Winter 2013
Chem 356: Introductory Quantum Mechanics
Chapter 3 – Schrodinger Equation, Particle in a Box .................................................................................. 34
Introduction to the Schrodinger Equation .............................................................................................. 34
Linear Operators ..................................................................................................................................... 36
Quantization of energy ........................................................................................................................... 39
Interpretation of Wave Function ............................................................................................................ 40
Determination of Constant C .................................................................................................................. 41
Useful integrals for particle in the box ................................................................................................... 43
Demonstration of Uncertainty Principle ................................................................................................. 44
Particle in a 3 dimensional box ............................................................................................................... 46
Chapter 3 – Schrodinger Equation, Particle in a Box
Introduction to the Schrodinger Equation
De Broglie suggested one can associate a wave with a particle and take p 

e ikx
2
k
p
h

h
k k
2
Generalization to 3 dimensional wave
eikx
p k
In chapter 2 we saw that waves in general satisfy a wave equation.
Try to postulate a wave equation for “electron-waves” (a guess)
Provide some rational for Schrodinger equation:
Wave equation
2u
1
2u


x 2 V 2 t 2
Choose solution with particular  

2
u ( x, t )   ( x) cos(t )
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Chem 356: Introductory Quantum Mechanics
Winter 2013
d 2  2
 ( x)  0

dx 2 V 2
  V
  2 v
 (nu) frequency ; V
velocity
d 2 4 2
 2  ( x)  0
dx 2

4 2
h

p
2
2
4 2
 p
 2  p2   
h
 
2
 2
 p 2 ( x)  0
dx 2
Now substitute p 2 :
p2
V  E
2m
 2 p 2

 ( x)  0
2m x 2 2m
2
 2
 ( E  V ) ( x)  0
2m x 2
2
 2
E ( x)  
 V ( x) ( x)
2m x 2
2
Or
Hˆ  ( x )
We obtain a differential equation for function  ( x )
Hˆ  ( x)  E ( x)
E is a constant, the energy
Ĥ is “operator” that acts on a function.
Summarizing:
 2
(using de Broglie + classical wave equation)
x 2
2) Substitute p 2  2m( E  V ( x))
1)
p 2 ( x)  
2
 2

 V ( x) ( x)  E ( x)
2m x 2
Hˆ  ( x)  E ( x)
2
p2
Hˆ 
 V ( x)
2m
‘energy operator’ (see later)
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
We need to discuss 2 mathematical items
p̂ , Ĥ , p̂ 2 ….?
a) Operators
b) Eigenvalue equations
Ĥ  E
E , p : numbers
p̂  p
Operators will be indicated by “^” hat or carot
Linear Operators
(in 1 dimension first)
ˆ ( x)  g ( x)
Af
Acting with an operator on a function yields a new function.
ˆ ( x)  g ( x)
Af
f ( x)
Â
d2
dx 2
 d2

d
 2  2  3
dx
 dx

d
x
dx
d
x
dx
d
i
dx


d2
 V ( x) 

2
 2m dx

2x
0
x3
6 x  6 x 2  3x3
x2
2 x2  x
d 2
(x )
dx
x2
d 2
( x )  3x 2
dx
e ikx
keikx
cos(kx)
 2k 2

 V ( x)  cos(kx)

 2m

The operators we consider are linear operators:
Aˆ (c1 f1 ( x)  c2 f 2 ( x))



ˆ ( x)  c Af
ˆ ( x)
= c1 Af
1
2
2

Where c1 , c2 are (complex) constants
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
( f ( x))2
Example of operator that is not linear: SQR( f ( x))
SQR( f ( x)  g ( x))  ( f ( x)) 2  ( g ( x)) 2  2 f ( x) g ( x)
=SQR( f ( x))  SQR( g ( x))  2 f ( x) g ( x)
Not linear therefore
We can act with operators in sequence
ˆ ˆ ( x)  Aˆ ( Bf
ˆ ( x))
ABf
In general:
ˆ ˆ ( x)  BAf
ˆ ˆ ( x)
ABf
d
  x , Bˆ 
dx
df
 d 
 x  f ( x)  x
dx
 dx 
d
df
d 
( xf ( x))  f ( x)  x
 x  f ( x) 
dx
dx
 dx 
Example
ˆ ˆ ( x)  BAf
ˆ ˆ ( x) , for any f ( x ) we write
If ABf
ˆ ˆ  BA
ˆˆ 0
AB
[ Aˆ , Bˆ ]  0
 and B̂ commute, the order does not matter
Eigenvalue equations (by example)
Aˆ ( x)  a ( x)
Acting with  on a function yields the same function multiplied by a constant
i
 ikx
e  (i )(ik )eikx
x
 keikx
 h 2  ikx


