Download Quiz #5: Physics 203

Modern Physics
Quiz #3
NAME: ________________________
1. The particle in a finite one-dimensional box.
Somewhat more realistic than the particle in an infinitely-walled box problem is the
particle in a finite one-dimensional box. If the potential energy well has length L and
the sides of the box have height U, then
2
 ( x)  U ( x) ( x)  E ( x)
2m x 2
0 if  L2  x  L2
where U ( x)  
L
U if x  2  where U  0

2
Before we begin it is useful to identify reasonable units in which to work. Joules and
meters don’t make sense here. However, the problem contains a natural unit of length
(L) and a natural unit of energy.
a. Show that by making the substitution x  L z we can rewrite Schrodinger’s
equation as
2
 2
 ( z )  U ( z ) ( z)  E ( z)
2 L m z 2
0 if  12  z  12
where U ( z )  
1
 U if z  2
2
b. That doesn’t look much better, but if we remember the particle in the infinite
walled box of length L, the ground state of that wave function had energy
Eunit 
2
2
2 L2 m
. So if E   Eunit , show that Schrodinger's equation becomes
 2

2
2
  2   u ( z )  ( z )     ( z )
 z

where u ( z ) 
U ( z)
Eunit
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Quiz 3
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This is much better. Simply by measuring position in units of L and energy in
units of Eunit we have simplified our algebra considerably. This is what physicists
refer to as using “natural units.” We now solve:
      u  z 
where U ( x)  u( z) Eunit .
2
The presence of  2 might look awkward, but it allows us to compare functions
and energies to something that we know (a particle in an infinitely deep box).
Inside of the finite one-dimensional box:
if z  12 , then u( z)  0 so     2 

The potential energy is never negative, so there can’t be a solution with a negative
value of  ( would curve away from the axis forever so such a solution would
not have finite probability). Since  is positive, the solution of this differential
equation involves sines and cosines.
c. Since the box is symmetric about the origin,  must be symmetric about the
origin as well. Explain why this means that possible wave functions include
2

odd

( z )  A sin   z
but not

neither

or


even


( z )  A cos   z



( z )  A sin   z  cos   z 


Outside the box:
if z  12 , then u( z)  u so     2 u    
d. Show that the solution to this equation is
exp  Cz  if z  12
 ( z)   BBexp
,
 Cz  if z   12
(a graph might help explain the signs!)

where C  
u 
Modern Physics
Quiz 3
Page 3
e. Now think of the edges the box (where the potential energy is discontinuous).
What is continuous and what is discontinuous at an edge of the box?
i. The wave function itself? (continuous or discontinuous?)
ii. The first derivative of the wave function? (continuous or
discontinuous?)
iii. The second derivative of the wave function? (continuous or
discontinuous?)
f. If we look at the edge where z   12
 or

x   L2 , show that if we apply your
answers from part e to the “odd” wave functions  odd we find:

 cos 
 cos 
A sin
A 

 u    , and
    B  u   exp 
u   ,
    u   sin   

  B exp
2



2

2


2
so

2
2
g. Using the same logic, show that if we look at  even we find:
 sin


2

   u   cos


2


Modern Physics
Quiz 3
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Now the fun part (or maybe just fun for your computer). We have absorbed the
width of the potential well into our units. The height of the potential well is given
(in units of Eunit ) by “u”. We now need to find the energy eigenvalues for this
system. That is the same as finding the values of  that satisfy:
  cos


either 

  sin



2

2

   u   sin

 or 
   u   cos




2
2




Which is the same as finding values of  that satisfy:
  cos



2

  u   sin


2
 
    sin


2

  u   cos


2

 0
This cannot be solved analytically. You will have to solve it numerically.
Different values of u will give you different solutions (and different numbers of
solutions). You can use a graphing calculator, Maple, or a program such as Excel
(look for functions with names like “Solver” or “Goal Seek”). Just a hint: if you
copy things down by hand, you will have to keep a fair number of significant
figures (five or six?) in your calculations.
h. First set u = 6, so the height of the potential wall is six times higher than the
ground state energy would be if the walls were infinitely high.
i. Figure out how many eigenvalues of  (or roots of the last equation)
there are. A graph might help. How many are there?
ii. Find all of them and record them in Table I on the following page.
i. Now set u = 20.
i. How many roots are there now?
ii. Find all of them and record them in Table I on the following page.
j. Now set u = 100. You don’t need to find all of the eigenvalues. Just find the
first six.
k. Now compare your results with the values we would get if u =  . The first
one has been written into the table. Fill in the next five.
l. Now fill in Table II on the following page. For this table, instead of recording
the energy levels, you just record the ratio of each energy level to the ground
state energy level for that value of u. Thus for Table II, the entire first row is
filled with the number 1.0000 (the ground state  value divided by itself). In
the column for u = 6, fill in the cell beneath the 1.0000 with the value of  for
the first excited state divided by the value of  for the ground state. Fill in the
rest of the table.
Modern Physics
Quiz 3
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Table I:
Values of  : Energy eigenvalues (in units of Eunit) for the particle in a finite walled
box where the wall height is given by u (in units of Eunit) . Note: there are more cells
than eigenvalues here, so some cells will be left blank.
Value of u :
6
20
100
Ground energy:  0

