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PHYS 20101INTRODUCTION TO QUANTUM MECHANICS

Prof W R Flavell

I. Wave Mechanics: the Schrödinger Wave Equation and theWavefunction

I.a) The Schrödinger Wave Equation

The initial ideas behind wave mechanics were formulated independently byHeisenberg and Schrödinger in 1926.

The Schrödinger Wave Equation may not be deduced from classicalmechanics, or proved in any other way, other than retrospectively as itspredictions are confirmed by experiment.

Schrödinger began with the De Broglie equation:

€

λ =hp

(1)We assume that the relationship

€

E = hν = hω

h =h2π,

ω = 2πν(angular frequency)

(2)

holds for particles as well as for photons. If we represent a moving particleby a progressive wave, it will have

wavevector

€

k =2πλ

=2πph

=ph

angular frequency

€

ω =Eh

2

For a non-relativistic particle

€

E =p2

2m for a free particle, i.e. all the energy of the particle is kinetic energy.

€

hω =h2k 2

2m(3)

Let us consider what sort of wave has these properties.

As a trial solution, let us try for a freely propagating particle:

€

Ψ(x, t) = Aei(kx−ωt )

then the relation (3) holds if the wave equation is of the form:

€

−h2

2m∂ 2Ψ∂x2

= ih∂Ψ∂t

(4)This is the time-dependent Schrödinger equation for a free particle of massm.

Equation (4) holds because:

€

∂ 2

∂x2ei(kx−ωt ) = (ik)2ei(kx−ωt )

∂∂t

ei(kx−ωt ) = (−iω)ei(kx−ωt )

so that

€

−h2

2m∂ 2Ψ∂x2

= −h2

2m(ik)2Ψ =

h2k 2

2mΨ

ih∂Ψ∂t

= ih(−iω)Ψ = hωΨ

More generally, the particle will be subject to some potential, V (e.g. for anelectron in the hydrogen atom, the attractive potential from the positively-charged nucleus). The total energy, E will be the sum of the kinetic energy

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(T) and the potential energy, V. If the potential varies with time (V(x,t)), thetotal energy will not be conserved.

For a particle moving in a possibly time-dependent potential, V(x,t), theequation corresponding to (4) is:

€

−h2

2m∂ 2Ψ∂x2

+V (x,t)Ψ = ih∂Ψ∂t

(5)(The 1D time-dependent Schrödinger equation, 1D TDSE)

I.b) The wavefunction

Ψ (x,t) is called the wavefunction of the particle.

In its general form it is complex, so it cannot be directly identified with anyphysical property of the system (but see later for the crucially importantinterpretation due to Born). However, the wavefunction contains all theinformation that we can know about the system.

I.c) The TDSE in 3 dimensions

We are more usually dealing with a three-dimensional wavefunction (e.g. ifwe are considering the properties of an electron in an atom). The mostgeneral wavefunction would be a function of three spatial coordinates andtime,

€

Ψ(x, y,z,t). Equation (5) then becomes:

€

−h2

2m∂ 2Ψ∂x 2

+∂ 2Ψ∂y 2

+∂ 2Ψ∂z2

+V (x, y,z,t)Ψ = ih∂Ψ

∂t

which is often expressed:

€

−h2

2m∇2Ψ+V (r,t)Ψ = ih∂Ψ

∂t

or

€

−h2

2m∇2 +V (r,t)

Ψ = ih∂Ψ

∂t(6)

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which is a statement of the TDSE in 3 dimensions

Here the dependence of V on spatial coordinates is expressed by the vectorr, used and

€

∇2 (called ‘del squared’) is the ‘Laplacian operator’,

€

∂ 2

∂x 2+∂ 2

∂y 2+∂ 2

∂z2

which will come back to haunt us later.

I.d) Temporal evolution: the time-independent SchrödingerEquation and ‘Stationary States’

If the potential V is not a function of time,

€

V (x, t)→V (x) (in one dimension),or more generally,

€

V (r,t)→V (r), then classically the total energy, E (=T+V)is conserved (time-independent).

In this special case, we have in the one-dimensional case

€

−h2

2m∂ 2Ψ∂x2

+V (x)Ψ = ih∂Ψ∂t

(7)

When the potential field V(x) does not explicitly include the time t, theequation may be solved by separation of variables, writing:

(8)

then

€

∂ 2Ψ(x, t)∂x 2

=d2ψdx 2

T(t)

∂Ψ(x,t)∂t

=ψ(x) dTdt

and if we substitute this into the 1D TDSE (7), we get:

(9)

€

Ψ(x, t) =ψ(x)T(t)

€

−h2

2md2ψdx 2

+V (x)ψ(x)

T(t) = ihψ(x) dT(t)

dt

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Rearranging to separate the time and space variables:

(10)The LHS depends only on x, while the RHS depends only on t. The twosides are therefore independent and must be equal to the same constant, E(which in fact has dimensions of energy, as we will see in (12) that 2πEt/hhas to be dimensionless):

(11)

Solution of (11) gives:

€

T(t) = e− iEt

h = e− iωt

or, more properly

(12)where

€

E = hω is the total energy of the system.

Assimilating the constant T0 into the form of ψ(x), this means the generalform of the wavefunction is

(13)In addition, from (10), we have:

(14)

This is the one-dimensional time-independent Schrödinger equation (TISE)

Solutions of the TISE are called ‘stationary states’ (see later). They arestates with a definite energy.

€

1ψ(x)

−h2

2md2ψdx 2

+V (x) =ihT(t)

dT(t)dt

€

ihT(t)

dT(t)dt

= E

€

−h2

2md2ψdx 2

+V (x)ψ = Eψ

€

T = Toe−iEt / h

€

Ψ(x, t) =ψ(x)e−iEt / h =ψ(x)e− iωt

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I.e) Interpretation of the wavefunction (due to Born)

There is no direct interpretation of Ψ. This ‘particle wave’ replaces theclassical trajectory of a particle.

