Chem 430
Particle on a ring
09/22/2011
Richard Feynman
I think I can safely say that nobody understands quantum mechanics
Quantum mechanics is based on
assumptions and the wave-particle duality
The nature of wave-particle duality is not known
To explain and predict experimental results:
(A) A quantum system has many possible states;
(B) Each state has a well defined energy;
(C) At anytime, the system can be in one or more states;
(D) The probability in each state is determined by energy
and other factors.
What is energy ?
In many cases, define probability
The energy of each state will not change
The system energy can change
s  c 0 ( t ) 0  c1 ( t ) 1  ...  c n ( t ) n
Energy value (frequency) is obtained
from the oscillation of the coefficients
Oscillating dipole generates electromagnetic radiation
0.10
CH3NCO
-1
CH3NCO/2288 cm
0.8
0.08
0.6
PP signal
Absorbance (normalized)
1.0
30cm-1
0.4
0.06
0.04
0.2
0.02
2200
2220
2240
2260
2280
2300
2320
2340
0.5
-1
Frenquency (cm )
1.0
1.5
2.0
2.5
Delay time (ps)
T 

c

1
3000  3  10
8
 1.1  10
 12
s  1.1 ps
3.0
3.5
4.0
Polar Coordinates (2D)
x  r cos 
y
x
r

O
y  r sin 
P (x,y)
y
x
  0 ~ 2
Cylindrical Coordinates (3D)
z
y
x
r
P(x,y,z)
y
x
x  r cos 
y  r sin 
z z
Why use the new coordinates rather than the
Cartesian Coordinates?
Fewer variables, easier to calculate
Variables can be separated because of symmetry
Rotation
a rotation is a rigid body movement which keeps a point fixed.
a progressive radial orientation to a common point
In Cartesian coordinates, two variables
In polar coordinates, only one variable
x  r co s 
y  r sin 
r is constant
Particle on a ring
Particle mass : m
Potential:0
Radius: r=a=constant
Angle: the only variable
a
General procedure
Write down Hamiltonian
Simplify Math with symmetry
Use boundary conditions to define energy levels
Particle on a ring
Hamiltonian only contains the kinetic energy part
2
H 
2
 
2
(

2 m x
2m

2

2
2
y
)
2
In the polar coordinate system
 
2

2
x
2


2
y
2


2
r
2

1 
r r

1

2
r 
2
2
 
2

2
x
2



2
y

2
1 
2
r
2


r r
2

x



2

r x
r
2
2

r
r x
 r
2
r r x

)
x
 r
r x

(
x
(
r
(


2
(
Need to eliminate

2



x

r x
 r
r x


 
 x
 
 x
)
  r

 r
  
)

(

)
 x x  r x  x x
 r

2

r r x

(
 r

x

x
2
The chain Rule
y  r sin 

2
r 
x  r co s 
Apply it twice

1

 r
r  x
2
r x
 

2
 x

2
0


2

2
x



 
2
)
r
 r x x

 
2
)

 x x
x  r cos  ; y  r sin 
dx  cos  dr  r sin  d  ;
dy  sin  dr  r cos  d 
dr  cos  dx  s in  dy ;
d  
sin 
cos 
dx 
r
2
x

r

2
cos 

2

r
sin   
cos 
 co s  ;
 s in  ;

;

.
x
y
x
r
y
r
r

dy
2
r
2

2 sin  cos 

r
r 
2

sin  cos  
r
2
sin  
2


r
r

sin  
r
2
2
2

2
y
2

sin 
2

2
r
2

2 sin  cos 

r
r 
2

sin  cos  
r
2

cos  
2

r
r

cos  
r
2
2
2

2
2
 
2

2
x
2


2
y
2


2
r
2

1 
r r

1

2
r 
2
2
http://en.wikibooks.org/wiki/Partial_Differential_Equations
/The_Laplacian_and_Laplace's_Equation
ra

