Transcript Physical Chemistry 2nd Edition

Chapter 13
The Schrödinger Equation
Physical Chemistry 2nd Edition
Thomas Engel, Philip Reid
Objectives
• Key concepts of operators, eigenfunctions,
wave functions, and eigenvalues.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Outline
1. What Determines If a System Needs to Be
Described Using Quantum Mechanics?
2. Classical Waves and the Nondispersive Wave
Equation
3. Waves Are Conveniently Represented as
Complex Functions
4. Quantum Mechanical Waves and the
Schrödinger Equation
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Outline
1. Solving the Schrödinger Equation: Operators,
Observables, Eigenfunctions, and
Eigenvalues
2. The Eigenfunctions of a Quantum Mechanical
Operator Are Orthogonal
3. The Eigenfunctions of a Quantum Mechanical
Operator Form a Complete Set
4. Summing Up the New Concepts
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.1 What Determines If a System Needs to Be Described Using
Quantum Mechanics?
•
•
Particles and waves in quantum mechanics are
not separate and distinct entities.
Waves can show particle-like properties and
particles can also show wave-like properties.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.1 What Determines If a System Needs to Be Described Using
Quantum Mechanics?
•
•
In a quantum mechanical system, only certain
values of the energy are allowed, and such
system has a discrete energy spectrum.
Thus, Boltzmann distribution is used.
ni g i ei  e j / kT

e
nj g j
where n = number of atoms
ε = energy
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.1
Consider a system of 1000 particles that can only have
two energies,    2  1, with 1 and  2 . The difference
in the energy between these two values is 1   2 .
Assume that g1=g2=1.
a. Graph the number of particles, n1 and n2, in states
1 and  2 as a function of kT /  . Explain your
result.
b. At what value of kT /  do 750 of the particles have
the energy  1 ?
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
We can write down the following two equations:
n2 / n1  e / kT and n1  n2  1000
Solve these two equations for n2 and n1 to obtain
1000e   / kT
n2 
1  e   / kT
1000
n1 
1  e   / kT
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Part (b) is solved graphically. The parameter n1 is
shown as a function of kT /  on an expanded
scale on the right side of the preceding graphs,
which shows that n1=750 for kT /   0.91.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.2 Classical Waves and the Nondispersive Wave Equation
•
13.1 Transverse, Longitudinal, and Surface Waves
•
•
•
A wave can be represented pictorially by
a succession of wave fronts, where the
amplitude has a maximum or minimum value.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.2 Classical Waves and the Nondispersive Wave
Equation
•
The wave amplitude ψ is:
x t 
x, t   A sin 2   
 T 
•
It is convenient to combine
constants and variables to
write the wave amplitude as
x, t   A sin kx  wt 
where k = 2πλ (wave vector)
ω = 2πv (angular frequency)
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.2 Classical Waves and the Nondispersive Wave
Equation
•
13.2 Interference of Two Traveling Waves
•
For wave propagation in a medium where
frequencies have the same velocity (a
nondispersive medium), we can write
 2   x, t  1  2   x, t 
 2
2
x
v
t 2
where v = velocity at which the wave propagates
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.2
The nondispersive wave equation in one dimension
is given by
 2  x, t  1  2  x, t 
 2
2
x
v
t 2
Show that the traveling wave  x, t   A sinkx  wt   
is a solution of the nondispersive wave equation.
How is the velocity of the wave related to k and w?
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
The nondispersive wave equation in one dimension is given
by
 2  x, t  1  2  x, t 
 2
2
x
v
t 2
Show that the traveling wave  x, t   A sinkx  wt    is a
solution of the nondispersive wave equation. How is the
velocity of the wave related to k and w?
( x, t )  A cos(t  kx   )  1 ( x) 2 (t )!!!
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.2
We have
 2 x, t  1  2 x, t 

2
x
v t 2
 2 A sin kx  wt     w
 2 A sin kx  wt   
2
t
v
Equating thesetwo results gives v  w / k
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.3 Waves Are Conveniently Represented as Complex
Functions
•
It is easier to work with the whole complex
function knowing as we can extract the real
part of wave function.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Standing (stationary) wave
( x, t ) 1 ( x) 2 (t )  Asin  kx)cos(t   
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.3
a. Express the complex number 4+4i in the
form re i .