One consequence of the uncertainty principle is that we can no longer talk ofcertainties, in terms of where a particle (such as an electron) is, but only interms of probabilities. Born suggested that the wavefunction is related to theprobability of finding the particle.

In 1D:

€

Ψ(x,t) 2dx = Ψ* (x,t)Ψ(x, t)dx = P(x, t)dx(15)

= the probability to find the particle in the interval

€

x→ x + dx at time t.

P(x,t) is called the ‘probability density’ to find the particle at x at time t.(sometimes written as ρ(x,t)).

So in 3 dimensions, Ψ2dv (or more properly Ψ*Ψdv if Ψ is complex) is theprobability of finding a particle in a given volume of in space (in volume dv)at time t.

(c.f. relationship between the amplitude and intensity of a wave.)

z

dv1 r1 y r2 x

dv2

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€

P1P2

=Ψ∗ (r1,t)Ψ(r1,t)dv1Ψ∗ (r2,t)Ψ(r2,t)dv2

=Ψ(r1,t)

2dv1

Ψ(r2,t)2dv2

(16)where P1,P2 are the probabilities of finding the particle in regions dv1 and dv2respectively.

But the particle must be somewhere, so for a ‘normalised wavefunction’:

€

Ψ(r,t)0

∞

∫2

dv =1

(i.e. total probability of finding 1 particle in all space =1)

or in one dimension:

€

Ψ(x,t)−∞

∞

∫2

dx =1

(17)The interpretation of the wavefunction in terms of probabilities is criticallyimportant, as it imposes a series of conditions on the wavefunction, Ψ.These in turn restrict the possible solutions of the Schrödinger equation.

1. The wavefunction must be finite at any point. If Ψ becomes infinite,the probability density is infinite, which corresponds to a certainty offinding the particle.

x

ψ

x

ψx

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2. Ψ must be single-valued – there must be only one probability offinding the electron at a particular point.

3. Ψ must be continuous, and must become zero at infinity. If theprobability of finding the particle at x has a particular value, that at(x+dx) must have a similar value as

€

x→ 0.

4. The derivative

€

∂Ψ ∂x must be continuous, except where there is aninfinite discontinuity in the potential energy function V(x). (Thisfollows from the SE.)

5. Note that if our wavefunction has the general form given in (13):

then the probability density is given by

€

Ψ(x, t) =ψ(x)e−iEt / h =ψ(x)e− iωt

x

ψ x

x

ψ x

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(18)which does not depend on t. For a stationary state the time-dependent partof the wavefunction has no physically observable quantitities. Thus fromthe point of view of calculating the probability of finding a particle, we mayat least begin by ignoring the time-evolution of the solutions, and using thetime-independent form of the Schrödinger equation (TISE) shown in (14).For most purposes, as we shall see, this is a reasonably good starting point.

Considerations such as the ones above mean that only certain solutions ofthe TISE are possible – the solutions have to satisfy certain ‘boundaryconditions’ (see later). We will see that this has very importantconsequences as it often means that only a discrete set of wavefunctions,each with a discrete energy is allowed.

Was Born right?You will see as the course progresses that Born’s interpretation isfundamental to how we interpret wave mechanics. So we should check thatequation (15) is correct.

Artist’s impression takenfrom Physics World, 23,September 2010, page 4.

An important experiment that shows that he was, as far as we can tell, hasbeen published recently. It involves passing single photons through aYoung’s slit experiment with 3 slits. If equation (15) (or its equivalent in3D) holds, interference between the wavefunctions passing through each slitalways occurs in pairs of possibilities (no third order interference terms, aconsequence of the square exponent in (15)); and none is seen in theexperiment.

To read more see: Physics World, 23, September 2010, page 4U Sinha, C Couteau, T Jennewein, R Laflamme and G Weihs, Science, 329,414 (2010) (23rd July 2010).

€

Ψ* (x, t)Ψ(x, t) =ψ * (x)ψ(x)

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II. Operators, expectation values and uncertainty

II. a) Observables, operators and expectation values

Quantum-mechanically, observables (things that can be measured, e.g.position, momentum, kinetic energy) are represented mathematically byoperators (which ‘operate’ on the wavefunction, ψ)

We can rewrite the TISE (14) as:

€

−h2

2md2

dx 2+V (x)

ψ(x) = Eψ(x)

(19)

We can interpret the lefthand side of this in terms of a mathematical‘operator’,

€

ˆ H , operating on the wavefunction:

€

ˆ H = −h2

2md2

dx 2 + V (x) , known as the Hamiltonian operator

Then the TISE reads (in its most famous form):

€

ˆ H ψ = Eψ

As we saw in the last section, the limitations on the choices of wavefunctionmean that solutions of the TISE have to satisfy certain ‘boundaryconditions’. This often restricts the set of solutions to a discrete set ofallowed wavefunctions, ψ1(x), ψ2(x)….. (called ‘eigenfunctions’), each withan energy (or ‘eigenvalue’) E1, E2,……. etc.

i.e.:

€

ˆ H ψn (x) = Enψn (x)(20)

for these solutions, where

€

ˆ H is the Hamiltonian operator, ψn(x) is aneigenfunction of

€

ˆ H , and En is an eigenvalue of

€

ˆ H . Equations of the generaltype of (20), together with boundary conditions are called ‘eigenvalueequations’.

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The fact that we typically obtain only a set of allowed wavefunctions, with acorresponding set of allowed energies leads us to the very importantdemonstration that quantisation of energy arises naturally in wavemechanics.

