Motion of Two Bodies
Hˆ  
2
2m1
ˆ 2

1
2
2m2
w
ˆ 2  Vˆ (r , r )

2
1 2
Each type of motion is best represented
in its own coordinate system best
suited to solving the equations involved
Rotational Motion
Motion of the C.M.
2

ˆ 
ˆ
Hˆ 

2m
Cartesian
Translational Motion
Internal motion (w.r.t CM)
 ˆ ˆ
ˆ
H
L  L Internal
coordinates
2I
2

k
ˆ
ˆ
ˆ
H
   (r  req )2
2
2
18_12afig_PChem.jpg
2
r2
Center of Mass
r
k
r1
r
2
Vibrational Motion
Rc
r
1
Origin
Motion of Two Bodies
Hˆ  Hˆ (R)  Hˆ (r)  
Centre of Mass
2
2M
ˆ 2

R
2
2
ˆ 2  Vˆ (r)

r
Weighted average of all positions
m1r1  m2r2
R
  X , Y , Z  r1   x1 , y1 , z1  r2   x2 , y2 , z2 
m1  m2
m1 x1  m2 x2
X
m1  m2
Internal Coordinates:
r  r1  r2
r  ( x, y, z)
m1 y1  m2 y2
Y
m1  m2
Z
m1 z1  m2 z2
m1  m2
In C.M. Coordinates:
r1  r1  R & r2  r2  R
r  r1  R   r2  R   r1  r2
Kinetic Energy Terms
Kˆ  Kˆ (R)  Kˆ (r)  
2
2M
ˆ 2

R
2
2
ˆ 2

r
Tanslational Motion: In C.M. Coordinates:
Kˆ (R)  
2
2M
ˆ 2

R
 d2
d2
d2 

 2  2

2
2M  dX
dY
dZ 
2
?
?
?
Rotation and Vibration: Internal Coordinates:
2
2
2


d
d
d
2
ˆ
ˆ
K (r)  
r  
 2  2

2
2
2  dx dy dz 
2
2
?
?
?
Centre of Mass Coordinates
d
dX d
dx d


dx1 dx1 dX dx1 dx
m1
m1
dX
d  m1 x1  m2 x2 




dx1 dx1  m1  m2  m1  m2 M
dx
d

 x1  x2   1
dx1 dx1
m1 d
d
d


dx1 M dX dx
Centre of Mass Coordinates
d
dX d
dx d


dx2 dx2 dX dx2 dx
m2
m2
dX
d  m1 x1  m2 x2 




dx2 dx2  m1  m2  m1  m2 M
dx
d

 x1  x2   1
dx2 dx2
m2 d
d
d


dx2 M dX dx
Centre of Mass Coordinates
d 2  m1 d
d   m1 d
d 

 
 
2
dx1
 M dX dx   M dX dx 
m21 d 2
m1 d d
d m1 d
d2
 2


 2
2
M dX
M dX dx dx M dX dx
m21 d 2
2m1 d
d2
 2

 2
2
M dX
M dXdx dx
Similarly
m22 d 2
2m2 d
d2
d2
 2

 2
2
2
dx2
M dX
M dXdx dx
Centre of Mass Coordinates
2
2
2
2
2






1
d
1
d
1x
2x
ˆ
Kx   


 

2
2  m1
m2 
2  m1 dx1 m2 dx2 2 
2
m1 d 2
1 d2
2 d
1 d2
 2


2
2
m1 dx1
M dX
M dXdx m1 dx 2
m2 d 2
1 d2
2 d
1 d2
 2


2
2
m2 dx2
M dX
M dXdx m2 dx 2
m1 d 2
m2 d 2
1 d2
1 d2
1 d2
1 d2

 2
 2


2
2
2
2
2
m1 dx1 m2 dx2
M dX
M dX
m2 dx
m1 dx 2
Centre of Mass Coordinates
 m1
m2  d 2  1
1 d2
1 d2
1  d2

 2  2
   2
2
2
2
m1 dx1 m2 dx2
M  dX
M
 m1 m2  dx
1 d2
1 d2


2
M dX
 dx 2
m1m2
1
1



 m1 m2
m1  m2
1
Reduced mass
2
2


1
d
1
d
ˆ
 Kx   

2
2  M dX
 dx 2 
2
 Kˆ  
 1 2 1 2 ˆ
 R  r   KCM  Kˆ internal

2 M
 
2
2
2
2
d
d
d
 2R 
 2 2
2
dX
dY
dZ
2
2
2
d
d
d
r2  2  2  2
dx
dy
dz
Hamiltonian
Hˆ  Hˆ (R)  Hˆ (r)  
2
2M
ˆ 2

R
2
2
ˆ 2  Vˆ (r)

r
Hˆ  (R, r)   Hˆ (R)  Hˆ (r)  (R) (r)   ER  Er  (R) (r)
Hˆ (R ) (R )  ER (R )

2
2M
Separable!
C.M. Motion
3-D P.I.B
ˆ 2 (R )  E  (R)

R
R
Hˆ (r ) (r )  Er (r )

2
2
ˆ 2 (r )  Vˆ (r )  E  (r )

r
r
Internal Motion
Rotation
Vibration
Rotational Motion and Angular Momentum
ˆ  (r )  
K
2
2
r2 (r )
rotational motion requires internal coordinates
Linear momentum of a rotating Body
p   mi vi
p(t1)
p(t2)
i
ds
vi 
dt
d
vi  ri
dt
Ds  riD
d
w
dt
Angular Velocity
vi  riw
Parallel to moving body
p   mi riw
i
Always perpendicular to r
Always changing direction with time???
Ds
f
Angular Momentum
L  rp
L  r p
 r p sin f  r p
p
v
f
m
w
r
L
Perpendicular to R and p
Orientation remains constant with time
Rotational Motion and Angular Momentum
L   ri  p i
As p is always perpendicular to r
i
  mi ri w
2
 r  r  r  r  r
2
i
r
r1
R
r2
Center
of mass

