N.9 Explanation of the band gaps

Chapter 6.21 explained the band gaps in spectra qualitatively as the remnants of the discrete energy states of the individual atoms. However, if you start from the free-electron gas point of view, it is much less clear why and when addition of a bit of crystal potential would produce band gaps. This note explores that question based on the Kronig & Penney model.

Figure N.1: Spectrum for a weak potential.
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To allow an easier comparison with the free-electron gas solutions, the drops in potentials will be greatly reduced compared to figure 6.23. That results in the spectrum shown in figure N.1. The figure also shows the corresponding free-electron spectrum at the right.

You would expect that the relatively small potential drops would have little effect at high energies. Therefore, at high energies the two spectra should agree. And so they do, mostly. But periodically there is still a very narrow band gap, however high the energy. And at those gaps the electron velocity plunges sharply to zero.

Figure N.2: The 17 real wave functions of lowest energy for a small one-di­men­sion­al periodic box with only 12 atomic cells. Black curves show the square wave function, which gives the relative probability of finding the electron at that location.
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To understand why this happens, it is necessary to revert from the Bloch waves to the more basic underlying real energy eigenfunctions. These more basic eigenfunctions can be obtained from the real and imaginary parts of the Bloch waves. For example, for free electrons the Bloch waves can be written as some unimportant constant times

\begin{displaymath}
e^{{\rm i}k_x x} = \cos(k_x x) + {\rm i}\sin(k_x x)
\end{displaymath}

In this case the real eigenfunctions are the cosine and sine functions.

In the presence of a crystal potential, the real eigenfunctions are no longer sines and cosines. Figure N.2 shows the first few eigenfunctions for a 12 atom periodic crystal. The crystal potential is shown in green. The black curves show the real energy eigenfunctions. More precisely, they show the square magnitudes of these functions. The square magnitude is shown since it gives the probability of finding the electron at that location.

Consider first the eigenfunction with $k_x$ zero. It is shown in the very bottom of figure N.2. The free-electron eigenfunction would be $\cos(0)$, which is a constant. In the presence of the crystal potential however, the eigenfunction is no longer a constant, as figure N.2 shows. It develops a ripple. The wave function now has peaks at the locations of the drops in potential. That makes it more likely for the electron to be at a drop in potential. The electron can lower its potential energy that way. If the ripple becomes too strong however, it would increase the kinetic energy more than the potential energy decreases.

Next consider the eigenfunctions for which $k_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$$\raisebox{.5pt}{$/$}$$d_x$. For that wave number, the free-electron eigenfunctions $\cos(k_xx)$ and $\sin(k_xx)$ have a period that is twice the atomic period $d_x$. In particular, the energy eigenfunctions cross zero every atomic period. For a 12 atom crystal, that means that the eigenfunction has 12 zeros. In the presence of a crystal potential, the eigenfunctions are no longer a cosine and sine, but the period stays equal to 2$d_x$. The number of zero crossings does not change.

It can be shown from the mathematical theory of equations like the one-di­men­sion­al Hamiltonian eigenvalue problem that the energy eigenfunctions remain arranged by zero crossings. The more crossings, the higher the energy.

The solutions in the presence of a crystal potential are marked as $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 in figure N.2. Note that what used to be the sine, the antisymmetric energy eigenfunction, has all its peaks at the drops in potential. That lowers the potential of this eigenfunction considerably. On the other hand, what used to be the cosine has all its zeros at the drops in potential. That means that for this eigenfunction, the electron has very little chance of being found at a location of low potential. Therefore the cosine solution has a much higher energy than the sine solution.

That produces the band gap. The energy of the sine-type solution gives the top energy of the lowest band in the spectrum figure N.1. The energy of the cosine-type solution is the bottom energy of the second band.

The next lower value of $k_x$ corresponds to wave functions with 10 zeros instead of 12. That makes a big difference because there is no way to align 10 zeros with 12 atomic periods. The two eigenfunctions are now just shifted versions of each other and have the same energy. The latter is unavoidable. If you take any eigenfunction and shift it over an atomic period, it is still an eigenfunction, with the same energy. But it cannot be just a multiple of the unshifted eigenfunction because that would require the shifted and unshifted eigenfunction to have the same zeros. If you distribute 10 zeros over 12 atomic periods, some periods end up with no zeros and other with at least one. Shift it over a period. and some periods must change their number of zeros. And if the shifted and unshifted eigenfunctions are different, then they are a complete set of eigenfunctions at that energy. Any other is just a combination of the two.

So no energy difference exists between the two $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10 solutions and therefore no energy gap. But you might still wonder about the possibility of an energy gap between the two $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10 solutions and the $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 solution of lowest energy. It does not happen. The $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10 energy has to stay below the $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 one, but the eigenfunctions struggle to achieve that. By sitting right on top of every potential drop, the $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 eigenfunction is highly effective in lowering its potential energy. The $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10 solutions cannot do the same because 10 zeros cannot properly center between 12 potential drops. The $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10 solutions instead slowly modulate their amplitudes so that the amplitude is high where the peaks are on top of the potential drops, and low where they are not. That has the same effect of increasing the probability that the electron can be found at low energy. The modulation however forces the energy eigenfunctions to give up some of their kinetic energy advantage compared to the $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 solutions. So the energies become closer rather than further apart. Since the changes in energy are a measure of the propagation velocity, the velocity plunges to zero.

