A few principles of classicaland quantum mechanics

The classical approach:

In classical mechanics, we usually (but not exclusively) solveNewton’s 2nd law of motion relating the acceleration a of thesystem of mass m to the external forces F:

2

2

dt

xd

dt

dvawithmaF

The kinetic energy K of the system expresses the energyinvolved in the motion of the system. Hence, K depends on themomentum p (or speed v = p/m) by:

m2

pmv

2

1K

22

The potential energy V of a system expresses the energyassociated with the position of the system within a force field F.Hence, V depends on the position x and is related to the forcefield by:

dx/dVF

The total energy E of a system is the sum of its kinetic energyK and of its potential energy V: VKE

The quantum mechanical approach:

In quantum mechanics, we need to solve the appropriateSchrödinger equation:

EH

where H is the Hamiltonian operator and is thewavefunction describing the state of our system at energy E.

We shall see that only certain wavefunctions and theircorresponding energies are acceptable. Hence, quantization ofenergies is a natural consequence of the equation and theconditions imposed on it (boundary conditions).

For a single particle of mass m, the Hamiltonian operator takesa simple form:

)x(Vdx

d

m2H

2

22

Very importantly, we notice that the first term defining H canbe related to the kinetic energy of the particle. It follows thatthe momentum operator is defined by:

dx

diP

Translational motion

The classical approach:

The classical description of free motion (V = 0) for a particleof mass m and momentum p in one dimension is given by:

m2

pE

2

Furthermore, its position is perfectly defined at all times:

000 xtvxvv0a0F

The quantum approach:

We solve the Schrödinger equation: EH , with

2

22

dx

d

m2H

hence:

E

dx

d

m2 2

22

The general solutions are (indexed with k):

xixi BeAe kkk

with:m2

E22k

k

All values of k are permitted. It follows that translational

motion of a free particle is not quantized.2

k is

independent of x position of the particle is completelyunpredictable. This is consistent with Heisenberg’suncertainty principle since the momentum is certain:

xhence0p2/xp

Vibrational motion(Characteristic of the harmonic oscillator)

The classical approach:

A particle undergoes harmonic motion if itexperiences a restoring force proportionalto its displacement (no dissipation):

kxF (k = force constant)

Potential energy is derived from forcefield:

2kx2

1V

dx

dVF

We solve the problem with Newton’s law:

0kxdt

xdmmaF

2

2

This is a well known differential equation whose solutions are:

tcosA)t(x 0 withm

k0

Total energy of the system: 222 kA2

1kx

2

1mv

2

1E

We notice that the total energy of a given vibration (orharmonic oscillation is a constant, independent of time (henceof frequency = 20).

Classically, as expected, the total energy can take any value(since A can take ‘any value’).

The quantum approach:

Schrödinger’s equation:

Ekx2

1

dx

d

m2

22

22

This is a standard differential equation whose solutions areknown (see later). We solve for En by forcing the wavefunctions

xat0n (boundary conditions):

2

1E nn with:

m

k and: n = 0, 1, 2, …

Separation in energy for twoadjacent levels:

nn EE 1

for any given level n.

Therefore, the energy levelsform a ladder of spacing .

Zero-point energy: 2

1E0

The detailed solutions for the wavefunctions have the form:

4/12

2/y

mk;

xy:withe)y(HNx

2

nnn

Nn = normalized constantHn(y) = Hermite polynomial

2/y2e = Gaussian function

Expressions for the ground state and the 1st excited state:

2/x1

2/y11

2/x00

22

22

ex2

Nye2Nx

eNx

Probability of locating the ground state particle in position x:

22 /x2

020 eNx

An example of application:

Atoms vibrate to one another in molecules with the bond actinglike a spring.

Consider an X-H chemical bond, where a heavy X atom formsa stationary anchor for the very light H atom. That is, only theH atom moves, vibrating as a simple harmonic oscillator.

The force constant of a typical X-H chemical bond is around500 Nm-1; for example, 516.3 Nm-1 for the 1H35Cl bond.

Because the mass of the proton is about 1.7·10-27 kg, using500 Nm-1 we get = 5.4·1014 s-1 (5.4·102 THz). It follows thatthe separation of adjacent levels is = 5.7·10-20 J (about0.36 eV). This energy separation corresponds to 34 kJ mol-1,which is chemically significant.

The zero-point energy of this molecular oscillator is about2.85·10-20 J, or 0.18 eV, or 17 kJ mol-1. The excitation of thevibration of the bond from one level to the level immediatelyabove requires 5.7·10-20 J.

Therefore, if it is caused by a photon, the excitation requiresradiation of frequency = E/h = 86 THz and the wavelengthis = c/ = 3.5 m. The associated wavenumber is about =1/ = 2900 cm-1. It follows that transitions between adjacentvibrational energy levels of molecules are stimulated by or emitinfrared radiation.

Rotational motion

The classical approach:

The rotational motion of a particle ofmass m around a central point at a fixeddistance r (radius) is described by its

angular momentum, J

(a vector!). Its

magnitude, J = J

is given by:

J = I = angular velocity (rad/s).I = moment of inertia (kg m2).

For a point of mass m moving in a circle of radius r, themoment of inertia about the axis of rotation is given by:

I = mr2

Note the analogous roles of m and I, of v and , of p and J inthe translational and rotational cases, respectively, with = v/rand J = pr.

