Chem 430

Particle on a ring

09/22/2011

Richard Feynman

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.

I think I can safely say that nobody understands quantum mechanics

What is energy ?

In many cases, define probability

The energy of each state will not change

The system energy can change

0 1( ) 0 ( ) 1 ... ( )ns c t c t c t n

Energy value (frequency) is obtained from the oscillation of the coefficientsOscillating dipole generates electromagnetic radiation

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.02

0.04

0.06

0.08

0.10

PP

sig

na

l

Delay time (ps)

CH3NCO/2288 cm-1

2200 2220 2240 2260 2280 2300 2320 2340

0.2

0.4

0.6

0.8

1.0

Ab

sorb

an

ce (

no

rma

lize

d)

Frenquency (cm-1)

CH3NCO

30cm-1

128

11.1 10 1.1

3000 3 10T s ps

c

Polar Coordinates (2D)y

x

P (x,y)

r

y

xcosx r

0 ~ 2 siny r

O

Cylindrical Coordinates (3D)

x

yz

P(x,y,z)r y

x

cos

siny

z

x

r

z

r

Why use the new coordinates rather than the Cartesian Coordinates?

Fewer variables, easier to calculateVariables 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 variablesIn polar coordinates, only one variable

cos

siny

x r

r

r is constant

General procedure

Write down Hamiltonian

Simplify Math with symmetry

Use boundary conditions to define energy levels

Particle on a ring

a Particle mass : mPotential:0Radius: r=a=constantAngle: the only variable

Particle on a ring

Hamiltonian only contains the kinetic energy part2 2 2 2

22 2

( )2 2

Hm m x y

In the polar coordinate system

2 2 2 22

2 2 2 2 2

1 1

x y r r r r

2 2 2 22

2 2 2 2 2

1 1

x y r r r r

cos

siny

x r

r

r

x r x x

The chain Rule

2

2

2 2 2 2

2

2 2 2 2

2

2 2

( ) ( )

( ) ( )

( )

( )

x x x x

r

r x x

r r r

r x r r x r x r x x

r r

r x r x x x x

r

r r x

r

r x x

r r

r x x r x x

0

x

Apply it twice

Need to eliminate

x

cos sin ; sin cosdx dr r d dy dr r d

; cossin

in scosdr dx dy d dx dyrr

cs o; ; ;co

siins .

ns

r r

x y r y rx

2

2

2 2 2 2 22

2 2 2 2

2sin cos sin cos sin sincos

x

r r r r r r r

2

2

2 2 2 2 22

2 2 2 2

2sin cos sin cos cos cossin

y

r r r r r r r

cos ; sinr yx r

2 2 2 22

2 2 2 2 2

1 1

x y r r r r

http://en.wikibooks.org/wiki/Partial_Differential_Equations/The_Laplacian_and_Laplace's_Equation

2

2 0 a

rr

r

2 2 2 22

2 2 2

1 1( )

2 2H

m m r r r r

Moment of inertia2I mr

2 2

22

dH

I d

2 2

22

dH E

I d

2 2

2 22mr

d

d

2 2

2 2IEd

d

2

1

2( )2

l lim im

l

Ae Be

mIE

On the ring, Points P = Q ( ) ( 2 )P Q

Cyclic boundary condition

( ) ( 2 ) l lim imAe Be

2 2

2lmEI

2 2l l l lim im im imAe e Be e

2 2l l l l l lim im im im im imAe Be Ae e Be e

2 cos 2 sin 2 1liml le m i m

0, 1, 2....lm 2 2

; 2

l

l

lm

immE e

IA

Why only choose one portion for the wave function?

Normalized2 2

0* 2 1

1

2

l lim imd d Ae Ae A

A

1

2lime

1*

2

Constant for all angles

Implications: (1) probability is same at any point

(2) Position can’t be determined at all

Why? Why is it different from in the 1D box?

Consequence of arbitrary position

No zero point energy

0, 0lm E

0; p

0, lE m l

Particle can rotate clockwise or counterclockwise

Double degeneracy

y

xa

The Circular Square Well

( ) 0

( )

V r a

V r a

2 2 2

2 2 2

1 1( )

2H

m r r r r

2 2 2

2 2 2

1 1( )

2E

m r r r r

Circular symmetry

can be separated into angular and radial parts

( , ) ( ) ( )r R r

The angular part is known from the above

2 2 2

2 2 2

2 2 2 2

2 2 2

1 1( )

2

1 1( )

2 2

R R RER

d d

m r r

d

d d dr r

d R dR dER

m dr r dr m r d

R

2

2

22 2

2 2

1 2( )

1d R dR mEr r r

R dr dr

d

d

22

2

1;

2lim

l

de m

d

2 2 2 2

2 2 2

1 1( )

2 2

d R dR dE

mR dr r dr m r d

Divided by R

2lm

Radial angular

22 2 2

2 2

2l

d R dR mEr r r R m Rdr dr

22

2 2

22

1(1 ) 0

2;

lmd R dRR

dz z dz zwith

mEk 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 stateto obtain its spatial distribution of probability