3
e
b. Express the complex number
in the
form a+ib.
i3 / 2
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
a. The magnitude of 4+4i is 4  4i 4  4i 1/ 2  4
The phase is given by
cos 
4
4 2

1
1 
or   cos1

2
2 4
Therefore, 4+4i can be written
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4 2ei  / 4 
2
Solution
b. Using the relation ei  expi   cos  i sin  , 3ei3 / 2
can be written
3
3 

3 cos  i sin
  30  i   3i
2
2 

Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.2 Some Common Waves (Traveling waves)
 2  r,t   2 (r,t ) 2 (r,t )



2
2
2
r

r
r
v t
 2 (r   r,t )  2 (r (r,t ))

2
2
r 2
v t
   x, t     x , t 

2
t
x 2
2
2
  r, t  
Planar wave
x t 
  x, t   A sin 2   
 T 
Chapter 13: The Schrödinger Equation
Physical Chemistry 2 Edition
 A sin(t  kx   )
© 2010 Pearson Education South Asia Pte Ltd
nd
0
r
Spherical wave
r t

cos 2      
 T

 2  r,t  1  (r,t )

(r
)
r
v 2t 2 r r
0
r
cos  kr  t   
Cylindrical wave
  r , t   r0 cos  kr  t   
13.4 Quantum Mechanical Waves and the Schrödinger
Equation
•
The time-independent Schrödinger
equation in one dimension is
h 2 d 2 x 

 V x  x   E x 
2
2m dx
•
It used to study the stationary states of
quantum mechanical systems.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.4 Quantum Mechanical Waves and the Schrödinger
Equation
•
An analogous quantum mechanical form of
time-dependent classical nondispersive wave
equation is the time-dependent Schrödinger
equation, given as
x, t 
h 2  2 x, t 
ih

 V x, t x, t 
2
t
2m x
•
where V(x,t) = potential energy function
This equation relates the temporal and spatial
derivatives of ψ(x,t) and applied in systems
where energy changes with time.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.4 Quantum Mechanical Waves and the Schrödinger
Equation
•
For stationary states of a quantum mechanical
system, we have
ih
•
  x, t 
 E  x , t 
t
Since
, we can show that that
wave functions whose energy is independent of
time have the form of
x, t    x e-iE/ht
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.5 Solving the Schrödinger Equation: Operators, Observables,
Eigenfunctions, and Eigenvalues
•
•
We waould need to use operators, observables,
eigenfunctions, and eigenvalues for quantum
mechanical wave equation.
The time-independent Schrödinger equation is
an eigenvalue equation for the total energy, E
 h2  2

 V x  n x   E n  n x 

2
 2m x


where {} = total energy operator or H
•
It can be simplified as Hˆ  n  En n
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd

O n  nn
Eigenequation, eigenfunction, eigenvalue

O n  nn
The effect of an operator acting on its eigenfunction is the same as
a number multiplied with that eigenfunction.
There may be an infinite number of eiegenfunctions and eigenvalues.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.5
2
2
d
/
dx
and
d
/
dx
Consider the operators
. Is the
function  x  Aeikx  Beikx
an eigenfunction of
these operators? If so, what are the eigenvalues?
Note that A, B, and k are real numbers.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
To test if a function is an eigenfunction of an
operator, we carry out the operation and see if the
result is the same function multiplied by a
constant:


d Aeikx  Beikx
 ikAeikx  ikBikx  ik Aeikx  Beikx
dx


In this case, the result is not  x  multiplied by a
constant, so  x  is not an eigenfunction of the
operator d/dx unless either A or B is zero.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
This equation shows that  x  is an eigenfunction
of the operator d 2 / dx2 with the eigenvalue k2.


d 2 Aeikx  Beikx
2
2 ikx
ikx
2
ikx
ikx
2





ik
Ae


ik
B


k
Ae

Be


k
 x 
2
dx
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd


Matrix as Quantum Mechanical Operator
1 0 1 0   0 1  0 i 

,
,
,

0
1
0

1
1
0

i
0

 
 
 

1 0 0 1 0 0  0 1 0  0 i 0

 
 
 

0
1
0
,
0
0
0
,
1
0
1
,

i
0
i

 
 
 

