Cohen-Tannoudji Problem 1.6
Zachary Burell, Vibha Gill, Anthony Morgan
January 20, 2009
Problem.
Consider a square well potential of width a and depth
V0
. We intend
to study the properties of the bound state of a particle in this well when its width
a approaches zero.
•
a) Show that there indeed exists only one bound state and calculate its
energy
E.
•
b) Show that
ρ−→ 0
and that
A0
2=A2wB1
2
. Deduce from this that,
in the bound state, the probability of nding the particle outside the well
approaches 1.
•
c) How can the preceding considerations be applied to a particle placed,
as in excercise 2, in the potential
V(x) = −αδ (x)
?
The nite square well, a quick review:
V(x) = (−V0,−a/2< x < a/2
0,|x|> a/2
(1)
Where
V0
is a positive constant. We consider the bound states (
−V0< E < 0)
( with eigenenergies given by,
Ek=~2κ2
2m
) .
In region
I
the potential is zero so the S.E. is :
−~2
2mψxx =Eψ
or
ψxx =κ2ψ
(2)
where
κ=√2m|E|
~
is real and positive. The general solution is ( in region I
x < −a/2
)
ψI(x) = Ae0−κx +Aeκx
but the rst term diverges as
x→ −∞
so
A0= 0
Therefore the solution in
region I is :
ψI(x) = Aeκx
(3)
1
