Lesson 9
I.
Particle In A Finite Well
We return to solving simple systems that can be used to model real
physical systems like the nucleus.
The first system considered is a particle of energy E in a finite well of
width L and height Uo as finite as shown below.
Uo
E
0
U 0
Ux   
0
A.
L
x
elsewhere
0 x L
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.
d2
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
2



III
where


dx 2
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.
The spatial derivative of the wave function must be continuous
at all points in space where the potential function is finite.
F.
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 L   III L  and
II
x

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

L   sin 
L 
  cos


 




 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
0.6
0.7
0.8
0.9
1
Left Hand Side
-1
-2
Eta
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


2 




1

sin
kL

cos
kL

B
sin kx   coskx  dx



2 2  k

k

0
L
2
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 and 2nd Allowed Energy States

x
0
L
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.