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Module 10 – Infinite and Finite Wells
I.
Particle In A Box With Infinite Walls
A.
Why Solve An Unphysical Problem?
There is no quantum system in the real world with infinite walls (implies an
infinite force). So why should we care about the solution to this problem?
All but the simplest of quantum systems are too complicated to solve by
hand. Thus, we shall start by solving idealized quantum problems that can
both be solved by hand and be used as models to approximate real systems in
the same way that we modeled real projectiles by neglecting air friction in
classical physics. In quantum physics this practice of modeling is even more
important as it builds our intuition for the quantum world which differs markedly
with our everyday experience. This modeling also allows us to gain familiarity
with the mathematical techniques needed to solve harder quantum problems.
B.
Solving the T.I.S.E.
We start by considering a particle of total energy E that is trapped in a region of
space from x = 0 to x = L. Inside the box, the potential is zero while outside the
box the potential is infinite.
U(x)
E
x
U=0
x=0
x=L
 0 0 x  L 
U

 otherwise 
The problem can be broke into three regions of constant potential:
Region I (- to 0), Region II (0 to L), and Region III (L to ).
Because the particle has no probability of being in either Region I or Regions III
(i.e. outside the well), we know that the wave function in each of these two
regions is zero!!
Φ I x   Φ III x   0
The problem is therefore reduced to solving the Time Independent Schrodinger
Equation in Region II where the particle’s kinetic energy (E-U) is positive. Using
the fact that U = 0 in this region, the Time Independent Schrodinger Equation
becomes:
d2
Φ II x    k 2 Φ II x 
2
dx
where
k2 
2mE
 constant  0.
2
Thus, we have a traveling wave with a real linear momentum and a real wave
number whose solution we considered in Lesson 9. Since we are dealing with a
bounded problem, we choose to use the cosine-sine solution form from Lesson 9
as it will make the math easier when applying boundary conditions.
Φ II (x)  A sin(kx)  B cos(kx) where k 
2mE

Next, we must consider the boundaries where we broke our problem into constant
potential regions. The wave function must be continuous through out all of space.
Thus, we have the following boundary conditions at x =0 and x=L:
Φ I (0)  Φ II (0)  0
and
Φ II(L)  Φ III (L)  0 .
Using the first condition, we get
Φ II (0)  A sin(0)  B cos(0)  B  0
Thus, our wave function in Region II has been reduced to only sine terms.
Φ II (x)  A sin(kx) where k 
2mE

We now apply our second boundary condition and obtain
Φ II (L)  A sin(kL)  0
Since the wave function must be non-zero in some region of space for the particle
to exist, the constant A and the constant k can not be zero!! Note that this
condition on k implies that the particle must have some kinetic energy (i.e. a
stationary particle is not allowed). This is consistent with the uncertainty
principle. If we confine the particle’s location then we can’t know its momentum
is exactly zero!! In order to make the equation true, k L must take on non-zero
values that make the sine function zero.
k L  n π where n is any positive integer
Thus, we see that only certain wave numbers are allowed. This is exactly the same
result that you would get by considering a bounded guitar string in classical
physics (only certain notes can be played). It also means from deBroglie’s
relationship that the particle has a limited number of linear momentum values (i.e.
it is quantized).
k
nπ
where n is any positive integer
L
p k 
nπ nh

where n is any positive integer
L
2L
By rearranging the relationship between wave number and particle energy and
inserting our new quantization condition for particle’s wave number, we see that
the particle’s energy is likewise quantized.
2 k 2 p2 n 2h 2
En 


where n  1, 2, 3, ....
2 m 2 m 8 m L2
We can multiply top and bottom of the equation by c2 to obtain
n 2 h c 
En 
n2
2
2
8 m c L
2
 1240 MeV fm 2 

 where n  1, 2, 3, ....
2


8
E
rest L


Again, the product hc sets the basic size of the energy levels. If you squeeze the
box (perhaps by placing atoms closer together), the energy levels get larger. This
simple model explains why nuclear levels where the box has a length of
femtometers is much greater than atomic energy levels where the width is on the
order of tenths of nanometers. We also see how the rest mass energy of the
particle affects the particle’s energy levels.
It is also important that you realize that each energy eigenvalue is associated with
a specific wave function called an energy eigenfunction (remember from previous
lesson that eigen is german for “proper” and means that only certain solutions are
possible). From our work, we have that the wavefunction corresponding to the nth
energy level is
nπx

