Lecture 18:
Delta-function Scattering
Phy851 Fall 2009
Delta-Function Scatterer
• Any very narrow barrier can be approximated
by a delta function:
V ( x) = gδ ( x)
• The coefficient g is then the area under V(x):
g= hxw
= Energy x Length
g ≈ V0 w
• Conditions for validity of delta-function
approximation:
– Incoming wave characterized by k, which gives
a length-scale: λ =2!/k
Thus we surely must require: w << λ

k w << 1
– But the delta-function must have another
length scale associated with it (from V0)
• Based on units only, we find a second
length scale, let’s call it `a’:
h2
h2
g=
=
a
2
ma ma
h2
a=
mg
`scattering length’
Do we also
Need:
k a << 1 ?
In the limit w → 0, scattering is governed by
the scattering length
Delta-Function Scatterer
• Scattering by the delta-function will be
handled by applying boundary conditions to
connect the wavefunctions on the left and
right sides
II
I
ψ 1 ( x) = Aeikx + Be − ikx
ψ 2 ( x) = Ce ikx + De −ikx
• RECALL: a delta-function in the potential
means that ψ_(x) is discontinuous
– But ψ(x) remains continuous
• PRIMARY GOAL: Determine the proper
boundary conditions for _ and _´ at the
location of a delta function scatterer
– Be able to solve `plug and chug’ problems
• Secondary Goal: find Mδ for the delta
potential:
C 
  = M δ
 D
 A
 
 B
Delta-function Boundary Condition
• All boundary conditions are derived from
Schrödinger's Equation:
h2
Eψ ( x) = −
ψ ′′( x) + gδ ( x)ψ ( x)
2m
• For the delta-potential, the trick is to
integrate both sides from −ε to +ε
– Then take limit as ε → 0
h2
E ∫ dxψ ( x) = −
dxψ ′′( x) + g ∫ dx δ ( x)ψ ( x)
∫
2 m −ε
−ε
−ε
ε
ε
ε
h2
(ψ ′(ε ) −ψ ′(−ε ) )+ gψ (0)
Eψ (0)2ε = −
2m
– Take ε 0:
ψ ′(ε ) → ψ 2′ (0)
ψ ′(−ε ) → ψ 1′(0)
h2
(ψ 2′ (0) −ψ 1′(0) )+ gψ (0)
0=−
2m
ψ 2′ (0) = ψ 1′(0) +
R
L
2mg
ψ (0)
2
h
R or L
Example
• Lets solve the delta-potential scattering
problem via `plug and chug’ method:
– Q: Let V(x) = g _(x). For a single incident
wave with momentum k, what are the
reflection and transmission amplitudes
and Probabilities?
ψ 1 ( x) = eikx + re − ikx
ψ 2 ( x) = te ikx
x=0
Drat, I can’t remember the
boundary condition…
x
Solution:
2
1
ka)
(
R=
T=
2
2
1+ ( ka)
1+ ( ka)
Transfer Matrix for Delta function
I
a
c
b
d
II
x
ψ I ( x) = Aeikx + Be − ikx
ψ II ( x) = Ce ikx + De −ikx
C 
  = M δ
 D
ψ 2 (0) = ψ 1 (0)
2
ψ 2′ (0) = ψ 1′(0) + ψ (0)
a
 A
 
 B
h2
a=
mg
C + D = A+ B
2
ik(C − D) = ik(A − B) + (A + B)
a
2
C − D = A − B − i ( A + B)
ka
€
1 1  C   1
2

  =
1
−
i
1 − 1 D  
ka
1
 A 
2  
− 1 − i  B 
ka 
Continued
1 1  C   1
2

  =
1 − 1 D  1 − i ka
1
 A 
2  
− 1 − i  B 
ka 
1 1   1
2

M δ (ka ) = 
1 − 1 1 − i ka
−1
1

2
−1− i 
ka 
1 
1 1 1  1 2
2
Mδ ( ka) = 
1− i
−1− i 
2 1 −1
ka
ka 

i
i 
1− ka − ka 
Mδ ( ka) = 
i
i 

1+ 
 ka
ka 
€
Vδ (x) = gδ (x)
h2
a=
mg
€
€
Summary of Transfer Matrix Results:
• Basic Elements:
 e ikL
M free (kL) = 
 0
0 

−ikL 
e 
1  k 2 + k1

M step (k 2 , k1 ) =
2k 2  k 2 − k1
k 2 − k1 

k 2 + k1 
1 
1 ika + 1
Mδ ( ka) =


ika −1
ika  −1
•
€ For n regions (n-1 boundaries):
M = M [ n ,n −1] M [fn −1]M [n −1,n − 2 ]K M [3, 2 ]M [f2 ]M [2,1]
R=
M12
M 22
€
€
2
T = 1− R