Interpretation of "spin-1/2" in classical Dirac field

Question

I emphasize that the proceeding is purely classical physics. Consider the Grassmann-valued field (where $\mathcal{N}$ is a Grassmann number), which is a solution to the Dirac equation, given by $$\psi(x) = \mathcal{N}\begin{pmatrix} \sqrt{\omega - k} \\ 0 \\ \sqrt{\omega + k} \\ 0 \end{pmatrix} e^{-i\omega x^0 + ikx^3}.$$ The $0$th component of the Noether current corresponding to the continuous symmetry of rotations around the $z$-axis yields $$l_z = \psi^\dagger \frac{1}{2} \begin{pmatrix} \sigma_3 && 0 \\ 0 && \sigma_3 \end{pmatrix} \psi$$ where the "orbital angular momentum density" contribution is clearly $0$, so it has been omitted. We identify the remaining contribution to $l_z$ as "spin angular momentum density". Hence, the spin angular momentum density for this particular $\psi(x)$ is given by $$l_z = \mathcal{N}^* \mathcal{N}\omega.$$ Hence, the "spin density" of the field is dependent on frequency (energy) $\omega$. How do I physically make sense of the spin density being variable and depending on $\omega$? Is this perhaps the fact that once we quantize a Dirac field we may have multiple "spin-$1/2$ excitations", i.e. particles, corresponding to said field, so the spin density should take into account the spin of all such excitations?