The corresponding solutions to the TDSE (see (13)) have the form:

€

Ψn (x, t) =ψn (x)e− iEnt

h

(21)These solutions are called ‘stationary states’ (states of definite energy).They are not time-independent, BUT all the measurable quantities, such asthe probability density |Ψ(x,t)|2 are time-independent, as we saw in (18).

• If the system is in the stationary state Ψn(x,t), it has a definite energy,En.

• A measurement of the energy in the state Ψn will always give thevalue En.

• After the measurement, the system will still be in the state Ψn. Thestate Ψn is an ‘eigenstate’ of energy En.

The momentum operator:

The operator for the x component of momentum is

€

ˆ p x = −ih ∂∂x

(22)For example, take a simple wavefunction

€

Ψ(x, t) = ei(kx−ωt ) (a sinusoidaltravelling wave):

€

ˆ p xei(kx−ωt ) = −ih ∂

∂xei(kx−ωt )

= hkei(kx−ωt )

(23)In other words, the wavefunction

€

Ψ(x, t) = ei(kx−ωt ) is an eigenfunction of

€

ˆ p xwith eigenvalue

€

hk . So the state

€

Ψ(x, t) = ei(kx−ωt ), (or

€

ψ(x) = eikx) represents aparticle with a definite momentum:

€

px = hk(24)

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(If px is measured in this state, you will always get the value

€

hk , and thesystem is in the same state after the measurement.)

The position operator:

€

ˆ x = x(25)

(i.e.

€

ˆ x ψ = xψ , the effect of the operator is to multiply by x.)

The kinetic energy operator:

Classically,

€

T =p2

2m (kinetic energy)

so it follows from (22) that:

€

ˆ T =ˆ p 2

2m= −

h2

2md2

dx2

(26)(In one dimension)

Now the total energy, E=T+V (the sum of kinetic and potential energy), sothe total energy operator (in one dimension) is:

€

ˆ H =ˆ p 2

2m+ V ( ˆ x ) =

−h2

2md2

dx 2 + V (x)

(27)

(where

€

ˆ H is the Hamiltonian operator, which we introduced in eqn (19). Soif our wavefunction is an eigenfunction of

€

ˆ H , then the action of the energyoperator on the wavefunction is to generate an eigenvalue for the energy,multiplied by the wavefunction,

€

ˆ H ψ = Eψ , as we saw at (20).

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Let us also note for future reference the comparison between (27) and theTDSE in eqn (5), which would suggest that another way of stating a generalenergy operator,

€

ˆ E is:

€

ˆ E = ih ddt

= −h

iddt

(28)We will return to this point later.)

General operator:

For any observable A (something that can be measured, such as energy,momentum, position) there is an operator

€

ˆ A whose eigenvalues an give thepossible values of A.

€

ˆ A ψn = anψn

n =1,2,............(29)

where ψn are the eigenfunctions of the operator

€

ˆ A , and an are the eigenvaluesof

€

ˆ A .

• The eigenvalues, an, of

€

ˆ A are the only possible values of A.• If the system is in the state ψn, the value an is always returned.• The system stays in the state ψn after the measurement.

Expectation values

If the system is in some arbitrary state Ψ (x,t), which is not necessarily aneignfunction of the operator,

€

ˆ A , the expectation value of A is

€

A = Ψ* ˆ A −∞

∞

∫ Ψdx

(30)If A is measured an infinite number of times in the same state, Ψ,

€

A is thevalue obtained. In the general situation where Ψ (x,t) is not an eignfunctionof the operator,

€

ˆ A , then at different times, or in different experiments aimedat measuring A, a range of different eigenvalues will be measured.

The operator that corresponds to the deviation of

€

ˆ A from

€

A is:

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€

Δ ˆ A = ˆ A − A

and we define the mean square deviation of

€

ˆ A from its mean value as:

€

(Δ ˆ A )2 = ( ˆ A )2 − A 2

(31)So the square of the uncertainty, or the ‘variance’ in e.g. the position andmomentum of a particle are given by:

€

Δx( )2= (Δ ˆ x )2 = ( ˆ x )2 − x 2

= x2 − x 2

Δpx( )2= (Δ ˆ p x )2 = ( ˆ p x )2 − px

2= px

2 − px2

(32)

If the wavefunction is an eigenfunction of the operator

€

ˆ A (and is alsonormalised), then we have:

€

A = Ψ* ˆ A −∞

∞

∫ Ψdx = a Ψ*−∞

∞

∫ Ψdx = a

(33)i.e. with complete certainty, only one value will be measured, and that valueis a.

Note:We saw earlier in (18) that for a stationary state, the probability density doesnot depend on t. More generally, the time-dependent part of thewavefunction has no physically observable quantities. For example theexpectation value of a general operator is:

€

A = dxΨ* (x,t) ˆ A Ψ(x,t)−∞

∞

∫

= dxψ * (x) ˆ A ψ(x)−∞

∞

∫because the time-dependent parts cancel,

€

eiEt / he− iEt / h =1

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II.b) Commutators, ‘compatible’ observables and uncertainty

Consider the operators

€

ˆ A and

€

ˆ B . The ‘commutator’ is defined as:

€

ˆ A , ˆ B [ ] = ˆ A ̂ B − ˆ B ̂ A

(34)If

€

ˆ A ̂ B = ˆ B ̂ A , i.e.

€

ˆ A , ˆ B [ ] = 0, the operators are said to ‘commute’, i.e.:

€

ˆ A ̂ B Ψ = ˆ B ̂ A Ψ for any Ψ.

If operators commute, they can share the same eigenfunctions (they have‘simultaneous’ eigenfunctions):

Suppose

€

ˆ A Ψn = anΨn and

€

ˆ B Ψn = bnΨn , then if the operators commute, it followsthat:

€

ˆ A ̂ B Ψn = ˆ B ̂ A Ψn = anbnΨn

(35)However, if the operators do not commute, they cannot have simultaneouseigenfunctions.

