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A simple tight binding picture shows how allowed states in a solid can be quantized with allowed wavenumber $$k_i = i \frac{2 \pi}{N a}$$ where $N$ is the number of atoms in a circular chain. For large system size, the spacing $k_{i+1}-k_i$ tend to zero and the spectrum is nearly continuous: the electronic bands.

An alternative explanation for the formation of (nearly) continuous energy levels (i.e., the bands) goes as follows: "... if a large number N of identical atoms come together to form a solid, such as a crystal lattice, the atoms' atomic orbitals overlap. Since the Pauli exclusion principle dictates that no two electrons in the solid have the same quantum numbers, each atomic orbital splits into N discrete molecular orbitals, each with a different energy. Since the number of atoms in a macroscopic piece of solid is a very large number ($N\approx10^{22}$) the number of orbitals is very large and thus they are very closely spaced in energy (of the order of $10^{−22}$ eV). The energy of adjacent levels is so close together that they can be considered as a continuum, an energy band." [copied from Wikipedia] ].

How to reconcile the two concepts?

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The volume (i.e., the size) of a solid is proportional to the number of particles in it, since the density is constant (otherwise it becomes a different solid) - so, if we say that the energy level spacing is small to the system size, it is also correct to say that it is small due to the large number of particles.

In an infinitely large system the energy spectrum is continuous - as it is the case for a free electron. The discrete energy levels appear when the system is bounded and the spacing between the levels increases as the bounds come closer. This is a manifestation of the Heisenberg uncertainty pronciple: localization in space means that we have more uncertainty in momentum and hence heigher energy. In this respect the solid differs from free electron by a periodic potential being superimposed.

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