Pauli principle for semiclassical electrons

Question

In the discussion of electrons in metals, often the semiclassical model is used. There, the electrons are treated as occupying localized wave packets $|k,x_i\rangle$ which have momentum $k$ and position $x$ with small fluctuations.

If this is really true, it would seem to be possible for many electrons (say 1000) to occupy the same momentum state, as long as they are localized at different points. (you can have an antisymmetric wavefunction like that if I'm not mistaken: $|\psi\rangle = \sum_P sgn(P) |k,x_{P(1)}\rangle |k,x_{P(2)}\rangle |k,x_{P(3)}\rangle... $).

In the experiment we observe the Fermi distribution: Only two electrons at most can have the same momentum. This seems to be in direct conflict with the semiclassical model.

What is going on here?

To clarify: The question is why the semiclassical model doesn't contradict the fact that electrons in a metal are distributed according to the Fermi distribution.

Edit: I guess what must happen is that the "semiclassical" description yields the same momentum distibution (at least in some approximation) because of the spacial constraint of the material which manifests in both models in a different way. In the wavepacket model we must fill the space with no overlapping $x_i$, while in the Bloch approach we have $\Delta k \sim \frac{2 \pi}{L}$. Is this true? If so, can you help me with the details?

Answer 1

The Pauli exclusion principle only refers to particles in exactly the same state. If they are local wavepackets localised at different locations then they aren't in exactly the same state. So it doesn't apply here. In fact, since Pauli is a (somewhat weak) corollary to the fact that fermion wavefunctions must be anti-symmetric under exchange, that you wrote down an anti-symmetric wavefunction shows that Pauli clearly doesn't apply here.

It's worth noting that neither the momentum nor the position in this case can be exactly certain (as per Heisenberg) - this is obvious for the positions since the wavepacket by definition has a certain spread, but its also the case for the momentum (on general grounds via Heisenberg, or you could pick your favourite wavepacket and convert to momentum space and check).