Lecture 11 Particle on a ring (c) So Hirata, Department of Chemistry, University of Illinois at...

24
Lecture 11 Particle on a ring (c) So Hirata, Department of Chemistry, University of Illinois at Urbana- Champaign. This material has been developed and made available online by work supported jointly by University of Illinois, the National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the sponsoring agencies.

Transcript of Lecture 11 Particle on a ring (c) So Hirata, Department of Chemistry, University of Illinois at...

Lecture 11Particle on a ring

(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has been developed and made available online by work supported jointly by University of Illinois, the

National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not

necessarily reflect the views of the sponsoring agencies.

The particle on a ring

The particle on a ring is the particle in a box curved with its both ends tied together.

This is a model for molecular rotation, a part of an atom’s electronic wave function, and even a part of a crystal’s electronic wave function.

We introduce a new operator, angular momentum operator, and a new boundary condition, cyclic boundary condition.

Rotational motion

We review the relevant classical physics and mathematical concepts: Angular momentum Cylindrical coordinates

We then switch to quantum mechanics: Angular momentum operator Hamiltonian in cylindrical coordinates Cyclic boundary condition The particle on a ring

The angular momentum

Which is the most effective way of applying force to rotate the gear?

(b); the right angle to the radius

(a) and (c) are wasteful because the effective component is |F| sin θ; (d) is not using leverage.

a

bc

θ

d

The angular momentum

For a given r (position) and p (momentum), its rotational motion is proportional to |r| and |p| sin θ.

Angular momentum J (or l) is a vector vertical to the plane formed by r and p, with length |r||p|sinθ.

θrp

The angular momentum

The mathematical definition of the angular momentum is a vector outer product:

Without losing generality, we can assume r = (rx, 0, 0) and p = (px, py, 0). Then l = (0, 0, lz).

( , , ) ( , , )x y z y z z y z x x z x y y xl l l r p r p r p r p r p r p

l r p

(0,0, ) (0,0, ) (0,0,| || | sin )z x yl r p r pNote that the angular momentum is a vector parallel to z-axis

Cylindrical coordinates

zz

ry

rx

sin

cos

The angular momentum operator

Classical angular momentum

The quantum mechanical operator can be obtained by the conversion

Thereforex

imvp xx

rrx

sincos

rry

cossin zz

ry

rx

sin

cos

The Hamiltonian

zz

ry

rx

sin

cos

The particle on a ring

Cyclic boundary condition

In the cylindrical coordinates, the wave function is Ψ(r,φ,z) at fixed r and z.

The wave function must satisfy the cyclic boundary condition:

After a complete revolution, the function must have the same value and also must trace the function in the previous cycle. Otherwise it would not be single-valued.

The particle on a ring

)()(2 2

2

2

2

Emr

22

2

22

2

kxkx

ikxikx

d ek e

dx

d ek e

dx

Promising

This will have problems with the cyclic boundary condition

ll imlim e

mr

me

mr 2

22

2

2

2

2

22

Energy

The particle on a ring

The cyclic boundary condition

Unlike the particle in a box, we have m = 0. Unlike the particle in a box, we have negative m’s. Energy is doubly degenerate for ml ≥ 1. No zero-point energy; no violation of uncertainty.

,2,1,0 lm

2

22

2mr

ml

The particle on a ring

Box Ring

Doubly degenerate

No zero-point energy

Energy differences in

the microwave range for

molecular rotations

Cyclic boundary condition

Imagine bending the particle in a box and connect the both ends to make a ring. Notice the difference between the boundary conditions used in two problems.

Quantum in nature

How does a microwave oven work?

Rotation of H2O

Quantum in nature

Why is grass green?

Rotation of electrons in chlorophyll

The particle on a ring

Normalization

2

11

2/12

0

2/12

0

2

ddeN lim

The angular momentum operator

Classical angular momentum

The quantum mechanical operator can be obtained by the conversion

Thereforex

imvp xx

rrx

sincos

rry

cossin zz

ry

rx

sin

cos

The angular momentum

Let us act the angular momentum operator on the wave function

The wave function is also an eigenfunction of angular momentum operator (H and lz commute and have the simultaneous eigenfunctions).

Double degeneracy comes from the two senses of rotations.

The uncertainty principle

The complementary observables

The angle and angular moment are the complementary observables.

x xipx

ˆ

t tiE

il x

The particle on a ring

For a state with a definite angular momentum, angle must be completely unknown.

The probability density along the ring is,

completely uniform regardless of ml.

2

1

22*

ll

ll

imim

mm

ee

Cyclic boundary condition and crystals

The crystalline solids have repeated chemical units. We impose cyclic boundary condition on the wave functions of crystals because infinite periodic crystals are mathematically isomorphic to rings of large radius. Crystals’ wave functions as a result have particle on a ring eigenfunctions as their part.

Summary

In the particle on a ring, we have learned a number of important concepts: angular momentum operator, cylindrical coordinates, cyclic boundary condition.

Similarities and differences from the particle in a box: the ring has no zero-point energy, doubly degenerate (|ml| ≥ 1), uniform probability density.