化學數學（一）
The Mathematics for Chemists
(I)
(Fall Term, 2004)
(Fall Term, 2005)
(Fall Term, 2006)
Department of Chemistry
National Sun Yat-sen University
Chapter 5 Differential Equations
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Simple Ordinary Differential Equations (ODE)
Kinetics of Chemical Reactions
Partial Differential Equations (PDE)
Chemical Thermodynamics
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Gamma Functions
Beta Functions
Hermite Functions
Legendre Functions
Laguerre Functions
Bessel Functions
Contents Covered in Chapters 11-14
Assignment
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P.260: 38,40,42
P.286: 24,27,28
P.302: 7,9,12,15
PP.323-324: 2, 8, 10
Overview of Differential Equations (DE)
DE: Equations that contains (partial) derivatives.
dy d 2 y
f ( x, y,
,
, )  0
2
dx dx
u  2u u  2u u  2u
 ( x, y, z, , 2 ..., , 2 ..., , 2 , )  0
x x
y y
z z
• Ordinary DE (ODE): One variable
First-order ODE, Second-order ODE, …
Constant coefficient ODE,
Variable coefficient ODE
• Partial DE (PDE): Multi-variable
Examples
First order:
dx
 kx
dt
Second order: constant coefficients
d 2h
 g
2
dt
d2y
dy
 5  6y  0
2
dx
dx
Second order: variable coefficients
ODE
  2 d 2 1 2
 kx   E
2
2m dx
2
2


m
2



(1  x ) y  2 xy  l (l  1) 
y0
2 
(1  x ) 

 2
 2 m ( x2 
2
 2
y 2

 2
z 2
)  V ( x, y, z )  E
PDE
Some First- and Second-order ODEs
dx
 kx
dt
First order rate process (growth/decay)
dx
 k (a  x)(b  x)
dt
Second-order rate process
d 2h
 g
2
dt
Free falling of an object
d 2x
2



x
2
dt
Classical harmonic oscillator
  2 d 2 1 2
 kx   E
2
2m dx
2
One-dimensional Vibration of atomic bonds
Solving A DE
• Find the function(s) (of one or more
variables) that satisfy the ODE/PDE. This
step normally involves integration and/or
series expansion.
• Initial or boundary conditions are usually
required to specify the solution.
Therefore, both equations and initial/boundary conditions
are equally important in solving a specific practical problem.
dy
I. First Order ODE  F ( x, y )
dx
• Examples:
dx
 kx
dt
First order rate process (growth/decay)
dx
 k (a  x)(b  x)
dt
Second-order rate process
dy
 2x
dx
dy
 dx dx   2 xdx
y  x c
2
y  x  10
2
Initial condition: y=10 when x=0
Classroom Exercise
Find the general and particular solutions of the
following equation with the given initial condition:
dx
  kx
dt
x(t  0)  a
x(t )  ce
 kt
x(t  0)  a
x(t )  ae
 kt
Solving First Order ODE
Separable Equations:
dy
 F ( x, y )
dx
dy f ( x)

dx g ( y )
dy
g ( y)
 f ( x)
dx
dy
 g ( y) dx dx   f ( x)dx  c
 g ( y)dy   f ( x)dx  c
First-order linear equations:
dy
 p( x) y  r ( x)
dx
+ initial conditions
Example: Separable First-Order ODE
dy
n
 ax
dx
y (0)  0
dy  ax dx
n
 dy  a  x
n  1
n 1
n
dx
ax
y ( x) 
c
n 1
ax n1
y ( x) 
n 1
Classroom Exercise:
Separable First-Order ODE
dy
2
 2  3 x dx
y
1
y (1) 
2
dy
  2  3 x 2 dx
y
1
3
 x c
y
1
y ( x)  3
x c
1
1
y (1) 

1 c 2
1
y ( x)  3
x 1
Reduction to Separable Form:
Homogeneous Equations
f (x, y,)   f ( x, y,)
For n=0:
F ( x,  y )  F ( x, y )
n
Example:
x y
 y
 y
 y
F ( x, y ) 
   1    f  
xy
x
x
x
2
2
2
 y 
F ( x,  y )  f   
 x 
dy
 F ( x, y ) 
dx
 y
f 
x
y

x
d
dx

f    
x
 y
f    F ( x, y )
x
dy d ( x )
d

  x
 f  
dx
dx
dx

d
 ln x  c
f    
Example: Separation of a
Homogeneous Equation
dy x 2  y 2

dx
xy
y

x
dy d x 
d
1

  x
 
dx
dx
dx

d 1
x

dx 
2
dx
 d   x
y2
 ln x  c
2
2x
2
y dy y
1


x 2 dx x 3 x
2
Check:
 ln x  c
d  y2  d
 2   ln x  c 
dx  2 x  dx
dy x 2  1 y 2  x y x 2  y 2
  3    

dx
y x x  y x
xy
Chemical Kinetics
aA  bB    pP  qQ  
2H 2  O2  2H 2 O
0   vB B
B
0  aA  bB  pP  qQ    v A A  v B B  v P P  vQ Q  
0  2H 2  O2  2H 2O
Rate of Reaction
nA  nA0  v A
d
r 
dt
d
1 dn A
r

dt v A dt
dn A
d
 vA
dt
dt

r
1 d A

V v A dt
1 d A
1 d B 1 d P 1 d Q





a dt
b dt
p dt
q dt
0  2H 2  O2  2H 2O
d O2  1 d H 2 O
1 d H 2 



2 dt
dt
2 dt
Rate Constant and Order
aA  bB    products
  k A B  

A  products
2A  products
A + B  products

first order
second order
second order
A  products: first order process
d A

 k A
dt
dx
  kx
dt
dx
 kdt
x
ln x  kt  c
ln[A]
ln A  kt  c
A  A0 e
ln[A]0
ln A  kt  ln A0
-k
 kt
0
t
2A  products: second order process
d A
2

 k A
dt
dx
 2  2kdt
x
1
 2kt  c
x
dx
 2kx 2
dt
1/[A]
1 1

 2kt
x x0

A0
A 
1  2kt A0
1
1

 2kt
A A0
2k
1/[A]0
0
t
A + B  products: second order process


d  A
d B 

 k AB 
dt
dt
 A0  a,  B0  b
d (a  x) dx

 k (a  x)(b  x)
dt
dt
dx
 k (a  x) 2
dt
1
1

 kt  c
a  x A
dx
 kdt
2
(a  x)
1
1

 kt
A A0
dx
 (a  x)(b  x)  kt  c
1
1  1
1 



 (a  x)(b  x) b  a   a  x b  x dx

dx
 kdt
(a  x)(b  x)
1
1  1
1 




(a  x)(b  x) b  a  a  x b  x 
1
 ln( a  x)  ln( b  x)   1 ln  b  x 
ba
ba a x
1
b
c
ln  
ba a
 A  a  x,  B  b  x
1
a(b  x)
ln
 kt
b  a b( a  x )
1
b x
ln 
  kt  c
ba a x
A B  kt
1
ln 0
B0  A0 B0 A
First-Order Linear Equations:
The Homogeneous Case
dy
 p( x) y  0
dx
dy
  p( x)dx
y
ln y    p( x)dx  c

y  ae
 p ( x ) dx
First-Order Linear Equations:
The inhomogeneous Case
dy
 p ( x) y  r ( x)
dx
F ( x)  e 
 p ( x) dx
d
p ( x)dx  p ( x)

dx
dF ( x)
 F ( x) p( x)
dx
dy
F ( x)  F ( x) p ( x) y  F ( x)r ( x)
dx
dy
dy dF ( x)
d
F  x  y 
F ( x)  F ( x) p( x) y  F ( x) 
y
dx
dx
dx
dx
d
F x  y   F x r x 
dx
F  x  y   F  x  r  x  dx  c
Example: Linear Equation
 p( x)dx   x
dy
 y  3e2 x
dx
p ( x ) dx