 A n sin(
) 0 x  L
Φ n (x)  
L

otherwise
0
We have one final requirement for the wavefunction that we can use to solve the
remaining constant An. The wave function must be normalized in order to relate
the square of the wave function to a probability density.Thus, the nth wave
function must meet the requirement that


1 Φ*n Φ n dx

L

1 A n
2
0
An  L
2

0
 nπ x 
sin 
 dx  A n
 L 
2
L
2
 nπ x 
 dx
L 
 sin 
2
0
1
 nπ x 
sin 2 
 dx
 L 
There is nothing to be gained by making the constant A complex. Thus, the
absolute value sign can be dropped.
An 
L

0
1
 nπ x 
sin 2 
 dx
 L 
You should have previously evaluated this integral in physics 2424 for AC
circuits and in Calculus II using integration by parts. Its value is the length of the
region over which you are integrating divided by 2. You could also of course look
it up in the Schaum’s Math Handbook. Thus, we find that the constant is the same
for all the wave functions (this is a special property of the square well and not true
for other potentials).
An 
2
L
This gives us our final results.
 2
nπx

Φ n (x)   L sin( L ) 0  x  L and E n  n 2
0
otherwise
 1240 MeV fm 2 


2


8
E
rest L


n 1,2,3,..
Know these results and be able to (1) apply them to solve problems both
analytically and graphically.
C.
Superposition
The Time Independent Schrodinger Equation also possesses the property that any
superposition of individual solutions is also a solution of the equation. This again
is a property that all waves share and therefore all wave equations must posses.
This means that the most general solution is an infinite series of the individual
energy eigenstates: Φ 

a
n
Φ n . The square of the individual constants, an,
n 1
give the probability that a measurement will find the particle with an energy En.
This probability requirement also means that the constants must obey the
requirements that
1

a
2
n
n 1
Let us consider a specific example in order to make this clearer. A particle in an
infinite square well might have a wavefunction of
  0.6  2  0.8  3
Such a wavefunction is called an ensemble of states and can be interpreted as the
particle being in multiple energy states at the same time!!
If we wanted to graph the wavefunction, we would need to graph the function:
 2
 2π x 
 3 π x 

0.6 sin 
 0.8 sin 

 0  x  L

 L 
 L 
 L 

otherwise
0
This would be somewhat painful to graph by hand, but easy using a graphing
calculator, MathCad, Excel, etc.
What value would we get if we tried to measure the particle’s energy? The answer
is that we can’t know for certain what energy value we would get!! In fact the
general interpretation of quantum mechanics (Copenhagen Interpretation) is that
the particle has no energy (i.e. energy has no reality) till we measure it. The
coefficients tell us that we have a 36% chance of measuring E2 and a 64% chance
of measuring E3. No other energy would be measured as the particle has no
probability of being in any state besides states 2 and 3. Even more amazing is that
the very act of measuring the particle’s energy changes the particle’s wave
function. This is called “collapsing the wavefunction.” If we measure the
particle’s energy to be E2 and then make a second measurement of the particle’s
energy we are guaranteed to get E2 again. This means that the particle must be
100% in the 2nd state and the wave function for the particle must be
  2
This is a very profound fact. It means that you can never observe the quantum
world without disturbing the system that you are trying to measure no matter how
good you make your instruments!! To see what strange consequences this fact has
upon our philosophy of physics, read In Search of Schrodinger’s Cat by
Gibbons or Quantum Reality by Nick Herbert. They are excellent non-technical
books on Quantum Physics and its impact on our way of thinking about science.
II.
Particle In A Finite Well
We next modify our infinite well to a more realistic potential that can be used to
model real physical systems like the nucleus.
We consider a particle of energy E trapped in a finite well of width L and finite
height Uo as shown below.
Uo
E
0
L
x
 U 0 elsewhere
Ux  
0 0  x  L
A.
Solution Method
1. Since the potential is not a function of time, we already know the time dependent
part of the solution and can solve the Time Independent Schrodinger Equation to
find the spatial part of the solution.
d 2
2m E  U 
2
2


