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I am having some trouble making the jump from single electrons to solids. In David Tong's notes, I have worked through the tight binding model section. Here, we solve for the energy levels that result from putting atoms on a lattice with probability of an electron moving between neighboring lattice points. The energies are $$E = E_0 - 2t\text{cos}(ka)$$ for $k$ restricted to the Brillouin zone.

I understand that, in quantum mechanics, we can solve for the orbitals an electron can occupy in an atom. However, I don't understand how to physically interpret the energy bands in the tight binding model. Are there just a large number of energies available to the electron at every point in the lattice?

I see pictures similar to this one often, but they don't help me see what is physically going on in the solid, so I think I am misunderstanding this model.

bands

Any help would be appreciated!

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I really think this is one of those things that really should not be done as an afterthought, as Tong is doing. You should be picking up a proper condensed matter / solid state physics text and learning step-by-step what is being meant.

The central idea that is not proved but is experimentally observed to be true for almost all systems, is that weakly interacting almost-isolated states exhibit a tiny repulsion from each other, forming QHO-like equally spaced energy states.

Consider an isolated H atom. Then its energy levels are given by the famous Rydberg formula to be of the form $E=-\frac R{n^2}$ that you must be familiar with. When you have a H$_2$ molecule, each energy level of the original H atom is pretty strongly perturbed because of the covalent bond, but even this covalent bonding can be seen as a bonding v.s. anti-bonding splitting of the original H energy levels, one pair of splitting per original energy state, i.e. a lot of perturbations from the $-\frac R{n^2}$ that you started with. These perturbations are, at least for near the ground state and first few excited states, very small compared to the original Rydberg energy, and so despite the fact that this perturbation is already pretty strong, on the level of $e$V, it is still considered a small perturbation.

When you bring a bunch of H$_2$ molecules together, their interactions will be van der Waals and not covalent. This makes the perturbation in their individual energy levels incredibly tiny. But this perturbation still exists, still makes for splittings in energy levels. If you have 3 H$_2$ molecules, then the simgle H$_2$ bonding orbital splits into 3 different energy levels, and similarly the 3 anti-bonding orbital splits into 3 different energy levels too. These are extremely tiny splittings, so basically you cannot see them.

However, when you are dealing with condensed matter with Avogadro's number of atoms, then this splitting will become essentially continuous. In the context of crystalline physics, the famous Bloch theorem tells us that the Hamiltonian commutes with translation operator and thus the solution wavefunctions can be expanded in a momentum basis multiplied by a periodic variation.

In particular, this means that the energy eigenstates can be plotted in momentum space, and they will form energy bands because there are so many energy states near each other in energy eigenvalue, this eigenvalue smoothly changing in momentum space. There could be band gaps, e.g. the huge jump from $1s$ to $2s$ energies will typically not be bridged by these perturbations, whereas the small difference in $2s$ and $2p$ original energy eigenvalues (assume some tiny perturbation in the H$_2$ system splits the degenerate H atom eigenenergies) might overlap and the band gap disappears.

On your picture on the left, if you ignore that the energy band is actually a thin hypersurface in momentum space, it will then appear as a thick band of allowed energies separated by band gaps. On the right, it is plotted with respect to interatomic distance $a$ rather than momentum, and it shows that at small $a$, you have two bands, at intermediate, you have the two bands overlap, and at infinitely large $a$, they go back to the single-atom-in-entire-universe energy eigenstates that are widely separated.

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You state:

I understand that, in quantum mechanics, we can solve for the orbitals an electron can occupy in an atom.

But do you understand what the word "orbital" means?

An electron does not occupy an orbital in the way , for example, earth occupies its orbit around the sun. Quantum mechanically, the word orbital is describing the locus in space where there is a **probability ** to find an electron; not an orbit for the electron. The probability comes from computing the $Ψ^*Ψ$ of the wavefunction solution for that particular problem including boundary conditions. Look at the results of this experiment https://physicsworld.com/a/quantum-microscope-peers-into-the-hydrogen-atom/

However, I don't understand how to physically interpret the energy bands in the tight binding model. Are there just a large number of energies available to the electron at every point in the lattice?

Analogously, the orbitals in the lattice are the probability loci where an electron can be found if measured. There will be energy levels depending on the boundary conditions of the problem, The band theory of solids is an approximation of this.

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  • $\begingroup$ I'm confused. I am well aware of the PDF nature of the wavefunction. Maybe occupy is a strange word to use. What I meant was that, in my physics coursework, I did a calculation of the solutions to the Schrodinger equation for the Hydrogen atom case. Plotting the PDFs of this, I showed that various linear combinations of these give rise to the molecular orbitals that we know and love from chemistry. In chemistry, we say these orbitals are where we can find electrons. This is my full understanding, is this incorrect..? $\endgroup$ Sep 25, 2023 at 4:31
  • $\begingroup$ Either way, here what I'm seeking is a physical picture of what this band model affords. In the lattice, how can I understand the physical placement of these new states. i.e. where will I find the electrons under this model, physically (in space)? $\endgroup$ Sep 25, 2023 at 4:32
  • $\begingroup$ " we say these orbitals are where we can find electrons. This is my full understanding, is this incorrect..?" if you go to an (r,theta,phi) of an orbital, there is a calculable** probability** of finding an electron there. $\endgroup$
    – anna v
    Sep 25, 2023 at 8:37
  • $\begingroup$ " i.e. where will I find the electrons under this model, physically (in space)? – " the electron has a probability to be anywhere in the lattice. The band theory allows to calculate the collective behavior of the orbitals , as the figure states, $\endgroup$
    – anna v
    Sep 25, 2023 at 8:43

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