CHAPTER V
One-Dimensional Potential Wells and Barriers
Part I. The Delta Function Potential
Introduction
So far, we have encountered two very different types of solution to the time
independent Schrödinger equation. In the case of the infinite square well the Schrödinger
equation
s <8 ÐBÑ œ I8 <8 ÐBÑ
L
(5.1)
gives us a set of normalizable, orthogonal wave functions (or eigenfunctions) <8 ÐBÑ, each
associated with a specific, discrete energy value (or eigenvalue) I8 . The most general
time-dependent solution to the Schrödinger equation in this case is a linear combination
of these eigenfunctions of the form
GBß > œ -8 /3=8 > <8 ÐBÑ
(5.2)
8
where =8 œ I8 Îh , and where the expansion coefficients, -8 , are determined from the
wave function at time > œ ! from the equation
-8 œ (
∞
∞
GBß !<8* ÐBÑ .B
(5.3)
These coefficients must also satisfy the normalization condition
l-8 l# œ "
(5.4)
8
The expectation value of energy of this system is given by
s œ l-8 l# I8
ØLÙ
(5.5)
8
In the case of the free particle, however, we encountered a solution to Schrödinger's
time-independent equation which was associated with a continuous range of possible
energy values (for I !). The most general normalizable solution to the Schrödinger
equation in this case is the integral
GÐBß >Ñ œ
∞
"
9Ð5Ñ/35B=> .5
È#1 (∞
(5.6)
where
9Ð5Ñ œ
∞
"
GÐBß !Ñ/35B .B
È#1 (∞
(5.7)
One-Dimensional Potential Wells and Barriers
2
Note: Although the “eigenfunctions” of the momentum operator are not
properly normalizable wavefunctions, we can express the expectation value
of the energy for a free particle using the equation
s œ(
ØLÙ
∞
∞
l 9 5 l#
h5 #
.5
#7
(5.8)
This is analogous to the way we have computed other expectation values,
except that, here, we are using the momentum dependent functions for the
free particle.
You should perform this operation starting with the free-particle
#
s œs
wave functions GÐBß >Ñ, and the Hamiltonian operator L
: Î#7
expressed in configuration space (where s
: œ 3h `Î`BÑ and show that
you obtain the result above. Notice that this expression is just the
expectation value of the energy (expressed in terms of the momentum
: œ h5) in momentum space.
The Delta-Function Potential Well Problem
We now want to consider a particularly simple (though somewhat unrealistic)
problem which exhibits both of these types of solutions - the delta-function potential
problem. In this problem, we assume that the potential energy is of the form
Z ÐBÑ œ α $ ÐBÑ
(5.9)
which acts something like an infinitely narrow, infinitely deep potential well in an
otherwise constant potential background. We will approach the solution to this problem
much like we did in the case of the infinite square well.
Schrödinger's time-independent equation for this problem is
h# ` 2
<ÐBÑ Z B<ÐBÑ œ I <ÐBÑ
#7 `B#
(5.10)
where Z B œ α $ B. This potential function is somewhat unique and looks
something like what is shown in the diagram below, with the effective depth of the well
going to negative infinity, but with the width of the well being so narrow that the area of
the well is unity! This means that for a $ -function potential, region II in the diagram
below has zero width. We simply draw this potential with a finite width to be able to
think clearly about what we are doing. Because region II has zero width, we only need to
be concerned about the wave function in regions I and III.
One-Dimensional Potential Wells and Barriers
3
V(x)
ΙΙ
Ι
ΙΙΙ
x
−ε
+ε
Figure 4.1 The delta-function Potential Well
We must now find the solution to Schrödinger's equation in regions I and III. In each of
these regions we must consider two different cases, one where I ! and one where
I !. We will first examine the case where I !.
The Delta-Function Potential Well Solution for I !
