Relationship between k and orbital phase of solid state?

[Alec Neeson]

I understand that quantum objects have wave and particle properties. I know that k = 2π / lamda. I am simply not understanding the x-axis of a bandstructure plot of E(k) vs. k. I've read parts of a book by Roald Hoffman on this subject. In the book it is shown that there are infinite chain of hydrogen atoms. K is related to the orbital phase in this book. I am not sure how to connect this k to orbital phase. I've heard k can be a measure of momentum. How exactly is k and the phase of an orbital related? What is the most realistic idea of the notion of k? I want to physically understand k. Does this momentum of k relate to the orbital angular momentum and that is how phase and momentum are related?

 

[TeethWhitener]

I imagine the book you’re referring to is “Solids and Surfaces” by Roald Hoffmann. It’s a really good book for approaching solid state physics from a chemical point of view, but it sounds like you’re struggling with some basic aspects of solid state physics.

##k## is the wave number (more generally, the wave vector) of a particle, which is related to the momentum of the particle by ##\mathbf{p} =\hbar \mathbf{k}##. Bloch’s theorem states that the energy eigenstates of a periodic lattice (such as a crystal) have the form
$$\psi(\mathbf{r})=\exp (i\mathbf{k}\cdot\mathbf{r})\phi(\mathbf{r})$$
The function ##\phi(\mathbf{r})## is periodic with the same period as the lattice, and the function ##\exp (i\mathbf{k}\cdot\mathbf{r})## is a plane wave (free particle state) with momentum ##\hbar\mathbf{k}##.

Solving the Schrodinger equation for an atom, for example, gives a discrete spectrum of energy eigenstates, but solving the Schrödinger equation for a periodic system gives a different eigenfunction for each value of ##\mathbf{k}##, which is a continuous variable. So if we were to draw the energy level diagram for a crystal as we would for an atom or molecule, we would simply draw a variable density continuum of states, which is not very enlightening. Instead, we use ##\mathbf{k}## as a kind of “continuous index” and plot ##E## vs. ##\mathbf{k}##. This plot gives us information on the continuous energy bands that are characteristic of a crystal.

 

[Alec Neeson]

In a bandstructure diagram of E(k) vs. k : each band can be occupied by 2 electrons similar to a molecular orbital or no? Also, do the orbitals change as you change the k value or just the phase of the orbitals?

 

[TeethWhitener]

each band can be occupied by 2 electrons similar to a molecular orbital or no?

No. Each band is a continuous collection of energy eigenstates. Each eigenstate can be occupied by 2 electrons.

Also, do the orbitals change as you change the k value or just the phase of the orbitals?

The orbitals themselves change. It might help to think of ##\mathbf{k}## as an index on the wavefunction, just as you’d think of ##n,l,m_l## as indices on atomic wave functions.