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Lecture 13: Angular Momentum-I.
The material in this lecture covers the following in Atkins.
Rotational Motion
Section 12.6 Rotation in two dimensions
Lecture on-line
Angular Momentum in 2D (PowerPoint)
Angular Momentum in 2D (PDF format)
Handout for this lecture
Tutorials on-line
Vector concepts
Basic Vectors
More Vectors (PowerPoint)
More Vectors (PDF)
Basic concepts
Observables are Operators - Postulates of Quantum
Mechanics
Expectation Values - More Postulates
Forming Operators
Hermitian Operators
Dirac Notation
Use of Matricies
Basic math background
Differential Equations
Operator Algebra
Eigenvalue Equations
Extensive account of Operators
Audio-visuals on-line
Rigid Rotor (PowerPoint)
(Good account from the Wilson Group,****)
Rigid Rotor (PDF)
(Good account from the Wilson Group,****)
Motion in 1D
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Confined translation
1D
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Free translation
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Harmonic Oscillation
1D
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Vibrating diatomic molecule
Thus
Quantum Mechanical
Rotation in 2D...
quantitative derivation
1
ml ( ) 
expiml
2
describes a state
2 m2
l ;
with precise energy E 
ml  0; 1; 2,..
2I
and angular momentum
component in z - direction
Lz  ml ; ml  0; 1; 2;....
Vibrational spectroscopy
Has the vibrational energy levels
A vibrating diatomic
molecule AB
E
v6
1
E  (  v)
2
v5
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11 
2
v4
9

2
v3
v2
v 1
A photon of energy h 
will be absorbed if
h  
v0
7

2
5

2
3 
2
1

2
x
   / 2

k

mAmB

mA  mB
Vibrational spectroscopy
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Molecules in general absorbs photons
of specific frequencies (fingerprints)
Motion in 2D
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Confined translation
2D
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Rotating Diatomics
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Rotation in 2D The rigid rotor
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Vibrating Diatomics
Quantum Mechanical Rotation in 2D
let us consider a particle of mass m
moving in the xy plane in a circle
of fixed radius a
The position of the particle
z
is given by
k
r = ix + j k
j
a
i
x
y
where
r = r r  x2  y 2  a
With the velocity given by
dr
dx
dy
= i
+ j
dt
dt
dt
v = i v x + j vy
Quantum Mechanical Rotation in 2D
We have
It is more informtive to
work in the spherical
r = i x + jy
coordinates (r, )
z
r = i rcos + j rsin
k
The velocity is given by
j

i
x
r
y
drcos
drsin
v=i
+ j
dt
dt
or since r is constant (r = a)
d
d
v = - i rsin
+ j rcos
dt
dt
Quantum Mechanical Rotation in 2D
z
k
j

i
y
r
We note that
|v | = vv 
d
a2 ( )2 [sin2   cos2 ]
dt
d
 a( )2
dt
v
x
also
v  r  v xrx  vy ry
r = i rcos + j rsin
dr
d
v = - i rsin
+ j rcos
dt
dt
d
 a sin 
cos  
dt
d
a sin 
cos   0
dt
Quantum Mechanical Rotation in 2D
Let us now evaluate
L = r p = m r v
z
k
We have
L = r  p = ( i x + j y)  ( i px  j p y )
j

i
x
y
r
v
= i  i (x px )  i  j (x p y )
 j  i (yp x )  j  j (yp y )
 kxpy  kyp x
L  m(xpy  yp x )k
Quantum Mechanical Rotation in 2D
L  m(xpy  yp x )k
or
L
z
d
2
2
L  [mr cos 

k
j

i
x
d
2
2
mr sin 
]k
y
r
dt
dt
v
d
2
Lmr
k
dt
L  m | r |  | v | k | r |  | p | k
r = i rcos + j rsin
dr
d
v = - i rsin
+ j rcos
dt
dt
x
Quantum Mechanical Rotation in 2D The total energy is
E = Ekin  Epot
L
2 d
Epot  0
z L  mr dt k
k
2


d

d

2
mr 2 ( )2 mr ( )
2
p

dt 
j
dt

y Ekin  2m 
r
2
mr 2

i
v
Lz 2
Ekin 
mr2
since I = mr 2 is moment of
inertia
d
2
| p | mr
dt
L z2
Ekin 
I
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization
We have
L z2
Ekin 
I
and
Lz  pr
I  mr 2
and
Lz  pr
I  mr 2
The angular momentum of a particle of mass m on a circular path
of radius r in the xy-plane is represented by a vector of magnitude
pr perpendicular to the plane.
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization
Consider a free paricle Its wavefunction is given by
moving to the right
 i 
 (x, t)  AExp Et Expikx


 i 
 AExp Et cos kx  sinkx


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

AExp


AExp

Its momentum is given by p  k
The wavelength  is
determined from the condition
 (x, t)   (x  , t)

k  2  k  2 / 
Etcos kx  sinkx 

p  (2 /  )  h / 
i 
Etcos kx    sinkx  
de Broglie

i
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization
Consider a particle moving at a ring
Its wavefunction () must satisfy
( )  (  2)
Same physical situation
acceptable
Not acceptable
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization

According to QM a particle of momentum p
h
de Broglie
moves as a wave with wave length  =
p

Standing wave must repete itself
thus wavelength must be a integer
fraction of the circumperence
2r

n
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization

We have
2r

m
de Broglie
h

p
m  0,1, 2, 3,...
Boundary cond.
Thus
2r h

m p
or
hm
p
2r
m =  0,1,2,3,4,...
Quantum Mechanical Rotation in 2D...
qualitative origin of quantization
hm
p
2r
m =  0,1, 2, 3,4,...
Lz  pr  m
The angular momentum of a particle confined to a
plane can be represented by a vector of length |ml|
units along the z-axis and with an orientation that
indicates the direction of motion of the particle. The
direction is given by the right-hand screw rule.
(pr)2 h2m2
E 

I
(2) 2 I
E
2
m2
I
m =  0,1, 2, 3, 4,...
We might also try to solve the problem
by standard procedures
The Hamiltonian is again
given by
2  d2
2 
d


H= 
2 dy 2 
2m 
dx

Next making use of
spherical coordinates
z

r
2 
 2 1 
1

ˆ 
H

 2 2 
2

2m  r r  r
r  
y
i
x
2
2
2 1 
1 2
 2 2
2 
2 
2 
x
y
 r r r
r 
2
k
j
Quantum Mechanical Rotation in 2D...
quantitative derivation
x  r cos  ; y = rsin
Since r is a constant, a
the first two derivatives
with respect to r can be
neglected and we have
2 
2 
1

ˆ 
H
2
2 
2m 
r



2 
2 
1

ˆ 


H
2
2
2m 
 r  

or since I = mr 2
and the Schroedinger
equation is
2 2 

 E
2
2I

Quantum Mechanical Rotation in 2D...
quantitative derivation
The normalized solutions are
1
 ml () 
expiml where
2
ml  
2IE
 here ml is just
at the moment a real number
or
2 
 2
2IE
 2 
2 
 2
 m2 
2IE
2
m 
2
We must have that
 ml ()   ml (  2)
since (r, ) and (r,   2)
represent the same point in
space
Quantum Mechanical
Rotation in 2D...
quantitative derivation
ml (   2 ) 
1
expiml (  2 )
2
1
=
exp im 2 exp im 
l
l
2

 

1
=
expi2ml ml ( )
2
1

12ml ml ()
2
Thus for ml (  2 ) to be
equal to ml ():
12ml  1
or
ml  0, 1; 2; 3;....;
Note :
Similar to
de Broglies
condition
Quantum Mechanical
Rotation in 2D...
quantitative derivation
We have that
2IE
ml  

Thus
2
ml2
E
;
2I
ml  0; 1; 2,..
Negative m values corrspond
to rotation in one direction
positive m values to rotations
in the other direction. For
m values of the same absolute
value but opposite signs the
energies are the same. The two
states are degenerate
Quantum Mechanical
Rotation in 2D...
quantitative derivation
2 m2
l ;
E
2I
ml  0; 1; 2,..
1
ml ( ) 
expiml
2
Note that the number of
nodes in the real
part of ml () increases
with ml .
The real parts of the wavefun ctions of a
particle on a ring. As shorter wavelengths
are achieved, the magnitude of the angul ar
momentum around the z-axis grows in steps
of
Quantum Mechanical
Rotation in 2D...
quantitative derivation
We have that the
angular momentum in
general is given by
L = r p = m r v
We have in the 2 - D
case
L = r  p = ( i x + j y)  ( i px  j p y )
= i  i (x px )  i  j (x p y )
 j  i (yp x )  j  j (yp y )
 kxpy  kyp x
L  m(xpy  yp x )k
L  (xpy  yp x )k  L zk
Lz  xpy  ypx
Quantum Mechanical
Rotation in 2D...
quantitative derivation
or quantum mechanically
Lˆ z  xˆ pˆ y  ypˆ x
 


Lˆ z  xˆ
 y 
i  y
x
or in spherical coordinates
Lˆ z 

i 
We note that
1
ml ( ) 
expiml
2
is an eigenfunction to Lˆ z
with eigenvalues
Lz  ml ; ml  0; 1; 2;....