# Why does the energy levels in electron bands lay so closely together?

I'm currently trying to figure out how electron bands form in solids. The Pauli exclusion principle states that:

two or more identical fermions (particles with half-integer spin) cannot occupy the same quantum state within a quantum system simultaneously

Here's another excerpt from the book University Physics with modern Physics:

Because of the electrical interactions and the exclusion principle, the wave functions begin to distort, especially those of the outer, or valence, electrons. The corresponding energies also shift, some upward and some downward, by varying amounts, as the valence electron wave functions become less localized and extend over more and more atoms.

As single atoms, the gaps are quite large between each subsequent energy level $$E_n$$. How come that the energy levels are so close to each other in solids, and what is the formula for determining the 2nd or 3rd atom's energy level?

How do bands form? How come that the energy levels are so close?

When you consider a single atom, this will have some number of distinct energy levels available for its electrons.

Now, when you solve the Hamiltonian for two of these atoms, you will see that the energy levels will split, with two near energy levels for each of the original levels. (Keep in mind that degeneracies in quantum mechanics are typically special; for example they appear in topological systems). You now have twice as many energy levels and all the electrons can sit on the lowest energy levels: one (two if there's spin degeneracy) per level, to satisfy Pauli's exclusion principle.

When you bring together a larger number of atoms, the splitting of the energy levels continues up to when you have, in the thermodynamic limit, an infinity of energy levels deriving from the splitting of a single initial atomic energy level. And you will find one band appearing for each of the initial atomic energy levels.

Does this violate Pauli's exclusion principle?

No, because quantum mechanics does not break down if the Hamiltonian of your system has a continuous spectrum. All the rules apply in the same way. Pauli's principle says that you can't have two fermions in the same quantum state $$|\psi_k\rangle$$. But they can be in $$|\psi_k\rangle$$ and $$|\psi_{k^\prime}\rangle$$ for each $$k\neq k^\prime$$, with $$k,k^\prime$$ discrete or continuous labels.

• Thanks for the answer. Unfortunately I have too low of a reputation to upvote it, but this made it a bit clearer. One more question: are the energy bands identical at all positions in the solid?
– Void
Mar 1, 2019 at 16:04
• I'd say no. In fact, if you fix an energy (or a small window of energy) you can ask, for example, where are the states with that energy (de)localised. And typically what you find is that they are not homogeneously spread over the whole system. For two dimensional systems, this local density of states (LDOS) can be measured with scanning tunneling microscopy (STM) experiments. Mar 1, 2019 at 16:11
• Ok, does that mean that as the volume tends to infinity, the bands wouldn't continue to grow, but rather look slightly different in different parts of the volume?
– Void
Mar 1, 2019 at 16:20
• I'm not sure I understand your question. However, suppose you have a system with a single band (for simplicity). And suppose that for a system size $N$ the width of the band is $\Delta E$. When you bring $N\rightarrow \infty$ this does not make $\Delta E$ go to $\infty$. Typically if $N$ is "big enough", the size of $\Delta E$ won't increase by increasing $N$. On the other hand, the number of states within the band will increase linearly with $N$, yielding a continuous spectrum of energies within the band. Mar 1, 2019 at 16:33