2

   mi ri  w
 i

L  Iw
I   mi ri
2
i
Moment of inertia
Proxy for mass in rotational motion
Moment of Inertia and Internal Coordinates
I  m1 r1  m2 r2  m1 r1  R  m2 r2  R
2
2
2
2
m1r1  m2r2  1

m
m
M
r1  R  r1 
  m1r1  m2r1  m1r1  m2r2   2 r1  r2   2 r
M
M
M
M
M
r2  R 
m1
m1
r

r


r
 2 1
M
M
r  r1  r2  r1  r2
2
2
m  2
m  2
I  m1  2  r  m2  1  r
M 
M 
r
r1
R
r2
Center
of mass
1
1   m1m2  2
   
 r
 m1 m2   M 
2
1
2
2
    r
 
 r
2
Angular Momentum and Kinetic Energy
K .E.  
i

R
r2
Center
of mass
Classical Kinetic Energy
2mi
mi vi
2
i
r
r1
pi
2
2

i
mi  ri w 
2
2

2w
2 w
   mi ri 
r
2
 i
 2
2
2
L
Iw 

Iw



2
2I
2I
2
2
2
Rotational Motion and Angular Momentum
L
K .E. 
r
r1
R
2I
ˆ2
L
ˆ  (r )    (r )
K
2I
r2
ˆ  (r )  
K
Center
of mass
r  r
2
2
ˆ
L 
r 


I
2
2
2
2
1 ˆ2
ˆ
K L
2I
r2 (r )
r2  i 2
2
 i  r  r   i  r   r 
 L̂  i r   r
2
ˆ 
K
2
2
r2
 r  r   r  r 
Since r and p are
perpendicular
Momentum Summary
Classical
Linear
QM
dr
Momentum p  mv  m
dt
2
Energy
Rotational Momentum
(Angular)
Energy
pˆ  i 
2
p
md r
K

2m 2 dt 2
L  r p
1
2
K  r  p
2I
ˆ 
K
2
2
2
ˆ  i r  
L
ˆ 
K
2
2I
r  
2
Angular Momentum
L  rp
L  x
y
z    px
py
i
j
k
L x
px
y
py
z
pz
pz 
Angular Momentum
L  Lx i  Ly j  Lz k
Lx  ypz  zp y
Ly  zpx  xpz
Lz  xp y  ypx
Angular Momentum in QM
ˆ ˆ z  zp
ˆ ˆ y  (r )
Lx  Lˆx (r )   yp
ˆˆ z  (r )
ˆ ˆ x  xp
Ly  Lˆ y (r )   zp
ˆˆ y  yp
ˆ ˆ x  (r )
Lz  Lˆz (r )   xp
 d
d 
ˆ
Lx  i  y  z 
dy 
 dz
d
 d
ˆ
Ly  i  z  x 
dz 
 dx
 d
d 
ˆ
Lz  i  x  y 
dx 
 dy
ˆ  Lˆ i  Lˆ j  Lˆ k
L
x
y
z
Angular Momentum
ˆ
L  rˆ  pˆ  i rˆ  
L  x
y
d
z   i 
 dx
i
L  i x
d
dx
j
y
d
dy
k
z
d
dz
d
dy
d

dz 
Angular Momentum
ˆ
ˆ
ˆ
ˆ
L  Lx i  Ly j  Lz k
 d
d 
ˆ
Lx  i  y  z 
dy 
 dz
d
 d
ˆ
Ly  i  z  x 
dz 
 dx
 d
d 
ˆ
Lz  i  x  y 
dx 
 dy
Two-Dimensional Rotational Motion
x  r cos(f )
y  r sin(f )
d
d 
   i  j
 dx dy 
2
2


d
d
2
  2  2
dy 
 dx
Polar Coordinates
y
r
f
x
How to we get:
d

d
   (r ,f ) i  (r ,f ) j
dy
 dx

2
2


d
d
2
   2 (r ,f )  2 (r ,f ) 
dy
 dx

d dx d dy d
d
d


 cos(f )  sin(f )
dr dr dx dr dy
dx
dy
d
dx d dy d
d
d


 r sin(f )  r cos(f )
df df dx df dy
dx
dy
Two-Dimensional Rotational Motion
d
d
d
 cos(f )  sin(f )
dr
dx
dy
r
d
d
d
d
d
 r cos(f )  r sin(f )  x  y 
dr
dx
dy
dx
dy
1 d
x d
y d
 2
 2
r dr r dx r dy
Consider
1 d d x d
y d  d
d 
r  2
 2  x  y 
r dr dr  r dx r dy   dx
dy 
x d d
x d d
y d d
y d d
 2
x  2
y  2
x  2
y
r dx dx r dx dy r dy dx r dy dy
product rule
x
 2
r
product rule
d d  xy d d yx d d
y
 dx d
 dx dx  x dx dx   r 2 dx dy  r 2 dy dx  r 2
 dy d
d d 
 dy dy  y dy dy 


x d x 2 d 2 xy d
yx d
y d y2 d 2
 2
 2 2 2
 2
 2
 2 2
r dx r dx
r dxdy r dydx r dy r dy
Two-Dimensional Rotational Motion
d
dx d dy d
d
d
d
d


 r sin(f )  r cos(f )   y  x
df df dx df dy
dx
dy
dx
dy
Consider
d2 
d
d 
d
d 


y

x

y

x
df 2  dx
dy   dx
dy 
d d 
d d
d 
d 
d d
 y y  y  x   x  y   x x
dx dx
dy  dx 
dy dy
 dx dy 
product rule
product rule
2
d2
dx d
d d
d
d d
2 d
y
y
 yx
 x  xy
x
2
dx
dx dy
dx dy
dx
dy dx
dy 2
2
2
d2
d
d
d
d
2 d
y
 y  yx
 x  xy
x
dx 2
dy
dxdy
dx
dydx
dy 2
2
1 d2
y2 d 2
y d yx d
x d xy d
x2 d 2
 2 2 2
 2
 2
 2
 2 2
2
2
r df
r dx
r dy r dxdy r dx r dydx r dy
Two-Dimensional Rotational Motion
1 d d x2 d 2
x d xy d
yx d
y d y2 d 2
r  2 2 2
 2
 2
 2
 2 2
r dr dr r dx
r dx r dxdy r dydx r dy r dy
1 d2
y2 d 2
x d yx d
xy d
y d x2 d 2
 2 2 2
 2
 2
 2
 2 2
2
2
r df
r dx
r dx r dxdy r dydx r dy r dy
1 d d
1 d2
x2 d 2 y 2 d 2 x2 d 2 y 2 d 2
r  2
 2 2 2 2 2 2 2 2
2
r dr dr r df
r dx
r dx
r dy
r dy
2
1 2
d2 
2  d
 2  x  y   2  2 
r
dy 
 dx
2
2
d
d
 2  2  2
dx
dy
Two-Dimensional Rotational Motion
2
2
2
d 1 d
1 d
d
d
r
 2
 2 2
2
dr r dr
r df
dx
dy
2
d
1
d
1
d
2
 r
 2
dr r dr r df 2
2


d
1
d
1
d
2
ˆ
H
 
r
 2

2
2
2  dr r dr r df 
2
2
Two-Dimensional Rigid Rotor
ˆ  (r ,f )  
H
2
2
 2 (r ,f )  E (r ,f )
2


d
1
d
1
d
2
ˆ
H
 
r
 2

2
2  dr r dr r df 2 
2
2
Assume r is rigid, ie. it is constant
2
2
1
d
ˆ 
H
 2r  
2
2  r 2 df 2
ˆ2
L
ˆ  z
H
2I
2
d
ˆ
L z  i  r  i
df
As the system is rotating about the z-axis
Two-Dimensional Rigid Rotor
2
2
d
ˆ  (f )  
H
 (f )  E (f )
2
2 I df
 2 d2

