dEdt
- 286
- 2
I don't understand how it's possible for the intrinsic parity of any elementary particle to be anything other than one. The parity operator makes the transformation [itex]\mathbf{x} \rightarrow -\mathbf{x}[/itex], so the only thing about a state that can change after the parity operator is applied to it are the components of its wavefunction, right? ie for a spin 1/2 particle, [tex]P \left( <br />
\begin{array}{cc} <br />
\psi_{+}(\mathbf{x}) \\ <br />
\psi_{-}(\mathbf{x})<br />
\end{array} <br />
\right)=\left( <br />
\begin{array}{cc} <br />
\psi_{+}(\mathbf{-x}) \\ <br />
\psi_{-}(\mathbf{-x})<br />
\end{array} <br />
\right).[/tex] It's elementary to show that if a state has orbital angular momentum [itex]l[/itex], then its parity is [itex](-1)^l[/itex].
But apparently some particles - specifically the antiparticles of fermions - have some sort of "intrinsic parity" equal to [itex]-1[/itex], which means that [tex]P \left( <br /> \begin{array}{cc} <br /> \psi_{+}(\mathbf{x}) \\ <br /> \psi_{-}(\mathbf{x})<br /> \end{array} <br /> \right)=-\left( <br /> \begin{array}{cc} <br /> \psi_{+}(\mathbf{-x}) \\ <br /> \psi_{-}(\mathbf{-x})<br /> \end{array} <br /> \right).[/tex]
How is this possible? Where does that extra factor of [itex]-1[/itex] come from?
But apparently some particles - specifically the antiparticles of fermions - have some sort of "intrinsic parity" equal to [itex]-1[/itex], which means that [tex]P \left( <br /> \begin{array}{cc} <br /> \psi_{+}(\mathbf{x}) \\ <br /> \psi_{-}(\mathbf{x})<br /> \end{array} <br /> \right)=-\left( <br /> \begin{array}{cc} <br /> \psi_{+}(\mathbf{-x}) \\ <br /> \psi_{-}(\mathbf{-x})<br /> \end{array} <br /> \right).[/tex]
How is this possible? Where does that extra factor of [itex]-1[/itex] come from?