Why? If operators

€

ˆ A and

€

ˆ B have a simultaneous eigenfunction, φ, then

€

ˆ A φ = aφˆ B φ = bφ

then

€

ˆ A ̂ B φ = ˆ A bφ = abφˆ B ̂ A φ = ˆ B aφ = baφ = abφ

so

€

ˆ A ̂ B φ = ˆ B ̂ A φ

which implies

€

ˆ A ̂ B = ˆ B ̂ A

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‘Compatible observables’ in quantum physics are those which can bedetermined without mutual interference or contradiction. The commutatorhas an important role in telling us which physical observables can bedetermined simultaneously and precisely.

Some examples

€

ˆ x and

€

ˆ p x:

€

ˆ x , ˆ p x[ ]Ψ = x −ih ddx

Ψ− −ih d

dx

xΨ = ihΨ

for any Ψ.

So

€

ˆ x , ˆ p x[ ] = ih(36)

This means that a particle cannot simultaneously have precise values ofx and px.

(in fact this is a statement of the Uncertainty Principle: see later)

However,

€

ˆ x , ˆ p y[ ]Ψ = x −ih ddy

Ψ− −ih d

dy

xΨ = 0

So

€

ˆ x , ˆ p y[ ] = 0 ,i.e. the particle can simultaneously have precise values of x and py.

II.c) Hermitian operators

Definition: An operator

€

ˆ A is Hermitian if

€

φ * ˆ A ψ( )dx = ˆ A φ( )∫∫ *ψdx

(37)for all normalisable φ and ψ.

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This is a particularly important class of operator, as all operators thatrepresent physical observables are Hermitian. These operators have anumber of important properties:

1. All eigenvalues of a Hermitian operator are real.

Proof:Suppose that φ is an eigenfunction of the Hermitian operator

€

ˆ A witheigenvalue

€

a, i.e.:

€

ˆ A φ = aφ

Then:

€

φ * ( ˆ A φ)dx =∫ ( ˆ A φ) *φdx∫φ * (aφ)dx =∫ (aφ) *φdx∫

a φ *φdx =∫ a * φ *φdx∫

so

€

a must be real (it is equal to its complex conjugate).

2. Eigenfunctions of

€

ˆ A with different eigenvalues are orthogonal if

€

ˆ A isHermitian

If 2 functions, φ and ψ, are orthogonal, we mean that the integral of theirproduct is zero:

€

φ *ψdx = ψ *φdx = 0∫∫(38)

Proof:Let φ1,φ2 be the eigenfunctions of

€

ˆ A with eigenvalues

€

a1,a2Then:

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€

ˆ A φ1 = a1φ1

ˆ A φ2 = a2φ2

φ1∫ * ˆ A φ2dx = φ1∫ * a2φ2dx = a2 φ1∫ *φ2dx =

( ˆ A φ1) *φ2dx∫ = (a1φ1∫ ) *φ2dx = a1 φ1∫ *φ2dx

a2 − a1( ) φ1∫ *φ2dx = 0

φ1∫ *φ2dx = 0(provided

€

a1 ≠ a2.)

In deriving this result we have used the definition of Hermiticity, andpostulate 1 above.

Using the properties of Hermitian operators, one may derive (albeit a littletortuously) a relation between the mean square deviation of two Hermitianoperators and their commutator:

€

(Δ ˆ A )2 (Δ ˆ B )2 ≥14

ψ * Δ ˆ A ,Δ ˆ B [ ]ψdx−∞

∞

∫2

(39)So returning to the example of the uncertainties in position and momentum,expressed in (36), we can write:

€

(Δˆ x )2 (Δˆ p x )2 = (Δx)2(Δpx )2 ≥14

ψ * Δˆ x ,Δˆ p x[ ]ψdx−∞

∞

∫2

(Δx)2(Δpx )2 ≥h2

4

ΔxΔpx ≥h

2(40)

(the more familiar form of the Uncertainty Principle.)

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II.d) The Superposition Principle

The TDSE (from (5)) may be written:

€

ih∂Ψ∂t

= ˆ H Ψ

(41)

This is linear in the wavefunction, so if Ψ1 and Ψ2 are solutions, then so is(AΨ1 + BΨ2), where A and B are possibly complex constants.

The Principle of Superposition states that:

If

€

Ψ1,Ψ2,Ψ3 .............Ψn are n solutions of the TDSE for a system, then a linearsuperposition

€

Ψ = c1Ψ1 + c2Ψ2 + c3Ψ3..........+ cnΨn is also a solution (where theconstants c1…… may be complex).

Does this imply that a particle can be in more than one energy state at thesame time?

Suppose:

€

ˆ H Ψ1 = E1Ψ1

ˆ H Ψ2 = E2Ψ2

The states Ψ1 and Ψ2 are states of definite energy E1 and E2 (stationarystates) with

€

E1 ≠ E2.

As we saw in (13), the time-dependent wavefunctions are given by:

and the TISE are:

€

ˆ H ψ1 = E1ψ1

ˆ H ψ2 = E2ψ2

€

Ψ1(x, t) =ψ1(x)e− iE1t / h

Ψ2(x, t) =ψ2(x)e− iE2t / h

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Solutions of the TISE are ‘stationary states’, with definite energy E1 and E2.However, the superposition:

€

Ψ(x, t) = Aψ1(x)e− iE1t / h + Bψ2(x)e

− iE2t / h

(42)is not a stationary state – it is a superposition of states with different E, or astate of uncertain energy. An experiment to measure the energy will yield aparticular value, E1 or E2 (but a series of experiments will not yield the samesingle definite result for each measurement; this is discussed further insection III, where we discuss ‘non-stationary’ states.) The probability toobtain E1 is |A|2, and to obtain E2 is |B|2 (assuming Ψ(x,t) is normalised.)