  2 I df 2  E  (f )  0


 d 2 2I 
 df 2  2 E  (f )  0


 d2
2
 df 2  k  (r )  0


k2 
2I
18_05fig_PChem.jpg
2
k2 2
EE
2I
Two-Dimensional Rigid Rotor
 d2
2
 df 2  k  (f )  0


d
 d

 df  ik   df  ik  (f )  0



d
 df  ik


d
 (f )  0 &  df  ik


  (f )  N  e
 ikf
18_05fig_PChem.jpg

 (f )  0

  (f )  N  eikf
Two-Dimensional Rigid Rotor
  (f )  N  e
ikf
Periodic - Like a particle in a circular box
  (f  2 )  N  eik f 2     (f )  N  eikf
ik  2 
e
 1  2 k  2 m  k   m  k 
2
2
2
2
k
m
E

 Em
2I
2I
m = quantum number
 m (f )  N  e
imf
18_05fig_PChem.jpg
 m  (f )  N  eimf
Two-Dimensional Rigid Rotor
  m (f )  N  e
2 2
m
ˆ  (f ) 
H
  m (f )
m
2I
2
2
2
d
ˆ 
H
 2r  
2
2 I df 2
m2 2
Em 
2I
ˆ  (f )   m (f )
L
z m
m
d
ˆ
L z  i  r  i
df
ˆ 2 (f ) 
L
m
2
Lz m  m
m 2  m (f )
2
d
ˆ2   2
L
df 2
L2m  m2
18_05fig_PChem.jpg
2
 imf
Two-Dimensional Rigid Rotor
Hˆ   m (f )  Em (f )
m2 2
Em 
2I
m
E
 imf
Lˆz  m (f )   m  m (f )   m (f )  N  e
Lz m  m
 m E
 2
 
m I 
Lz m
6
 6
18.0
6
5
 5
12.5
5
4
 4
8.0
4
3
2
1
 3
 2
 1
4.5
2.0
0.5
3
2

Only 1 quantum number is require to determine the state of the system.
Normalization
2
 
2
(f )  m (f ) df  1
*
m
&
0
 
(f )   m (f ) df  1
*
m
0
 m (f )  A eimf
 m (f )   A e
*
imf
*
  A*e  imf
 m (f )  A eimf
 A*  A
 m (f )
*
  m (f )
Normalization
2
1
1
 
2
(f )   m (f ) df    m (f ) m (f ) df
*
m
0
0
2
2
 
(f )   m (f ) df    m (f ) m (f ) df
*
m
0
0
2
2
2
0
0
0
1  N  N   e imf e  imf df  N  N   eimf e imf df  N 2  1df  N 2 [2 ]
1
N
2
1 imf
1  imf
 m (f ) 
e
 m (f ) 
e
2
2
Orthogonality
2
 
(f )  m (f ) df   m,m
*
m
0
2
 1 imf 
0  2 e 
*
For m = m’
1
1 imf
e df 
2
2
1

2
For m ≠ m’
2
1
1
i mm f

e
d
f

2 0
2
2
2
 imf imf
e
 e df
0
2
2
0 1df  2  1
2
 cos( m  m f )  i sin( m  m f ) df
0
2
1
i


cos(
m

m
f
)
d
f

sin( m  m f ) df




2 0
2 0
1
i

0
0  0 since  m  m    0
2
2
18_06fig_PChem.jpg
Spherical Polar Coordinates
r  xi  y j zk
d
d
d
  i  j k
dx dy
dz
r  x ( r , , f ) i  y ( r ,  , f ) j
 z ( r , , f ) k
r  r sin cosf i  r sin sin f j  r cos k
d

d
d
   ( r , , f ) i  ( r ,  , f ) j  ( r ,  , f ) k 
dy
dz
 dx

14_01fig_PChem.jpg
?
Spherical Polar Coordinates
x  r sin cosf y  r sin sin f
d dx d dy d dz d



dr dr dx dr dy dr dz
z  r cos
d
d
 .... &
 ....
d
df
d
d
d
d
 cosf sin 
 sin f sin 
 cos
dr
dx
dy
dz
d
d
d
d
 r cos f cos
 r sin f cos
 r sin 
d
dx
dy
dz
d
d
d
 r sin f sin 
 r cosf sin 
df
dx
dy
14_01fig_PChem.jpg
The Gradient in Spherical Polar Coordinates
d 
 dr 
   cos f sin 
 d    r cos f cos
 d  
   r sin f sin 
d 
 df 
Gradient in Spherical Polar
coordinates expressed in
Cartesian Coordinates
sin f sin 
r sin f cos
r cos f sin 
d 
 dx 
cos   
d 


r sin 
  dy 
0   
d 
 dz 
 SP  W Cart .
14_01fig_PChem.jpg
The Gradient in Spherical Polar Coordinates
1
Gradient in Cartesian
coordinates expressed in
Spherical Polar Coordinates
Cart
Cart .  W  SP
d  
 dx  cos f sin 
  
d  


 sin f sin 
 dy  
  
 d   cos
 dz  
cos f cos
r
sin f cos
r
sin 

r
14_01fig_PChem.jpg
sin f   d 

r sin    dr 
 
cos f   d 
r sin    d 
 d 
0  
  df 
The Gradient in Spherical Polar Coordinates
ˆ  i r  
L
Cart .
Cart .
 d  cos f sin 
 dx  
  
d

     sin f sin 
 dy 
  
d  
 dz  
d cos f cos d
sin f d 


dr
r
d r sin  df 

d sin f cos d
cos f d 
1



W
 SP

dr
r
d r sin  df

d sin  d

cos


dr
r d
1
ˆ
L  i r  W  SP
14_01fig_PChem.jpg
The Gradient in Spherical Polar Coordinates
ˆ  i r  
L

cos f sin 
 r cos(f )sin( )  

 i  r sin(f )sin( )    sin f sin 
 r cos( )  