In region I and III, the Schrödinger equation is
h# ` 2
<ÐBÑ œ I <ÐBÑ
#7 `B#
(5.11)
`2
#7I
<ÐBÑ œ # <ÐBÑ œ 5# <ÐBÑ
#
`B
h
(5.12)
or
which has a solution of the form
<ÐBÑ œ E/-B
(5.13)
Plugging this back into the Schrödinger equation reveals that
- œ „ 35
(5.14)
In the case where I !, 5 is imaginary, so we write
5œÊ
#7I
#7lIl
Ê
œ
œ 3,
h#
h#
(5.15)
and the solution becomes
<M ÐBÑ œ E/,B F/,B
for B !
(5.16)
<MMM ÐBÑ œ J /,B K/,B
for B !
(5.17)
One-Dimensional Potential Wells and Barriers
4
Now, in region I, where B !, the solution blows up as B p ∞, unless F œ !.
Likewise, in region III, where B !, the solution blows up as B p ∞, unless J œ !.
This means that the solutions in region I and III are
<M ÐBÑ œ E/,B
for B !
(5.18)
<MMM ÐBÑ œ K/,B
for B !
(5.19)
As we have discussed earlier, the wave function must be continuous to have
meaning as a probability amplitude. This means that for a $ -function potential where
region II in the diagram above has no real width, the wave function in region I must have
the same value as the wave function in region III at the point where B œ !. This requires
that E œ K, so the solution to Schrödinger's equation is
<M ÐBÑ œ E/,B
for B !
(5.20)
<MMM ÐBÑ œ E/,B
for B !
(5.21)
or
<MßMMM ÐBÑ œ E/,lBl
(5.22)
The second condition on the wave function solution to Schrödinger's equation is
that the first derivative of the wave function must be continuous for piecewise-continuous
potentials. However, in the case of the delta-function potential, the potential is not
piecewise continuous - it is infinite. Let's look again at the requirement imposed on the
first derivative of the wave function by the Schrödinger equation. As you will remember,
we integrated the Schrödinger equation with respect to B over a small interval ?%
Bo %
Bo %
h # Bo % ` 2 <ÐBÑ
.B (
Z B<ÐBÑ .B œ I (
<ÐBÑ .B
(
#7 Bo % `B#
Bo %
Bo %
(5.23)
In the limit as ?% p ! the integral over the wave function must go to zero, since it is a
continuous, single-valued function. The integral of the second derivative of the wave
function is just the first derivative, so that we are left with
` <ÐBÑ Bo %
#7 Bo %
lim
œ lim # (
Z B<ÐBÑ .B
º
?%Ä! `B
?%Ä! h
Bo %
Bo %
(5.24)
As pointed out in the last chapter, the integral of the product of the potential energy
function and the wave function can be represented by the diagram shown for piecewisecontinuous potential energy functions i.e., those having a finite number of finite steps.
One-Dimensional Potential Wells and Barriers
V(x)
5
ψ(X)
x
∆ε
However, for potential energy functions which have infinite steps, such as the
infinite square well and the delta-function potential, the first derivative of the wave
function may not be continuous at a boundary. In fact, using the definition of the deltafunction, we can evaluate the integral of the product of the wave function and the
potential energy function to obtain
` <ÐBÑ
#7 %
#7α
lim
œ
α$
ÐBÑ
<
ÐBÑ
.B
œ
<Ð!Ñ
(
º
?%Ä! `B
?%Ä! h # %
h#
%
%
lim
(5.25)
which we write as
lim Œ
?%Ä!
` <ÐBÑ
` <ÐBÑ
#7α
#7α
º º  œ # <Ð!Ñ œ # E
`B %
`B %
h
h
(5.26)
Now the first partial derivative is evaluated in the region B !, while the second partial
is evaluated in the region B !, giving
lim Œ,E/,B º
?%Ä!
%
,E/,B º  œ %
lim #,E/,% œ ?%Ä!