However, until such a measurement is made, the description of the state isgiven by a wavefunction built up from different energy eigenfunctions. Sothe answer to our question is basically, yes, in this sense, it is possible for aparticle to be in more than one energy state at the same time! (And in fact,this is the basis for quantum computing…. more later in Section III)

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III. One-dimensional Infinite Potential Wells

III.a) The ‘Infinite Square Well’

Let us consider the simple, rather abstract case of a particle in a box (1D).

The particle is confined to the box by an infinite potential at each wall. Letus imagine that the potential inside the box is zero:

Step 1: We need the 1D TISE:

€

d2ψdx 2

+8π 2mh2

(E −V )ψ = 0

(43)

Step 2: introduce the potential, V:

In regions I and III, where

€

V =∞,

€

d2ψdx2

+8π 2mh2

(E− ∞)ψ = 0

(44)In region II, where the particle is constrained to be,

I II III

0 L

V=0

€

V =∞

€

V =∞€

∞

€

∞

22

€

d2ψdx 2

+8π 2mh2

(E − 0)ψ = 0

(45)

Step 3: Solve the equations subject to the limitations on

€

ψ .

In regions I and III,

€

ψ must be =0, because for any finite value of

€

ψ , theLHS of eqn (44) will be infinite, not zero.

In region II, rearranging (45) we have:

€

−h2

8π 2md2ψdx2

= Eψ

(46)

or

€

d2ψdx 2

+ k 2ψ = 0,

k 2 =2mEh2

(47)We need to find ‘well-behaved’ solutions to this. Two independentsolutions to this equation are sinkx and coskx (or eikx and e-ikx)

So the general solution is:

€

ψ(x) = c1 sin(kx) + c2 cos(kx)(48)

What limits do the boundary conditions impose?We know that ψ=0 in regions I and III, and we know that ψ must becontinuous. This means that

€

ψ(0) = 0ψ(L) = 0

(49)at x = 0,

€

0 = c2

23

at x=L

€

0 = c1 sin(kL) + c2 cos(kL)= c1 sin(kL)so

€

sin(kL) = 0(50)

(as if c1=0, the wavefunction is zero everywhere, i.e. no particle in the box).

Hence

€

kL = nπ(n =1,2,3.....)

so the allowed values of k are restricted to

€

kn =nπL

n =1,2,.......(51)

which means that only certain values of the energy are allowed, belonging tothe set

€

En =h2

2mkn2 =

n2π 2h2

2mL2

(52)with the wavefunction given by

€

ψ(x) = An sinnπxL

(53)where An is the normalisation constant

(n=0 gives a mathematical solution, but gives ψ=0 for all x, i.e. there is noparticle in the box.)

The complete set of stationary states (energy eigenfunctions) for a particle ina 1D infinite ‘square-well’ potential is therefore

24

€

Ψn (x, t) = An sinnπxL

e

− iEnth ,

0 ≤ x ≤ L with n=1,2,3…..

(54)and

€

Ψ(x, t) = 0x < 0,x > L

(55)i.e. outside the box.The probability density for the position, x, in the nth state is

€

Ψn (x,t)2

= An2 sin2 nπx

L

0 < x < L(and Ψ=0 elsewhere)

(56)The normalisation condition becomes

€

An2 sin2 nπx

L

0

L

∫ dx =1

(57)(since Ψ=0 outside the well), giving:

€

An2L2

=1,

An =2L

so the normalised energy eigenstates are

€

Ψn (x, t) =2Lsin nπx

L

e

− iEnth ,

0 ≤ x ≤ L

where

€

En =n2π 2h2

2mL2(58)

25

The solutions look like this:

(from R L DeKock and H B Gray, Chemical Structure and Bonding, Benjamin/Cummings, Menlo Park, Califorrnia, 1980.)

General points:• We have quantised energy levels, corresponding to different values of n. Note also

that we have a 1D box, and one ‘quantum number’, n.• The energy is independent of x, but the probability of finding the electron in the box

is not the same throughout the box. Positions of zero probability (ψ2=0) are called‘nodes’.

• The state with n=1 is known as the ground state, and has a non-zero energy,

€

E1 =h2

8mL2

• Higher n values give states with progressively higher energy (excited states). Thestate with n=2 is the ‘1st excited state’. Note that the spacing between levels increasesas n increases.

• If V(x) is symmetric about some point, the eigenfunctions are either even or odd withrespect to that point.

• The eigenvalues are real, so the operator is Hermitian, and we can easily show thatthe wavefunctions in (53) are orthogonal:

€

ψn0

L

∫ * (x)ψm(x)dx =2L

sin nπxL

sin

0

L

∫ mπxL

dx

=1,n = m= 0,n ≠ m

26

III.b) Non-stationary (time-dependent) states

The states given in (58):

€

Ψn (x, t) =2Lsin nπx

L

e

− iEnth ,

0 ≤ x ≤ Lwith n=1,2,3………….

(58)are the normalised energy eigenfunctions (stationary states) for a particle inan infinite square well. In each of these states, a measurement of the energygives a definite value (i.e. if the system is in the state Ψn, the measuredenergy is En).

These functions are described as ‘orthonormal’, which means mutuallyorthogonal and normalised) in the interval 0,L (in the box), i.e.:

€

dx0

L

∫ Ψn * (x, t)Ψm(x, t) =1,m = n

=0, otherwise(59)

This follows from the result on the previous page, i.e.

€

ψn0

L

∫ * (x)ψm(x)dx =2L

sin nπxL

sin

0

L

∫ mπxL

dx

=1,n = m= 0,n ≠ m

(60)(We normalised the eigenfunctions, and they are in addition orthogonal, aproperty of the eigenfunctions of a Hermitian operator.)