e
 2  

e
We say
ikx
h

e
i
eikx  peikx
2

x
periodic with period 

x
 px eikx
pˆ x  i
pˆ x eikx
Number
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
The wave function e ikx is an eigenfunction of operator pˆ x
d
, with eigenvalue
dx
pˆ x  i
k
h

pˆ ( x)  p ( x)
A particle with definite momentum px is described by eigenfunction of operator pˆ x
Consider kinetic energy operator
 

i


2
2
pˆ
d2
x 


2m
2m
2m dx 2
2
Eigenfunctions of Kinetic energy:

2
2 2
d 2 ax
a
e


2
2m dx
2m
 0 !!
(if areal)
Not physical

2
2
2
d
sin(ax)  +
a 2 sin(ax)
2
2m dx
2m
Constant Eigenvalue
Or
2
2
d2

cos(ax) 
a 2 cos(ax)
2
2m dx
2m
Also
2
eiax
2m
a 2 eiax
Or Hamiltonian operator:
2
pˆ 2
2
ˆ
H
 V ( x) = 
 V ( x)
2m
2m x 2
Hˆ  ( x)  E ( x)
: particle described by eigenfunction ( x ) has the definite energy E
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
Quantization of energy
We saw that a fundamental feature of ‘new’ quantum mechanics was that energy cannot take
on any value, but only certain values. Why is that?
Let us consider a particle in a box problem:
V ( x)  0
0 xa
V ( x)  
elsewhere
We wish to solve

2
d2
 ( x)  V ( x) ( x)  E ( x)
2m dx 2
Constant
Outside the box V ( x )   we want finite values of E , the only possibility is  ( x)  0 outside the box.
We also wish  ( x ) to be continuous:
Inside the box we have V  0

d 2
 E ( x)
2m dx 2
2
Boundary Condition:
 (0)   (a)  0
We considered before this equation
General Solution:
x0
c sin(kx)  b cos(kx)
 0
b0
E
2
k2
2m
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
xa
Chem 356: Introductory Quantum Mechanics
n
, n  1, 2,3
a
k
c sin(ka)  0
Any c , c not equal to 0
 n x 

 a 
 ( x)  c sin 
n 2 2 h 2 n 2
E

2ma
8ma
2
-
n  1, 2,3.....
Quantization: Combination of wave equation + Boundary conditions
n  1, 2, 3 also possible, but yields “same” solutions
n x
 n x 
c sin  
  c sin
a
 a 
-
c can be anything (still)
ˆ  ( x)  cAˆ ( x)
Ac
 ca ( x)
 a (c ( x))
If  ( x ) is an eigenfunction of operator  then also c ( x ) is eigenfunction. ( c is constant)
Interpretation of Wave Function
In Mathchapter B we discussed probability distribution p ( x ) dx :
p( x)  0
x

 p( x)dx  1


x 
 xp( x)dx
etc.

The absolute square of the wave function  ( x)   * ( x) ( x) is to be interpreted like a probability
2
distribution.
p( x)dx   ( x) dx
2
Probability to find particle between x and x  dx
 ( x)  f ( x)  ig ( x)
complex
f ( x ) , g ( x ) real
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
 * ( x)  f ( x)  ig ( x)
 * ( x) ( x)  [ f ( x)  ig ( x)][ f ( x)  ig ( x)]
 f ( x)  g ( x)  i[ f ( x) g ( x)  g ( x) f ( x)]
2
So
2
 ( x)  0
2
everywhere
Probability distribution

  ( x)
Moreover:
2
dx  1

Normalization
Multiply  ( x ) by constant c , choose c such that c ( x)   new ( x) is normalized
Particle in the box (later)
2
 n x 
sin 

a
 a 
 n ( x) 
Further Interpretation
xhigh
  ( x)  ( x)dx
*
xlow
Probability to find particle between xlow and xhigh
And
x   x ( x)* ( x)dx
Determination of Constant C
We will impose that the wave functions are normalized