F  x  e
 e x
e x y   e x  3e2 x dx  c  3 e x dx  c  3e x  c


y  e 3e  c  3e
x
x
2x
 ce
x
Classroom Exercise:
Linear Equation
dy 2
 y  3x3
dx x
 2 x  dx

F  x  e
 e2 ln x  x 2
6
x
2
2
3
x y   x  3x dx  c   c
2
x4
c
y
 2
2 x
Chemical Kinetics Example
k1
k2
A 
B 
C
d  A
d B 
 k1 A  k 2 B   A0  a,  B0  0, C 0  0.
  k1  A
dt
dt
 A  a  x,  B  y, C   x  y.
dy
d a  x 
 k1 a  x   k 2 y
 k1 a  x 
dt
dt
dy
dy
 k1t
 k1t
 k 2 y  ak1e  k1t
a  x  ae
 ak1e  k 2 y
dt
dt
p(t )  k 2
r (t )  ak1e  k1t
F t   e
 p t dt
e
k 2t
F (t ) y 
 F ( x)r (t )dt  c
ak1  k2 k1 t

c
 k 2  k1 e
k 2t
k 2  k1 t
e y  ak1  e
dt  c   ak1t  c

if k1  k 2
if k1  k 2
 k ak1k e k1t e k2t 
2
1
y


if k1  k 2
 ak1te k1t

 Ak1 e  k1t e  k 2t 

 k1t

k 2  k1
A  A0 e
B    Ak1tek1t

C  A0  A  B 
  k ak1k
c  0 2 1

A
A
C
k1  1, k2  10
B
if k1  k2
C
k1  k2  1
B
if k1  k2
if k1 k2
if k1  k 2
if k1  k 2
A
B
k1  1, k2  0.1
C
Example: Electric Circuit
Three sources of electric potential drop ( drop of voltage):
E R  RI
dI
EL  L
dt
Q
Ec 
C
R
E
L
dI
dI
L  RI  E
 ( R / L) I  E / L
dt
dt
F x   e
  R L dt
e
I t   e
Rt L
 Rt L
 Rt L E

dt  c 
  e
L

For constant electromotive force: E=E0
Initial condition, I(0)=0:
E
 E
I t   e  Rt L  0  e Rt L dt  c   0  ce  Rt L
L
 R

E0
I t  
1  e  Rt L
R

Inductive time constant:
L  L R
II. Second-Order ODE: Constant Coefficients
Inhomogeneous, linear, variable coefficients:
d2y
dy
 p( x)
 q( x) y  r ( x)
2
dx
dx
Inhomogeneous, linear and constant coefficients:
d2y
dy

a
 by  r ( x )
2
dx
dx
Homogeneous and linear, variable coefficients:
d2y
dy
 p ( x)
 q ( x) y  0
2
dx
dx
Homogeneous, linear and constant coefficients:
d2y
dy

a
 by  0
2
dx
dx
Principle of Superposition: Example
d2y
dy
 p ( x)
 q ( x) y  0
2
dx
dx
Linearly independent (not related by a proportional coefficient)
d2y
dy
 5  6y  0
2
dx
dx
y1  e
2x
y2  e
3x
Particular solutions
2
2
dy1
d
y1 dy1
2x
2x
2x
2x

5

6
y

4
e

5

2
e

6
e
0
 2e
1
2
dx
dx
dx
d y1
2x

4
e
dx 2
dy 2
d 2 y2
3x
3x

3
e
 9e
2
dx
dx
d 2 y2
dy 2
3x
3x
3x

5

6
y

9
e

15
e

6
e
0
2
2
dx
dx
2
y  c1 y1  c2 y2
d y
dy
5
 6y  0
2
dx
dx
Principle of Superposition
(for Homogeneous Linear DEs)
y  c1 y1  c2 y2
d 2 y1
d 2 y2
d2y
 c1
 c2
2
2
dx
dx
dx 2
dy1
dy 2
dy
 c1
 c2
dx
dx
dx
 d 2 y1
d 2 y2 
d2y
dy
 p( x)  q( x) y   c1 2  c2 2 
2
dx
dx
dx 
 dx
dy2 
 dy1
 p( x)  c1
 c2
  q( x)  c1 y1  c2 y2 
dx 
 dx
 d 2 y1

 d 2 y2

dy1
dy2
 c1  2  p ( x)
 q ( x) y1   c2  2  p( x)
 q ( x ) y2   0
dx
dx
 dx

 dx

The linear combination of two (particular) solutions of a homogeneous DE is also
a solution of the DE.
The general solution (constant coefficients)
d2y
dy
 a  by  0
2
dx
dx
d x
e   e x
dx
x
x
 e  ae  be
2
x
  a  b  0
1
   a  a  4b 
2
2
2
1
y1  e
 0

2
a  4b  0
 0

1x
ye
guess
x
2
d x
2 x
e


e
2
dx
0


e x 2  a  b  0
(characteristic equation or auxiliary equation)

1
2   a  a 2  4b
2
2 x

y2  e
two real roots
y1 and y2 are linearly independent
real double root y1  y2 , a new solution is needed!
two complex roots y1 and y2 are linearly independent
Example
2
d y
dy
 5  6y  0
2
dx
dx
  5  6  0
2
  5  6  (  2)(  3)  0
2
y1  e
1 x
e
2x
y2  e
2 x
e
3x
The two particular solutions being linearly independent, the general solution is
y  c1 y1  c2 y 2  c1e
2x
 c2 e
3x
Three Cases
 0

2
a  4b  0
 0

two real roots
y1 and y2 are linearly independent
real double root y1  y2 , a new solution is needed!
two complex roots y1 and y2 are linearly independent
1 x
y  c1 y1 ( x)  c2 y2 ( x)  c1e  c2e
y1 ( x)  e
1x
e
y( x)  c1e
ax 2
y2 ( x)  xy1 ( x)  xe
2 x
 ax 2
 (c1  c2 x)e
 ax 2

y( x)  c1 y1 ( x)  c2 y2 ( x)  e  ax 2 c1eix  c2 e ix
y( x)  e  ax 2 d1 cos x  d 2 sin x 

 c2 xe
 ax 2
 ax 2
  i a 2 / 4  b
d1  c1  c2 d2  i(c1  c2 )
Example: Double root
y   4 y   4 y  0
  4  4  (  2)  0
2
1
y( x)  (c1  c2 x)e
2
2x
Example: Complex roots
  2  2  0
y  2 y  2 y  0
1
  (2  4  8)
2
y1 ( x)  e
1 x
2
1
2  1*  1  i
1  1  i
e
1i  x
y 2 ( x)  e 2 x  e 1i x
y ( x)  c1e 1i  x  c2 e 1i x

 e c1e  c2 e
x
ix
ix

e
 ix
 cos x  i sin x
y ( x)  e x c1 cos x  i sin x   c2 cos x  i sin x 
 e x c1  c2  cos x  ic1  c2 sin x
 e x d1 cos x  d 2 sin x 
d1  c1  c2
d 2  i(c1  c2 )
Classroom Exercise
Find the general solution of the following ODE:
y  12 y  36 y  0
a  12, b  36
12
    62  36  6
2
y ( x)  (c1  c2 x)e
6x
Classroom Exercise
Find the general solution of the following ODE:
y  6 y  25 y  0
a  6, b  25, a 2 / 4  b  0
32  25  i ,   4
y ( x)  e
y ( x)  e
3 x
3 x
(c1e
i x
 c2e
 i x
)
(d1 cos  x  d 2 sin  x)
d1  c1  c2
d2  i  c1  c2 
Particular Solutions
Solutions with initial or boundary conditions.
y( x)  c1 y1 ( x)  c2 y2 ( x)
y ( x0 )  y 0
y (0)  0
y  y  6 y  0
y ( x0 )  y1
y (0)  5
    6  (  2)(  3)  0
2
y( x)  c1e
2x
 c2 e
3 x
2x
3 x

y ( x)  2c1e  3c2e
y( x)  e 2 x  e 3 x
y(0)  0  c1  c2
y(0)  5  2c1  3c2
Boundary Conditions
y  y  6 y  0
y( x)  c1e
2x
 c2 e
x 
y ( 0)  1
y ( x)  0
y(0)  1  c1  c2
c1  0
y ( x)  e
3 x
3 x
Example: The particle in a 1D box
 2 d 2