Since the TDSE is linear, any linear superposition of stationary states willalso be a solution, i.e.:

€

Ψ(x, t) = c1Ψ1(x, t) + c2Ψ2(x, t),(61)

is a solution, where c1, c2 are arbitrary possibly complex constants, andΨ(x,t) will also be normalised if

27

€

c12

+ c22

=1(62)

BUT this wavefunction does not represent a state of definite energy.

The new wavefunction is (see (42)):

€

Ψ(x, t) = c1ψ1(x)e− iE1t / h + c2ψ2(x)e

− iE2t / h

(63)The position probability density (in the case of real c1, c2) is:

€

Ψ(x,t) 2 = c1ψ1(x)eiE1th + c2ψ2(x)e

iE2th

c1ψ1(x)e

−iE1th + c2ψ2(x)e

−iE2th

= c12ψ1

2(x) + c22ψ2

2(x) + 2c1c2ψ1(x)ψ2(x)cosE2 − E1( )t

h

(64)So the probability density has a part that oscillates with angular frequency

€

ω =E2 − E1

h

This means that the probability to find the particle in a given position ischanging with time.

If the particle is a charged particle, such as an electron, then

€

Ψ(x,t) 2 alsorepresents a probability density for charge. The state Ψ(x,t) represents acharge distribution which is oscillating with frequency

€

ω =E2 − E1

h, i.e.:

€

hω = E2 − E1,(65)

which is the energy of the photon that would be emitted in a transition fromstate 2 to state 1.

Quantum states of uncertain energy are called non-stationary states, becausethey have some observable properties that change with time.

28

Measurement Theory

What happens if we measure the energy of the particle in state Ψ(x,t)?

We do not get some kind of average of E1 and E2 – the state does not have adefinite energy (it is not a stationary state).

Instead we get either E1 or E2!

• The probability to obtain the results E1 and E2 are |c1|2 and |c2|2respectively (assuming that Ψ(x,t) is normalised, eqn. (62)).

• Immediately after the measurement, the system is in the stationary stateΨ1(x,t) or Ψ2(x,t) corresponding to the energy measured (E1 or E2).

• Afterwards the system stays in the new stationary state until a time-dependent force (or potential) disturbs it.

The instantaneous change of

€

Ψ(x, t)→Ψ1(x, t) or

€

Ψ(x, t)→Ψ2(x, t) when ameasurement is made is called ‘collapse of the wavefunction’.

Quantum bits or ‘qubits’

A quantum bit or qubit is a unit of quantuminformation formed from a 2-level quantum-mechanical system, which in principle may beused to great effect in quantum computing.Take for example, a quantum property such asthe ‘spin’ of an electron. The electron may beregarded as spinning on its axis either clockwiseor anticlockwise, giving 2 states called ‘spin up’and ‘spin down’. ‘Spin-up’ and ‘spin-down’states have different wavefunctions (see later…).It is thus possible to create a ‘horizontal’ spinstate (for example using circularly polarisedlight), which is in effect a coherent superpositionof the wavefunctions for spin-up and spin-downstates.

[From Awschalom, Flatte and Samarth, Scientific American, June 2002, 52]

29

This is a ‘quantum bit’ or ‘qubit’: if each ‘up-spin’ or ‘down-spin’ representsa bit of information (1 or 0), then a qubit can represent both 0 and 1simultaneously. (In fact from (42), the qubit may represent arbitrarycombinations of both values - i.e. an infinite number of possibilities between0 and 1.)

In a conventional computer, every bit has a value of 0 or 1. A series of 8bits can represent any number from 0 to 255, but only one number at a time.Eight qubits can represent every number from 0 to 255 simultaneously! Anelectron with horizontal (or other tilted spin) is a natural qubit (so is anuclear spin). A quantum computer has a large number of qubits insuperpositions of alternative states. This gives the computer a massiveparallelism (the ability for algorithms to operate on many different numberssimultaneously).

Quantum Entanglement

Imagine a large box in which we have a pair of electrons (A and B),produced in such a way that the total spin of the pair is zero. Before wemeasure, we don’t know which electron has which spin. A possiblewavefunction before the measurement is:

€

ψ =12[ψ↑ (A)ψ↓ (B) −ψ↓ (A)ψ↑ (B)]

(where /

€

1/ 2 is the normalisation factor. I’m afraid a full explanation ofwhy we have this wavefunction rather than any other possibility will have towait until the end of the course!)

But if we measure an up-spin (

€

ψ↑) on A, then if the overall spin is zero, wemust have

€

ψ↓ on B, i.e. a measurement on A causes the wavefunction tocollapse to

€

ψ↑ (A)ψ↓ (B) and determines the result of the measurement forB – even though electron B may be a long distance from electron A.This is called quantum entanglement: the spin components of the twoelectrons are indeterminate prior to any measurement, but produce resultsthat are completely correlated when they are measured.

In a quantum computer, an initial spin (or other quantum) state imposed onthe system is allowed to evolve in time - the spin states may becomeseparated spatially, but in order to retain information, the spins states of the

30

electrons must remain correlated (aligned and in phase with each other;‘entangled’).

Qubits can easily be destroyed by interactions with their surroundings, beingconverted into random ordinary bits (through collapse of the wavefunction -disaster).

