Quantum Mechanics for Engineers 

© Leon van Dommelen 

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 that gives the values of
at given the values at 0:
Simply take the initial conditions to be 1,0 and 0,1 respectively, and
find the solutions at to find the two columns of .
Since the potential is the same in all atomic cell, matrix
describes the change over any cell, not just the first one. And for a
periodic solution for a box with atoms,
after
applications of the original values of must be
obtained. According to linear algebra, and assuming that the two
eigenvalues of are unequal, that means that at least one
eigenvalue of raised to the power must be 1.
Now matrix must have unit determinant, because for the two basic
solutions with 1,0 and 0,1 initial conditions,
for all . The quantity in the left hand side is called the
Wronskian of the solutions. To verify that it is indeed constant,
take times the Hamiltonian eigenvalue problem for
minus times the one for to get
According to linear algebra, if 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 , so there are no
periodic solutions. Energies that produce such matrices are in
the band gaps.
If the eigenvalues of are complex conjugates, they must have
magnitude 1. In that case, the eigenvalues can always be written in
the form
for some value of . For either eigenvalue raised to
the power to produce 1, must be a whole multiple of
. That gives the same wave number values as for the
freeelectron gas.
To see when the eigenvalues of 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 must be greater than 2 in magnitude. (One way of seeing
that for positive eigenvalues is to multiply out the expression
0. For negative ones, add two
minus signs in the square roots.) Conversely, when the eigenvalues
are complex conjugates, their sum equals according to
the Euler formula (2.5). That is less than 2 in magnitude.
So the condition for valid periodic eigenfunctions becomes
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
because the cosine is at its extrema at the gap energies. So
the velocity becomes zero at the ends of the bands.
Identification of the eigenfunctions using the growth matrix 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.