= Solution
Let $\chi=(\sqrt5/3,2/3)^{\mathsf T}$, so $\chi^\dagger\chi=1$. The <Pauli matrices>[Pauli identity] $(\boldsymbol\sigma\cdot\mathbf p)^2=|\mathbf p|^2I$ gives
$$
\psi^\dagger\psi=|N|^2\left(1+\frac{|\mathbf p|^2}{(E+m)^2}\right)=|N|^2\frac{2E}{E+m},
$$
using $E^2=|\mathbf p|^2+m^2$. Thus unit <spinor> <probability density> is obtained with
$$
\boxed{N=\sqrt{\frac{E+m}{2E}}.}
$$
An arbitrary constant <phase> is immaterial. For the alternative <Relativistic normalization of a one-particle state>[Lorentz normalization] $u^\dagger u=2E$, choose $N=\sqrt{E+m}$. Neither convention makes an infinite <plane wave> <Square-integrable function>[square-integrable]; a box-normalized state includes an additional $V^{-1/2}$, or one forms a normalizable <wave packet>.
Back to article page