To encode 2 bits of information on two quantum particles:

‘Conventional computing’:4 possibilities: 0102, 0112, 1102, 1112

Quantum computing:An infinite number of superpositions possible!e.g. the so-called ‘Bell states’:

ψ+ =

€

12

(0112 + 1102)

ψ- =

€

12

(0112 - 1102)

φ+ =

€

12

(0102 + 1112)

φ- =

€

12

(0102 - 1112)

Key points:• The single particle states are ‘entangled’ in these superpositions• All the information is distributed among 2 qubits, and none of the

individual particles carries any information• If two particles in an entangled state are widely separated, a

measurement on one will still influence the quantum state of the other• Performing an operation on one of the qubits above switches from one

of the 4 states to any other

31

The result:

Bob can send two bits of information toAlice using just one qubit, providedAlice has access to both qubits and isable to determine which of the four Bellstates they are in. This can be used as abasis for quantum cryptography.

Quantum computation requires that thequantum states once created remaincoherent (undisturbed by interactionswith the outside world) for long times.The quantum states also must becontrolled very precisely.

From Physics World, March 1998

Further readingYou might find it interesting to read:

On the Einstein-Podolsky-Rosen (EPR) argument and Bell’s Theorem/Bell’sInequality:‘Quantum Physics: an Introduction’ ed. J Manners (IOP in association withthe Open University), 2000, shelfmark 530.145/M132

On quantum cryptography:‘Less reality, more security’, Artur Ekert, Physics World, September 2009,pg 29

and for a not altogether serious take, try ‘Schrödinger’s Dog’, LateralThoughts, Luiz Henrique Alves Monteiro, in the same edition of PhysicsWorld, pg 52.

32

IV. Tunnelling and finite potential wells

IV.a Barrier penetration

In section III, we saw that when a particle is confined by an infinitepotential well, then (from (44)), the wavefunction ψ = 0 outside the box,and therefore there is no probability of finding the particle outside thebox.

In classical mechanics, if the particle is confined by a finite potential,with V>E, then the particle cannot penetrate into this region (as thisimplies a negative kinetic energy). Such a region is called the‘classically forbidden region’.

In quantum mechanics, unless the confining potential is infinite, there is afinite probability of finding the particle in classically forbidden regions –i.e. for a particle in a box where the walls are created by finite potentials,there is some probability of finding the electron outside the box! Theprobability depends upon V-E.

[Taken from B Bleaney, Contemp Phys, 25, 320 (1984).]

33

A quantum particle can penetrate and tunnel through a classically forbiddenregion. This effect is known as tunnelling (or barrier tunnelling).

Consider a particle encountering a square barrier of height VB:

i.e.

€

V (x) = 0,−∞ < x < 0V (x) =VB ,0 < x < aV (x) = 0,a < x < +∞

(66)The figure shows the situation where the energy of the particle, E<VB.

A classical particle approaching from the left would be reflected if its energyis below VB and transmitted if its energy is above VB. For a quantumparticle, the outcome is uncertain; it may be reflected or transmitted (andreflection and transmission persist as possible options until the particle isdetected, when the wavefunction collapses to one or other possibility). Thisis a non-stationary state.

We need to find a suitable wavefunction Ψ(x,t) which describes thesepossibilities, and as the problem is dynamic, we may need the TDSE:

I II III

0 a

V(x)=VB

€

V (x) = 0

€

V (x) = 0

E

34

€

−h2

2m∂ 2Ψ∂x2

+V (x,t)Ψ = ih∂Ψ∂t

(5)Fortunately, provided the uncertainty in the energy of the particle is smallcompared with the variations in V(x), we can calculate the probabilities ofreflection and transmission by considering a stationary state, and using theTISE:

(67)where ψΕ(x) is an eigenfunction with energy E satisfying the eigenvalueequation:

(68)The procedure is to find the solutions of equation (68) in each differentregion of x, and then to join the solutions up to give acceptable choices ofwavefunction.

In region I:

Assume the particle approaches the barrier from the left. On the lefthandside of the barrier, V(x)=0, so we have:

€

d2ψE

dx 2= −k 2ψE

E =h2k 2

2m

(69)We need a suitable ‘trial solution’ of this that has both an incident and areflected wave.

€

Ψ(x, t) =ψE (x)e− iEt / h

€

−h2

2md2ψE

dx 2+V (x)ψE = EψE

35

A suitable form would be:

€

ψE (x) = AIeikx + ARe

− ikx

(70)This represents an incident wave of intensity |AI|2 (travelling left to right withwavevector k), and a reflected wave of intensity |AR|2 (after reflection,travelling right to left, with wavevector –k) – these are indicated by thearrows in the figure.

In region II:

Inside the barrier, there are two possible sets of solutions, depending onwhether the energy of the particle, E is >VB or < VB. We are interested in thecase where E < VB (the classically forbidden region, shown in the figure),but note that if E > VB, the particle is in a classically allowed region, and inregion II, the eignfunction is governed by:

€

d2ψE

dx 2= −kB

2ψE

E =h2kB

2

2m+VB

(71)

This has a general solution consisting of transmitted and reflected waveswith wavevector kB (because there will be some probability of reflection atthe righthand edge of region II), both with arbitrary amplitudes:

€

ψE (x) = AeikBx + A'e− ikB x

(72)

(Note that as we are using TISE and assuming a stationary state, energy isconserved at the boundaries. So from (69) and (71), we can see that kB<k, sothe wavelength of the wavefunction becomes longer in the barrier region.)

36

When E < VB, region II is classically forbidden. We can write:

€

d2ψE

dx 2= β 2ψE

E = −h2β 2

2m+VB

(73)(defined thus so β is positive)

which has a general solution

€

ψE (x) = Be−βx + B'e+βx

(74)where B and B’ are arbitrary constants.

In Region III

On the righthand side of the barrier, the eigenfunction must again satisfyequation (69). The solution must represent a transmitted wave travellingfrom left to right (indicated by the arrow in the figure), of intensity |AT|2, i.e.:

€

ψE (x) = ATe+ ikx

(75)

Probabilities for reflection and transmission

The probabilities of reflection, R, and transmission, T, are given by:

€

R =AR

2

AI2

T =AT

2

AI2

(76)and in addition, R + T = 1.