Nothing to do with rotation

2

0

1( )1 2( ) ( ) ( 1)

2 !( )!l

l

n

nm

n

m

l

krJ kr kr

n n m

Boundary condition

;

( ) 0lm

r a

J ka

22

2mEk

The condition gives allowed k and therefore energy

2

0

1( )1 2( ) ( ) ( 1)

2 !( )!l

l

n

m nm

n l

krJ kr kr

n n m

0lm 2

00

1( )2( ) ( 1)

! !

n

n

n

krJ kr

n n

2.405ka 2 2 2

25.783

2 2

kE

m ma

2

00

1 2.405( )2( ) ( 1)

! !

n

n

n

raJ krn n

0 1-1

0

1

r/a

2 2 2

230.471

2 2

kE

m ma

0.0 0.5 1.0-1

0

1

r/a

2

00

1 5.520( )2( ) ( 1)

! !

n

n

n

raJ krn n

2

20

1 5.520( )2.749 2( , ) ( 1)

! !

n

n

n

rar

a n n

2

20

1 2.405( )1.181 2( , ) ( 1)

! !

n

n

n

rar

a n n

normalized

22

0 0( , ) 1

ar rdrd

0 2 4 6 8 10-1

0

1

kr

(A) (C)(B)

0 2.405 0J (A) 0 ( ) 0J ka

0 5.520 0J 5.520ka (B)

0.0 0.5 1.0-1

0

1

r/a

0 8.654 0J 8.654ka 2 2 2

274.887

2 2

kE

m ma

2

00

1 8.654( )2( ) ( 1)

! !

n

n

n

raJ krn n

2

20

1 8.654( )4.320 2( , ) ( 1)

! !

n

n

n

rar

a n n

(C)

Many more states are possible if kr is bigger

1lm

2

10

1( )1 2( ) ( 1)

2 !( 1)!

n

n

n

krJ kr kr

n n

0 2 4 6 8 10-1.0

-0.5

0.0

0.5

1.0

kr

1 3.832 0J 3.832ka 2 2 2

214.682

2 2

kE

m ma

2

10

1 3.832( )1 2( ) ( 1)

2 !( 1)!

n

n

n

raJ kr kr

n n

0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

r/a

2

20

1 3.832( )1.963 1 2( , ) ( 1)

2 !( 1)!

n

i n

n

rar e kr

a n n

1 7.016 0J 7.016ka 2 2 2

249.218

2 2

kE

m ma

2

10

1 7.016( )1 2( ) ( 1)

2 !( 1)!

n

n

n

raJ kr kr

n n

0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

r/a

2

20

1 7.016( )3.534 1 2( , ) ( 1)

2 !( 1)!

n

i n

n

rar e kr

a n n

2

0

1( )1 2( ) ( ) ( 1)

2 !( )!l

l

n

m nm

n l

krJ kr kr

n n m

2lm

2

22

0

1( )1 2( ) ( ) ( 1)

2 !( 2)!

n

n

n

krJ kr kr

n n

0 2 4 6 8 10-1.0

-0.5

0.0

0.5

1.0

kr

2 5.136 0J 5.136ka 2 2 2

226.375

2 2

kE

m ma

2

22

0

1 5.136( )1 2( ) ( ) ( 1)

2 !( 2)!

n

n

n

raJ kr kr

n n

0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

r/a

2

2 22

0

1 5.136( )2.759 1 2( , ) ( ) ( 1)

2 !( 2)!

n

i n

n

rar e kr

a n n

2 8.417 0J 8.417ka 2 2 2

270.850

2 2

kE

m ma

2

22

0

1 8.417( )1 2( ) ( ) ( 1)

2 !( 2)!

n

n

n

raJ kr kr

n n

0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

r/a

2

2 22

0

1 8.417( )4.322 1 2( , ) ( ) ( 1)

2 !( 2)!

n

i n

n

rar e kr

a n n

2

0

1( )1 2( ) ( ) ( 1)

2 !( )!l

l

n

m nm

n l

krJ kr kr

n n m

0 2 4 6 8 10-1.0

-0.5

0.0

0.5

1.0

kr

0 ( )J kr

1( )J kr

2 ( )J kr

Energy level

energy

0 0 5.78

1 1 14.68

2 2 26.38

3 0 30.47

4 3 40.71

5 1 49.22

lm

Angular Momentum

y z x yx zx y z

i j ky z x yx z

l r p x y z i j kp p p pp p

p p p

componant ( )z k

z y xl xp yp

( ) ( ) ( )zl x y x yi y i x i y x i

1

22

l

l

im

imlz l

em

l e mi i

Spherical Polar Coordinates

P(x,y,z)r

x

y

z

x

y

z

sin cos

sin sin

cos

x r

y r

z r

22 2

2 2 2 2 2

1 1 1sin

sin sinr

r r r r r