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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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The proton and neutron are defined by convention to have wavefunctions that don't change sign under parity transformations, or positive parity. An antifermion has the opposite intrinsic parity of its matter counterpart, so the antiproton and antineutron have negative parity. Other baryon states have measured parities which are tabulated by the Particle Data ...

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There isn't parity violation in classical physics; it's a phenomenon unique to the weak charged current. There are axial vectors (a.k.a. psuedovectors) in classical physics, like the magnetic field or the angular momentum vector, which would change sign if computed using a "left-hand rule" instead of the usual right-hand rule. However whenever you compute ...

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To clarify what's going on here, let's start by considering an example with an electric field. Suppose you have some complicated, asymmetric shape like a guitar, and it has positive charge distributed on it. At some nearby point $\mathbf{r}$, we have an electric field $\mathbf{E}$. Under a full parity inversion $(x,y,z)\rightarrow(-x,-y,-z)$, the entire ...

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