2
H 

r
2
 
2
2m
(



r
2
2 m r
2

2
H 
H  
d 
0
2
1 
r r
I  mr
Moment of inertia
2
2
d
2 I d
2
2
r 
2
2
2
) 
2mr
2
2
2 I d
2
2
 E


1
d 
2

d
2

2 IE
2

d
2
2
d
2
  Ae
ml  (
im l 
2 IE
2
 Be
 im l 
1
)
2
 E 
ml
2
2
2I
Cyclic boundary condition
P ( )  Q (  2  )
On the ring, Points P = Q
 ( )   (  2 )
Ae
im l 
 Be
 im l 
 Ae
im l 
e
im l 2 
 Be
 im l 
e
 im l 2 
Ae
im l 
 Be
 im l 
 Ae
im l 
e
im l 2 
 Be
 im l 
e
 im l 2 

e
im l 2 
 cos m l 2   i sin m l 2   1

m l  0,  1,  2....
E ml 
ml
2
2
;
  Ae
im l 
2I
Why only choose one portion for the wave function?
 d 
Normalized
*   d  A e
Ae
2
im l 
 2 A  1
2
0
1
 A
 * 
 im l 
2
1
2
 
1
2
e
im l 
Constant for all angles
Why? Why is it different from in the 1D box?
Implications: (1) probability is same at any point
(2) Position can’t be determined at all
Consequence of arbitrary position
No zero point energy
m l  0, E  0
   p   0;
  
E  0, m l   l
Double degeneracy
Particle can rotate clockwise or counterclockwise
The Circular Square Well
y
a

V (r  a )  0
V (r  a )  
x
2
H 
2

2
2 m r

2
(
2m
(

r
2

2

1 
r r
1 
r r


1
2
r 
2
1 
2
)
2

r 
2
2
)  E
C ircular sym m etry
  can be separated into angular and radia l parts
 ( r ,  )  R ( r ) ( )
The angular part is known from the above
2

d R
2
(
2m
dr
2


2
1 d R
2

r dr
 d R
2
(
2m
1 d R
dr

2
1 dR
r
d
2
2
)
r dr
)  E R
2
R 1 d 
2
2m r
2
d
2
 E R
Divided by R 
2
2

(
d R
2mR
1
dr
2

2
(r
R
 
2
d R
dr
1
2
1 dR
r
)
2
e
im l 
;
d 
2
2 m r
2mE
dr
Radial
2
)
r dr
dR
1 d 
2
2
d
 E
1 d 
2
r 
2
 d
2
angular
2
2
d
2
 m 
2
l
m
2
l
2
r
2
d R
dr
2
r
dr
2
d R
dz
2
dR

1 dR

2mE
2
 (1 
z dz
r Rm R
2
m
z
2
l
2
2
l
)R  0
w ith
k 
2
2mE
2
; z  kr
Bessel’s Equation
Chem 430
Particle in circular square well and
09/27/2011
If particle is confined in a ring,
At 0K, what is the most probable location to find it?
How about at high temperature?
How to explain these in terms of QM?
1. Find out possible states
2. Find out the energy of each state
3. Find out the wave function of each state
to obtain its spatial distribution of probability
Nothing to do with rotation
J m l ( kr )  (
1
(

kr )
ml
2

(  1)
n
n0
1
kr )
2n
2
n !( n  m l ) !
Boundary condition
r  a;
J m l ( ka )  0
The condition gives allowed k and therefore energy
k 
2
2mE
2
J m l ( kr )  (
1
(

kr )
ml
2
 (  1)
n0
n
1
kr )
(A)
2n
2
n !( n  m l ) !
J 0  2 .4 0 5   0
(

J 0 ( kr ) 
 (  1)
1 2.405 r
2
n
normalized
 (r , ) 
(

J 0 ( kr ) 
 (  1)
n0
n
1
kr )
1.181
a
2n
2
2
k
2
 5.783
2m
2ma
2
1
2n
(
(  1)
1 2.405 r
2
n
)
2n
a
n !n !
n0
-1
2
n !n !
0
1
r/a
(B)
J 0  5 .5 2 0   0
E 
k
2
2
 30.471
2m
(

J 0 ( kr ) 

(  1)
1 5.520 r
2
n
)
2n
2ma
2
1
a
n !n !
n0
(A)
2
ka  5 .5 2 0
1
0
E 
0


)
ka  2 .4 0 5
a
n !n !
n0
ml  0
2
J 0 ( ka )  0
(C)
0
(B)
 (r , ) 
2.749
a
2
(

 (  1)
1 5.520 r
2
n
n0
)
2n
a
n !n !
-1
0.0
0.5
1.0
r/a
-1
0
2
4
6
8
10
(C)
J 0  8 .6 5 4   0
(