The second $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 12 solution is unusually effective in avoiding the regions of low potential energy, and the $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 14 solutions have to keep up with that.

If you actually want to show mathematically that the propagation velocity is indeed zero at the band gaps, you can do it using a linear algebra approach. Define the “growth matrix $G$ that gives the values of $\psi,\psi'$ at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $d_x$ given the values at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

\begin{displaymath}
\left(\begin{array}{c} \psi(d_x) \\ \psi'(d_x) \end{array}...
...\left(\begin{array}{c} \psi(0) \\ \psi'(0) \end{array}\right)
\end{displaymath}

Simply take the initial conditions to be 1,0 and 0,1 respectively to find the two columns of $G$.

For a periodic solution for a box with $N_x$ atoms, after $N_x$ applications of $G$ the original values of $\psi,\psi'$ must be obtained. According to linear algebra, and assuming that the two eigenvalues of $G$ are unequal, that means that at least one eigenvalue of $G$ raised to the power $N_x$ must be 1.

Now matrix $G$ must have unit determinant, because for the two basic solutions with 1,0 and 0,1 initial conditions,

\begin{displaymath}
\psi_1\psi_2'-\psi_1'\psi_2 = \mbox{constant} = 1
\end{displaymath}

for all $x$. The quantity in the left hand side is called the Wronskian of the solutions. To verify that it is indeed constant, take $\psi_1$ times the Hamiltonian eigenvalue problem for $\psi_2$ minus $\psi_2$ times the one for $\psi_1$ to get

\begin{displaymath}
0 = \psi_1\psi_2'' - \psi_2 \psi_1''
= \left(\psi_1\psi_2' - \psi_2 \psi_1'\right)'
\end{displaymath}

According to linear algebra, if $G$ has unit determinant then the product of its two eigenvalues is 1. Therefore, if its eigenvalues are unequal and real, their magnitude is unequal to 1. One will be less than 1 in magnitude and the other greater than 1. Neither can produce 1 when raised to the power $N_x$, so there are no periodic solutions. Energies that produce such matrices $G$ are in the band gaps.

If the eigenvalues of $G$ are complex conjugates, they must have magnitude 1. In that case, the eigenvalues can always be written in the form

\begin{displaymath}
e^{{\rm i}k_x d_x} \quad\mbox{and}\quad e^{-{\rm i}k_x d_x}
\end{displaymath}

for some value of $k_x$. For either eigenvalue raised to the power $N_x$ to produce 1, $N_xk_xd_x$ must be a whole multiple of $2\pi$. That gives the same wave number values as for the free-electron gas.

To see when the eigenvalues of $G$ have the right form, consider the sum of the eigenvalues. This sum is called the trace. If the eigenvalues are real and unequal, and their product is 1, then the trace of $G$ must be greater than 2 in magnitude. (One way of seeing that for positive eigenvalues is to multiply out the expression $(\sqrt{\lambda_1}-\sqrt{\lambda_2})^2$ $\raisebox{.3pt}{$>$}$ 0. For negative ones, add two minus signs in the square roots.) Conversely, when the eigenvalues are complex conjugates, their sum equals $2\cos(k_xd_x)$ according to the Euler formula (2.5). That is less than 2 in magnitude. So the condition for valid periodic eigenfunctions becomes

\begin{displaymath}
\mbox{trace}(G) = 2 \cos(k_xd_x) \qquad k_xd_x = \frac{n_x}{N_x} 2\pi
\end{displaymath}

From the fact that periodic solutions with twice the crystal period exist, (the ones at the band gaps), it is seen that the values of the trace must be such that the cosine runs through the entire gamut of values. Indeed when the trace is plotted as a function of the energy, it oscillates in value between minima less than -2 and maxima greater than 2. Each segment between adjacent minima and maxima produces one energy band. At the gap energies

\begin{displaymath}
v^{\rm p}_x =
\frac{{\rm d}{\vphantom' E}^{\rm p}_x}{{\...
...{{\rm d}\mbox{trace}(G)}{{\rm d}{\vphantom' E}^{\rm p}_x} = 0
\end{displaymath}

because the cosine is at its $\pm1$ extrema at the gap energies.

Identification of the eigenfunctions using the growth matrix $G$ is readily put on a computer. A canned zero finder can be used to find the energies corresponding to the allowed values of the trace.

Since the eigenfunctions at the band gap have zero propagation velocity, the electrons in these states cannot move through the crystal. If you train an electron beam with such a wave number onto the crystal, the beam will be totally reflected. This can be verified using the so-called Bragg reflection theory of wave mechanics. Indeed, the fact that the crystal spacing is a half-integer multiple of the wave lengths that are reflected is a standard Bragg result. It can be easily derived if you wave your hands a lot, chapter 10.7.2. It then provides an intuitive justification for some of the features of the band structure, in particular for the fact that the velocity is zero at the band gaps.