ˆ  Lˆ i  Lˆ j  Lˆ k
L
x
y
z
d cos f cos  d
sin f d 


dr
r
d r sin  df 

d sin f cos  d
cos f d 


dr
r
d r sin  df 

d sin  d

cos 


dr
r d

d cos
d 
ˆ
Lx  i   sin f

cos f

d

sin

d
f



d cos
d 
ˆ
Ly  i cos f

sin f

d

sin

d
f


d
ˆ
Lz  i
df
14_01fig_PChem.jpg
The Laplacian in Spherical Polar Coordinates

2
Cart
  W  SP    W  SP 
1
1
2
1
d
d
1
d
d
1
d
2  2
r2
 2
sin 
 2 2
r dr dr r sin  d
d r sin  df 2
Radial Term
OR
Angular Terms
2
1
d
2 
r  .....
2
r dr
OR
14_01fig_PChem.jpg
2
d
2 d
2
  2
 ....
dr
r dr
Three-Dimensional Rigid Rotor
2
2
2


d
d
d
2
ˆ
H
Cart .  
 2  2

2
2
2  dx
dy
dz 
2
ˆ 
H
2
2
 2SP
2
1 d d
1
d
d
1
d2 

r  2
sin 
 2 2

2  r dr dr r sin  d
d r sin  df 2 
2
Assume r is rigid, ie. it is constant. Then all energy is from rotational motion only.
2

 L2
1
d
d
1
d
ˆ
H
sin 
 2
2 
2 r  sin  d
d sin  df 2   2I
2
2


1
d
d
1
d
2
2
ˆ
L  
sin 
 2
2
sin

d

d

sin

d
f


Three-Dimensional Rigid Rotor
2


1
d
d
1
d
ˆ
H ( ,f )   
sin 
 2
 ( ,f )  E ( ,f )
2
2 I  sin  d
d sin  df 
2
 1 d
d
1 d2 
 
sin 
 2
( )(f )  E( )(f )
2
2 I  sin  d
d sin  df 
Separable?
2
d
d
d2
sin 
sin 
( )(f )  2 ( )(f )    sin 2 ( )(f )
d
d
df

2 IE
2
E
18_05fig_PChem.jpg
2
2I
Three-Dimensional Rigid Rotor
d
d
d2
(f )sin 
sin 
( )  ( ) 2 (f )    sin 2 ( )(f )
d
d
df
2
1
d
d
1
d
sin 
sin 
( )   sin 2   
 (f )
2
( )
d
d
 (f ) df
Two separate
independent
equations
1 d2
2

(
f
)


k
 (f ) df 2
k2= Separation
Constant
1
d
d
sin 
sin 
( )   sin 2   k 2
( )
d
d
Three-Dimensional Rigid Rotor
1 d2
2
 (f )   k
2
 (f ) df
d
2
 df 2  k  (f )  0


2
k  m  k 
Recall 2D Rigid Rotor
 m (f ) 
1 imf
e
2
 m (f ) 
18_05fig_PChem.jpg
1  imf
e
2
Three-Dimensional Rigid Rotor
1
d
d
sin 
sin 
( )   sin 2   m2
( )
d
d
d
d
sin 
sin 
( )   sin 2 ( )  m 2( )
d
d
d
d

2
2
sin

sin



sin


m
( )  0


d
d

This equation can be solving using a series expansion, using a Fourier Series:
l ,m ( )  C P
m m
l
l
Where
 cos( ) 
  l (l  1)
18_05fig_PChem.jpg
Legendre polynomials
El  
2
2I

2
l (l  1)
2I
Three-Dimensional Rigid Rotor
ˆ  ( ) (f )  E  ( ) (f )
H
l ,m
m
l l ,m
m
l ,m ( )  Clm Pl (cos( ))
m
 m (f ) 
El 
1 imf
e
2
 1

m
ˆ
H
Clm Pl (cos( ))eimf  
 2

2
2
l (l  1)
2I
l (l  1)  1
m m
imf 
C
P
(cos(

))
e
l
l

2 I  2

ˆ Y ( ,f )  E Y ( ,f )
H
l ,m
l l ,m
1
m
Yl ,m ( ,f ) 
Clm Pl (cos( )) eimf
2
Spherical Harmonics
The Spherical Harmonics
1
m
Yl ,m ( ,f ) 
Clm Pl (cos( )) eimf
2
1/2
 (2l  1)(l  m )!
Clm  (1) m 

2(
l

m
)!


m  l , l  1,..., l  1, l
1/2
for m  0
 (2l  1)(l  m )!
Clm  

2(
l

m
)!


for m  0
m
 d

Pl (cos )  sin  
Pl (cos )

 d (cos ) 
m
m
l

1  d
2
l
Pl (cos )  l 
(cos


1)
2 l !  d (cos ) 
0
For l=0, m=0
Pl (cos ) 

1  d
2
0
(cos


1)
1


0
2 0!  d (cos ) 
0
 d

0
P0 (cos )  sin 0  
11

 d (cos ) 
The Spherical Harmonics
1
m
Yl ,m ( ,f ) 
Clm Pl (cos( )) eimf
2
Y
0,0
For l = 0, m = 0
P00 (cos( ))  1

1
2 
 r  ro ,  (0,  ), f  (0,2 )
Everywhere on the surface of
the sphere has value 1
2 
1/2
 (2*0  1)(0  0)!
Clm  (1)0 

2(0)!



1
2
1 1
1
i0
Y0,0 ( ,f ) 
(1) e 

2 2
2 
what is ro ?
r = (ro, , f
The Spherical Harmonics
Normalization:
 Y  Y
*
0,0
S
0,0
x f  ro y f  f ( x j ) z f  g ( x f , y f )

dV 


xi  ro yi  f ( xi ) yi  g ( xi , yi )
*
 1  1
dxdydz  1


2   2 
In Spherical Polar Coordinates dV  dxdydz
 2
as ro  1
 ro2 sin( ) d df  sin( ) d df
*
 1  1
0 0  2   2  sin( ) d df 

0 0

Z
 2
2
1
1
sin( ) d df 
sin  d  df

4
4 0
0

Y
1
1

2

cos

f

2(2 )  1

0  0
4
4
The wavefunction is an angular
function which has a constant value
over the entire unit
r = (1, , f
circle.
X
1
2 
The Spherical Harmonics


