Why is the gradient of Energy vs. wavenumber(k) zero at the edge of Brillouin zones?

Question

My question relates to semiconductor physics. The relationship between energy E and wavenumber k for the Kronig-Penny model looks like this, where a is the period of the potential function:

I am referring to a statement made in the book Advanced Semiconductor Fundamentals, 2nd Edition by Robert F. Pierret, page 61. In reference to the E-k diagram of the Kronig-Penney model it says:

...the energy band slope dE/dk is zero at the k-zone boundaries.

and referring to the gradient:

This is a feature common to all E-k plots, even those characterizing real materials.

Why is the gradient zero at the k-zone boundaries (i.e. edges of Brillouin zones)? What does it signify?

Answer 1

The x axis on the plot is proportional to momentum. The k-zone boundary in your diagram, is the point with zero momentum (no net motion of a carrier.); So, in the plot you show, that's why it's the minimum point as all other points have a net motion greater than it. You will notice, that the graph is symmetrical about the k=0 point in the kronig-penny model. The gradient is zero because of that symmetry. From basic calculus, when you take the derivative of a function, it is always zero at an extremium.

Answer 2

A detailed mathematical analysis of Kronig-Penney model yields a mathematical equation comprising of two parameters k and q.

$$\cos(qa)+\frac{mg}{{\hslash}^2 q \sin(qa)} = \cos(ka)$$ where $$q = \sqrt{\frac{2mE}{\hslash}}$$ q is the parameter involved the solution of wavefunction of Schrodinger equation.

Here the k parameter denotes the crystal momentum not the electron momentum. Now, not going into detailed math, in this model (1-D), the Bragg law of reflecetion must be restated. It is essential to note this law in order to explain why the dE/dk is 0 at top of the curve and the negative differential effective mass of electron.

Bragg law: wavelength - w; angle - t;

$$w = 2a\sin{t}$$

Note 1-D model, so only back scattering so t = 90 degrees.

Also by de Broglie formula:

$$w = \frac{h}{p}$$

$$p=\hslash q$$

$$\hslash=2 \pi h$$

On solving above equations we get \$q=\frac{n \pi}{a}\$. For lowest degree, n=1. So \$q=\frac{\pi}{a}\$, Bragg reflection takes place even when the k vector is in the opposite direction.

What this results in is a standing wave due to the creation of two waves in two directions (Bloch wave functions). dE/dk is a measure of group velocity and this being zero implies there is a presence of standing wave at Bragg Reflection points