 E
2
2m dx
Two distinct regions:
well and wall
The microscopic entity cannot
be outside of the well:
  0 for ( x  0 or x  L)
Within the well, the particle is
a free particle:
 k ( x)  C sin kx  D cos kx
for 0  x  L
Ek
k 2 2

2m
Boundary Conditions
 k ( x)  C sin kx  D cos kx
 k (0)  0
 k ( L)  0
 k (0)  C * 0.0  D  0  D  0
 k ( L)  C * sin kL  D cos kL  0
To ensure
C0
sin kL  0  kL  n , n  0,1,2,...
k
2L

n
2

 n ( x )  C sin( nx / L)
Quantization of Energy
E
k
p

2
k 2 2
 2m 
2m
2 2
n h
2
8m L
Only some energies are allowed:
E 
1
h
2
;E
2
8m L
2

4h
2
;E
2
8m L
3

9h
2
...
2
8m L
Where there is constraint, there is
quantization
n: quantum numbers
Normalization
 n ( x )  C sin( nx / L)
L
1   |  | dx  C
2
0
L
2
dx
sin

2 n x
L
 LC
1
2
2
0
C 
2
L
2
 n (x)   
L
1/ 2
 nx 
sin 

 L 
for
0xL
First five normalized wavefunctions
n
1/ 2
 2


L
1/ 2
1  2



2i  L 
n
k
L
 nx 
sin 

 L 
e
ikx
 e ikx

2L

n
Standing wave
Where there is constraint, there is
quantization
Orthogonality

*
 n  n ' d
1/ 2
2
 3 ( x)   
L
 x 
sin  
 L
 dx ( x) 3 ( x) 
*
1
L
2
L
 dx[sin
0
0
2
 n ( x)   
L
 3x 
sin 

 L 
1/ 2
2
 1 ( x)   
L
0
1/ 2
x
L
sin
3x
L
]
 nx 
sin 

 L 
Example: The particle in a ring
d 2
2 IE
2







2
2
d
i
 i
1
2
 ( )  c e
c e
 (  2 )   ( )
in 
 in 
 n ( )  c1e  c2 e
2
 n
En 
2I
n  0,  1,  2,
2
Choosing c2=0 (because n can take both positive and negative
values) and normalizing the wavefunction:
2
1    ( ) n ( )d  c1
0
*
n
 n ( ) 
2

2
0
e
in in
e d  c1
1 in
e
2
2

2
0
d  2 c1
2
Probability distribution
 ml 
 ml 
*
1
2
1
2
e
 ml  ml 
*

e
iml
Re
 iml
1
2
e
iml
1
2
e
iml
1
2
Re( )  Re(cosm  i sin m )
 cos m
Im( )  Im(cosm  i sin m )
 sin m
Im
Orthogonality

2
0

2
0
 ( ) m ( )d  0
if n  m
*
n
1
 ( ) 2 ( )d 
2
*
1
1

2
2

2
0
1
e e d 
2
 i
2 i
1
1 i 
2 i

e

e
 1  0

 i 
2 i
0

2
0
ei d
Quiz
• Solve the following ODEs:
dy
 x2 y  2x2
dx
y-8y+2y  0
y ( 0)  1
x 
y '( x) 
0
2
1 3
x
dx
dy
x
2
2

3
e
 x y  2x F  x   e
dx
e
1 x3
3
y  e
1 x3
3
 2x dx  c  2e
2
y-8y+2y  0
 2  8 1  2  0   
y  c1e(2
14 ) x
 c2 e(2
4  64 8
2
1 x3
3
c
y  2  ce
y ( 0)  1
- 13 x3
x 
y '( x) 
0
 2  14
14 ) x
y (0)  1  c1  c2  1
x 
y '( x) 
 0  lim(c1 (2  14)e(2
x 
y  e(2
14 ) x
14 ) x
 c2 (2  14)e(2
14 ) x
)  0  c1  0
Inhomogeneous, Linear ODE
2
d y
dy
 a  by  r ( x)
2
dx
dx
y   3 y   2 y  2 x
y  a1  2a2 x
y  a 0  a1 x  a 2 x 2
2
y   2a2
y   3 y   2 y  (2a 2  3a1  2a 0 )  (6a 2  2a1 ) x  2a 2 x 2
(2a2  3a1  2a0 )  0
(6a2  2a1 )  0
y
7
 3x  x 2
2
a2  1
General Solution
d 2 yp
dx
2
a
dy p
dx
 by p  r ( x)
y h ( x)  c1 y1 ( x)  c2 y 2 ( x)
d 2 yh
dy h
a
 by h  0
2
dx
dx
y ( x)  y h ( x)  y p ( x)
2

dy p
 d 2 yh
  d yp
dyh
d 2 y dy
 a  by   2  a
 byh    2  a
 by p   r ( x)
dx
dx
dx
dx2
dx
dx


 
The general solution of an inhomogeneous linear ODE is the sum of the general
solution of the corresponding homogeneous equation and a particular solution of the
inhomogeneous equation.
Example
y   3 y   2 y  2 x
2
y   3 y   2 y  0
yh ( x)  c1e
yp
x
 c2 e
2 x
7
( x) 
 3x  x 2
2
y ( x)  y h ( x)  y p ( x)  c1e
x
 c2 e
2 x
7
  3x  x 2
2
Some Important Particular Solutions
term in r ( x)
Choice of y p
ce ax
ke ax
cx n (n  0,1, 2,...)
c cos  x(or c sin  x)
a0  a1 x  a2 x  ...  an x
2
n
k cos  x  l sin  x
ax
ax
e
(
k
cos

x

le
sin  x)
ce cos  x(or ce sin  x)
ax
ax
The determination of the coefficient(s) in yp is obtained by substituting it back to the
inhomogeneous equation. However, if yp is already in yh then the general solution
should be: y ( x )  y ( x )  c ( x ) y ( x )
h
p
where the choice of c(x):
If the characteristic equation of the corresponding homogeneous equation has two
(real or complex) roots, then c(x) =x, or else, c(x)=x2 .
If r(x) is the sum of terms given in above table, the total yp(x) is the sum of
respective yp of all terms. [This leads to a method of series expansion for general r(x) ]
d2y
dy

a
 by  r ( x)
2
dx
dx
term in r ( x)
Choice of y p
ce ax
keax
cx n (n  0,1, 2,...)
c cos  x(or c sin  x)
a0  a1 x  a2 x  ...  an x
2
n
k cos  x  l sin  x
ax
ax
e
(
k
cos

x

le
sin  x)
ce cos  x(or ce sin  x)
ax
ax
The determination of the coefficient(s) in yp is obtained by substituting it back to the
inhomogeneous equation. However, if yp is already in yh then the general solution
should be: y ( x )  y ( x )  c ( x ) y ( x )
h
p
where the choice of c(x):
If the characteristic equation of the corresponding homogeneous equation has two
(real or complex) roots, then c(x) =x, or else, c(x)=x2 .
If r(x) is the sum of terms given in above table, the total yp(x) is the sum of
respective yp of all terms. [This leads to a method of series expansion for general r(x) ]
The Method of Undetermined
Coefficients
y   3 y   2 y  3e
yh ( x)  c1e
x
2 x
 c2 e
2 x
2 x
y p ( x)  kxe
yp ( x)  ke2 x  2kxe2 x
yp  3 yp  2 y p  kxe2 x
yp ( x)  4ke2 x  4kxe2 x
y p  3xe2 x
y( x)  yh ( x)  y p ( x)  c1e  x  c2 e 2 x  3xe2 x
Classroom Exercise
2x