Its rotation energy is - as function of : 2I2

1E

- as function of J:I2

JE

2

(V = 0). Classically, we expect the rotational energy to take anyarbitrary value.

The quantum approach:

The treatment is broken down in two parts: motion in 2D andmotion in 3D.

1) Rotation in 2D: a particle on a ring

We recall from classical mechanics:

I2

JE

2z

Because Jz = ±pr, and, from the de Broglie relation, p = h/,the angular momentum about the z-axis is: Jz = ±hr/.

For meaningful sense, the only allowedwavelengths are:

lm

r2 ml = 0, ±1, ±2,…

When: 0lm

Hence, the angular momentum is quantized, that is:

,...2,1,0Jz ll mm

ml > 0 (< 0) corresponds to CW (CCW) rotation around the z-axis.

It follows that the energy is limited to the values:

,...2,1,0I2I2

JE

222z l

l mm

Formally we solve Schrödinger’s equation (in 2D, V = 0):

2

2

2

22

yxm2HEH

Transformation to cylindrical coordinates:

)sin(ry

)cos(rx

0rwithr

1

r

1

rr

1

ryx

2

2

2

2

2

22

2

2

2

2

2

Hence:2

22

2

2

2

2

I2mr2H

Schrödinger’s equation becomes:

22

2 EI2

d

d

Normalized general solutions are:

2/1

i

2

e)(

l

l

m

m with

2/1EI2lm .

Cyclic boundary conditions: )()2( .

By substitution it becomes: lll

mmm

21)()2(

Hence: 11 2 lmml = 0, ±1, ±2,…

The ground state ( 0lm ): 2/10 2/1)( has the

same value at all points on the circle.

We arrive at the following conclusions:(1) energy quantized (ml) and independent

of sense of rotation ( 2lm ).

(2) levels doubly degenerate for lm >0

(except for ml = 0).(3) wavefunctions have increasing # of

nodes with increasing Jz.(4) decreases as lm increases.

(5) probability density is 2/1*ll mm ,

independent of ml and of hence thelocation of the particle is completelyindefinite (since Jz or p are preciselyknown!)

2) Rotation in 3D: a particle on a sphere

A second cyclic boundary condition (thewavefunction should match as a path istraced over the poles as well as round theequator of a sphere of radius r) will introducea second quantum number, l.

Again, we solve Schrödinger’s equation (in 3D, V = 0) with:

Em2m2

H 22

22

where2

2

2

2

2

22

zyx

= laplacian (for convenience)

Transformation to spherical coordinates:

)cos(rx

)sin()sin(ry

)sin()cos(rx

2

2

2

22

22

r

1

r

1

rr

2

r

with

sin

)sin(

1

)(sin

12

2

2

2

Schrödinger’s equation becomes: 2

2 EI2

Separation of variables , and substitution in

Schrödinger’s equation leads to two equations:

222

22

2

)(sinEI2

d

d)sin(

d

d)sin(

d

d1ll mm

The first is exactly what we found for the 2D case; hence it hasthe solutions described earlier. The second is much morecomplex to solve. Cyclic boundary conditions on result in theintroduction of a second quantum number, l, identifyingacceptable solutions and restricting the values of ml by thevalue of l.

Quantization: l = 0, 1, 2,… and ml = l, l-1, …, -l+1, -l.l is named the orbital angular momentum quantum number.ml is named the magnetic quantum number.

Energy is:I2

)1(E2

ll , l = 0, 1, 2,… independent of ml, hence

a level with quantum number l is (2l + 1)-fold degenerate.

Normalized wavefunctions ),(Y lml, = spherical harmonics.

It follows that: )1(J ll l = 0, 1, 2,…

And we saw that: lmzJ , ml = l, l-1,…, -l

Note that: (1) Number of angular nodes increase with l.(2) No angular nodes around z-axis for ml = 0.(3) For a given l, the probability of migratingtowards the x-y plane increases with ml.

An example of application:

Under some circumstances, the particle on a sphere is areasonable model to describe the rotation of diatomicmolecules.

Consider the rotation of a 1H127I molecule: because the largedifference in atomic masses, it is appropriate to picture the Hatom as orbiting a stationary I atom at a distance r = 160 pm,the equilibrium bond distance. The moment of inertia of HI isthen: I = mHr2 = 4.288·10-47 kg m2

It follows that: meV810.0J10297.1I2

222

This energy corresponds to 78.11 Jmol-1. The table belowreports a few values for the first four levels: (1) rotationalenergy levels (in meV), (2) the degeneracies of the levels, and(3) the magnitudes of the angular momentum of the molecule:

l (1) (2) (3)0123

01.6194.8579.712

1357

021/261/2

121/2

It follows from our calculation that the l = 0 and l = 1 levels areseparated by E = 2.594·10-22 J = 1.619 meV. A transitionbetween these two rotational levels of the molecule can bebrought about by the emission or absorption of a photon with afrequency given by the Bohr frequency condition:

GHz5.391Hz10915.3h

E 11

Radiation with this frequency belongs to the microwave regionof the EMR spectrum, so microwave spectroscopy is aconvenient method for the study of molecular rotations.