Simple Harmonic Simple Harmonic Oscillator (SHO)Oscillator (SHO)

Quantum Physics II

Recommended Reading:

Harris: chapter 4 section 8

The Simple Harmonic Oscillator (Classical)The Simple Harmonic Oscillator (Classical)

0xdt

xd

0kxdt

xdm kx

dt

xdm

maF

22

2

2

2

2

2

0

ω (1)

One of the most fundamental problems of physics is the simple harmonic oscillator. This problem occurs in areas as diverse as molecular vibrations, lasers, quantum descriptions of the electromagnetic field, phonon propagation, calssical mechanics ( pendulum, mass on spring)

Consider a particle of mass m subject to a restoring force F = -kx.

Classically (Newton 2nd Law) the equation of motion is

where mk

0 ω

(2)

0 is the natural frequency of the oscillator, it depends on the mass m of the particle and the strength of the spring k

tcosBtsinAx 00 ωω A general solution to equation (1) is

Quantum S.H.O. Quantum S.H.O. The Schrodinger equation uses the Potential rather than the force

dx FUx U

F

x

xRecall that force and potential are related

Knowing F we can calculate U

kxF Hooke’s Law Where k = spring constant N.m-1.

2kx21

dx kx U (3)

In quantum mechanics solving the simple harmonic oscillator reduces to solving the Schrodinger equation for the potential 2kx

21

U

Schrodinger Equation for S H OSchrodinger Equation for S H OThe Schrodinger equation for the potential given in equation (3) is

(4) Ex21

dx

dm2

22

22φφκ

φ

Difficult differential equation to solve. So we start by guessing its solutions and require that they satisfy boundary conditions.

The wave-function must approach zero as x and in this limit the solutions of eqn (4) behave like exponentials of -x2.

and comparing with the solutions for the infinite potential well, we expect just one peak in the lowest energy wave function.

x

2bxexpAx φ (5)

Where A and b are constants to be determined by requiring that this wave-function satisfies equation (4). Take first and second derivative of (5) and plug into (4)

xbx2 bxexpbxA2dx

xd 2 φφ (6)

bxexpA bx2 bx2 bxexpA b2dx

xd 222

2φ

(7)

Since the potential is symmetric about the origin expect the wave-functions to have alternating even and odd parity

So, for the ground state we try a solution of the form

x b2xb4bxexpA b2xb4dx

xd 222222

2φ

φ

differentiate again

collect terms together

Now plug equations (5) and (7) for (x) and (d2 / dx2) into equation (4) and see if there is a solution.

222 bxbx2bx222

Ae E Aekx21

Ae b2xb4m2

(8)0Em

b x

mb2

k21 2

222

Equation (8) must hold for all values of x the coefficients of each power of x must be equal to zero, this then gives

0m

b2k

21 22

and 0Em

b

2

So we can now solve for b and E. From (9) we find

(9) (10)

2m

2km

b 0ω and from eqn (10)

021

mk

2E ω (11) (12)

The constant A must be found from the Normalisation Condition

collect terms together

The NORMALISATION Condition is

1b2

AdxAedxt,x 22

bx2 2

πΨ

NormalisationNormalisation

So the constant A is therefore 81

22

41 mkπb2

A

πPutting this together with equation (5) we find

2

81

22x

2mk

expmk

xπ

φ0

Equation (12) gives the quantised lowest energy level while equation (14) gives the lowest wavefunction of a particle of mass m confined in a potential kx2/2. This is the ground state solution of the oscillator (Lowest energy) = ZERO POINT ENERGY

(13)

(14)

00 21

mk

2E ω (12)

Ground State (n=0) wavefunctionGround State (n=0) wavefunction

2

81

22x

2mk

expmk

xπ

φ0

Note that the oscillator extends beyond the classical turning points of the oscillator. (Compare this to the finite potential well where the wavefunction penetrated into the classically forbidden region.

classically forbiddn regions

Complete SolutionComplete SolutionThe full solution of the Schrodinger equation for Harmonic Oscillator potential (equation 4) is difficult. To find the wavefunctions and allowed energies, we assume that the general solution is of the form

2nnn bxexp)x(HAx φ

where Hn(x) are polynomial functions of x in which the highest power of x is xn. That is

n

0p

pp

nn

2210n xaxaxaxaa)x(H

We then find these functions by substituting equation 16 into the TISE and solving for the coefficients an, a lot of algebra

(15)

(16)

The polynomials that satisfy the Schrodinger equation for the Harmonic Oscillator potential are called the HERMITE Polynomials

Recall 2m

2km

b 0ω and define x b2z

then the first few Hermite Polynomials are:

Hermite PolynomialsHermite PolynomialsThe wave functions all have the form of a Hermite Polynomial Hn(x) multiplied by a Gaussian exp(-bx2) function . So we first look at the shape of the Hermite polynomials

H4(x)

12z48z16)x(H 244

H0(x)

1)x(H0

H1(x)

z2)x(H1

H2(x)

2z4)x(H 22

H3(x)

z12z8)x(H 33

H5(x), H6(x)...

