Vibrational Motion

Harmonic motion occurs when a particle experiences a restoring force that is proportional to its displacement.

F=-kx Where k is the force

constant. The stiffer the spring, the greater the value of k.

Force is also the gradient of the potential energy V. In 1-D: F=-dV/dx

For F=-kx V=1/2kx2

Vibrational Motion

For two masses connected by a spring: The force on particle 1 will be equal and opposite of the force

on particle 2. This force will only depend on the relative distance between

the particles. Define this relative distance as x = x1 – x2.

If we are to speak in terms of this relative coordinate, then we must also use a relative mass to validate F = ma = -kx.

21

21 mass reducedmm

mmu

QM of the harmonic oscillator

What is the SWE for a vibrational motion in terms of our relative coordinates x?

Mathematicians solved this problem long before quantum mechanics.

The solution has eigenvalues:

Notice that Ev is never zero. For v = 0, Ev= 1/2h This is defined as the zero point energy (ZPE)

Without ZPE the uncertainty principle would be violated. We’d simultaneously know both the position (0 displacement) and

momentum (0) of the particle. The energy level spacing is uniform:

,2,1,0 )21( vhvEv

h1 vv EEE

The Harmonic Oscillator Energy level diagram

The energy level diagram

,2,1,0 )21( vhvEv

h1 vv EEE

The harmonic oscillator wavefunctions

The wavefunctions (eigenfunction of the SWE) for a harmonic oscillator have the form:(x) = N x (polynomial in x) x (bell-shaped Gaussian function)

12.1. Tablein given spolynomial Hermitte theare )(

!2

1 ; where

)()(

2/1

41

2

22/12

xH

vN

k

exHNx

v

vv

x

vvv

The harmonic oscillator wavefunctions

41

2

22/1

!2

1 ; where

)()(2

vN

k

exHNx

vv

x

vvv

The probability distributions

Note: correspondence principle – the results are more classical as v increases

Properties of oscillators

Show that 0 and 1 are orthogonal. Calculate the average displacement of a harmonic

oscillator in its first vibrationally excited energy level (v = 1).

Vibrational Motion

What are the turning points of a classical harmonic oscillator in its ground vibrational state (v = 0)?

What is the probability of finding the particle outside the classical turning points?

Tunneling – penetration through classically forbidden zones.

Vibrational Motion

Show that the zero-point level of an harmonic oscillator is in accord with the uncertainty principle.

2

; 2222

px

pppxxx

2/1

1210

2

2

)12(312

nnbxn

b

nex

Useful Integrals

Rotational Motion in 2-D

Consider a particle of mass m constrained to a circular path of radius r in the xy plane (A particle on a ring).

Rotational Motion in 2-D

The de Broglie relation gives us the wavelength of a particle with momentum p.

We must place boundary conditions on such that condition (b) is met.

The circumference of the ring must be an integer multiple of the wavelength

l

l

m

r

rm

2

2

Rotational Motion in 2-D

An acceptable wavefunction for this problem is:

The probability density is independent of the angle so we know nothing about the particles location on the ring.

The sign on ml indicates the direction of travel, just as the sign on eikx indicated direction for our 1-D free particle.

There’s also an angular momentum operator that can operate on Y to give the angular momentum of our particle.

2/12)(

l

l

im

m

e

2

1

22)()( 2/12/1

* ll

ll

imim

mm

ee

Rotational Motion in 3-D

A particle on a sphere must satisfy two cyclic boundary conditions; this requirement leads to two quantum numbers needed to specify its angular momentum state.

Spherical polar coordinates

We can discard any term than involves differentiating with respect to r since r is constant.

sinsin

1

sin

1

legendrian theis squared)-(lamba where

12

2

2

22

2

222

22

rrrr

Rotational Motion in 3-D

The SWE for rotational motion is:

The wavefunctions are the spherical harmonics (see table 12.3).

The energy levels are:

There are 2l + 1 different wavefunctions (one for each value of ml) that are degenerate.

A level w/ quantum number l is (2l+1)-fold degenerate.

Angular Momentum for Particle on a sphere

Since energy is quanitized it follows that angular momentum J should also be quantized.

So the magnitude of the angular momentum = {l (l +1)}1/2ħ

I

JE

2

2

Putting it all together

Our particle on a ring in 2-D gave us the z-component of angular momentum = mlħ where ml = l,l-1,…-l

Think of ml as the angular momentum quantum number for the tip of a top recessing around the z-axis. The angular momentum of the spinning top will be specified by l.

Conclusion

Quantum mechanics says that a rotating body may NOT take up any arbitrary orientation with respect to some specified axis. This orientation restraint is called space quantization.

The quantum number ml is referred to as the orbital magnetic quantum number because it indicates the orientation of a magnetic field caused by the rotation of a charged body about an axis.

The Stern and Gerlach experiment verified the idea of space quantization.

(a) A magnet provides an inhomogeneous field that Ag atoms must pass through.

(b) Classically, the atoms should be deflected uniformly since they should have arbitrary configurations as they pass through the magnetic field.

(c)The observed behavior agrees with space quantization imposed by QM.

Space Quantization

Spin

The angular momentum of a particle due to motion about its own axis is called spin. Spin of a particle is sophisticated and actually comes from the theory

of relativity. DO NOT think of it as an actual spinning motion. For an electron, only one value of s is allowed, s = ½. There are thus

2s + 1 = 2 different orientations. ms = +1/2 (denotes as a or spin-up) ms = -1/2 (denoted as b or spin-down)

Fermions are particles with half-integer spin (electrons & protons). Bosons are particles with integer spin (photons & neutrons).

Summary of Angular Momentum Quantum #’s

Name Symbol Range of values

Orbital angular momentum quantum number

l 0, 1, 2, …..

Orbital magnetic quantum number ml 0,±1,…, ±l

Spin angular momentum quantum number

s ½ for an electron

Spin magnetic quantum number ms ±1/2 for an electron

Methods of Approximation

Every application we have encountered thus far has had an exact solution of the SWE.

Most ‘real’ problems do NOT have exact solutions.

There are two approximation methods for treating these ‘unsolvable’ problems1. Perturbation theory2. Variation theory

Perturbation Theory

Assume Hamiltonian of our ‘unsolvable’ problem is a sum of:1. A simple Hamiltonian, Ĥ(0), which has known

eigenvalues and eigenfunctions.2. A contribution Ĥ (1) which represents the extent to

which the true Hamiltonian differs from the simple Hamiltonian.

Perturbation Theory Example

Find the 1st order correction to the ground-state energy for a particle in a box where the bottom of the box is well-shaped with a variation in the potential of the form V = - sin(x/L)

Variational Method

Variation principle: If an arbitrary wavefunction is used to calculate the energy, the value calculated is never less than the true energy.

The arbitrary wavefunction is the trial wavefunction. The variational method allows us to calculate an upper bound

to the energy eigenvalue.

We can include an adjustable parameter in our trial wavefunction and adjust the parameter to minimize the variational energy.

Wd

dH

*

ˆ*The variational integralW ≥ Etrue