Particle on a RingParticle on a Ring

An introduction to An introduction to Angular MomentumAngular Momentum

Quantum Physics II

Recommended Reading:

Harris, Chapter 6

Particle confined to a circular ringParticle confined to a circular ringIn this problem we consider a particle of mass m confined to move in a horizontal circle of radius. This problem has important applications in the spectroscopy of molecules and is a good way to introduce the concept of ANGULAR MOMENTUM in quantum mechanics.Can specify the position of the particle at any time by giving its x and y coordinates. r

m

BUT this is a one dimensional problem and so we only need one coordinate to specify the position of the particle at any time t.The position is specified by the angle (t), the angle made by the vector r with the horizontal.If orbit is horizontal then there potential energy of the particle is constant and can be taken as zero: U = 0What is the potential if the orbit is vertical? Then we must include gravity, much more difficult problem.

Classical TreatmentClassical Treatment

r

m

v

s

m = mass of particlev = instantaneous velocity = ds/dtr = radius vectors = arc length = angle = s/rCan then define angular velocity r

vdtds

r1

dtd

The kinetic energy is constant and equal to ½mv2

In terms of the angular velocity this can be expressed as 22mr21

E

The quantity I = mr2 is the moment of inertia of the particle, then 2I21

E

The angular momentum L of the particle is defined as

Irm vmrL 2

Can write the energy of the particle in terms of the angular momentum

I 2L

I21

E2

2

Note that the angular momentum L is perpendicular to both r and v (since L = mr v ).

can have two directions for the velocity, clockwise or anti-clockwise,

r

L

v

rv

L

Magnitude of L is the same in both situations, but direction is different. Must remember that L is a vector

TISE for particle confined to a circular ringTISE for particle confined to a circular ring

E0dy

d

dx

dm2 2

2

2

22

The time independent Schrodinger equation is (in (x,y) coordinates)

(1)

If we express this in terms of the angle then

x = r.cos(), y = r.sin()or tan() = y/x

2

2

22

2

2

2

d

d

r

1

dy

d

dx

d

φ

and Schrodinger’s equation becomes

Ed

d

r

1m2 2

2

2

2

φ

(2)

(3)but mr2 = I, the moment of inertia of the particle, So TISE is

ψφ

Ed

dΙ2 2

22

(4)want to solve this equation for the wavefunctions and the allowed energy levels

3 ,2 ,1 ,0egerintm

1m2exp

)(imexp)exp(im

L

L

LL

π

2π2π

(5)

Solutions to TISE for particle on a circular Solutions to TISE for particle on a circular ringring

rearrange (4)

0md

d 0

E2

d

d 2L2

2

22

2

I

A general solution of this differential equation is

LL im-BexpimAexp (6)

where 21

2LIE2

m

Check this

(7)

The wavefunction must be single valued for all values of . In particular if we rotate through an angle 2 the wavefunction must return back to the same value it started with:

from (7) we then get 2

22L

22L

r2

m

2

m E

mI

Energies are quantised. Allowed values are given by eqn (8)

(8)

Energy SpectrumEnergy Spectrum

2L

2m

2 E

I

where mL =0, 1, 2, 3, ….

mL =0mL =1

mL =2

mL =3

mL =4

mL =5

E0 =0

E1 =1

E2 =4

E2 =9

E2 =16

E2 =25

Energies in units of

I2

2

Energy SpectrumEnergy Spectrum

2L

2m

2 E

I

where mL =0, 1, 2, 3, ….

Energies in units of

I2

2

mL =0mL =1

mL =2

mL =3

mL =4

mL =5

E0 =0

E1 =1

E2 =4

E2 =9

E2 =16

E2 =25

En

erg

y

all states are doubly degenerate except the ground state (m = 0) which is singly degenerate.

- mLc

mL

+ mL

Degeneracy of SolutionsDegeneracy of SolutionsEquation (6) shows that there are two solutions for each value of mL, (except mL = 0) [ exp(imL) and exp(-imL)] doubly degenerate system. The two solutions correspond to particles with the same energy but rotating in opposite directions

rm

φφψ LimAexp

rm

φφψ Lim-Aexp

Compare this with linear motion of a free particle, where the solutions are also doubly degenerate; Aexp(ikx) (+x direction) and Aexp(-ikx) (-x direction)

NormalisationNormalisationFrom normalisation condition

1A2 1d A

1d imLexpAimLexpA 1d

22

0

2

2

0

2

0

πφ

φφφφψ*ψ

π

ππ

Hence

21

π21

A

And the normalised wavefunctions are

φπ

φπ L

21

L2

1

im-exp21

and imexp21

(9)

(10)

Probability DistributionProbability DistributionWhat is it probability P, that the particle will be found at a particular angle ?

21

imexp21

im-exp21

P L2

1

L2

12

πφ

πφ

πφψφ*ψφψ

Note that this is independent of the angle and the quantum number mL equal probability of finding the particle anywhere on the ring no matter what state it is in!

0 2

Angle

Pro

bab

ilit

y

1/2

However, note that the wavefunctions are complex. They have a real and an imaginary part

The WavefunctionsThe Wavefunctions isincos e and isincos e i-i θθθθ θθ Recall that

so the wavefunctions can be written as

φm ins im cos 21

e 21

LLmi L φ

ππφ

Real part of the wavefunction = cos(mL)

Imaginary part of the wavefunction = sin(mL) and again we see that

π

φφπ

φφπ

φφπ

φψφ*ψφψ

21

msinmcos21

msinimcos 21

msinimcos21

P

L2

L2

LLLL

2

We can visualise the real and imaginary parts of the wavefunction as follows.

sin(1.) cos(1.) 1

11

0 2

Angle

sin(2.) cos(2.)

22

0 2

Angle

33

sin(3.) cos(3.)

0 2

Angle

sin(1.) cos(1.) 1

22