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 \mathbf{x} \rightarrow -\mathbf{x}, 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, 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). It's elementary to show that if a state has orbital angular momentum l, then its parity is (-1)^l.
But apparently some particles - specifically the antiparticles of fermions - have some sort of "intrinsic parity" equal to -1, which means that 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).
How is this possible? Where does that extra factor of -1 come from?
But apparently some particles - specifically the antiparticles of fermions - have some sort of "intrinsic parity" equal to -1, which means that 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).
How is this possible? Where does that extra factor of -1 come from?