*
( x) ( x)dx  1
For reasons discussed before

 * ( x) : complex conjugate of functions
 ( x)  f ( x)  ig ( x)
f ( x ) , g ( x ) real
 ( x)  f ( x)  ig ( x)
*
 ( x) ( x)  [ f ( x)  ig ( x)][ f ( x)  ig ( x)]
*
 [ f ( x)]2  [ g ( x)]2  i[ f ( x) g ( x)  g ( x) f ( x)]
 f ( x)  g ( x)
2
2
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
  ( x)
 0 everywhere
2
If  ( x ) is real then  ( x)   ( x) 2
2
Consider particle in the box wave functions:
 n ( x) 
n x
a
Cn sin
0 xa
0

2
n x 

  ( x) dx  0 Cn  sin a  dx
a
2
2
 Cn 2
a
1
2
2 i
e
a
Choose Cn 
2 ik
e would work too.
a
We can always choose the function  ( x ) to be normalized
Simplest,
A physically meaningful wave function would be normalized
If Aˆ ( x)  a ( x) eigenfunction of  , eigenvalue a

 * ( x) Aˆ ( x)
Then

  * ( x)a ( x)  a * ( x) ( x)

And:

*


( x) Aˆ ( x) dx


 a   * ( x) ( x)dx

 a 1
IF  ( x ) is normalized
We define:
Aˆ   * ( x)Aˆ ( x)dx
Called the expectation value of operation  , depending on ( x ) , also called the
average value of Â
If  ( x ) is normalized, then  would be the average value measured for quantity A
If  ( x ) is an eigenfunction of  , then one would always measure a , and the average
value A  a IF  ( x ) is normalized
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
If  ( x ) is not an eigenfunction of  , then many values could be obtained if A is measured.
The average value would be Â
One more definition:
 Aˆ  Aˆ 
2
: The standard deviation from the average. The spread of the measured
values
 Aˆ  Aˆ  Aˆ  Aˆ 
 Aˆ 2  2 Aˆ Aˆ  Aˆ
2
 Aˆ 2  2 Aˆ Aˆ  Aˆ
 Aˆ 2  Aˆ
2
2
Depends on wave function  ( x )
  A2
Useful integrals for particle in the box
 sin
2
axdx 
x sin 2ax

2
4a
2
 x sin axdx 
x 2 x sin 2ax cos 2ax


4
4a
8a 2
2
2
 x sin axdx 
x3  x 2
1 
cos 2ax
   3  sin 2ax  x
6  4a 8a 
4a 2
Definite Integrals (Most important)
a
 sin
0
a
2
n x
a
dx 
a
2
2
 x sin
0
a
n x
a2
dx 
a
4
2
2
 x sin
0
a
 sin
0
n x
a3
a3
dx 
 2 2
a
6 4n 
n x
n x
cos
dx  0
a
a
Chapter 3 – Schrodinger Equation, Particle in a Box
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Chem 356: Introductory Quantum Mechanics
Winter 2013
Demonstration of Uncertainty Principle
Using the above integrals, we can calculate the following
a) Normalize  n  Cn sin
n x
a
n x 

2 a
Cn   sin
1
 dx  Cn
a 
2
0
2
a
2
Normalized particle in the box eigen states:
b)
2
a
Cn  C 
2
 n x 
sin 

a
 a 
Calculate x for normalized  n ( x) :
2
n x
n x
x   sin
x  sin
dx
a0
a
a
a

2 a2 a


a 4 2
center of the box
c) Calculate x 2
2
n x 2
n x
sin
x  sin
dx

a0
a
a
a
x2 

2  a3
a3
  2 2
a  6 4n 
 a2
a2



2 2
 3 2n 
d) Standard deviation in x :
 x 2  x2  x
2
2 2

a2
a2
a2
a2
a
 a  n 
 2 2   
 2 2 
 2
 
3 2n 
12 2n 
2
 2 n   3

2

2
2
n x 
d 
n x
sin
dx
 i
 sin

a0
a 
dx 
a
a
e)
Px 
2
n 
n x
n x
cos
dx  0
 i
  sin
a
a 0
a
a
a

f)
Px
2
a
2
n x 
  sin

a0
a 
2

a
d 2  n x
 sin
a
dx 2 
n 2 2 
n x 
 2   sin
 dx
a 
a 0
a
2
2
2
a
2
Chapter 3 – Schrodinger Equation, Particle in a Box
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Chem 356: Introductory Quantum Mechanics
Winter 2013
n 2 2 h 2 n 2


a2
4a 2
hn
 ( Px ) 
2a
2
(  2mEn , of course!)
We can test the Heisenberg Uncertainty Principle
1
  2 n2
 2 hn
 x p 

