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OPTI 507 - Solid-State Optics
HW 2 Solutions
Jason M. Auxier
Copyright ©c 2004
All rights reserved.
August 30, 2004
Homework 2 consists of problems 2.5 - 2.8 in Introduction to Semiconductor Optics by Peyghambarian et al [1].
1 Problem 2.5 - 10 Points
Calculate the degeneracy of the lowest three energy bands at the Γ point for a face-centered cubic lattice in the free-electron approximation. The Γ point is the point k = 0 in momentum space. (Degeneracy is defined as the independent number of the wave functions for the same energy. For simplicity, ignore the spin degeneracy.)
1.1 Solution
First, we must understand what the authors meant by the question. Unfortu- nately, this is not entirely clear and so I grade this question accordingly. He meant to ask for the degeneracy of the bands and not the states. That is, when you plot energy in reciprocal space, how many bands are on top of each other. The free-electron approximation means that we ignore the lattice potential and spin-orbit coupling, so
E =
¯h^2 k^2 2 me
Since k is scalar, it does not specify a direction in 3D-reciprocal space. Starting from the Γ point and going in any direction, we travel along three bands (one for each dimension in space). Because of this, we say that we have 3-fold degeneracy (ignoring spin).
1.1.1 Note
This solution is really an oversimplification. To handle this problem properly, one would need to calculate the band structure for a face-centered cubic lattice
Homework 2 Solutions 2
using perturbation theory to at least 1st-order (assuming some crystal size). This is not trivial and beyond the scope of this course, so I did not show it here. This would also be required to calculate the state degeneracy. Now, if the electron is truly free (i.e., in free space), then the degeneracy of states is infinite. This is because in free space, we have a continuum of states. For any value of E = ¯h
(^2) k 2 2 me , there is a spherical shell of k-states which are degenerate.^ Now, if you have a lattice potential for a finite crystal, then the story is quite different.
2 Problem 2.6 - 10 Points
Show that the wave function for the tight-binding model
ψk(r) =
n
eik·rn^ φk (r − rn) , (2)
satisfies the Bloch theorem, Eq. (2.37).
2.1 Solution
We want to show that Eq. 2 obeys Bloch’s Theorem:
ψk(r + R) = eik·Rψk(r). (3)
In Eq. 2, consider r → r + R:
ψk(r + R) =
n
eik·rn^ φk (r + R − rn). (4)
But, the (r − R) site in the lattice is the (n − 1) lattice site, so (rn − R) ≡ rn− 1. Therefore,
ψk(r + R) =
n
eik·(rn−^1 +R)φk (r − rn− 1 )
= eik·R^
n
eik·rn−^1 φk (r − rn− 1 ).
Now, since we are summing over all of the lattice sites of a large lattice (N → ∞), we can write
ψk(r + R) = eik·R^
n′
eik·rn′^ φk (r − rn′^ )
= eik·Rψk(r).
Thus,ψk(r) obeys Bloch’s theorem. Now, this large crystal approximation could be replaced by using periodic boundary conditions (PBC). The problem lies at the lattice site n = 0. Here, you cannot translate to the n = −1 site if the crystal is finite and it starts at n = 0 and goes to n = N − 1. With PBC, we say that lattice sites n = −1 and n = N − 1 are the same (the lattice wraps around on itself). So, you break up the sum and handle this site separately and then everything works out.
Homework 2 Solutions 4
translation r → r + N ai is performed, where N is the total number of primitive cells in the crystal, ai are the three primitive lattice vectors. Consider a one- dimensional solid with N unit cells each of length a. Use the periodic boundary condition and show that the number of allowed states is equal to the number of unit cells.
4.1 Solution
The periodic boundary condition (PBC) is
ψk (r + N ai) = ψk(r). (13)
You can write Bloch’s theorem as
ψk (r) =
eik·r √ V
u(r),
u(r) = u(r + R),
where u(r) is the lattice periodic part of the wavefunction. In 1D, Rn = na, so RN = N a. Thus, substituting this Bloch wavefunction into Eq. 13, we find that
eikxxeikxN a^ = eikxx.
⇒ kx =
2 πnx N a
where nx is an integer. Therefore, the PBC gives us quantized states. Now, for a crystal of any size, N is very large (say 10^8 sites for each direction), so these states are very close together. This is why we refer to the collection of states as a band. But, we’re not finished yet! Since nx is an integer, we need to limit its values. To do this, we use Bloch’s theorem (Eq. 3) and the fact that the direct and reciprocal lattices are periodic. The period is Pk = (^) R^2 minπ = (^2) aπ. Now, Bloch’s theorem can also be written as:
ψk(r + R) = eik·Rψk(r). (16)
Now, letting k′^ = k + G, where G is a reciprocal lattice vector. Noticing that G · R = 2πn, where n is an integer, we have
ψk(r + R) = eik
′·R e−iG·Rψk(r) = eik
′·R ψk(r).
This is Bloch’s thm. again, which means that k′^ and k are equivalent states, that is, they are really the same state. Thus, we only need to talk about states within the 1st Brillouin zone (since we can always pick the right G to get us there). Again, the period is Pk = (^2) aπ , but by convention, we choose the range kmin = − πa to kmax = πa.
REFERENCES 5
So, from Eq. 15, you find that there are only N allowed states within this range. All other values of k are redundant.
For you EE’s, the Fourier transform of a comb function is a comb function. Translating into the 1st Brillouin zone is exactly the same as saying all the information about a periodic signal is contained within a limited bandwidth (period) in Fourier space.
References
[1] Nasser N. Peyghambarian, Stephan W. Koch, and Andre Mysyrowicz. In- troduction to Semiconductor Optics. Prentice Hall, Englewood Cliffs, NJ,
