One explanation on bandgap formation uses nearly free electron theory and Bragg reflection. The diagram used for the explanation appears:

enter image description here

taken from here.

I also learnt that in free electron theory, not all k are allowed. For a cubic lattice with side length a,

$$k= \pm\sqrt{k_x^2+k_y^2+k_z^2}, {\space} k_x=\frac{n_x\pi}{a}, {\space} k_y=\frac{n_y\pi}{a},{\space} k_z=\frac{n_z\pi}{a}$$

$$k = \pm \frac{n\pi}{a}, {\space} n=\sqrt{n_x^2+n_y^2+n_z^2}$$

where $n_x, n_y, n_z$ all must be integers.

That is, not all k are allowed. However, those k allowed ($\frac{\pm \pi}{a}, \frac{\pm 2\pi}{a}$...) will always satisfy Bragg condition.

Doesn't that mean that all states will then satisfy Bragg condition and no band will form? To put my question simply,

enter image description here

why is there allowed energy/ k in the red-circled regions when those k are not allowed from the solution of SE with periodic boundary conditions?


1 Answer 1


You are confusing the size of the entire crystal with the size of a single unit cell. Furthermore, you have used reflecting boundary conditions which require $n_x, n_y, n_z$ to be positive integers, but in the diagram $k$ can be negative so we need periodic boundary conditions which include negative integers.

The diagram considers a one-dimensional crystal for simplicity, which I'll assume to be in the $x$ direction. In such a crystal, if the unit cell has length $a$ and there are $N$ of them, the crystal has length $L=Na$. This $L$ is what should be used to find the allowed $k_x$. Periodic boundary conditions require that the plane wave $\psi(x)=e^{ik_x x}$ is equal on either side of the crystal, so $\psi(0)=\psi(L)$ giving

$$1=e^{ik_x L}\Rightarrow k_x=\frac{2\pi n}{L}=\frac{2\pi n}{Na}$$

where $n$ is any integer, not just a positive integer. Note that what I call $k_x$ is what the diagram calls $k$, which is quite confusing because $k$ can also mean the magnitude of the wavevector (which would be positive).

As $N>1$ there are values of $k_x$ which lie inside the first Brillouin zone (i.e. have $-\pi/a<k_x<\pi/a$) and do not satisfy the Bragg condition. In fact, usually $N\gg 1$ as there are a macroscopic number of atoms in a crystal, so the spacing between allowed $k_x$ states is much smaller than the size of the Brillouin zone, and we can consider there to be a continuum of $k_x$ states as is shown on the diagram.

  • $\begingroup$ The periodic boundary condition, $\psi(0) = \psi(L)$, the L is the length of the bulk crystal, not of the unit cell? What is the rationale for that? I thought it should be that of unit cell because same values of wavefunction repeat at same position in different unit cells. $\endgroup$
    – TheLearner
    Dec 30, 2021 at 13:48
  • $\begingroup$ I can follow your argument, just couldn't get the meaning of L in the boundary condition. I read $\psi(x) = \psi(x+L)$, if L = length of the bulk crystal, x + L already outside of the crystal, then what will be the interpretation on the equation? $\endgroup$
    – TheLearner
    Dec 30, 2021 at 13:52
  • 1
    $\begingroup$ You can think of it like this: instead of a one-dimensional crystal with actual ends, you join the ends together and make it into a circle. The physics must repeat itself when you move around the circle so $\psi(0) = \psi(L)$. What you call $x=0$ is arbitrary so this is the same as $\psi(x) = \psi(x+L)$. $\endgroup$ Dec 30, 2021 at 14:04
  • 1
    $\begingroup$ You might then ask how can we consider it to be a circle even when it is not. The answer is that most physical quantities are measured far from the boundary and so are independent of the choice of boundary conditions. $\endgroup$ Dec 30, 2021 at 14:07

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