12z48z16)x(H 244

1)x(H0

z2)x(H1

2z4)x(H 22

z12z8)x(H 33

Hermite PolynomialsHermite Polynomials

200 bxexpN)x( φ

211 bxexpz2N)x( φ

2222 bxexp2z4N)x( φ

2333 bxexpz12z8N)x( φ

bxexp12z48z16N)x( 22444 φ

The first five Hermite polynomials are defined as

where Nn is the Normalisation Constant which is determined from the normalisation condition. In general the normalisation constant for the nth wavefunction is given by 21

nn! n2

1N

π

What do the wavefunctions look like??

Ground State Wave functionGround State Wave functionAll of the SHO wave functions have a Gaussian term exp(-z2/2) which looks like this

Normalisation Normalisation FactorFactor 21

nn! n2

1N

π

Which for the first 5 wave functions is

n Nn

0 0.75111 0.53112 0.26563 0.10844 0.03835 0.0121

Putting everything together we find that the wavefunctions look like

SHO Wave FunctionsSHO Wave Functions

n=0

n=1

n=2

n=3

n=4

n=5

SHO Wave FunctionsSHO Wave Functions)x(φ 2)x(φ

Note PARITY of wavefunctions: Alternating even and odd parity

n = 0: even

n = 2: even

n = 4: even

n = 1: odd

n = 3: odd

SHO WavefunctionsSHO WavefunctionsFirst four wavefunctions and corresponding probability distributions for the SHO potential.

SHO Energy LevelsSHO Energy LevelsWe have already seen above that the ground state solution of the oscillator (Lowest energy = ZERO POINT ENERGY) is given by

00 21

mk

2E ω (12)

What about the energies of the higher states. When we solve the Schrodinger equation we find that the energies are quantised (just like the particle in an infinite potential well). However, the energy levels (spectrum) of the SHO are given by

0n 21

nE ω

n=0

n=5

n=4

n=6

n=3

n=2

n=1

Note that the energy levels are evenly spaced

Tim

e

req

Atom oscilating backwards and forwards

Application: Molecular VibrationApplication: Molecular Vibration

Application: Molecular VibrationApplication: Molecular Vibration

Chemical bonds are not rigid, the atoms in a molecule vibrate about an equilibrium position.

The force required to stretch or compress the bond length to r is proportional to (r-req)

The energy required to stretch the bond is 22

eq kx21

rrk21

E

Just like a spring which obeys Hooke’s Law.

where x = (r-req)

k = force constant of bond, units N.m-1

req

r

m1 m2

r

Can consider molecule as a single oscillating particle with an effective (reduced) mass

21

21mm

mm

In a diatomic molecule both masses oscillate relative to the centre of mass of the molecule.

centre of mass.

m1 m2

This id equivalent to a single oscillating particle of mass (reduced mass)

21

21mm

mm

Potential Energy CurvePotential Energy CurveConnection between shape of potential energy curve and strength of bond k.

22

02

2kx

21

xdx

Vd21

xV

...xdx

Vd21

xdxdV

0VxV 2

02

2

0

Potential energy curve is approximately parabolic and the force constant k is related to the potential by

02

2

dx

Vdk

2

2

dx

VdBig steep potential large k

Small shallow potential small k

2

2

dx

Vd HCl, k = 516 Nm-1, weak.

CO, k = 1902 Nm-1, strong.

Energy Levels in a Molecule Energy Levels in a Molecule Example: 1H35Cl has a force constant of 519 cm-1 find its oscillation frequency 0.

kg1061.1amu351351 27

μ

Hz100.9

kg1061.1

m.N 51621

2

13

27

10

0

πν

And therefore

m3.3

m103.3

sec100.9

s.m103c

6

113

18

μ

νλ

We can excite the 1H35Cl vibration with radiation of this wavelength infra-red

First we calculate the reduced mass of the molecule

we can then calculate the vibrational frequency of the molecule

This gives rise to the whole area of Infra red (IR) Spectroscopy of molecules

0ν Hz

n=0

n=2

n=1

n=3

n=4

Selection Rules:

n = 1 +1 absorption of photon

-1 emission of photon

Vibrational Spectrum Vibrational Spectrum

The energy of a photon required to cause a transition from the n = 0 to the n = 1 state is

μλω

ωωνω

khc

21

23

EEh

0

0012

Spectrum has only one line. Usually in the infra red region

all transitions will occur at the same frequency. Do you see why?

Schrodinger Equation Schrodinger Equation

xExkx21

x

x2

22

22ψψ

ψ

μ

Where is the effective mass of the molecule 21

21mm

mm

μ

2ynnn

2eyHN ψ

Hn(y) = Hermite Polynomials, and

ω

21

nEnμk

ω

vibrational quantum number, n = 0, 1, 2, 3, …

See Applet at following website: http://www.falstad.com/qm1d/