 2 
2 n  3
 2a
a
1
  2 n2
2
 
 2
2 3

Note 1:
x 
a
12

2
as n   is the same as uncertainty in uniform distribution:
1 x2
x 
a 2
a

0
a
2
a
x
2
x

Px
11 3
1

x  a2
a3 0 3
uniform

a2 a2 a2


3
4 12
grows with n. Why?
Pn   (2mEn )  
n 2 2
a2
2
This
Spiked distribution
Large Uncertainty
represents the classical limit of particle of
energy En
bouncing back and forth in the box
Note 2:
be calculated for any wave function
 ( x)  Cx(a  x)
for example:
x , x 2 , Px , Px 2
Can
also satisfies the boundary conditions
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
Particle in a 3 dimensional box
Consider rectangular box of length a, b, c
3D Schrodinger Equation:
 2
2
2 




 ( x, y, z )  E ( x, y, z )
2m  dx 2 dy 2 dz 2 
2
Boundary Conditions:
 (0, y, z )   (a, y, z )  0
 ( x, 0, z )   ( x, b, z )  0
 ( x, y, 0)   ( x, y, c)  0
y, z
x, z
x , y
“The wave function at the faces of sides of a box is zero”
Technique to solve: Separation of variables.
 ( x, y, z )  X ( x)Y ( x) Z ( z )
Try
Substitute in Schrodinger equation and divide by  ( x, y, z ) (as we did for vibrating strings)

2
2
2
1 d2X
1 d 2Y
1 d 2Z


E
2m X ( x) dx 2 2m Y ( y ) dy 2 2m Z ( z ) dz 2
This can only be true if each term itself is constant: E x , E y , E z
We get 3 equations
a)

2
d2X
 Ex X ( x )
2m dx 2
2
d 2Y
 EyY ( y)
b) 
2m dy 2
c)

2
d 2Z
 Ez Z ( z )
2m dz 2
Ex  E y  Ez  E
X (0)  X (a)  0
Y (0)  Y (b)  0
Z (0)  Z (c)  0
Chapter 3 – Schrodinger Equation, Particle in a Box
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Winter 2013
Chem 356: Introductory Quantum Mechanics
This is just 3 times the 1D particle in the box equation! We know the (normalized)
solution:
X ( x) 
2
 k x 
sin 

a
 a 
Ex 
h2  k 2 
 
8m  a 2 
Y ( y) 
2
 l y 
sin 

b
 b 
Ey 
h2  l 2 
 
8m  b2 
Z ( z) 
2
 n z 
sin 

c
 c 
Ey 
h2  n2 
 
8m  c 2 
Or
n n n 
x y z
8
n 
sin  x x
abc
 a
 ny  y

  sin  b


2
nz 2 
h 2  nx 2 n y
E
 2  2 

8m  a 2
b
c 

 nz  z 
  sin 

 c 

nx , ny , nz  1, 2,3....
Degeneracies for Cubic box
Consider the special case of a Cubic box a  b  c . Then the energy takes the form
h2
E
n 2  n y 2  nz 2 
2  x
8ma
For each triplet nx , n y , nz we get a different wave function, but different values of nx , n y , nz may yield
the same energy.
Such energy levels are called degenerate. Eg.for atoms we know there are 1 s-orbital, 3 p-orbitals, 5 dorbitals.
Table of energies
E
h2
8ma 2
14
12
11
9
6
3
n , n , n 
Degeneracy
(1, 2,3), (1,3, 2), (2,1,3), (2,3,1), (3,1, 2), (3, 2,1)
(2, 2, 2)
(1,1,3), (1,3,1), (3,1,1)
(2, 2,1), (2,1, 2)(1, 2, 2)
(1,1, 2), (1, 2,1), (2,1,1)
(1,1,1)
6
1
3
3
3
1
x
y
z
Chapter 3 – Schrodinger Equation, Particle in a Box
47