D.84 Band gap ex­pla­na­tion de­riva­tions

To see math­e­mat­i­cally how the re­sults of note {N.9} were ob­tained re­quires knowl­edge of lin­ear al­ge­bra. If you are un­aware of it, def­i­nitely skip the be­low de­riva­tion.

First 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:

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

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

Since the po­ten­tial is the same in all atomic cell, ma­trix $G$ de­scribes the change over any cell, not just the first one. And 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,

$$\psi_1\psi_2'-\psi_1'\psi_2 = \mbox{constant} = 1$$

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

$$0 = \psi_1\psi_2'' - \psi_2 \psi_1'' = \left(\psi_1\psi_2' - \psi_2 \psi_1'\right)'$$

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

$$e^{{\rm i}k_x d_x} \quad\mbox{and}\quad e^{-{\rm i}k_x d_x}$$

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

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

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

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

be­cause the co­sine is at its $\pm1$ ex­trema at the gap en­er­gies. So the ve­loc­ity be­comes zero at the ends of the bands.

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.