J 0 ( kr ) 
 (  1)
n
1 8.654 r
2
 
0
0
2
 74.887
)
2ma
2
1
2n
0
2
 ( r ,  ) rdrd   1
2
a
n !n !
n0
a
E 
k
2m
kr
2
2
ka  8 .6 5 4
 (r , ) 
4.320
a
2
(


n0
(  1)
n
1 8.654 r
2
a
n !n !
Many more states are possible if kr is bigger
)
2n
-1
0.0
0.5
r/a
1.0
J m l ( kr )  (
1
(

kr )
ml
2

(  1)
1
kr )
J 1  3.832   0
2n
2
ka  3 .8 3 2
2
n !( n  m l ) !
n
n0
E 
k
2
2
 14.682
2m
2ma
2
1.0
ml  1
J 1 ( kr ) 
1
2
(

kr  (  1)
n
n0
1
J 1 ( kr ) 
kr )
2n
1
2
(

kr  (  1)
1 3.832 r
)
2n
0.5
2
a
n !( n  1) !
n
n0
2
n !( n  1) !
0.0
 (r , ) 
1.963
a
1.0
2
e
 i

1
2
(

kr  (  1)
1 3.832 r
)
2n
-0.5
2
a
n !( n  1) !
n
n0
-1.0
0.0
0.5
1.0
r/a
0.5
J 1  7.016   0
0.0
ka  7 .0 1 6
k
2m
-0.5
-1.0
2
E 
2
2
 49.218
2ma
2
1.0
0
2
4
6
8
10
1
J 1 ( kr ) 
kr
2
(

kr  (  1)
n
n0
0.5
1 7.016 r
2n
)
2
a
n !( n  1) !
0.0
-0.5
 (r , ) 
3.534
a
2
e
 i

1
2
(

kr  (  1)
n0
n
1 7.016 r
)
2n
2
a
n !( n  1) !
-1.0
0.0
0.5
r/a
1.0
J m l ( kr )  (
1
(

kr )
ml
2

(  1)
n
n0
1
kr )
J 2  5 .1 3 6   0
2n
2
n !( n  m l ) !
J 2 ( kr )  (
(

kr )
2
2
 (  1)
n0
n
1
J 2 ( kr )  (
kr )
E 
k
2
2
 26.375
2m
ml  2
1
2
ka  5 .1 3 6
2n
1
(

kr )
 (  1)
2
2
1 5.136 r
 (r , ) 
a
2
e
 2 i
n0
(
1
(

kr )
2
2
 (  1)
2n
2
a
n !( n  2) !
n
2
n !( n  2) !
2.759
)
n
n0
1 5.136 r
)
2ma
2
1.0
0.5
2n
2
a
n !( n  2) !
0.0
-0.5
1.0
-1.0
0.0
0.5
0.5
J 2  8 .4 1 7   0
0.0
ka  8 .4 1 7
2
E 
k
2m
-0.5
-1.0
1.0
r/a
2
2
 70.850
2ma
2
1.0
0
2
4
6
8
10
J 2 ( kr )  (
kr
 (r , ) 
4.322
a
2
e
 2 i
(
1
kr )
2
2
1
2
(

 (  1)
(

kr )
 (  1)
n0
n
)
2n
2
a
n !( n  2) !
n
n0
2
1 8.417 r
1 8.417 r
0.5
0.0
-0.5
2n
)
2
a
n !( n  2) !
-1.0
0.0
0.5
r/a
1.0
Energy
level
J 0 ( kr )
1.0
J 1 ( kr )
0.5
J 2 ( kr )
0.0
-0.5
-1.0
0
2
4
6
kr
8
10
0
1
2
3
4
5
ml
0
1
2
0
3
1
energy
5.78
14.68
26.38
30.47
40.71
49.22
Angular Momentum
i
j
l  r p  x
k
z i
y
px
py
pz
y
z
py
pz
lz  x (
l z 
i y

i 
)  y(


i
z
px
pz
k
x
y
px
py
l z  xp y  yp x
z com ponant ( k )

 j
x

i x
)
1
2

(x
i
e

y
 y

x
)

i 
im l 

ml
2
e
im l 
 m l
Spherical Polar Coordinates
z

P(x,y,z)
r
y
x
y

x
z
x  r sin  co s 
y  r sin  sin 
z  r co s 
 
2
1 
r r
2
r
2

r

1

r sin   
2
sin 




1
2
r sin   
2
2
2