#7α
E
h#
#7α
E
h#
(5.27)
(5.28)
which in the limit as % p 0 fixes the value of , (and therefore the energy I ) according to
the equation
,œ
7α
#7lIl
œÊ
#
h
h#
(5.29)
One-Dimensional Potential Wells and Barriers
6
where α is the “depth” of the delta-function potential. We see from this equation that
there is only one allowed energy for I ! given by
7 # α# h #
7 α#
Iœ % †
œ #
h
#7
#h
(5.30)
The only thing remaining for the case where I ! is to normalize the wave function and
determine E
(
!
∞
E# /#,B .B (
∞
E# /#,B .B œ "
(5.31)
!
E# /#,B !
E# /#,B ∞
œ"
º
º
#, ∞
#, !
(5.32)
E# #,B
/
/#,B º — œ "
º
–
#,
∞
!
(5.33)
E#
E#
c " ! ! " d œ
œ"
,
#,
(5.34)
!
∞
Ê E œ È,
(5.35)
Thus, for the delta-function potential, we have only one eigenstate corresponding to an
energy I ! given by
<I ÐBÑ œ È,/,lBl
Iœ
7 α#
#h #
(5.36)
________________________________________________________________________
Problem 5.1 The Double Delta Function Potential
Consider the double delta function potential energy function given by
Z ÐBÑ œ αc$ ÐB +Ñ $ ÐB +Ñd
(a)
(b)
Sketch this potential energy function.
Show that there are at most two possible bound state energies (i.e., energies where
I !) allowed. Skematically plot the wave function corresponding to each of
these two possible solutions. Do both solutions always exist?
(c) In particular, find the allowed energies for the two cases α œ h # Î7+ and
α œ h # Î%7+, and sketch the corresponding wave functions.
________________________________________________________________________
One-Dimensional Potential Wells and Barriers
7
The Delta-Function Potential Well Solution for I !
In region I and III, the Schrödinger equation is, again
h# ` 2
<ÐBÑ œ I <ÐBÑ
#7 `B#
(5.37)
`2
#7I
<ÐBÑ œ # <ÐBÑ œ 5# <ÐBÑ
#
`B
h
(5.38)
or
which has a solution of the form
<ÐBÑ œ E/-B
(5.39)
Plugging this back into the Schrödinger equation reveals that
- œ „ 35
(5.40)
In the case where I !, 5 is real, given by
5œÊ
#7I
h#
(5.41)
and the solution becomes
<M ÐBÑ œ E/35B F/35B
for B !
(5.42)
<MMM ÐBÑ œ J /35B K/35B
for B !
(5.43)
The time-dependent form of these two equations have the form of two traveling
sinusoidal waves moving in opposite directions. Let's assume that we are dealing with a
particle (or group of particles) that are originating in the negative half-plane (i.e., in the
region where B !). The particles will move in from the left, encounter the deltafunction potential at B œ !, and will either continue moving in the B direction (i.e., they
are transmitted through the region of potential change), or will be reflected and move
back in a B direction. This means that in the region where B ! we must allow for the
possibility of two opposite going waves (or particles), but in the region where B ! there
is only one possibility - the particle (or wave) moves only in the B direction. This
means that we must require that K œ !, based upon the initial conditions (the assumption
that particles are originating in the negative half-plane). This leaves us with the equations
<M ÐBÑ œ E/35B F/35B
<MMM ÐBÑ œ J /35B
for B !
for B !
(5.44)
(5.45)
One-Dimensional Potential Wells and Barriers
8
Since the wave function must be continuous, these two equations must be equal at
B œ !, giving the condition
EF œJ
(5.46)
The second condition on the wave function solution to Schrödinger's equation is
that the first derivative of the wave function must also be continuous for piecewisecontinuous potentials. But again, the potential in not piecewise continuous - it is infinite.