1.0000
Next energy: 1
Next energy:  2
Next energy:  3
Next energy:  4
Next energy:  5
Table II:
Ratios of  to  0 : Energy eigenvalues of excited states divided by the ground state
energy eigenvalue for different values of u. Again, there are more cells than
eigenvalues here, so some cells will be left blank.
Value of u :
0 / 0
1 /  0
2 / 0
3 / 0
4 / 0
5 / 0
6
20
100

1.0000
1.0000
1.0000
1.0000
Modern Physics
Quiz 3
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2. The quantum mechanical harmonic oscillator.
The solution of the harmonic oscillator problem in quantum mechanics has
applications in all branches of atomic, solid state, and particle physics. To construct
the Schrodinger equation for the harmonic oscillator, the potential energy function is
U ( x)  12 kx 2 or
1
2
m 2 x 2
where  
k
m
, so
2


2 1

 m 2 x 2   E

2
 2m x 2

a. Finding the natural units: Step One. The only constants in the problem that we
can use to find dimensions of length are , m, and .
i. Show that the combination with dimensions of length is
1
Losc

2


 m 
ii. Show that if we make the substitution x  L z then Schrodinger’s
equation becomes:
 1

2 1


  z 2   E

2
z
2
 2

b. Now the easy part. Show that if we measure the energy in units of  then

1  2
  2  z 2     , where E   .
2  z

Modern Physics
Quiz 3
Page 7
The differential equation on the previous page can be tough. It doesn’t
immediately yield to any of the standard methods for solutions of differential
equations. But it can be made a lot simpler if one notices that part of it at least
looks quadratic:
 2
2
 2 z 
 z




  z   z .
 z
  z

"looks sort of like"
c. Check the similarity (see what “sort of like” means). Show that

1 
1  2




   z    z     2  z 2  z  z  1
2  z
2  z
z
z 
  z

Which means that if for some solution  ,
n

1  2
  2  z 2    n  , then
n
2  z
 n
1 


   z    z    n  12 
n
2  z
  z
 n


d. By the same logic show that

1 
1  2




   z    z     2  z 2  z  z  1
2  z
2  z
z
z 
  z

1 

so    z 
2  z



1
  z  n   n  2
 z


 n
Modern Physics
Quiz 3
Page 8
The last equation is crucial. Since the potential energy function is never negative,
the eigenvalues of E (and thus  ) must be positive. That means that we don’t
expect to find a solution where  = -1/2 , but we just might find a solution where
 = +1/2.
e. Show that if we can find a solution to the first order equation


  z  0  0, then it will also be a solution to
 z

1 


   z    z  0   0  12
2  z
  z



0
, where  0 
1
2
Now to solve the first order equation. Let  0 ( z)  A0 exp 0 ( z) where A0 is just
a constant.
f. Show that


  z  0  0, becomes
 z


0  z  0, or 0 ( z )   12 z 2
z
Modern Physics
Quiz 3
Page 9
The previous solution turns out to be the ground state of the quantum harmonic
oscillator. There is no state with lower energy. So we have found:
 m 2 
x 
 2

 0  A0 exp   12 z 2   A0 exp  
g. Go back to equations with x in them and show that this really is a solution to
2


2 1

 m 2 x 2  0 ( x)  E0  0 ( x), and find E0

2
 2m x 2


h. Now find A0. Use the requirement that

2
0
dx  1 to find A0.

-
note #1: That is an integral with respect to x, not z.

-
note #2: You may want to know that
 exp   y  dy 
2


Modern Physics


Now define  1  A1   z  0
 z

Quiz 3
where
Page 10
A1 is another constant.
i. Find  1 ( z ) . (You don’t need to find A1 but you can if you want. It turns out
to be easier to find A1 than it was to find A0.)
j. Show that

1  2
  2  z 2  1  1  1 , where 1 
2  z

3
2
so E1 
3
2
  E0  .
Modern Physics
Quiz 3
Page 11
It turns out that  1 is the first excited state (okay, there is another “A” sitting out
front that you might have to find eventually but you don’t need it now). There are
no states with energies between  0 and 1 . As a matter of fact, you can find all of
the solutions of the quantum harmonic oscillator using the “recurrence relation”

n 1


 An 1   z  n
 z

where
An 1 is another constant.
k. Extra Credit! Prove it! Show that for any  n

1  2


if   2  z 2  n   n  n , and  n 1  An 1   z  n , 
2  z
 z



1  2
then   2  z 2  n 1   n 1  n 1 , where  n 1   n  1,
2  z


so En 1  En    E0  (n  1)   n  1  12

.
The tools you need are:
1 

   z
2  z

1 

   z
2  z



1
  z  n   n  2
 z



1
  z  n   n  2

z



 n,

  n.
and
Apply those equations to the following to prove the result:

1  2
 1 

 1
  2  z 2  n1     z    z    n1
2  z
  z
 2
 2  z