For l =1, m = 0 P1 (cos )  11  d  (cos 2   1)1  cos
2 1!  d (cos ) 
0
 d

0
P1 (cos )  sin 0  
cos  cos

 d (cos ) 
Z
Y1,0 ( ,f ) 
1
3
3
i 0f

1
cos(

)
e


cos( ) Along z-axis
 
2
2
2 
r = (1, , f
Y1,0 ( )  
3
cos( )
2 
   (0,  ), f  (0, 2 )
Y
X
The wavefunction is an angular
function which has a value varying  3 cos( )
2 
as on the entire unit circle.
The spherical Harmonics
are often plotted as a
vector starting from the
origin with orientation 
and f and its length is
Y(,f)
The Spherical Harmonics
For l=1, m =±1 P 1 (cos )  sin1  
1

1

d
d
P
(cos

)

sin

cos  sin 
 1
d
(cos

)
d
(cos

)


Y1,1 ( ,f ) 
3
sin( ) e  if
2 2
Complex Valued??
1
Y1,1 ( ,f )  Y1,1 ( ,f )  
2
3
sin( ) e  if  e  if  
4 2
3
sin( )cos(f ) 
2 2
Along x-axis
1
Y1,1 ( ,f )  Y1,1 ( ,f )  
2i
3
sin( ) e  if  e  if  
4i 2
3
sin( )sin(f ) 
2 2
Along y-axis
18_05fig_PChem.jpg
The Spherical Harmonics
P20 (cos( )) 
1
3cos 2 ( )  1

2
P21 (cos( ))  3sin( )cos ( )
Y2,1 ( ,f ) 
Y2,1  Y2,1
2

15
sin  cos e  if
2 2
15
sin  cos f cos
8
XZ
Y2,0 ( ,f ) 
5
4 
(3cos 2 ( )  1)
P22 (cos( ))  3sin 2 ( )
Y2,2 ( ,f ) 
Y2,1  Y2,1
2i
18_05fig_PChem.jpg
15
sin 2  e 2if
4 2

15
sin  sin f cos
8
YZ
The Spherical Harmonics Are Orthonormal
 2
*
Y
  l ,mYl,m sin( ) d df   l ,m,l,m
0 0
 2

0 0
1
m
Clm  Pl (cos ) eimf
2
1
m
Clm Pl (cos ) eimf sin  d df
2

2
1
 imf imf
m m
m m

e
e
d
f
C
P
(cos

)
C
Pl (cos ) sin  d


l
l
l


2 0
0
Clm Clm
  m,m
2
Clm Clm
 When m  m
2
Example
 Pl (cos ) Pl (cos ) sin  d
m
m
0
 Pl (cos ) Pl (cos ) sin  d   l,l
m
m
0
P10 (cos( ))  cos( )
P00 (cos( ))  1
C10C11
2
2
2
2
C00C11 2
0 P0 (cos ) P1 (cos ) sin  d  2 0 cos sin  d  0
ODD
0
0
Yl,m are Eigenfuncions of H, L2, Lz
1
m
Yl ,m ( ,f ) 
Clm Pl (cos( )) eimf
2
ˆ Y ( ,f ) 
H
l ,m
2
2I
l (l  1)Yl ,m ( ,f )
El  
2


1
d
d
1
d
ˆ
H 
sin 
 2
2 I  sin  d
d sin  df 2 
2
2I
l (l  1)
2
ˆ 2Y ( ,f ) 
L
l ,m
2
l (l  1)Yl ,m ( ,f )
2


1
d
d
1
d
2
2
ˆ
L  
sin 
 2
2
sin

d

d

sin

d
f


Ll 2 
Ll 
2
l (l  1)
l (l  1)
ˆ Y ( ,f )  m Y ( ,f )
L
z l ,m
l ,m
d
ˆ
L z  i
df
Lz  m
Dirac Notation
Continuous Functions
   j ds  i, j
*
i
S
N
 i   ci ,mfm
m
*
fm 

f
f




m
m
m
,
m

S
Vectors
v  v j  i, j
*
i
Dirac
v i  ci*,1
ci*,2
ci*, N 1
ci*, N 
is complete
 c j ,1 


c
 j ,2 

vj  


 c j , N 1 
 c j ,N 


*

 i  j ds   i |  j   i , j
 i  vi
Bra
S
j  vj
Ket
 v i  v j   ij
ˆ  *O
ˆ  ds   O
ˆ   O
ˆ 
O
i
j
i
j
 i j
S
ˆ |  E 
H
j
j
j
Dirac Notation
ˆ  Y |H
ˆ |Y 
H
l ,m
l ,m
2 
*
ˆ Y ( ,f )sin  d df
Y
(

,
f
)
H
l
,
m
l ,m

0 0
2 


0 0
2 
1
m
ˆ
Clm Pl (cos( )) e  imf H
2
1
m
Clm Pl (cos( )) eimf sin  d df
2
 1 d
d
1 d2 
 sin  d sin  d  sin 2  df 2 


 1
m m
imf 
C
P
(cos(

))
e
l
l

 sin  d df
 2

2
 1

m
  
Clm Pl (cos( )) e  imf  
 2I
0 0  2
Dirac Notation
Example
ˆ  Y |H
ˆ |Y 
H
1,1
1,1
2 
*
ˆ Y ( ,f )sin  d df
Y
(

,
f
)
H
1,1
1,1

0 0
2
ˆ Y ( ,f )  E Y ( ,f ) 
H
l ,m
l l ,m
ˆY 
H
1,1
2
2I
1(1  1) Y1,1 

2I
l (l  1)Yl ,m ( ,f )
2
I
Y1,1
2
Y1,1 |
I
Y1,1 
ˆ |Y  Y | H
ˆ |Y  Y | H
ˆ |Y
Y1,1 | H
1,1
1,0
1,0
1,1
1,1 
2
I
Y1,1 | Y1,1 
2
Degenerate
I
2
I
Dirac Notation
Example
ˆ2 
L
2 
*
ˆ 2 Y ( ,f )sin  d df  Y | L
ˆ2 | Y
Y
(

,
f
)
L
1,1
1,1
1,1
  1,1
0 0
ˆ 2Y
 Y1,1 | L
1,1
ˆ 2Y ( ,f ) 
L
l ,m
ˆ2Y 
L
1,1
2
2
l (l  1)Yl ,m ( ,f )
1(1  1) Y1,1  2
2
Y1,1 
 Y1,1 | 2 2Y1,1
2
2
Y1,1 | Y1,1  2
2
L 2
ˆ 2Y
Y1,1 | L
1,1  2
2
ˆ 2Y
Y2,1 | L
2,1  2
2
ˆ 2Y
Y2,1 | L
1,1  ?
Dirac Notation
Example
ˆ 
L
z
2 
*
ˆ Y ( ,f )sin  d df  Y | L
ˆ |Y
Y
(