As before, we integrate the Schrödinger equation with respect to B over a small interval
?%
Bo %
Bo %
h # Bo % ` 2 <ÐBÑ
<
<ÐBÑ .B
.B
Z
B
ÐBÑ
.B
œ
I
(
(
(
#7 Bo % `B#
Bo %
Bo %
(5.47)
Again, in the limit as ?% p ! the integral over the wave function must go to zero, the
integral of the second derivative of the wave function is just the first derivative, and so
we are left with
` <ÐBÑ Bo %
#7 Bo %
lim
œ lim # (
Z B<ÐBÑ .B
º
?%Ä! `B
?%Ä! h
Bo %
Bo %
(5.48)
Using the definition of the delta-function, we can evaluate the integral of the product of
the wave function and the potential energy function to obtain
` <ÐBÑ %
#7 %
#7α
lim
œ lim # ( α$ ÐBÑ<ÐBÑ .B œ # <Ð!Ñ
º
?%Ä! `B
?%Ä! h
h
%
%
(5.49)
which we write as
lim Œ
?%Ä!
` <ÐBÑ
` <ÐBÑ
#7α
º º  œ # E F `B %
`B %
h
(5.50)
Now the first partial derivative is evaluated in the region B !, while the second partial
is evaluated in the region B !, giving
lim Œ35J /35B º
?%Ä!
%
ˆ35E/35B 35F/35B ‰º  œ %
lim Œ35J /35 % ˆ35E/35 % 35F/35 % ‰º  œ ?%Ä!
%
lim 35 ˆJ F /35 % E/35 % ‰ œ ?%Ä!
#7α
E F h#
#7α
E F h#
#7α
E F h#
(5.51)
Now, in the limit this reduces to
J FEœ
3#7α
E F œ #3" E F h#5
(5.52)
One-Dimensional Potential Wells and Barriers
9
where " œ 7αÎh # 5 . But from the condition on the continuity of the wave function we
know that J œ E F , so this last equation reduces to
F œ 3"E F (5.53)
or
Fœ
3"
E
" 3"
(5.54)
"
E
" 3"
(5.55)
Solving for J gives, since J œ E F ,
J œ
It should be obvious that the amplitude E is related to the probability of measuring
an incoming particle, while the amplitude F is related to the probability of measuring a
reflected particle and the amplitude J to measuring a transmitted particle. To determine
a precise relationship between these quantities we return to the concept of the probability
current density defined by the equation
4ÐBß >Ñ œ 3h
` GÐBß >Ñ
` G* ÐBß >Ñ
*
G
ÐBß
>Ñ
G
ÐBß
>Ñ
”
•
#7
`B
`B
(5.56)
As you will recall, this equation has the form
3h
t GÐ<ß
t G* Ð<ß
t4Ð<ß
t >Ñ œ t >Ñf
t >Ñ GÐ<ß
t >Ñf
t >Ñ“
’G* Ð<ß
#7
(5.57)
in three-dimensions, which satisfies the continuity equation
t
` 3Ð<Ñ
t † t4 œ !
f
`>
(5.58)
This equation states that the change in the probability density at some point is associated
with the probability current flowing away from that point. If we integrate the continuity
equation over some enclosed volume Z , we obtain
(
Z
`3
t † t4 .@ œ * t4 † 8
^ .=
.@ œ ( f
`>
Z
W
(5.59)
where the last integral is due to Stokes' theorem. Thus, if the probability of finding a
particle within an enclosed volume decreases in time, this corresponds to a probability
flux, the sum of all probability currents flowing out of an enclosed volume. Thus in
three-dimensions t4 corresponds to the relative number of particles per unit volume
flowing outward through the enclosing surface W , times the speed at which the particles
leave. This has units of the number of particles passing through a given cross-sectional
area per unit time. Thus, if we consider a stream of particles leaving an enclosed volume
and incident upon some surface area .W , we would designate 438-3./8> as the relative
number of particles per unit volume that were approaching the surface, 4</06/->/. as the
One-Dimensional Potential Wells and Barriers
10
relative number of particles per unit volume that were reflected from that surface and
4><+8=73>>/. as the relative number of particles per unit volume that were passing through
the surface. We define the transmission and reflection coefficients in terms of these
current densities according to the equations
V´º
4</06/->/.
º
438-3./8>
and
X ´º
4><+8=73>>/.
º
438-3./8>
(5.60)
The solutions we have obtained for our delta function potential are traveling plane
waves of the form GÐBß >Ñ œ E/35B=> . The probability current density for a plane wave
is given by
t4 œ 3h ”G* ÐBß >Ñ ` G ÐBß >Ñ G ÐBß >Ñ ` G* ÐBß >Ñ•
#7
`B
`B
(5.61)
t4 œ 3h ”E* /35B=> ` E/35B=> E/35B=> ` E* /35B=> •
#7
`B
`B
(5.62)
t4 œ 3h E* /35B=> 35E/35B=> E/35B=> 35 E* /35B=> ‘
#7
(5.63)
t4 œ 3h 35lEl# 35 lEl# ‘
#7
(5.64)
t4 œ 3h #35lEl# ‘
#7
(5.65)
t4 œ h5 lEl#
7
(5.66)
t4 œ : |E|# œ @|E|#
7
(5.67)
or
This has the form
and will be positive or negative depending upon the sign of 5 .
This means that the incident, reflected and transmitted probability current densities
for plane waves are givenß respectively, by
t4 38- œ h5 lEl#
7
(5.68)
t4</06 œ h5 |F |#
7
(5.69)
One-Dimensional Potential Wells and Barriers
11
t4><+8= œ h5 |J |#
7
(5.70)
Thus, the reflection coefficient for our delta function potential is given by
Vœ
"#
lFl#
"
œ
œ
#
#
lEl
""
" "Î" #
(5.71)
and the transmission coefficient is given by
X œ
lJ l#
"
œ
#
lEl
" "#
(5.72)
Notice that in this particular problem, the value of 5 œ È#7I Z Îh #
is the same in regions I and III of figure 4.1, but in other problems, the
value of 5 may be different. In those cases, the transmission and reflection
coefficients depend upon the value of 5 in the respective regions.
You should notice that the sum of the reflection and transmission probabilities equal
unity as it must. Now using the definition of " and 5 , we find that
"# œ
7 # α#
7 # α#
7 α#
œ
œ
h % 5#
h % #7IÎh # #h # I
(5.73)
So we can express the reflection and transmission coefficients for the delta-function
potential well in terms of the energy and mass of the incident particle:
Vœ
X œ
"
#h # IÎ7α# (5.74)
"
" 7α# Î#h # I (5.75)
"
Thus, for situations where the energy of the incident particle is small, there is a relatively
high probability for reflection at B œ !, but as the energy of the incident particle
increases, the probability for reflection goes down.
However, there is one small problem we have not yet addressed. Plane wave
solutions for the free particle are not normalizable, so are not really acceptable solutions!
At first glance, this might seem a bit problematic, but we have shown that an acceptable
solution can be formed from a combination of plane wave solutions using a Fourier
integral. This means that each component of the Fourier integral will be transmitted or
reflected with different probability, because each component has a slightly different
energy. We can, however, obtain an estimate of the approximate transmission or
reflection coefficient based upon the more prevalent component of the wave packet.
One-Dimensional Potential Wells and Barriers
12
The Delta-Function Potential Barrier Problem
For the case of a delta-function potential barrier, only the sign of the potential
energy function is changed, so that the potential energy function has the form
Z ÐBÑ œ α $ ÐBÑ
(5.76)
which acts something like an infinitely narrow, infinitely tall potential barrier in an
otherwise constant potential background. It should be obvious that there in no bound
state solution for this case (i.e., where I !). We only have to consider the case where
I !. For this case all we have to do in order to find the solution for the infinite barrier
is to change the sign of α. But the reflection and transmission coefficients are a function
only of the square of α, so that we obtain the same result as for the potential well! Thus,
whether we are dealing with a delta-function well or barrier, the transmission and
scattering coefficients are identical. This means that a quantum mechanical particle can
penetrate a potential barrier of infinite height - provided it is not too wide! This
phenomena is called tunneling. We will see how this same situation plays out for finite
width barriers and wells in the next chapter.