Piecewise Constant Potentials
We now want to turn again to solutions to the time independent Schrödinger’s equation
for other potentials. The next level of complication after dealing with the free particle
and the infinite square-well potential, is piecewise constant potentials, an example of
which I show below:
Now, the scale of the potential does not affect physical observables. This can be
demonstrated by considering the time dependent Schrödinger’s equation:


 
 ( x, t )  H  ( x, t )
i t
If we add a constant to H:


H  H  V0 
( x, t )  ( x, t ) e iV0 t
But this is an overall phase that does not affect any observable. So, let’s take the zero of
the potential to be the potential at infinity, and our picture above becomes:
Now, in all regions of space, the potential is a constant Vc so that our Hamiltonian
operator is:


p2
H
 Vc
2m
and the TISE becomes:

H  E ( x)  E E ( x) 
  2 d 2


 Vc  E ( x)  E E ( x) 
2
 2m dx

2
d  E ( x)   2m E  Vc  

 E ( x)  0
dx 2
2


where ψE are our energy eigenstates. Now, we can simplify this by making the
substitution:
kc 
 2m E  Vc 
2
such that the TISE in a region of constant potential Vc becomes:
d 2  E ( x)
 kc2  E ( x)  0
2
dx
There are two special cases for the solutions to this differential equation:
1) E > Vc implies that kc is real and then:
 E ( x)  A eik x  B eik x
c
c
which are our familiar plane wave solutions.
2) E < Vc implies that kc is imaginary and then let:
2m
Vc  E  
2
 E ( x)  A ec x  B ec x
 c  i kc 
We will now use these two classes of solutions to solve specific examples of piecewise
constant potentials:
Finite Square-Well
As our first example, let’s look at the finite square well potential:
0,


V ( x)   V0 ,

0,
x  a
2
a  x  a
2
2
xa
2
(region I)
(region II)
(region III)
Generally, there are two classes of solutions for this problem:
1) E > 0 means that the particle is unbound. We will deal with this later.
2) E < 0 means that the particle is bound. Since we have a feel for this type of solution
already, let’s examine it first.
So, for E < 0, let E = -Eb, where Eb is positive and is what is called the binding energy
(the amount of energy that is required to bring E > 0 and thus have an unbound state).
Then, the TISE becomes:
d 2  E ( x)  2m  Eb  V ( x)  

 E ( x)  0
dx 2
2


Now, since the boundary conditions are different in the different regions, we must look
for solutions separately:
Region I
Here V(x) = 0 and we have:
d 2  I ( x)  2mEb  I
  2  ( x)  0
dx 2
  
2mEb
kI   

2
 I ( x)  A ex  B e x
Region II
Here V(x) = -V0 and we have:
d 2  II ( x)  2mV0  Eb   II

 ( x)  0
dx 2
2


2mV0  Eb 
k II  k 

2
 II ( x)  C eikx  D eikx
Region III
Here again, V(x) = 0 and we have:
d 2  III ( x)  2mEb  III
  2  ( x)  0
dx 2
  
2mEb
k III   

2
 III ( x)  F ex  G ex
Now, these are all three parts of a single wavefunction, and must obey boundary and
continuity conditions at +/- infinity and at the boundaries between the regions:
As x → -∞, Be-κx → ∞, so B = 0.
As x → ∞, Geκx → ∞, so G = 0.
So, we have:
 I ( x)  A ex
 III ( x)  F ex
 II ( x)  C eikx  D eikx
Now we must insure continuity of the wavefunction and its first derivative (no infinite
potentials here!) at the boundaries between the regions:
1)  I  a 2   II  a 2  C e
2)  II a 2   III a 2  C e
ika
 ika
2
2
 De
 De
 ika
ika
2
2
 Ae
 Fe
a
a
2
2
ika
ika
a
d  I  a 2 d  II  a 2

 ik  C e 2  D e 2    A e 2


dx
dx
II
III
ika
ika
a
d  a 2 d  a 2

 ik  C e 2  D e 2     F e 2
4)


dx
dx
3)
OK, now some algebra:
Multiply 2) by kappa and add to 4):
(ik   ) C e
ika
2
 (ik   ) D e
 ika
2
0
   ik  ika
C
 e
  
D
   ik 
In the same way, multiply 1) by kappa and add to 4):
   ik  ika
C
 e
  
D
   ik 
These two equations lead us to C = +/-D
This means that in region II, there are again two classes of solutions, one for each
relationship between C and D. The two classes are labeled as “Even Parity” (C = +D)
and “Odd Parity” (C = -D) solutions for reasons that will soon become clear. For even
parity solutions in region II,  II  ( x)  2C cos(kx) , and for odd parity solutions in
region II,  II  ( x)  2iC sin( kx) .
Let’s first look more at the even parity solutions:
We must ensure continuity of the wavefunction at the boundaries:
5)  I   a 2   II   a 2  2C coska 2  A e
6)  II  a 2   III  a 2  2C coska 2  F e
AF
a
a
2
2





and continuity of the first derivative:
a
d  I   a 2 d  II   a 2

 2Ck sin ka 2  A e 2
dx
dx
II 
III 
d  a 2 d  a 2

 same
8)
dx
dx
7)
Then dividing 7) by 5) we have:
k tan ka 2   
ka
a
tan ka 2  
2
2
For the odd parity solutions we get similar results:
ka
a
cot ka 2 
2
2
Our final constraint comes from the relation between k and kappa:
k
2m V0  Eb 
,
2
2m Eb

2
2m
k 2  2 V0   2 


 ka   a   k0 a 
    
 ,
 2  2   2 
2
2
2
k0 
2mV0
2
These transidental equations cannot be solved analytically, so we solve them numerically.
This last equation is just the equation for a circle of radius k0a/2. We can plot all these on
the same plot of ka/2 vs. κa/2 and look for the points of intersections. We can see then
these conditions give us discrete values of allowed k and kappa, and therefore give
discrete allowed energies.