- #1

- 288

- 2

## Main Question or Discussion Point

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(

\begin{array}{cc}

\psi_{+}(\mathbf{x}) \\

\psi_{-}(\mathbf{x})

\end{array}

\right)=\left(

\begin{array}{cc}

\psi_{+}(\mathbf{-x}) \\

\psi_{-}(\mathbf{-x})

\end{array}

\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(

\begin{array}{cc}

\psi_{+}(\mathbf{x}) \\

\psi_{-}(\mathbf{x})

\end{array}

\right)=-\left(

\begin{array}{cc}

\psi_{+}(\mathbf{-x}) \\

\psi_{-}(\mathbf{-x})

\end{array}

\right).[/tex]

How is this possible? Where does that extra factor of [itex]-1[/itex] come from?

\begin{array}{cc}

\psi_{+}(\mathbf{x}) \\

\psi_{-}(\mathbf{x})

\end{array}

\right)=\left(

\begin{array}{cc}

\psi_{+}(\mathbf{-x}) \\

\psi_{-}(\mathbf{-x})

\end{array}

\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(

\begin{array}{cc}

\psi_{+}(\mathbf{x}) \\

\psi_{-}(\mathbf{x})

\end{array}

\right)=-\left(

\begin{array}{cc}

\psi_{+}(\mathbf{-x}) \\

\psi_{-}(\mathbf{-x})

\end{array}

\right).[/tex]

How is this possible? Where does that extra factor of [itex]-1[/itex] come from?