,
f
)
L
z 1,1
1,1
z
1,1
  1,1
0 0
ˆ Y
 Y1,1 | L
z 1,1
ˆ Y ( ,f )  m Y ( ,f )
L
z l ,m
l ,m
ˆ Y 
L
z 1,1
Y1,1 
 Y1,1 | Y1,1

ˆ |Y  0
Y1,0 | L
z
1,0
ˆ |Y
Y1,1 | L
z
1,1  
Y1,1 | Y1,1 
ˆ |Y  ?
Y1,0 | L
z
1,1
3-D Rotational motion & The Angular Momentum Vector
Rotational motion is quantized not continuous. Only
certain states of motion are allowed that are determined
by quantum numbers l and m.
Lz  m
L
l determines the
L
l (l  1)
length of the angular
momentum vector
m
18_16fig_PChem.jpg
indicates the
orientation of the angular
momentum with respect to
z-axis
E
Three-Dimensional Rigid Rotor States
El 
l
2
2I
l (l  1)
m
32
3
2
2
1
0
l
 m ,..,m
Y
Ll 
l (l  1)
 2
 
l I 
Ll Lz m
E
Lz  m
3
2
10
-1 -2
-3
3
6.0
Y3,2,1,0,
1, 2, 3
12

2
2
0
2
10
-1 -2
1 0
-1
0
2
2,1,0,1,2
Y
1
Y1,0,
1
0
Y0
3.0
6

2
1.0
0.5
2

0
0
0
Only 2 quantum numbers are require to determine the state of the system.
Rotational Spectroscopy
19_01tbl_PChem.jpg
Rotational Spectroscopy
EJ 
2
2 r
2
o
J ( J  1)
EJ 
2
2I
J ( J  1)
J : Rotational quantum number
DE  EJ 1  EJ

2
2I
( J  1)( J  2)  J ( J  1)
DE 
19_13fig_PChem.jpg
2
I
 J  1
Rotational Spectroscopy
DE  h 
hc

 hc
Wavenumber (cm-1)
h( J  1)

4 2 Ic
  2 B( J  1)
h
8 2  ro2 c
Rotational Constant
Line spacing
Dv  cD  c  J 1  J 
B
 c  2 B( J  1  1)  2 B( J  1)   2cB
v
Frequency (v)
Dv
Rotational Spectroscopy
Predict the line spacing for the 16O1H radical.
r = 0.97 A = 9.7 x 10-11 m
mO = 15.994 amu = 2.656 x 10-26 kg
mH = 1.008 amu = 1.673 x 10-26 kg
1 amu = 1 g/mol = (0.001 kg/mol)/6.022 x 10-23 mol-1
= 1.661 x 10-23 kg
1


B
1
1
27




1.774

10
kg
26
27
2.656 10 kg 1.673 10 kg
h
8 2  ro2 c

8 1.774 10
2
6.626 1034 kgm 2 / s
27
kg  9.7 10
11
m  (2.99 108 m / s )
2
 1890m 1
Dv  2cB  2  (2.99 108 m / s)  (1890 m1 )  1.13 1012 Hz
Dv  2B  2  (1890 m1 )  3780 m1 37.8 cm1
Rotational Spectroscopy
The line spacing for 1H35Cl is 21.19 cm-1,
determine its bond length .
mCl = 34.698 amu = 5.807 x 10-26 kg
mH = 1.008 amu = 1.673 x 10-26 kg
1


1
1
27




1.626

10
kg
5.807 1026 kg 1.673 1027 kg
Dv (21.19 cm 1 )(100 cm / m)
B

 1059.5 m1
2
2
B
h
8 2  ro2 c
 ro 
h
B8 2  c
6.626 1034 kgm2 / s
10


1.257

10
m  1.257 A
1
2
27
8
(1059.5 m )8 1.626 10 kg  (2.99 10 m / s)
The Transverse Components of Angular Momentum
ˆ  i r    Lˆ i  Lˆ j  Lˆ k
L
x
y
z
Lz  m
L
L
l (l  1)
ˆL2  Lˆ2  Lˆ2  Lˆ2
x
y
z
 d
d 
ˆ
Lx  i  y  z 
dy 
 dz
d
 d
ˆ
Ly  i  z  x 
dz 
 dx
 d
d 
ˆ
Lz  i  x  y 
dx 
 dy
Yl ,m ( ,f )
L2  l (l  1)
?
?
Lz  m
Ylm are eigenfunctions of L2 and Lz but not of Lx and Ly
Therefore Lx and Ly do not commute with either L2 or Lz!!!
Commutation of Angular Momentum Components
 Lˆx , Lˆ y   Lˆx Lˆ y  Lˆ y Lˆx



2
 d
d  d
d  d
d  d
d 
 y  z   z  x    z  x   y  z  
dy   dx
dz   dx
dz   dz
dy  
  dz
FOIL
 d
d  d
d
d d
d d
d d
d d
y

z
z

x

y
z

y
x

z
z

z
x
 dz



dy   dx
dz 
dz dx
dz dz
dy dx
dy dz

product rule
d d
d d
d d
d d
 dz d
 y
z

yx

zz

zx
dz dx 
dz dz
dy dx
dy dz
 dz dx
d
d
d2
d
d
 y  yz
 yx 2  z 2
 zx
dx
dzdx
dz
dydx
dydz
Commutation of Angular Momentum Components
FOIL
d  d
d 
d d
d d
d d
d d
 d
 z  x   y  z   z y  z z  x y  x z
dz   dz
dy 
dx dz
dx dy
dz dz
dz dy
 dx
product rule
 dz d
d d
d d
d d
d d 
  zy
 zz
 xy
 x
z

dx dz
dx dy
dz dz
dz
dy
dz
dy


d
d
d2
d
d
2
  zy
z
 xy 2  x  xz
dxdz
dxdy
dz
dy
dzdy
Commutation of Angular Momentum Components
 d
d
d2
d
d 
2
 y dx  yz dzdx  yx dz 2  z dydx  zx dydz 

 Lˆx , Lˆ y    2 

2


 zy d  z 2 d  xy d  x d  xz d 

dxdz
dxdy
dz 2
dy
dzdy 

2
  d
 d
d
d 
 y  x   i  ih  y  x  
dy 
dy  
 dx
  dx

 d
d 
 i  ih  x  y    i Lˆz
dx  
 dy

 Lˆ y , Lˆx   Lˆ y Lˆx  Lˆx Lˆ y  i Lˆz


Commutation of Angular Momentum Components
 Lˆx , Lˆz   Lˆx Lˆz  Lˆz Lˆx


 d
d  d
d  d
d  d
d 
  2   y  z   x  y    x  y   y  z    i Lˆ y
dy   dy
dx   dy
dx   dz
dy  
  dz
 Lˆz , Lˆx   Lˆz Lˆx  Lˆx Lˆz  i Lˆ y