k

where
k

dx 2
2
2. We can simplify the problem by breaking the problem into regions where the
wave number is constant (i.e. regions of constant U).
In this case, we can break the problem into following three regions:
Region I:
  x 0
Region II:
0 x  L
Region III:
L x 
The advantage to this approach is that we know the solution to the problem in
each region by inspection. The disadvantage is that we have to use properties of
the wave function and its derivatives at the boundaries between the various
regions to merge our individual solutions into the total wave function solution.
B.
Solution for Region I
In this region, we have the differential equation
d 2 I
2mU o  E 
  2  I where  2 
2
dx
2
As we have discussed previously in this course, the general solution to this
differential equation is
 I x   A e  x  B e x
C.
Solution for Region III
In this region, we have the same differential equation as region I
d 2  III
2mU o  E 
  2  III where  2 
2
dx
2
Thus, the general solution for region III is of the same form as the solution for
region I.
 III x  E e  x  F e x
D.
Solution for Region II
In this region, we have the differential equation
d 2  II
2mE
  k 2 II where k 2  2
2
dx

As we have discussed previously in this course, the general solution to this
differential equation is
 II x   C sin kx   D cos kx 
E.
Applying Boundary Conditions To  and 

There are four boundaries in this problem. The boundaries are located at x    ,
x  0 , x  L , and x    .
To represent a quantum system, the wave function must have the following
required properties:
1.
2.
3.
The wave function must be continuous and finite.
The wave function must go to zero at x    .
The wave function must be normalizable.

   dx 1


4.
F.
The spatial derivative of the wave function must be continuous at all
points in space where the potential function is finite.
Applying Boundary Condition At x   
Using our wave function conditions and our solution in Region I, we have
 I   A e   B e    0
A e   B 0 0
A e  0
The only way that this condition can be obtained is if term involving A in the
general solution is not retained (i.e. A =0).
Thus, our wave function solution for region I is reduced to
 I x   B e x
G.
Applying Boundary Condition At x   
Using our wave function conditions and our solution in Region III, we have
 III   E e    F e   0
F e   E 0 0
F e  0
The only way that this condition can be obtained is if term involving F in the
general solution is not retained (i.e. F =0).
Thus, our wave function solution for region I is reduced to
 III x   E e  x
H.
Boundary Conditions at x = 0
At the boundary between region I and region II, we require that both the wave
function and the spatial derivative of the wave function is continuous.
 I 0    II 0  and
 I
x

x 0
 II
x
x 0
Using our wave solutions for regions I and II, continuity of the wave function
requires that
B e 0  C sin 0 D cos0
B D
Using our wave solutions for regions I and II, the continuity of the spatial
derivative of the wave function requires that
 B e 0  k C cos0 D sin 0
 B k C
C
B
k
Thus, our wave function in region II is


 II x   B  sin k x   cosk x 
k

H.
Boundary Conditions at x = L
At the boundary between region II and region III, we require that both the wave
function and the spatial derivative of the wave function is continuous.
 II
x
 II L    III L  and

xL
 III
x
xL
Using our wave solutions for regions II and III, continuity of the wave function
requires that


B sin kL   coskL   E e  L
k

Solving this equation for E, we obtain that


E  B e L  sin kL   coskL 
k

Using our wave solutions for regions II and III, the continuity of the spatial
derivative of the wave function requires that


B k  coskL  sin kL     E e - L
k

Using our equation for E, we have that




B k  coskL  sin kL     B  sin kL  coskL  e L e - L
k

k



k  coskL   sin kL    
k



 k sin kL   coskL 
Substituting our definitions of  and k, we obtain the following constraint on the
allowed energies of the system.
 Uo  E
 2mE 
 2mE 




cos
L

sin

 

  L 
E
E 




 1
U o  E  U o  E  2mE 
 2mE 
sin 
L   cos
L 

E







We can simplify this condition by writing the energy as E  U o where  is a
number greater than 0 and less than 1. Our energy constraint condition now
becomes
 1
 2mU o

 2mU o





cos
L


sin
L








  




