Applied Quantum Mechanics (3rd Edition, 2024) – Mid-Term Examples Solutions – Levi

4 Examples Solutions

,MIDTERM example 1
SOLUTION GUIDE


SI-MKS

c = 2.99792458  10 m s
8 –1
Speed of light in free space
 = 6.5821188926  10
– 16
Planck’s constant eV s
 = 1.054571596  10
– 34
Js
e = 1.602176462  10
– 19
Electron charge C
m 0 = 9.10938188  10
– 31
Electron mass kg
m n = 1.67492716  10
– 27
Neutron mass kg
m p = 1.67262158  10
– 27
Proton mass kg
k B = 1.3806503  10
–23 –1
Boltzmann constant JK
k B = 8.617342  10
–5 –1
eV K
 0 = 8.8541878  10
– 12 –1
Permittivity of free space Fm
 0 = 4  10
–7 –1
Permeability of free space Hm
Speed of light in free space c = 1  00
N A = 6.02214199  10
23 –1
Avagadro’s number mol
a B = 0.52917721 10 m
–10
Bohr radius
4 0 
2
a B = ----------------
-
m0e 2

Inverse fine-structure constant  –1 = 137.0359976
4 0 c
 –1 = -----------------
-
e2




Applied quantum mechanics

,Problem 1
An electron moves in the following one-dimensional periodic potential V(x) with
L = 1 nm ..
L
10




Potential energy, V(x) (eV)
8

6

4

2

0
0 1 2 3 4 5 6
Position, x (nm)

(a) Assuming periodic boundary conditions are applied to the potential defined in
the region 0  x  L  6 , write down the Hamiltonian for the system and show that the
Bloch wave function can be written
 k  x + L  =  k  x e ikL
where k is the Bloch wave vector. (20%)
(b) Ignoring electron spin, what is the value of k for the 1st (lowest) energy eigen-
value E b1 and what is the value of k for the 2nd energy eigenvalue E b2 ? Sketch the cor-
responding wave functions  k 1  x  and  k2  x  . (20%)
(c) The periodic potential has a missing-atom crystal defect and the new potential is
sketched in the following figure
L
10
Potential energy, V(x) (eV)




8

6

4

2

0
0 1 2 3 4 5 6
Position, x (nm)

Sketch wave functions  1  x  and  2  x  corresponding to the lowest and next lowest-
energy eigenvalues E c1 and E c2 . Determine if eigenenergy E c1 is greater or smaller
than E b1 in part (b). (40%)
(d) The periodic potential has a crystal defect corresponding to an additional atom
and the new potential is sketched in the following figure
L
10
Potential energy, V(x) (eV)




8

6

4

2

0
0 1 2 3 4 5 6
Position, x (nm)

Sketch the lowest-energy wave function  1  x  with eigenenergy E d1 and state its par-
ity. Determine if eigenenergy E d1 is greater or smaller than E b1 in part (b). (20%)

, Problem 1 solution
(a) The one-electron Hamiltonian is
– d
2 2
ˆ = -------
H - -------- + V  x 
2m dx 2
where V  x  = V  x + nL  for integer n. The electron wave function can be a Bloch
function of the form
 k  x  = U k  x e ikx
where U k  x + nL  = U k  x  has the same periodicity as the potential and n is an inte-
ger. The term e ikx carries the phase information between unit cells via what is called the
Bloch wave vector k. If we wish to know how the value of an electron wave function
changes from position x to x + L , then we need to evaluate
 k  x + L  = U k  x + L e ik  x + L 
 k  x + L  = U k  x e ikx e ikL
Hence,
 k  x + L  =  k  x e ikL
(b) k 1 = 0 and k 2 =   3L .




(c) In this case the defect appears as a potential barrier that breaks symmetry and
separates the periodic crystal in half. The barrier acts to localize the electron and so the
lowest bound state envelope function is similar to the lowest state of an infinite rectan-
gular potential well. Note, as expected, the presence of the cell periodic part of the
wave function. Independent of Bloch wave vector k, each cell periodic part contributes
  2
2
a maximum of U 0 = ---------  ------- to electron energy, where L eff is an effective well
2m 0  L eff
width.
For part (b) we have
max
E 1b = 6U 0 + 0
and
max 4U
E 2b = 6U 0 + ---------0 = 6.111U 0 .
36
For part (c) one may estimate k 1 = 2  10L =   5L and k 2 = 2  5L to give



Applied quantum mechanics