 Lˆ y , Lˆz   Lˆ y Lˆz  Lˆz Lˆ y



2
 d
d  d
d  d
d  d
d 
  z  x   y  z    y  z   z  x    i Lˆx
dz   dz
dy   dz
dy   dx
dz  
  dx
 Lˆz , Lˆ y   Lˆz Lˆ y  Lˆ y Lˆz  i Lˆx


Cyclic Commutation of Angular Momentum
 Lˆx , Lˆ y   i Lˆz


 Lˆ y , Lˆx   i Lˆz


 Lˆ y , Lˆz   i Lˆx


 Lˆz , Lˆ y   i Lˆx


 Lˆz , Lˆx   i Lˆ y


Lˆ z
 Lˆ y , Lˆx   i Lˆz


 Lˆx , Lˆz   i Lˆ y


ˆL
x
i
 Lˆx , Lˆ y   i Lˆz


ˆL
x
Lˆ y
Lˆ z
i
Lˆ y
Commutation with Total Angular Momentum
 Lˆ2 , Lˆz    Lˆ2x , Lˆz    Lˆ2y , Lˆz    Lˆ2z , Lˆz 

 
 
 

 Lˆ2x , Lˆz   Lˆ2x Lˆz  Lˆz Lˆ2x  Lˆx Lˆx Lˆz  Lˆz Lˆx Lˆx


Recall  Lˆx , Lˆz   Lˆx Lˆz  Lˆz Lˆx
 Lˆx Lˆz   Lˆx , Lˆz   Lˆz Lˆx
 Lˆz Lˆx  Lˆx Lˆz   Lˆx , Lˆz 

 

 Lˆx  Lˆx , Lˆz   Lˆz Lˆ x  Lˆ x Lˆ z   Lˆ x , Lˆ z  Lˆ x
 Lˆx  Lˆx , Lˆz   Lˆx Lˆz Lˆx  Lˆx Lˆz Lˆx   Lˆx , Lˆz  Lˆx
 i Lˆx Lˆ y  i Lˆ y Lˆx
Commutation with Total Angular Momentum
 Lˆ2y , Lˆz   Lˆ2y Lˆz  Lˆz Lˆ2y  Lˆ Lˆ Lˆ  Lˆ Lˆ Lˆ
y y z
z y y


Recall
 Lˆ y , Lˆz   Lˆ y Lˆz  Lˆz Lˆ y


 Lˆ y Lˆz   Lˆx , Lˆz   Lˆz Lˆ y
 Lˆz Lˆ y  Lˆz Lˆ y   Lˆx , Lˆz 

 

 Lˆ y  Lˆ y , Lˆz   Lˆz Lˆ y  Lˆ y Lˆz   Lˆ y , Lˆz  Lˆ y
 Lˆ y  Lˆ y , Lˆz   Lˆ y Lˆz Lˆ y  Lˆ y Lˆz Lˆ y   Lˆ y , Lˆz  Lˆ y
 i Lˆ y Lˆx  i Lˆx Lˆ y
Commutation with Total Angular Momentum
 Lˆ2z , Lˆz   Lˆ2z Lˆz  Lˆz Lˆ2z  Lˆz Lˆz Lˆ z  Lˆ z Lˆ z Lˆ z  0


 Lˆ2 , Lˆz    Lˆ2x , Lˆz    Lˆ2y , Lˆz    Lˆ2z , Lˆz 

 
 
 

 i Lˆx Lˆ y  i Lˆ y Lˆx  i Lˆ y Lˆx  i Lˆx Lˆ y  0
  Lˆ2 , Lˆz   0
Therefore they have simultaneous
eigen functions,
Yl,m
Also note that:
 Lˆ2 , Lˆx   0  Lˆ2 , Lˆ y   0




Therefore the transverse components do not
share the same eigen function as L2 and Lz.
This means that only any one component of angular momentum can be
determined at one time.
Ladder Operators
Consider:
Lˆ  Lˆx  iLˆ y
Lˆ  Lˆx  iLˆ y
Note:  Lˆz , Lˆ    Lˆz , Lˆx   i  Lˆz , Lˆ y 
 i Lˆ y  i 2 Lˆx  i Lˆ y  Lˆx
  Lˆx  iLˆ y   Lˆ
 Lˆz , Lˆ    Lˆz , Lˆx   i  Lˆz , Lˆ y 

 
 

 i Lˆ y  i(i ) Lˆx i Lˆ y  Lˆx
   Lˆx  iLˆ y    Lˆ
ˆ ,L
ˆ  L
ˆ
L

 z 
Like an eigen equation
but for an operator!
ˆˆ  ˆ 
L z L    Lˆ 
Super operator
ˆˆ  ˆ   ˆ ˆ 
L z O    L z , O 
ˆˆ  ˆ 
L z L     Lˆ 
Ladder Operators
What do these ladder operators actually do???
Lˆ Yl ,m  Lˆx Yl ,m  iLˆ y Yl ,m
Recall That:
 Lˆz , Lˆ   Lˆ


?
?
 Lˆz Lˆ  Lˆ Lˆ z  Lˆ
Lˆ Yl ,m   Lˆz Lˆ  Lˆ Lˆz  Yl ,m
 Lˆz Lˆ Yl ,m  Lˆ Lˆz Yl ,m
 Lˆz Lˆ Yl ,m  m Lˆ Yl ,m
 Lˆ Yl ,m  Lˆz Lˆ Yl ,m  m Lˆ Yl ,m
 Lˆz  Lˆ Yl ,m   (m  1)  Lˆ Yl ,m 
Lˆz Yl,m1  (m  1) Yl,m1  Lˆ Yl ,m  Yl,m1
Similarly
 Lˆ Yl ,m  Yl,m1
Raising Operator
Lowering Operator
Ladder Operators
Note:
Similarly:
Consider:
 Lˆ2 , Lˆ    Lˆ2 , Lˆx   i  Lˆ2 , Lˆ y   0  i0  0

 
 

 Lˆ2 , Lˆ   0


 Lˆ2 , Lˆ  Yl ,m  0


  Lˆ2 Lˆ  Lˆ Lˆ2  Yl ,m
 Lˆ2 Lˆ Yl ,m  Lˆ Lˆ2 Yl ,m
 Lˆ2 Lˆ Yl ,m 
Lˆ2  Lˆ Yl ,m  
2
l (l  1)  Lˆ Yl ,m 
Which implies that
2
l (l  1) Lˆ Yl ,m
Therefore Lˆ Yl ,m
is an
Eigenfunction of Lˆ2 & Lˆ z
with eigen values l and m+1
Lˆ Yl ,m  Cl,m Yl ,m1
Ladder Operators
Lˆ Yl ,m  Cl,m Yl ,m1
These are not an eigen relationships!!!!
Lˆ Yl ,m  Cl,m Yl ,m1
Cl,m
is not an normalization constant!!!
These relationships indicate a change in state, by Dm=+/-1, is caused by L+ and Ln
n
Can these operators be applied indefinitely??  Lˆ  Yl ,m  ?  Lˆ  Yl ,m  ?

