1. In a one dimensional crystal of lattice spacing a each atom is
represented by a Dirac delta potential, V(x)
= V0/a d(x/a). (Note: the Dirac delta function is non-zero
only at one point, and its integral is 1). Use the nearly free electron
approximation and determine the energy gap between the lowest and the next
higher energy band. Under what condition
will the nearly free electron approximation work? (V0
should be much less than what?)
Solution:
In the nearly free electron approximation we assume
that the wavefunction of the electron is still a free wave: y = 1/L1/2
exp(ikx),
where the 1/L1/2 factor
normalizes the function so that the total probability of the electron being in
the crystal of length L is one. Since the energies belonging to k and –k are
the same, the linear combination of two waves is also a solution. The gap occurs at k=p/a and at k=-p/a. We pick the y(+) = (2/L)1/2 cos (p/a) and y(-) = i(2/L)1/2 sin (p/a). Notice that the wavefunctions are still
normalized so that the integral |y|2 is 1. The energy gap is an integral over the whole length
L of
the function [Sn V(x-na)] [|y(+)|2-|y(-)|2]. Here Sn V(x-na) is the potential due to the N=L/a atoms. Since the potential is made of Dirac delta
functions, it is non-zero only at x= 0, a, 2a, .. na. At these points
the [|y(+)|2-|y(-)|2] function is always exactly 2/L.
Therefore the integral yields Eg=
N (2V0/a) /L =2V0. (Note Eq. 6 in Chapter 7 of the book yields
the same).
This works as long as the gap is much less than the
bandwidth, V0<<(hbar2/2m)(p/a)2.
2. Consider electrons in a two-dimensional square
lattice in the tight binding approximation.
The energy is E=E0 (2
– cos a kx –
cos aky).
a./ Make a sketch of the kx
– ky plane, indicate the
Brillouin zone boundaries..
b./ This band has a small
number of electrons. What is the
effective mass in terms of
E0 and
a?
c./ Draw the Fermi surface
(approximately) for 1.9 electron per atom.
Is the sign of the Hall coefficient positive or negative? Justify your
answer in one sentence.
Solution:
a./
b./ expand the cos function around
k=0. We get is E=
E0 [(akx)2
– (aky)2]/2
= E0 (ak)2/2. Compare this to E= hbar2 k2 /2m, to obtain m= hbar2/E0 a2
c./ Hall coefficient is
positive, since the conduction is by holes in a nearly full band.
3./ Determine the plasma
frequency of copper. Here are some data you may use. Resistivity: 1mWcm,
electron density: 8x1022 cm-3, mean free path
400Angstrom, Fermi velocity 1.6x108cm/s.
Solution.
The plasma frequency is wp=[4pne2/m]1/2. In
principle we should determine the mass from the electron density and Fermi
velocity: m=(hbar/vF)(3p2n)1/3=8.7x10-28g,
pretty close to the free electron mass m=9.1x10-28g. We get wp=[4pne2/m]1/2= 1.6x10161/s. (Be careful with the units, Coulomb does not work in
CGS!)