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The Dirac Comb:
A First Glance at
Periodic Potential
Jed Brody
Outline
•Review of basic quantum mechanics
•Solution to the Dirac comb problem
•Some semiconductor devices I made
2
The Schrödinger Equation
•Given V(x) and boundary conditions, find (x) and E
that satisfy
 2 d 2 ( x)

 V ( x) ( x)  E ( x)
2
2m dx
•We generally find an infinite number of solutions:
1(x) and E1
2(x) and E2
All energies except
3(x) and E3
.
.
.
E1, E2, E3, etc.
are forbidden!!!
3
Example: Infinite Square Well
n 
En 
,
2
2ma
n  1,2,3,...
2
2
Plot allowed energies in units
 
2
of
2
2ma
2
:
First 10 allowed energies
2
100
0
4
Continuity of  and d/dx
• is continuous everywhere
•d/dx is continuous everywhere except where V is
infinite
•If V(x)=-(x-a),
d
dx
a
d

dx
a
2m
  2  (a)

Outline of derivation:
•Integrate Schrödinger equation from a-e to a+e
•Let e0
5
Commuting Operators
If operators A and B commute ([A,B]=AB-BA=0), then
we can find eigenfunctions of A that are also
eigenfunctions of B.
Proof:
Let [A,B] operate on a1, an eigenfunction of A with eigenvalue a1
(A a1 =a1 a1):
AB  BA  0
 AB  BA  a1  0
AB a1  BA  a1  0
AB a1  Ba1 a1  0
A( B a1 )  a1 ( B a1 )  0
A( B a1 )  a1 ( B a1 )  B a1 ~  a1  B a1  b1 a1
6
The Dirac Comb
•Consider an infinite, one-dimensional chain of
regularly spaced atoms
•Suppose each positively charged atomic core acts
on electrons as a delta-function well (not realistic)
V(x)
-a
a
2a
3a
x
Observe:
V(x+a)=V(x)
7
Bloch’s Theorem
If V(x+a)=V(x), then the solutions to the Schrödinger
equation satisfy (x+a)=eiKa(x) for some constant K.
(x+a)
•Suppose we know (x) only
between 0 and a:
(x)=sin(x/a),
0<x<a
(x)
•Replace x with (x+a):
(x+a)=sin[(x+a)/a], 0<x+a<a
-a<x<0
•Bloch’s theorem gives
(x)=e-iKa(x+a),
(x)=e-iKasin[(x+a)/a], -a<x<0
-a
a
x
8
First Step of Proof: D and H commute
•Define the displacement operator: Df(x)=f(x+a)
•H is the Hamiltonian, (-ħ2/2m)d2/dx2+V(x), for a
periodic potential: V(x+a)=V(x)
•If D commutes with (-ħ2/2m)d2/dx2 and with V(x),
then D commutes with H
•D commutes with (-ħ2/2m)d2/dx2 because taking the
second derivative of a function and then shifting it to
the left is the same as shifting it to the left and then
taking its second derivative
•[D,V(x)]f(x)=[DV(x)-V(x)D]f(x)
=DV(x)f(x)-V(x)f(x+a)
=V(x+a)f(x+a)-V(x)f(x+a)
=0 because V(x+a)=V(x)
9
Completion of Proof
•D and H commute, so we can choose eigenfunctions
of H that are also eigenfunctions of D
•Let D act on one of these eigenfunctions:
D(x)=(x), where  is the eigenvalue of D
•Thus since D(x)=(x+a),
(x+a)=(x)
•Note that (x+2a)=2(x), (x+3a)=3(x), etc.
•To keep (x) from blowing up or vanishing upon
repeated applications of D, ||=1 (x+a)=eiKa(x)
10
Restrictions on K
•The electrical properties of a solid deep in its bulk
are independent of edge effects. To avoid edge
effects, use periodic boundary conditions
•Wrap the x axis around on itself, forming a loop of
length Na; N is the very large number of potential
wells
•After travelling a distance Na, you return to your
starting point: (x+Na)=(x)
•Bloch’s theorem gives (x+Na)=eiNKa(x), so eiNKa=1
NKa=2nK=2n/(Na), n=0,±1,±2,...
11
Solving the Dirac Comb Problem
The potential is a series of delta functions:
N 1
V ( x)     ( x  ja)
j 0
In between delta functions (e.g.,0<x<a), V(x)=0:
V(x)
-a
a
2a
3a
x
 2 d 2

 E
2
2m dx
2
d
2


k

2
dx
2mE
k

12
Conditions on (x) at x=0
 ( x)  A sin( kx)  B cos( kx), 0  x  a
Replace x with (x+a):
 ( x  a)  A sin[ k ( x  a)]  B cos[ k ( x  a)],  a  x  0
Apply Bloch’s theorem, (x)=e-iKa(x+a):
 ( x)  e
 iKa
{ A sin[ k ( x  a)]  B cos[k ( x  a)]},a  x  0
(x) must be continuous at x=0:
d
dx

0
d
dx
B  eiKa [ A sin( ka)  B cos(ka)]

0
2m
 (0) :
2

kA  e
iKa
2m
k[ A cos( ka)  B sin( ka)]   2 B

13
Equation To Find Allowed Energies
•Eliminating A and B from previous equations gives
m
cos( Ka)  cos( ka)  2 sin( ka)
 k
•On LHS, we know K=2n/(Na), where n=0,±1,±2,…
•On RHS, define zka=a(2mE)1/2/ħ and ma/ħ2
sin z
 2n 
cos
 f ( z)
  cos( z )  
z
 N 
Example: N=8
n
2n/N
cos(2n/N)
0
0
1
±1
±/4
21/2/2
±2
±/2
0
±3
±4
±3/4 ±
-21/2/2 -1
±5
±6
±5/4 ±3 /2
-21/2/2 0
14
N=8 Solution
sin z
 2n 
cos
 f ( z)
  cos( z )  
z
 N 
=5
15
N=50 Solution
sin z
 2n 
cos
 f ( z)
  cos( z )  
z
 N 
16