L̂ Y2,0  Y2,1
L̂ Y2,0  Y2,1
L̂ Y2,1  Y2,2
L̂ Y2,1  Y2,2
Lˆ Y2,2  0
Lˆ Y2,2  0



Not allowed
Recall: There is a max & min value for m, as it
represents a component of L, and therefore
must be smaller than l. ie.
m  l (l  1)   l (l  1)  m  l (l  1)
Why is
L l
?
More Useful Properties of Ladder Operators
Recall
Lˆ2  Lˆ2x  Lˆ2y  Lˆ2z
Lˆ2x  Lˆ2y  Lˆ2  Lˆ2z   Lˆ2x  Lˆ2y  Yl ,m   Lˆ2  Lˆ2z  Yl ,m
 Lˆ2 Y  Lˆ2 Y
l ,m
z
l ,m
  l (l  1)  m 2  Yl ,m
This is an eigen equation of a physical observable that is always greater
than zero, as it represents the difference between the magnitude of L and
the square of its smaller z-component, which are both positive.
l (l  1)  m 2  0   l (l  1)  m  l (l  1)
This means that m is constrained by l, and since m can be changed by ±1
 m  l , l  1, l  2,...., l  2, l  1, l.
More Useful Properties of Ladder Operators
Knowing that:
Lˆ Yl ,mmax  0
Lˆ Yl ,mmin  0
Lets show that mmin & mmax are l & -l.
Consider
Lˆ Lˆ Yl ,mmax  0
Lˆ Lˆ & Lˆ Lˆ
Lˆ Lˆ Yl ,mmin  0
have to be determined in terms of
Lˆ2 & Lˆz
Lˆ Lˆ   Lˆx  iLˆ y   Lˆx  iLˆ y   Lˆ2x  iLˆx Lˆ y  iLˆ y Lˆx  Lˆ2y
Lˆ Lˆ   Lˆx  iLˆ y   Lˆx  iLˆ y   Lˆ2x  iLˆx Lˆ y  iLˆ y Lˆx  Lˆ2y
1
  Lˆ Lˆ  Lˆ Lˆ   Lˆ2x  Lˆ2y
2
 Lˆ Lˆ  Lˆ Lˆ   2 Lˆ2  2Lˆ2z


More Useful Properties of Ladder Operators
Also note that:
 Lˆ , Lˆ    Lˆ , Lˆx   i  Lˆ , Lˆ y 

 
 

  Lˆx , Lˆx   i  Lˆ y , Lˆx   i  Lˆx , Lˆ y   i (i )  Lˆ y , Lˆ y 
 0  i (i ) Lˆz  i (i ) Lˆz  0  2 Lˆz
2 Lˆ2  2 Lˆ2z  Lˆ Lˆ  Lˆ Lˆ
 Lˆ Lˆ  Lˆ Lˆ   Lˆ , Lˆ 

Similarly
Lˆ Lˆ  Lˆ2  Lˆ z  Lˆ2z
Lˆ Lˆ  Lˆ2  Lˆz  Lˆ2z
Ladder Operators
Recall
Lˆ Lˆ Yl ,max  0
  Lˆ2  Lˆz  Lˆ2z  Yl ,max
 Lˆ2 Yl ,mmax  Lˆz Yl ,mmax  Lˆ2z Yl ,mmax


2
l  l  1 Yl ,mmax  mmax Yl ,mmax  mmax 2
2
 l  l  1  m
max

(mmax  1)  Yl ,m
l  l  1  mmax (mmax  1)  0
 mmax  l
2
Yl ,mmax
Ladder Operators
Recall
Lˆ Lˆ Yl ,mmin  0
  Lˆ2  Lˆz  Lˆ2z  Yl ,mmin
 Lˆ2 Yl ,mmin  Lˆz Yl ,mmin  Lˆ2z Yl ,mmin
l  l  1 Yl ,mmin  mmin Yl ,mmin  mmin 2

2

2
 l  l  1  m
min

(mmin  1)  Yl ,mmin
l  l  1  mmin (mmin  1)  0
2
mmin
 mmin  l  l  1  0  mmin  l  1,& l
Since the minimum value cannot be larger than the
maximum value, therefore .
mmin  l
2
Yl ,mmin
Spin Angular Momentum
Intrinsic Angular Momentum is a fundamental property
like mass,and charge.
S , ms  I , m
Iˆ 2 I , m 
2
I  I  1 I , m
Iˆ z I , m  m  I  1 I , m
2
3
Iˆ 2α 
α
4
2
3
Iˆ 2β 
β
4
Iˆ z α  α
2
Iˆ z β   β
2
1 1
α ,
2 2
1 1
β  ,
2 2
Coupling of Spin Angular Momentum
S3 , m3  S1 , m1 S2 , m2
&
where S1  S2  S3  S1  S2
 S3  m3  S3
1 1 1 1
S, m  ,
,
2 2 2 2
S3  0, m3  0
0,0    
S3  1, m3  1,0,1
1, 1  
1,0    
1,1  
Spin and Magnetic Fields
Paramagnetism
ESR (EPR),
NMR (NPMR)
NQR
Mossbauer
Precession
Zeeman Splitting
μ  I
EZ  μ  B   Iˆ  B
   Bo Iˆ z   wo Iˆ z
For B  (0,0, Bo )
Nuclear Magnetic Resonance
XeF5+SbF6-
Fig. 1. 19F-NMR spectrum (56.4 MHz, 26°C) of the XeF5
cation (4.87 M XeF5 SbF6 in HF solution): (A) axial fluorine and
(a) 129Xe satellites; (X) equatorial fluorines and (x) 129Xe
satellites [30].