Plane wave solution to Schrodinger's equation for periodic potential

Question

Schrodinger's equation for periodic potential is

$$\left( -\dfrac{\hbar^2}{2m}\nabla^2 + V(\vec{r}) \right) \psi(\vec{r}) = E \psi(\vec{r})$$

The term $-\dfrac{\hbar^2}{2m}\nabla^2$ represents the kinetic energy, the term $V(\vec{r})$ represents the potential energy, $E$ is the total energy of the electron, and $\psi(\vec{r})$ is the wavefunction of an electron.

In semiconductor physics, it is said that, according to Bloch's theorem, the solution is of the form $\psi(\vec{r}) = e^{i \vec{k} \cdot \vec{r}} u(\vec{r})$, where $u(\vec{r})$ is a periodic function with the same periodicity as the crystal.

I found this interesting, because in my optics textbook, Optics, Fifth Edition, by Hecht, the complex representation of a plane wave is given as

$$\psi(\vec{\mathbf{r}}) = Ae^{i \mathbf{\vec{k}} \cdot \mathbf{\vec{r}}}.$$

My research indicates that the term $u(\vec{r})$ in $\psi(\vec{r}) = e^{i \vec{k} \cdot \vec{r}} u(\vec{r})$ is the amplitude, and so it is then said that the plane wave is a periodically modulated plane wave.

Given all of this, I am now wondering about a few things:

1. Since $A$ is the amplitude in $Ae^{i \mathbf{\vec{k}} \cdot \mathbf{\vec{r}}}$, and $u(\vec{r})$ is the amplitude in $\psi(\vec{r}) = e^{i \vec{k} \cdot \vec{r}} u(\vec{r})$, is it the case that $A = u(\vec{r})$? Why is one a constant and the other a function?

2. What does it mean to say that a plane wave is "periodically modulated"? What is the role of the term $u(\vec{r})$ in this?

3. Is this just a simple case of the plane wave being used in both cases, (and/)or is there some deeper physical connection between these two domains/theories?

I would greatly appreciate it if some of the more knowledgeable people would please take the time to clarify these points. Please keep in mind that I am a novice who has just started learning this material, so explanations at a commensurate level would be appreciated.

Answer 1

So indeed, a free electron is described by plane wave $$\Psi(r) =A e^{ikr}$$ where $A$ is constant. This means the probability to observe the electron will be equal to $|\Psi(r)|^2=A^2$ for every $r$, i.e. the probability to observe a free electron in space is everywhere the same. Now if we take into account a periodic potential $U(r)$ we indeed get a Bloch function $$\Psi(r) =u(r) e^{ikr}$$ where $u(r)$ is a function of position with the same periodicity as the periodicity of potential/lattice and also satisfies the Schrödinger equation. Now we have that $|\Psi|^2 =|u(r)|^2$ i.e. this function gives an idea about how the charge distribution looks within a unit cell of the lattice. So what this function does is taking into account the ion core potential of your lattice. Whereas, the free electron had just a flat probability distribution to be found in space. The Bloch wave representing a nearly free electron in the lattice will have some spatial distribution which will have periodic recurrence, typically charge will pile up on the positive ion cores. There are some methods to obtain an expression for $u(r)$, one of them is performed by Wigner and Seitz.

So in your third comment you ask whether it is a special case of a free electron? Well, it is the description of a nearly free electron subjected to a periodic potential which will force the wavefunction to adopt a periodic modulation similar to that of the lattice. I hope this answers your questions more or less, to fill the gaps I would highly recommend reading Kittel 'introduction to solid state physics' CH7 (optional CH9)!

Answer 2

Strict definition of "plane wave" means that the phase of the wave is constant along parallel planes, and subsequently the phase velocity is perpendicular to these planes. It doesn't matter if the wave represents classical light of quantum electrons. Mathematically it means that the phase velocity of the wave is constant and along single direction. Phase of such wave is $\phi=\vec{k}\cdot\vec{r}-\omega t$ and thus the constant phase velocity $d\phi=\vec{k}\cdot\vec{dr}-\omega\cdot dt = 0$ means that $\frac{dr}{dt}=\frac{\omega}{|k|}\equiv c$ (here $c$ is the phase velocity in the medium, which is the speed of light for electromagnetic waves).

Notice that $V(x)=0$ case is also a periodic potential (with "periodicity" 0), thus the Bloch solution is also valid for this case. Bloch solution is valid for any system with periodic potential (not restricted to semiconductors, as you mentioned, though mostly used for crystalline structures).

Since the Schrodinger equation is linear, its solution can be a superposition of two distinct solutions. So superposition of $\psi_1 = Ae^{-ik_1r}$ and $\psi_2 = Be^{-ik_2r}$ will also be a solution. However, if $k_1$ is not parallel to $k_2$ the result will not be a plane wave. i.e. the phase of the solution is not constant along any plane.

$A=u(r)$ in case of $V(x)=0$. Constant is also a function of $r$, just independent of coordinate. For $V(x)\ne 0$ the wave is modulated - the amplitude of the wave changes periodically as $u(r)$, but the phase relations remain the same (it is still a plane wave). The probability of finding the particle $|u(r)^2|$ depends on position periodically, whereas for $u(r)=A$ the probability is position independent