y  4y  4y  e
 2  4  4  0    2
y h ( x)  (c1  c 2 x)e 2 x
Check above table, we find
y p  ke2x
Double roots of homo. eq.
But it’s one solution of homo eq.
y p  kx 2 e2 x
yp  2kxe2 x  2kx2e 2 x
2x
2x
2 2x


y p  2ke  8kxe  4kx e
y p  4 y p  4 y p  2ke2 x
y ( x )  y h ( x )  y p ( x )  (c1  c 2 x 
1 2
x )e 2 x
2
Forced Oscillations
Harmonic force
Damping force (friction)
d 2x
dx
F  m 2  kx  c  F0 cos t
dt
dt
External periodic force
d 2x
dx
m 2  c  dx  F0 cos t
dt
dt
When no friction is considered and the external force is electric force on a charge e:
F  kx  eE0 cos t
d 2x
2


o x  A cos t
2
dt
0  k m
General Solution
d 2x
2


ox  0
2
dt
xh (t )  d1 cos 0 t  d 2 sin 0 t
x p (t )  c cos t  d sin t
d 2xp
dt
2
dx p
dt
 c sin t  d cos t
  (c cos t  d sin t )
Using method of undetermined coefficients:
d  0, c  A ( 02   2 )
A cos t
x p (t )  2
0   2
A cos t
x(t )  x h (t )  x p (t )  d1 cos  0 t  d 2 sin  0 t  2
0   2
Resonance
A cos t
x(t )  x h (t )  x p (t )  d1 cos  0 t  d 2 sin  0 t  2
0   2
On resonance, xp is a solution of the homo. eq., therefore the correct xp should be
x p (t )  t (cos  0 t  d sin  0 t )
d  A 20
x p (t ) 
A
2 0
t sin  0 t
Inhomogeneous, linear, variable coefficients:
d2y
dy
o( x )
 p ( x)
 q ( x ) y  r ( x)
2
dx
dx
It’s hard or impossible to obtain the solution of a general second-order ODE
III. Second-Order ODE:
Special Cases of Variable Coefficients
Legendre: (1- x ) y  2 xy  l (l  1) y  0
2
2
2


Associated Legendre: (1- x ) y  2 xy  [l (l  1)  m /(1  x )] y  0
Hermite: y  2 xy  2ny  0
Laguerre: xy  (1  x) y  ny  0
Associated Laguerre: xy  (m  1  x) y  (n  m) y  0
2
Bessel: x 2 y  xy  ( x 2  n 2 ) y  0
The Power-Series Method
y  p ( x) y  q ( x) y  r ( x)

y ( x)  a 0  a1 x  a 2 x 2  a3 x 3    a m x m
m 0
y '  a1  2a2 x  3a3 x 2
y ''  2a2  6a3 x  12a4 x 
2
a0 , a1 , a2 , a3
am ,

  mam x m 1
m 1

  m(m  1)am x
m2
m2
can be determined.
dy
y0
dx
Example
y  ae
x

y ( x)  a 0  a1 x  a 2 x 2  a3 x 3    a m x m

m 0
dy
 a1  2a 2 x  3a3 x 2     mam x m 1
dx
m 1
dy
 y  (a1  2a 2 x  3a3 x 2  )  (a 0  a1 x  a 2 x 2  )
dx
 (a1  a 0 )  (2a 2  a1 ) x  (3a3  a 2 ) x 2 
a1  a0  0 ,
2a2  a1  0 , 3a3  a2  0 ,
a0
a0
 a1
a2
a1  a0 , a2 
  , a3     ,
2
2!
3
3!

a0 (1) m x m
(  x) m
y ( x)   a m x  
 a0 
m!
m!
m 0
m 0
m 0


m
e
x

x2 x3
(  x) m
 1 x 
   
2! 3!
m!
m 0
am
(1) m a 0

m!
d2y
y0
2
dx

y   am x m
m0
y  c cos x  d sin x
Example

y    mam x

y    m(m  1)a m x m  2
m 1
m2
m 1

y    m(m  1)a m x m  2  (2  1)a 2  (3  2)a3 x  (4  3)a 4 x 2  
m2

  (m  2)( m  1)a m  2 x m


y   y   a m x   (m  2)( m  1)a m  2 x m
m
m 0
m 0
m 0

am  (m  2)( m  1)am 2  0
  a m  (m  2)( m  1)a m  2 x m
m 0
a0
a
a
a
a
, a 4   2   0 , a 6   4   0 ,
2 1
43
4!
65
6!
a
a
a
a
a
a
a3   1   1 , a5   3   1 , a 7   5   1 , 
3 2
3!
5 4
5!
76
7!
a2  

 

y  a 0  a 2 x 2  a3 x 4    a1  a3 x 3  a5 x 5  




x2
x4
x6
x3
x5
x7
 a 0 1 


   a1  x 


 
2!
4!
6!
3!
5!
7!




The Frobenius Method
y  p( x) y  q ( x) y  0 b( x)  xp( x), c( x)  x 2 q( x)
b( x )
c( x)
y 
y 
y 0
2
x
x
y ( x)  x (a 0  a1 x  a 2 x  a3 x  )  x
r
r: indicial parameter.
2
3
x 2 y  b0 xy  c0 y  0

r
m
a
x
 m
m 0
(Euler-Cauchy Eq.)
y  x r a 0  a1 x  
y   x r 1 ra0  (r  1)a1 x  
y   x r  2 r (r  1)a 0  (r  1)ra1 x  
x r r (r  1)a0  (r  1)ra1 x    b0 x r ra 0  (r  1)a1 x    c0 x r a0  a1 x    0
r (r  1)  b0 r  c0  0
(Indicial equation)
Examples
1
1
x y  xy   y  0
2
2
2
x y  xy  y  0
2
x y  xy  y  0
2
1
1
r (r  1)  r   0
2
2
r (r  1)  r  1  (r  1)2  0
r (r  1)  r  1  0
Solutions from Frobenius Method
y( x)  c1 y1 ( x)  c2 y2 ( x)
Case 1: distinct roots not differing by an integer
y1 ( x)  x r1 (a 0  a1 x  a 2 x 2  )
y2 ( x)  x r2 ( A0  A1 x  A2 x 2 
)
Case 2: Double root
y1 ( x)  x r (a0  a1 x  a2 x 2 
)
y2 ( x)  y1 ( x) ln x  x r 1 ( A0  A1 x  A2 x 2 
)
Case 3: roots differing by an integer
y1 ( x)  x (a 0  a1 x  a 2 x  )
r1
2
y2 ( x)  ky1 ( x) ln x  x r2 ( A0  A1 x  A2 x 2  )
r1  r2
Examples
y( x)  c1 y1 ( x)  c2 y2 ( x)
1
1
x y  xy 
y 0
2
2
2
1
1
r ( r  1)  r 
 0, r  1, 12
2
2
y1 ( x)  x(a0  a1 x  a2 x 2 
)
y2 ( x)  x1/ 2 ( A0  A1 x  A2 x 2 
x y  xy  y  0
2
)
r (r  1)  r  1  (r  1)  0, r  1,1
2
y1 ( x)  x(a0  a1 x  a2 x 
2
)
y2 ( x)  y1 ( x) ln x  x 2 ( A0  A1 x  A2 x 2 
x y  xy  y  0
2
)
r (r  1)  r  1  0, r  1, 1
y1 ( x)  x(a0  a1 x  a2 x 2 
)
y2 ( x)  ky1 ( x) ln x  x 1 ( A0  A1 x  A2 x 2 
)
The Legendre Equation
(1  x ) y  2 xy  l (l  1) y  0
2
y   am x m  a0  a1 x  a2 x 2  
m
a m 2
m(m  1)  l (l  1)
(l  m)(l  m  1)

am  
am
(m  1)( m  2)
(m  1)( m  2)
l (l  1)
a2  
a0
2
(l  1)(l  2)
a3  
a1
3!
(l  3)(l  1)(l  2)(l  4)
(l  2)(l  3)
(l  2)l (l  1)(l  3)
a1
a4  
a2  
a0 a5  
3 4
4!
5!
The power-series solution of the equation is therefore
y( x)  a0 y1 ( x)  a1 y 2 ( x)
y1 ( x)  1 
l (l  1) 2 (l  2)l (l  1)(l  3) 4
x 
x 
2!
4!
y 2 ( x)  x 
(l  1)(l  2) 3 (l  3)(l  1)(l  2)(l  4) 5
x 
x 
3!
5!
The Convergence Condition
am
m(m  1)  l (l  1)

 1 as m  
a m 2
(m  1)( m  2)
For -1<x<1, the series is convergent.
The Legendre Polynomials
Pl ( x) 
1  3  5 (2l  1)
l!


l (l  1) l  2 l (l  1)(l  2)(l  3) l  4
 x l 
x 
x  
2(2l  1)
2  4(2l  1)( 2l  3)


P0 ( x)  1
P1 ( x)  x
1 2
P2 ( x)  (3x  1)
2
1
P4 ( x)  (35 x 4  30 x 2  3)
8
1
P3 ( x)  (5 x 3  3x)
2
1
P5 ( x)  (65 x5  70 x3  15 x)
8
Example
Show that the Legendre function of order 3 satisfies the Legendre equation
1
y  P3 ( x)  (5 x 3  3 x)
2
3
y   (5 x 2  1)
2
y   15 x
(1  x 2 ) y   2 xy   l (l  1) y
3 2
1 3
2
 (1  x )  15x  2 x  (5x  1)  3  4  (5 x  3x)
2
2
 (15  15  30) x3  (15  3  18) x  0
The Recurrence Relation of
Legendre Functions
(l  1) Pl 1 ( x)  (2l  1) xPl ( x)  lPl 1 ( x)  0
Example
Use the recurrence relation to derive
P2 ( x), P3 ( x)
(l  1) Pl 1 ( x)  (2l  1) xPl ( x)  lPl 1 ( x)  0
Take l=1:
2 P2  3xP1  P0  0
1
1
P2  (3 xP1  P0 )  (3 x 2  1)
2
2
3P3  5 xP2  2 P1  0
1
1
1
1
2
P3  (5 xP2  2 P1 )  (5 x  (3x  1)  2 x)  (5 x 3  3x)
3
3
2
2
The Associated Legendre Functions
The associated Legendre equation
2


m
2
(1  x ) y   2 xy   l (l  1) 
y0
2 
(1  x ) 

Under conditions:
l  0,1,2,3,
m  0,1,2, ,l
The particular solutions are associated Legendre functions:
Pl ( x)  (1  x )
m
2
m 2
d
m
dx
m
Pl ( x)
Example
Use
Pl ( x)  (1  x )
m
2
m 2
d
m
dx
m
Pl ( x)
to derive
P3m ( x) for m  1, 2,3
x  cos
3
2

y  (5 x  1)
2
1
y  P3 ( x)  (5 x 3  3 x)
2
y   15
y   15 x
3
P ( x)  (1  x 2 )1 2 (5 x 2  1)
2
3
P (cos  )  sin  (5 cos 2   1)
2
P ( x)  15(1  x ) x
P32 (cos  )  15 sin 2  cos 
P ( x)  15(1  x )
P33 (cos  )  15 sin 3 
1
3
2
3
3
3
2
2 32
1
3
Orthogonality: The Legendre Functions

1
1
Pl ( x) Pl  ( x)dx  0 if l  l '
P1 ( x)  x
1 2
P2 ( x)  (3x  1)
2
1 3
P3 ( x)  (5 x  3x)
2
4
2 1
1
1 1 3
1  3x x 
1 P1 ( x) P2 ( x)dx  2 1 (3x  x)dx  2  4  2 
1
1  3 1   3 1 
         0
2  4 2   4 2 


1 1 4
1 5 3 1 1
2
1 P1( x) P3 ( x)dx  2 1 (5x  3x )dx  2 x  x 1  2 0  0  0
1
Orthogonality and Normalization:
The Associated Legendre Functions

1
1
Pl ( x) P ( x)dx  0
m
 
1
1
m
l
(l  l ')
2 (l  m )!
Pl ( x) dx 
 2l  1 (l  m )!
m

2
(2l  1) (l  m )! m
l ,m ( x) 
Pl ( x)
2 (l  m )!

l ,m ( x)l ,m ( x)dx   l ,l   
1
1
1
0
if l l 
if l  l 
The Hermite Equation
y  2 xy  2ny  0
Hermite polynomials:
H n ( x)  2 x 
n
y( x)  a0 y1 ( x)  a1 y 2 ( x)
n(n  1)
n(n  1)( n  2)( n  3)
n2

(2 x) 
(2 x) n  4  
1!
2!
H1 ( x)  2x
H 0 ( x)  1
H 2 ( x)  4 x 2  2
H 4 ( x)  16 x  48 x  12
4
2
H 3 ( x)  8 x 3  12 x
H 5 ( x)  32 x 5  160 x 3  120 x
The recurrence relation:
H n1 ( x)  2 xH n ( x)  2nH n1 ( x)  0
Hermite Functions


y  1  x 2  2n y  0
Its solution:
y ( x)  e
y  e
 ( x)
 x2 2
e
 x2 2
The Hermite functions:
x2 2

2



  2 x  (1  x )
   2x   2n   0
yn ( x)  e
 x2 2
H n ( x)
Orthogonality:




ym ( x) yn ( x)dx   e

 x2
H m ( x) H n ( x)dx  2 n!  m,n
n

Classroom Exercise
Write down the normalized Hermite functions:




ym ( x) yn ( x)dx   e H m ( x) H n ( x)dx  2n n!  m,n
 x2

Yn ( x) 
e
 x2 2
Hn ( x)
2n n! 
which satisfy orthonormal condition:



Ym ( x)Yn ( x)dx 
1
2n n! 



e H m ( x) H n ( x)dx   m,n
 x2
Example
Show that the Hermite function
e
x 2 2
H n (  x)
  km  2
is a solution of the Schrödinger equation for the harmonic oscillator
 2 d 2 1 2

 kx   E
2
2m dx
2
Let
z  x
d
dx
   (  x)   ( z )
d dz
d

 
  
dz dx
dz
d 2
  
2
dx
2
k 2

  
z   E
2m
2
   (1  z  2n)  0
2
1
if E  ( n  ) 
2
The Laguerre Equation
xy  1  xy  ny  0
n: real number
Laguerre polynomials:
2
2
2

n
n
(
n

1
)
n
n
n 1
n2
n 
Ln ( x)  (1)  x 
x 
x    (1) n!
1!
2!


L0 ( x)  1
L1 ( x)  1  x
L2 ( x)  2  4 x  x 2
L3 ( x)  6  18 x  9 x 2  x 3
Recurrence relation:
Ln 1 ( x)  (1  2n  x) Ln ( x)  n Ln 1 ( x)  0
2
Associated Laguerre Functions
The associated Laguerre equation
xy  m  1  x y  (n  m) y  0
It’s solution is associated Laguerre polynomials:
m
d
Lmn ( x)  m Ln ( x)
dx
they arise in the radial part of the wavefunctions of hydrogen atom in the form of
associated Laguerre functions:
 x 2 l 2l 1
n ,l
n l
which satisfy:
f ( x)  e
f  
xL
( x)
2
 n l (l  1) 1 
f  
 f 0
x
x
4
x
and are orthogonal with respect to the weight function x2 in the interval [0,∞]:


0

f n ,l ( x) f n,l ( x) x dx   e  x x 2l L2nll1 ( x )L2nl1l ( x ) x 2 dx
2
0
2n ( n  1)!

 n , n
( n  l  1)!
3
Bessel Functions
The Bessel equation:
x 2 y  xy  ( x 2  n 2 ) y  0
y  y / x  (1  n / x ) y  0
2
2
Therefore, it can be solved by Frobenius method.
y ( x)  x (a0  a1 x  a2 x 
r  n
r
2
)
2
4


x
x
n
y1 ( x)  a0 x 1 

 
 2(2n  2) 2  4(2n  2)( 2n  4)

2
4


x
x
n
y 2 ( x)  a0 x 1 

 
 2(2n  2) 2  4(2n  2)( 2n  4)

Bessel Functions for Integer n
Bessel functions of the first kind of order n:
 x
J n ( x)   
2
Examples:
(1)
 x
 

m  0 m!( n  m)!  2 
n 
2
4
2m
6
1  x
1  x
1  x
J 0 ( x)  1 




 
2 
2 
2 
(1!)  2  (2!)  2  (3!)  2 
3
J 1 ( x) 
5
7
x 1  x
1  x
1  x



 
 
  
2 1!2!  2 
2!3!  2  3!4!  2 
Zeros of Bessel Functions
J 0 ( x)  0
x  2.405, 5.520, 8.654, 11.792, 14.931, 
J1 ( x)  0
x  0, 3.832, 7.016, 10.173, 13.324, 
1
J0
J1
0
10
x
An Approximate Expression
2
n 
J n ( x) ~
sin( x 
 ) for large x
x
2 4
Bessel Functions of Half-Integer Order
Bessel functions of half-integral order can be expressed
in terms of elementary functions.
2
J 1 ( x) 
sin x
2
x
2
J  1 ( x) 
cos x
2
x
All others can be obtained with the recurrence relation:
Examples:
2n
J n 1 ( x) 
J n ( x)  J n 1 ( x)  0
x
J 3 2 ( x) 
J 3 2 ( x) 
1
J 1 2 ( x)  J 1 2 ( x)  0
x
1
2  sin x

J 1 2 ( x)  J 1 2 ( x) 

cos
x


x
x  x

1
2  cos x
J 3 2 ( x)   J 1 2 ( x)  J 1 2 ( x)  
 sin

x
x  x

x

Spherical Bessel Functions
j l ( x) 

2x
J l 1 2 ( x)
(l  0)
Spherical Neumann Functions
 l ( x)  (1)
l 1

2x
J l 1 2 ( x) (l  0)
These functions are important in treating scattering processes (which are
always useful in dynamics of molecules, atoms, nucleons and more
elementary particles ).
IV. Partial Differential Equations
2 f
1 2 f
 2 2
2
x
 t
1-D wave equation
2 f
1 f

2
D t
x
1-D diffusion equation
2 f 2 f 2 f
2




f 0
2
2
2
x
y
z
 f  g ( x, y, z )
2
3-D Laplace equation
3-D Poisson equation
2 2

   V ( x, y, z )  E Time-independent Schrödinger equation
2m
2 2


   V ( x, y, z )  i
2m
t
Time-dependent Schrödinger equation
General Solutions
 f
1  f
 2 2
2
x
 t
2
2
f
dF u dF


x du x du
f dF u
dF


t du t
du
f ( x , t )  F ( x  t )
2 f
d 2F

2
x
du 2
2
2 f
d
F
2


t 2
du 2
The general solution of 1D wave equation:
f ( x, t )  F ( x  t )  G ( x  t )
Yeah! Both F and G are arbitrary functions!
The general solution of an ODE contains an arbitrary constant, the general solution of
a PDE may contain a number of arbitrary functions.
Example
Verify that the function
f ( x, t )  3x 2  2 xt  3 2t 2
is a solution of the 1D wave equation.
f
 6 x  2t
x
2 f
6
2
x
f
 2 x  6 2 t
t
2 f
2

6

t 2
2 f
1 2 f
6
 2
2
x
υ t 2
The above solution can be written as
f ( x, t )  ( x  t ) 2  2( x  t ) 2  F ( x  t )  G( x  t )
Classroom Exercise
Verify that the function

f ( x, t )  a exp  b( x  t )
is a solution of the 1D wave equation.

 
d2 f
2
2

2
ab

1

2
b
(
x


t
)
exp

b
(
x


t
)
dx 2

 

d2 f
2
2
2

2
ab


1

2
b
(
x


t
)
exp

b
(
x


t
)
dt 2
d2 f
1 d2 f
 2
2
dx
 dt 2

2

Separation of Variables
f f

0
x y
f   XY 
dX

Y
x
x
dx
f ( x, y )  X ( x )  Y ( y )
f
dY
X
y
dy
dX ( x)
dY ( y )
Y ( y)
 X ( x)
0
dx
dy
1 dX ( x)
1 dY ( y )

X ( x) dx
Y ( y ) dy
dX
 CX
dx
 1 dX ( x)   1 dY ( y ) 
 X ( x) dx    Y ( y ) dy   0

 

1 dY
 C
Y dy
1 dX
C
X dx
dY
 CY
dy
X ( x)  AeCx Y ( y)  Be Cy
f ( x, y )  X ( x )  Y ( y )
 Ae  Be
Cx
 Cy
 De
C ( x y )
Motion in two and high dimensions
 2   2  2 

  E


2
2 

2m  x
y 
Separation of Variables
2

2m
  2
 2 


 x 2  y 2   E


x , y   Xx Yy 
 2   2 X ( x)Y ( y )
 2 X ( x)Y ( y ) 




 EX ( x)Y ( y )
2
2


2m 
x
y

 2   2 X ( x)Y ( y )  2 X ( x)Y ( y ) 




 EX ( x)Y ( y )
2
2


2m 
x
y

 1  2 X ( x) 1  2Y ( y) 
2 mE
 X ( x )




Y ( y)
2
2

2
x
y 

2 d2X

 ExX
2
2m dx
E  Ex  Ey
2
d 2Y

 E yY
2
2m dy
Reason?
A 2D problem reduced to two 1D problems!
Eigenfunctions of a particle on a
surface
 n x  
2 d2X

 ExX
2
2m dx
2
1
n  y 
2
d Y

 E yY
2
2m dy
 n 1 , n 2 x , y  
E  Ex  Ey
2
2
1
L1 L 2 2


2
L1 
1
2
sin
2
 L2 
1
2
n1x
L1
sin
n2 y
L2
n 1 x
n 2 y
sin
sin
L1
L2
E n1 , n 2
 n 12
n 22  h 2
  2  2 
L 2  8m
 L1
Some wavefunctions of a particle on a
surface
n1  1, n2  1
n1  1, n2  2
n1  2, n2  1 n1  2, n2  2
n1  1, n2  1
 1,1 x, y  
2
L1L2 
maximum :
x
L`
2
;y 
1
2
x
L1
L2 `
2
sin

x

2
L1
sin
y
L2
; Ly2 

2
n1  1, n2  2
x 2y
 1, 2 x, y  
sin sin
1
L1L2 2 L1 L2
2
maximum :
x
L`
2
;y 
minimum
x
L`
2
;y 
x
L1




2
;
2y
L2


;
2y
L2

3
2
2
L2 `
4
:
x
L1
3 L2 `
4
2
n1  2, n2  1
 2,1 x, y  
2
L1L2 
maximum :
x
L`
4
;y 
3 L`
4


2
; Ly2 

2
L2 `
4
:
2x
L1
;y 
L2 `
2
minimum
x
2x
L1
1
2
2x y
sin
sin
L1
L2

3
2
; Ly2 

2
n1  2, n2  2
 2 , 2  x, y  
L1L2 
1
2
maximum 1 :
x 
L`
4
;y 
maximum
x 
3 L`
4
2x
L1
3 L`
4
2:
;y 
x
L`
4
;y 

2
;
2y
L2


2
2x
L1

3
2
;
2y
L2

3
2
3 L2 `
4
;y 
minimum

L2 `
4
minimum 1 :
x
2x
2y
sin
sin
L1
L2
2
2x
L1

3
2
;
2y
L2


2
L2 `
4
2:
2x
L1
3 L2 `
4


2
;
2y
L2

3
2
Picture of the Interactions in an Atom
H 

2  2 
i
2 mi
i ( nuc, elec)


V
i
j ( nuc elec pair)
U ij
mn(elec  elec pair)
i
2
2
2
2
2
2



Ze
e



; Vj  
; U mn 
2
2
2
4

r
4 0 rmn
0 j
xi
yi
zi
Hamiltonian of Hydrogenic Atoms


 2  2 
2mi i
i ( nuc
,electwo
) terms
Only

Vi

jOnly
( nucone
electerm
pair)

U ij  E
mn(elec  elec pair)
2
2
2
2



Ze
i  2  2  2 ; V j  
4 0 r j
xi
yi
zi
2
2
e
U mn 
4 0 rmn
2
2
2



N  
 e 2  VN  e   Etotal
2m N
2 me
N
2
2
2
2
2
2
2
2









; e 


;
2
2
2
2
2
2
x N
y N
z N
xe
y e
z e
2
Ze
VN  e  
4 0 rN e
Magic! 玩個小魔術
2
2
2


H 
N 
 e 2  VN  e
2m N
2 me
N
2
2
2
2
2
2
2
2









; e 


;
2
2
2
2
2
2
x N
y N
z N
xe
y e
z e
2
Ze
VN  e  
4 0 rN e
2
2
2
Ze

H 
 
 H center of  mass
2
4 0 rN e

2
2
2
2






2
2
x
y
z 2
1  1  1
 me m N
Reduced mass
How a two-body problem is turned into a one-body
problem
Center of mass
m  mN  me ; 1  1  1
 me m N
mN
me
X
xN 
xe ; x  x N  xe
m
m
Classical Treatment
m  m1  m2
x1  X 
X 
m2
x
m
m1
m
X1  2 X 2
m
m
x2  X 
p12
p 22
E

V
2m1 2m2
m1m2

p1  m1 x  m1 X 
x
m
m1
x
m
m1m2

p2  m2 x 2  m2 X 
x
m
p12
p 22
1 2 1 2

 mX  x
2m1 2m2 2
2
E
P2
2m

p2
2
V
Free motion of the center of mass
One-body (with reduced mass) in a potential
Quantum Treatment
me
mN
x, xe  X 
x
m
m
mN
  
  X   x 
X x N
x x N
m X
x
xN  X 

x N
   X   x  me   
xe
X xe
x xe
m x
x
2
2
2
mN
2m N
m2 N
2
2





(

) 


m X
x
m
Xx
x 2 N
m 2 X 2
x 2
2
 2  ( me    ) 2  m e  2  2me
2  2
m x
x
m Xx
x 2 e
m 2 X 2
x 2
2
2  2
2  2
2
2
 2 mN

2 m N x 2 N
2 m 2 X 2
m Xx
2 m N x 2
 2   2 me
2  2
2  2
2
2
2 me x 2 e
2 m 2 X 2
m Xx
2 me x 2
2
 2   2 mN
 2   2 me
2
2
2

2 m N x 2 N
2 me x 2 e
2 m 2 X 2
2 m 2 X 2
2
 
2 me
2  2
2
2 m N x 2
x 2
2
2
2
2






2 m X 2
2  x 2
2
2
2



N 
e 2 
2m N
2me
2
2
2
2
2
2
2
2









(


)
( 2 

)
2
2
2
2
2m N x 2
2
m
e xe
y N
z N
ye
z e
N
2
2
2
2
2
2
2
2









(


)
( 2  2  2)
2
2
2m X 2
2

Y
Z
x
y
z
2
2
2



1 
 22 
2m1
2 m2
2
2
2
2
2
2
2
2









( 2  2  2 )
( 2
 2)
2
2m1 x
2m2 x
y1
z1
y 2
z 2
1
2
2
2
2
2
2
2
2
2









( 2  2  2)
( 2  2  2)
2m X
2  x
Y
Z
y
z
Motion of the center of mass
Relative motion
2
2
2



N  
 e 2   VN  e   Etotal
2m N
2 me
N
2
2
2
2
2
2
2
2









; e 


;
2
2
2
2
2
2
x N
y N
z N
xe
y e
z e
2
Ze
VN  e  
4 0 rN e
   c.m.
Etotal  E  Ec.m.
2
2
2
2





( 2  2  2 ) c.m.  Ec.m. c.m.
2m Free
X particle
Y sucks,
Z let’s forget it!
2
2
2
2





( 2  2  2 )  V  E
2  x
y
z
Two-body problemFree particle (Center of mass)
+ one-body problem
2
2
2
2
2



Ze


( 2  2  2 ) 
  E
2 x
4 0 rN e
y
z
 
2

2
r 2
 
2
2
x 2


2 
r r
2
y 2

2
1
sin2   2

1
r2
2
z 2

2
 sin1


sin 


 (r, , )  R(r )Y ( ,  )
2
2
2



2
1
 ( 2
 2  ) RY  VRY  ERY 
2 r
r r r
2
2


 [Y R2  2Y R  R2 2Y ]  VRY  ERY
2 r
r r r
Radial Wave Equation
2
2
2
2
2
2

R

R

r
2
r
1

[

  Y ]  Vr  Er
2
2  R r
R r Y
2

1 2Y  const

2 Y
2
2
2

R

R

r
2
r

[

]  Vr  Er  const
2
2  R r
R r
2
2
2
2
2

1


 Y  const  
[l (l  1)]
2 Y
2
2
2
2
2
2 2

R

R

r
2
r

[

]  Vr  Er 
l (l  1)
2
2  R r
R r
2
2
2


2 R ]  V R  ER

[ R

eff
2 r 2
r r
2
2
Ze

Veff  

l (l  1)
2
4 0 r 2r
Effective potential energy
2
2
Ze

Veff  
 2 l (l  1)
4 0 r 2r
Non-s orbitals
2
Veff |l  0   Ze
4 0 r
Very different close to the nucleus
but similar far from it
S orbitals
Solutions of wavefunction and energy
for the two cases are very different close
to the nucleus but similar to each other
at far distances.
Laguerre Equation and Laguerre Polynomials
2
2