 0 0 1   0 0 1  0 1 0   0 i 0 

 
 
 


O n  nn
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
 ak 
0 1
 0 1   ak 

 k  kk  
  b   k  b 
1 0
 1 0  k 
 k
0 1 0
 0 1 0  ak 
 ak 



 
 
1
0
1





1
0
1
b


k k
k  bk 

 k

 k 
0 1 0
 0 1 0  c 
c 



 k 
 k
Eigenvalues and eigenvector of the Matrix Operator

O n  nn
 ak 
0 1
 0 1   ak 

 k  k k  
  b   k  b 
1
0
1
0



 k 
 k
1  ak 
1 
 0  k
 0  k


0

det
 

0
0  k  bk 
0  k 
 1
 1
 (0  k ) 2  1  0  k  1  1  1, 2  1.
 a2 
 0 1   a1   a1   0 1   a2 

1
,


1
 

    
 
 1 0   b1   b1   1 0   b2 
 b2 
b1  a1
b2  a2
 1  1: 
2  1: 
a1  b1
a2  b2
 a1 
Using normality condition:  a1 b1     1,  a2
 b1 
a1  b 1  12 , a2  b2  12 .
 a1  
1  1, 1     
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
 b1  
© 2010 Pearson Education South Asia Pte Ltd
 a2 
b2     1
 b2 
1

 a2   2 
 , 2  1,  2      1  .
1 
 b2    2 
2 
1
2
Exercise

O n  nn
 ak 
 0 i
 0 i   ak 

 k  kk  
  b   k  b 
 i 0 
 i 0  k 
 k
 ak 
   ? k  ?
 bk 
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Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Review of Orthogonal decomposition of
vectors and functions:
Vector in 3D space: V  Ax i  Ay j  Az k
Orthogonality: i j  0, i k  0, j k  0,
{m , m  0,1, 2,3...}   *n m = nm ,
Normality:i i  1, j j  1, k k  1.
Denoting i =i1 , j  i 2 , k  i 3
   ann ,
  1 if j  k
 i j i k   jk  
 0 if j  k

n 0
Extend to functions of continuous variables:
{m ( x), m  0,1, 2,3...} 
 ( x)   ann ,
n 0
n
Vector in nD space: V  A1i1  A2i 2  A3i 3  ...  An i n   Aj i j
j 1
Orthogonality: i1 i 2  0, i1 i 3  0,... i 2 i 3  0,..., i n 1 i n  0.
Normality:i1 i1  1, i 2 i 2  1, i 3 i 3  1,..., i n 1 i n 1 , i n i n  1.


 dx *
n
( x)m ( x)= nm ,


Chapter 13: The Schrödinger Equation
  1 if j  k
Physical

i j i k Chemistry
 jk  2nd Edition
© 2010 Pearson Education
 0 South
if j Asia
k Pte Ltd
an   *n 

an 
 dx *
n

( x) ( x)
13.6 The Eigenfunctions of a Quantum Mechanical Operator
Are Orthogonal
•
•
Orthogonality is a concept of vector space.
3-D Cartesian coordinate space is defined by
x y  x z  y z  0
•
In function space, the analogous expression
that defines orthogonality is




 dx *i ( x) i ( x) 
Orthogonormality:
2
dx
|

(
x
)
|
1
i

0 if i  j
 dx *i ( x) j ( x)  ij  1 if i  j

Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia
Pte Ltd
Example 13.6
Show graphically that sin x and cos 3x are
orthogonal functions. Also show graphically that

 sin mxsin nxdx  0 for n  m  1


  sin mx sin nx  dx  0

1
2
for n  m

  sin nx sin nx  dx  1 for n  integers.


 sin mx  cos nx  dx  0
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition 
© 2010 Pearson Education South Asia Pte Ltd
for any n, m
Solution
The functions are shown in the following
graphs. The vertical axes have been offset to
avoid overlap and the horizontal line indicates
the zero for each plot.
Because the functions are periodic, we can
draw conclusions about their behaviour in an
infinite interval by considering their behaviour
in any interval that is an integral multiple of the
period.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
The integral of these functions equals the sum
of the areas between the curves and the zero
line. Areas above and below the line contribute
with positive and negative signs, respectively,

and indicate that  sin m xcos3xdx  0 and
 sin xsin xdx  0 . By similar means, we could
show that any two functions of the type sin mx
and sin nx or cos mx and cos nx are orthogonal
unless n=m. Are the functions cos mx and sin
mx(m=n) orthogonal?


Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.6 The Eigenfunctions of a Quantum Mechanical Operator
Are Orthogonal
•
•
•
3-D system importance to us is the atom.
Atomic wave functions are best described
by spherical coordinates.
0 if i  j
 dx *i ( x) j ( x)   ij  1 if i  j

0 if i  j
  dxdydz *i ( x, y, z ) j ( x, y, z )   ij  
1 if i  j
V
0 if i  j
  r sin  drd d *i (r , ,  ) j (r , ,  )   ij  
1 if i  j
V
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 13.8
r
Normalize the function e over the interval
0  r  ; 0     ; 0    2
Solution:
Volume element in spherical coordinates is
r 2 sin drdd , thus
2


0
0
0
N 2  d  sin d  r 2 e  2 r dr  1

4N 2  r 2 e  2 r dr  1
0
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Using the standard integral ,


0
x n eax dx  n!/ a n1 a  0, n is a positiveinteger
2!
1
we obtain 4N 23  1 so that N  
2
The normalized wave function is
1

e r
Note that the integration of any function involving
r, even if it does not explicitly involve  or  , e  r
requires integration over all three variables.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
13.7 The Eigenfunctions of a Quantum Mechanical Operator
Form a Complete Set
•
•
•
The eigenfunctions of a quantum mechanical
operator form a complete set.
This means that any well-behaved wave
function, f (x) can be expanded in the
eigenfunctions of any of the quantum
mechanical operators.
13.4 Expanding Functions in Fourier Series

 ( x)   a  ,
Chapter 13: The Schrödinger Equation n n
Physical Chemistry 2nd Edition n 0
© 2010 Pearson Education South Asia Pte Ltd

an 
 dx *  ( x)
n

13.7 The Eigenfunctions of a Quantum Mechanical Operator
Form a Complete Set
•
Fourier series graphs
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Fourier series

cosnx
  




n  0,1, 2,
cos mx cos nxdx  0


sin nx
m n



n  1, 2,3,
sin mx sin nxdx  0 m  n
cos mx sin nxdx  0 all m, n
2 ,
cos nx cos nxdx  
 ,
if n  0
if n  0



sin nx sin nxdx   if
a0
f ( x) 
 a1 cos x  a2 cos 2 x  a3 cos 3 x 
2
 b1 sin x  b2 sin 2 x  b3 sin 3x 

a0

  (an cos nx  bn sin nx)
2 n 1

1 
1
Chapter
13:
The Schrödinger
Equation
a
f
( x) cos nxdx
bn   f ( x)sin nxdx
n

Physical Chemistry 2 
Edition

 
© 2010 Pearson Education South Asia Pte Ltd
nd
n0
The function f(x) (blue line) is approximated by the summation of
sine functions (red line):
a0
f ( x) 
 a1 cos x  a2 cos 2 x  a3 cos 3 x 
2
 b1 sin x  b2 sin 2 x  b3 sin 3x 

a0

  (an cos nx  bn sin nx )
2 n 1
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Fourier Transforms (FT)

f ( x)   u( y) cos xy   ( y)sin xy dy
0
1
u( y) 


 ( y) 
1






Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
f ( x) cos xydx
f ( x) sin xydx
FT in exponential form

f ( x)   u( y) cos xy   ( y)sin xy dy
0
u( y) 
1



f ( x) cos xydx
 ( y) 
1 ixy
cos xy  (e  e  ixy )
2
1



f ( x) sin xydx
1 ixy ixy
sin xy  (e  e )
2
1
w( y )  u ( y )  i ( y ) 
2

f ( x)   w( y)eixy dy

1 
 ixy
w( y ) 
f ( x)e dx

2 
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
For students who are not familiar with orthogonal
expansion of functions, you may find the following ppt
tutorial helpful:
• http://140.117.34.2/faculty/phy/sw_ding/teac
hing/chem-math1/cm07.ppt
You should also read a chemistry math book and do some exercises
for better understanding.
Chapter 13: The Schrödinger Equation
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd