Long Answer
This 'physical meaning' that people speak of is the ability to build up a field strength to measure at the classical level. To build up a macroscopic strength of any field you need to be able to have a large number of quanta (particles) with identical quantum numbers. If this is possible then you can get the field strength that is observed classically. These states are
coherent states which are superpositions of multiparticle states:
|\alpha,\,\mathbf{p}\rangle=e^{|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}\,|n,\,\mathbf{p}\rangle\,,
with |n,\,\mathbf{p}\rangle=a^\dag_{\mathbf{p}}\ldots a^\dag_{\mathbf{p}}|0\rangle being defined as the n-particle state of momentum \mathbf{p}. The parameter \alpha sets the amplitude (and phase) of the coherent state (plane wave).
Now to answer your question: we never speak of the
electron field or the
dirac field in undergraduate chemistry and physics courses because we can never build up macroscopic field strength. This is because as Demystifier mentioned, the Dirac (electron) field is quantized using
anticommutation relations: \{a^\dag_{\mathbf{p}},\,a^\dag_{\mathbf{p'}}\}\,=0. What does this mean? It means we can create a 1-particle state by acting the creation operator on the vacuum state: a^\dag_{\mathbf{p}}|0\rangle=|1,\,\mathbf{p}\rangle, but the 2-particle state vanishes: a^\dag_{\mathbf{p}}a^\dag_{\mathbf{p}}|0\rangle=-a^\dag_{\mathbf{p}}a^\dag_{\mathbf{p}}|0\rangle=0, where in the first step, I've used the anticommutation relation (hard to tell since the a^\dag_\mathbf{p}'s are the same). All higher-particle states vanish for the same reason.
So, look up at the formula for the coherent state. The sum abruptly terminates after n=1 for the Dirac field, preventing us from building up that macroscopic strength we were looking for. This is why the Dirac field is said to be 'classically unphysical.'