2 R ]  V R  ER

[ R

eff
2 r 2
r r
2
2
Ze

Veff  

l (l  1)
2
4 0 r 2r
 l
Rn.l (r )  N n,l ( ) Ln,l (  )e   / 2n
n

2 Zr ; a 
0
a0
4 0  2
me e
Normalization factor
En,l  
2
Laguerre polynomails
Bohr radius=0.053 nm
Z 2 e 4
32
2 2 2 2
0  n
; n  1,2,3,...
Bound state

 l
 n,l ,ml (r, ,  )  Rn,l (r )Yl ,ml ( ,  )  N n,l ( ) Ln,l e 2nYl ,ml ( ,  )
n
Hydrogenic radial wavefunctions
Orbital
1s
2s
n
l
ml
1
0
2
0
Rnl
2(
Z 3 / 2  / 2
)
e
a0
1 ( Z ) 3 / 2 (2  1
2
2 2 a0
 )e   / 6
1 ( Z ) 3 / 2 e   / 4
4 6 a0
2p
2
1
3s
3
0
1 ( Z ) 3 / 2 (6  2   1
9
9 3 a0
1
1 ( Z ) 3 / 2 (4  1
3
27 6 a0
2
Z 3 / 2 2  / 6
1
(
)
 e
a
81 30
0
3p
3d
3
3
R n ,l ( r )  N n ,l (

n
l
) Ln ,l e

2n
 2 )e   / 2
 )e   / 6
The radial wavefunctions of the first few hydrogenic
atoms of atomic number Z Rnl
1s radial wavefunction
Z 3 / 2  / 2
2(
)
e
a0
2s radial wavefunction
1 ( Z ) 3 / 2 ( 2  1  )e   / 6
2
2 2 a0
3s radial wavefunction
1 ( Z ) 3 / 2 (6  2   1
9
9 3 a0
 2 )e   / 2
2p radial wavefunction
1 ( Z ) 3 / 2 e   / 4
4 6 a0
3p radial wavefunction
1 ( Z ) 3 / 2 ( 4  1  )e   / 6
3
27 6 a0
An Illustration
Calculate (1) the probability density for a 1selectron at the nucleus and (2) the probability
of finding a 2s-electron in a sphere with the
nucleus at the center and radius of 0.053 nm.

 l
 n,l ,ml (r, ,  )  Rn,l (r )Yl ,ml ( ,  )  N n,l ( ) Ln,l e 2nYl ,ml ( ,  )
n
For a 1s-electron: n=1,l=0,ml=0, the wavefunction is
 n,l ,ml (r, ,  )  R1,0 (r )Y0,0 ( ,  )
Probability density is
| 1,0,0 (r, , ) |2 | R1,0 (r )Y0,0 ( , ) |2
At the nucleus, r=0,   2Zr
a0
| 2(
Z 3/ 2  / 2 2 1
)
e
|
4
a0
3
Z
| 1,0,0 (0, ,  ) | 

a03
2
2.15 106 pm3 ( wit h Z  1)
 2,0,0 (r, ,  )  R2,0 (r )Y0,0 ( ,  )
For a 2s-electron,
Z 3/ 2
R20 
(
)
( 2  1  )e   / 6
2
2 2 a0
1
|  2,0,0 (r , ,  ) |2 | R2,0 (r ) |2 1
4
Z 3
1
1

( ) (4  2   1
4 8 a
4
0

 2 )e   / 3

0
n
x e
 ax
dx  ann!1
2 2
Z 3
1
4
Zr
Z
|  2,0,0 (r , ,  ) | 
( ) (4 
 2r )e 2 Zr / 3a0
32 a
a0
a0
0
2
The probability of finding a 2s-electron inside the sphere is
2

a0
 d  sin d 
0
0
0
a0
Z 3
1
r dr |  2,0,0 (r , ,  ) |  ( )
8 a
0
2
2

0
2 2
4
Zr
Z
r dr(4 
 2r )e  2Zr / 3a0
a0
a0
2
Structure of a Hydrogenic Atom
Principal quantum number n determines energy
En,l  
Z 2 e 4
32 2 02  2 n 2
; n  1,2,3,...
Orbital quantum number l gives the angular
momentum
| L | l (l  1) ; l  0,1,2, n  1
Magnetic quantum number ml gives the
“z”-component of angular momentum
Lz  ml ; ml  0,1,2m...  l
Energy Levels
 1
1

  RH  2  2
n2
 n1
~




unbound state=free state
H e
hc H 
2 2 2
32  0 
4
H 
H
me

me e 4

8 02 h 3 c
Ionization energy
Rydburg constant
Ionization energy
E1  hc H
I  hc H
For hydrogen atom, E1=13.6 eV
Ground state
Spectroscopic Measurement of Ionization
Energy
~
1 En
RH
  RH   I  2
n
12 hc
Shells and subshells
Principal quantum numbersshells
Orbital quantum numberssubshells
Shells and subshells
Atomic Orbitals:General
Considerations
1s 
1
(a03 )1 2
e  Z r a0
Appropriate balance between potential
and kinetic energy (c)
(a): electron tends to escape
(b): Electron tends to fall into the nucleus
Some typical atomic orbitals
1s 
1
(a03 )1 2
e  Z r a0
1 ( Z ) 3 / 2 ( 2  1  )e   / 6
2
2 2 a0
2a0
1
(2   )  0    4  r 
2
Z
Node(節點/節面)
Boundary Surface
The probability of finding the electron
inside the sphere is 90%
Mean radius of hydrogenic atoms
 r 

2


 
d
0

 *rd 
r |  |2 d

sin d
0


2
r 2 dr[rRnl
| Yl .ml |2 ] 
0


2
r 3 Rnl
dr
0
2
 sin d  dY
0
* '
l .m 'l ( ,  )Yl .ml ( ,  )   ll '  m m '
l
l
0

 r 

2
r 3 Rnl
dr
0
For 1s
R1,0 

 r 

0
r3
1
3
a0
1
(a03 )1 2
3a0
e  2 Z r a0 dr 
2Z
e  Z r a0
Radial Distribution Functions

P robability rr  dr 
2


 
d sin d
0
0
r  dr


r
P (r )dr
 *d 
voxel
r  dr
2

r

|  |2 d
voxel
2
r dr[Rnl
| Yl.ml |2 ] 
r  dr

2
r 2 Rnl
dr
r
2
2
P ( r )  r Rnl (r )
2
2
P ( r )  r Rnl (r )
For 1s,
R1,0 
1
(a03 )1 2
e  Z r a0
3
2 2Zr / a0
4
Z
P( r )  3 r e
a0
Its maximum:
3
dP( r )
4
Z
 0  3 d (r 2e 2Zr / a0 ) / dr  0
dr
a0
3
2
a
4
Z
2
Zr
 3 ( 2r 
)e  2 Zr / a0  0  r |P  max  0
a0
Z
a0
Most probable radius
The behavior differences between s, p, d and f
orbitals near the nucleus
s: has big probability
amplitude near the nucleus
p: probability amplitude ~r
near the nucleus
d: probability amplitude ~r
near the nucleus
2
f: probability amplitude ~r
near the nucleus
3
2p Orbitals
2 p z   2,1,0 (r , ,  )  R2,1(r )Y1,0 ( ,  ) 
1 ( Z )5 / 2 r cose  Zr / 2a0
4 2 a0
 r cosf (r )
2 p z  zf ( r )
Nodal plane
2 p 1   2,1,1 (r , ,  )  R2,1 (r )Y1,1 ( ,  )  
1
8 
( Z )5 / 2 r sin e  Zr / 2a0 e i
a0
  1 r sin e  i f (r )
2
2 px  
2 py 
1
( 2 p1  2 p1 )  r sin  cosf ( r )  xf ( r )
21 2
i
( 2 p1  2 p1 )  r sin  sin f ( r )  yf ( r )
21 2
3d Orbitals
d z 2  1 (3z 2  r 2 ) f (r )
2 3
d x 2  y 2  12 ( x 2  y 2 ) f (r )
d xy  xyf (r )
d zx  zxf (r )
d yz  yzf (r )
Where are the nodal planes?
Quiz
1. Write the recurrence relation for Legendre,
Hermite and Laguerre polynomials,
respectively.
2. Write the zeroth order and first order
Lagendre and Hermite polynomials.
3. Give the approximate expression of the
state of a 3p electron in an atom.