 1
1   1    2mU o

 2mU o

L   cos
L 
  sin 







The only possible  and therefore energies are those where the plot of the
left-hand side are equal -1.
For an electron trapped in a well with a width of 0.1 m and a height of 4 eV, I
have plotted the constraint as a function of energy using Excel. We see that only
one energy state is allowed (E = 3.32 eV, eta = 0.83).
Finite Well
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
Value
-2
-3
Numerator
Denominator
Left Hand Side
-4
-5
-6
-7
-8
-9
Eta
If we increase the height of the well to 40 eV and the width to 0.5 nm, we get the
following graph
Finite Well
2
Value
1
Numerator
0
Denominator
0
0.1
0.2
0.3
0.4
0.5
-1
-2
Eta
0.6
0.7
0.8
0.9
1
Left Hand Side
We see that both the numerator and denominator are oscillatory leading to several
places where the left-hand side passes through -1. Thus, we have several allowed
energy states for this system. These energies are approximately 5.84eV, 12.6eV,
19.5eV, 26.4eV, 33.2eV, and 39.4eV.
The graph is somewhat misleading because of the denominator going to zero. The
vertical lines are an artifact of using Excel to graph points for discrete values of
eta. No line should be drawn between points on two sides of the asymptote where
the denominator went to zero.
I have placed this spreadsheet on the website for you to download and investigate
the effect of the height and width of the well upon the energy eigenstates.
How are the results of your investigation similar to what we found for the
particle in an infinitely deep well?
How are the results of your investigation different from what we found for
the particle in an infinitely deep well?
J.
Wave Function and Normalization
We can now write our wave function solution for all three regions in terms of a
single constant B.


B e x
  x0




 x   
B sin kx   coskx 
0 x  L
k



Be L  sin kL   coskL  e   x L  x  
 k


Once the width and height of the potential well is specified, this constant can be
found by normalizing the wave function for an given energy state!
0
1 B
2
e
2 x

2 2 L
dx  B e
α

 k sin kL   coskL 
2
e
 2 x
dx
L


 B 2  sin kx   coskx  dx
k

0
L
2

2
B2 B2  α



1

sin kL   coskL   B2  sin kx   coskx  dx

2 2  k

k

0
2
L

B2 
1
2
1
α



(1   sin kL   coskL  )   sin kx   coskx  dx
2
k

k

0
2
L

For a given well, the normalization constant, B, is different for each energy state
since the particle’s energy effects both alpha and k. Although this is somewhat
painful to do b y hand, it is relatively easy to do numerically with a computer.
K.
Graphing the Wave Function
Although computational work is difficult without a computer, it is still easy to
graph the energy eigenfunctions for a particle trapped in a finite well by hand if
we take into account that our results must converge with the infinite well as we
increase the depth of the well.
1. In region I & III, we have a wave function that is decreasing exponentially.
2. In region II, we have a wave function exhibiting oscillatory behavior.

x
0
L
1st Allowed Energy States

x
0
L
2nd Allowed Energy States
The existence of a particle in regions I & III is prohibited in classical physics
as the kinetic energy in this region would be negative according to
conservation of energy. However in quantum mechanics, the particle does
have a non-zero probability of being located in such regions according to our
graphs above. This leads to a strange and important phenomenon called
tunneling which is common to all waves.
In regions I & III according to quantum mechanics it is possible for a particle to
violate conservation of energy without experimental detection due to the
uncertainty principle.
E t 

2
One can picture a particle borrowing a small amount of energy for a very short
amount of time that is less than h-bar over 2. If a particle must borrow more
energy, then the particle can travel less distance into the forbidden region.
L.
Final Thoughts
Although our work may seem mathematical and tedious, the problem is one of the
simplest that we could have attempted. For instance, the well was of constant
height in regions I & II instead of being a function of position as in the case of the
Coulomb potential. Also, we made the well symmetrical in height and isolated
from any other well. Real systems are much more complicated. Thus, you should
see the difficulty in performing detailed quantum mechanics calculations on real
systems, the importance of numerical techniques in physics and engineering, and
the importance of knowing the results of simple models to guide you to quick
qualitative results.
The finite well is an essential starting block to understanding band theory for
electrical engineers and radioactivity for nuclear physicists.