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.

  • $\begingroup$ A minor detail: the Schrödinger equation you use ist the Schrödinger equation. V(r) can be any potential, including periodic or non-periodic potentials. The solutions of the equation depend on that potential. $\endgroup$ Jan 17, 2020 at 11:12
  • $\begingroup$ @HartmutBraun Oh, ok. Thanks for making me aware of this. As is obvious from my post, I've only just started studying the Schrödinger equation, so small details like this are appreciated. $\endgroup$ Jan 17, 2020 at 11:14

2 Answers 2


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)!

  • 1
    $\begingroup$ Thanks for the answer! Out of curiosity, what norm is used for $|\Psi|^2 =|u(r)|^2$? $\endgroup$ Jan 17, 2020 at 22:39
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    $\begingroup$ Generally it's unity over the unit cell of the crystal, but also can be normalized to number of conduction electrons in that cell. $\endgroup$
    – Alexander
    Jan 18, 2020 at 0:27

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

  • $\begingroup$ Thanks for the answer. Can you please explain how you got that $\frac{d\phi}{dt}=\vec{k}\cdot\vec{dr}-\omega\cdot dt = 0$? Is $\vec{r}$ a function of $t$? And if so, then shouldn't we have $\frac{d\phi}{dt}=\vec{k}\cdot \dfrac{d\vec{r}}{d\vec{t}}-\omega = 0$? $\endgroup$ Jan 17, 2020 at 23:05
  • $\begingroup$ It should be $d\phi$. I corrected the post. $r$ and $t$ are space and time coordinates (not a property of the wave). $k$ and $\omega$ are properties of the wave. In a general case $\omega$ is a function of $k$ (it's called dispersion relations) You can ask what is the phase of the wave at specific $(r,t)$ and it is $\phi=kr-\omega t$ or what is the phase difference from a reference point if you move $dr$ in space and $dt$ in time. Then you constrict it to be zero and get the relation for the phase velocity (if you move $kdr$ in space and $\omega dt$ in time then you're get the same phase) $\endgroup$
    – Alexander
    Jan 18, 2020 at 0:22
  • $\begingroup$ Thanks for the clarification. Can you please explain why the $V(x)=0$ case is a periodic potential with periodicity $0$? I didn't understand this part. $\endgroup$ Jan 18, 2020 at 9:52
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    $\begingroup$ Periodic function satisfies $f(x)=f(x+a)$ with periodicity being the smallest $a$ that satisfies the relation for any $x$. Constant function satisfies this relation too. However for constant function $a$ can be infinitesimal, as small as you want it. $\endgroup$
    – Alexander
    Jan 18, 2020 at 17:36

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