoscalar particle?


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I expand my comment into an answer. The idea is to fix $\alpha, \beta \in \mathbb R$ in order that, if $P:= U(\alpha, \beta)$ (which is automatically unitary), we have (i) $PP=e^{ik}I$ for some $k\in \mathbb R$, (ii) $P\hat{x}P^\dagger = -\hat{x}$, (iii) $P\hat{p}P^\dagger = -\hat{p}$ Since $\hat{x}$ and $\hat{p}$ has to be treated symmetrically, we ...


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$$[P,H]f(x)=(PH-Hp)f(x)$$ But $$H=P^2/2m+E(x)$$ $$ =PE(x)-Hf(x)$$ $$ =E(-x)-E(-x)$$ $$ =0 $$ The parity operator therefore commutes with Hamiltonian.



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