37

Treatment of the Boundary Conditions

We can calculate R and T by joining the solutions for regions I, II and IIItogether at x=0 and x=a. In general, as we saw for the 1D infinite box,solution of the TISE is achieved by matching the solutions achieved oneither side of the boundaries in the problem, i.e. by consideration of theboundary conditions.

There are two important considerations that need to be met (see section I)• The wavefunction (in this case ψΕ(x)) must be continuous• The first derivative of the wavefunction, dψΕ(x)/dx, must be

continuous (except where there is an infinite discontinuity in thepotential energy function V(x)).

From the considerations above for regions I, II and III, the wavefunction of aparticle with E < VB is given by:

€

ψE(x) = AI eikx + ARe

−ikx , − ∞ < x < 0ψE(x) = Be−βx + B'eβx , 0 < x < aψE(x) = ATe

ikx , a < x <∞

(77)We can find the constants by applying the boundary conditions.

Continuity at x=0

Continuity of ψΕ(x) when x=0 requires

AI + AR = B + B’ (I)

Continuity of dψΕ(x)/dx when x=0 requires

ikAI -ikAR = -βB + βB’

or

€

ikAI = β(−B + B' ) + ikAR (II)

but (from (I)) we know that

€

ikAI + ikAR = ikB + ikB'

So, adding (II), we find: 2ikAI = −(β−ik)B + (β + ik)B’(78)

38

Continuity at x=a

Similarly, continuity of ψΕ(x) and dψΕ(x)/dx when x=a requires

Be−βa + B’e βa = ATeika (III) and -βBe−βa + βB’eβa = ikATeika (IV)

Substituting for ATeika from (III) in (IV) we obtain:

-βBe−βa + βB’eβa = ik Be−βa + ikB’e βa

B’(βeβa - ikeβa)=B(βe-βa + ike-βa)

or

€

B'= Be−2βa (β + ik)(β − ik)

then, substituting for B’ in (III):

€

ATeika = Be−βa + Be−2βa (β + ik)eβa

(β − ik)

=Be−βa

(β − ik)(β − ik) + (β + ik){ }

€

ATeika =

2β(β − ik)

Be−βa

So overall, we obtain:

€

ATeika =

2β(β − ik)

Be−βa

B'= Be−2βa (β + ik)(β − ik)

(79)Tunnelling through wide barriers

If the particle encounters a wide barrier, a is large, so e−2βa<<1. In this case,the constant B’<<B (see equation (79)), which means we can simplify (78)to:

€

2ikAI ≈ −(β − ik)B(80)

39

Now, combining (80) with (79), we obtain:

€

ATeika ≈

−4ikβe−βa

(β − ik)2AI ,

ATAI

≈−4ikβe−(β + ik )a

(β − ik)2

So we can now obtain an expression for the transmission (or tunnelling)probability from (76):

€

T ≈ 16k 2β 2

(β 2 + k 2)2

e−2βa

We can re-express this in terms of E and VB by using (69) and (71), i.e.:

€

k =2mEh

β =2m(VB − E)

hwhich gives:

€

T ≈ 16E(VB − E)VB

2

e−2βa

(81)This expression is valid for a wide barrier (e−2βa<<1). In this situation, thewavefunction inside the barrier is dominated by the exponentially decayingterm Be−βx (see (77), B’ is small), as shown in the figure overleaf.

As can be seen, there is a finite oscillatory wavefunction on the righthandside of the barrier, with the small amplitude of the oscillations correspondingto the small probability that tunnelling has taken place. The form of thewavefunction on the lefthand side is a superposition of incident and reflectedwaves characterised by wavevectors +k and –k respectively.

β is known as the penetration parameter, and it is the e−2βa part of (81) that iscritical in determining the tunnelling probability. This tells us that thetunnelling probability is dependent on:

• the width of the barrier• the mass of the particle• the deficit VB-E in the energy compared with the value required for a

classical particle to pass the barrier.

40

[adapted from P A Cox, Introduction to Quantum Theory and Atomic Structure, Oxford University Press, Oxford, 1996.]

For an electron with an energy deficit of 1 eV and a barrier of 2 Å wide (justover an atom!), e−2βa is 0.13, showing that in this situation there is a highprobability of tunnelling through the barrier. However, for a proton in thesame situation, the probability is as small as 10-38. So tunnelling is onlylikely to be observed for very light particles.

The tunnelling probability is also incredibly sensitive to the width of thebarrier. For example, for an electron tunnelling across a vacuum gapbetween two metal surfaces with an energy deficit of 4 eV, there is a 2%change in tunnelling probability when the gap changes by as little as 0.01 Å(less than 10% of an atomic diameter.) This also means that tunnelling onlyoccurs over very short distances.

Examples of tunnelling

• Scanning tunnelling microscopy (STM)This is an experimental technique for imaging the individual atomsmaking up surfaces. It makes use of the extreme sensitivity of thetunnelling probability to a. A metal tip is positioned very close to thesurface of a conducting solid, and a small potential difference is applied.If the gap between the tip and the solid is small enough, electrons can

a

41

tunnel through the vacuum, and a current passes. The tip is scannedacross the surface, its height being controlled by a feedback mechanismdesigned to keep the current constant. Because the tunnelling probabilityis such a sensitive function of distance, the height can be controlled towithin a fraction of an atomic diameter.

Schematic diagram of STM, taken fromwww.ieap.uni-kiel.de/ surface/ag-kipp/stm/stm.htm

The STM tip can also be used to makeatomic structures. Here is a picture of theChinese characters for ‘atom’ (originalchild), constructed of Fe atoms on a Cusurface.

(From www.almaden.ibm.com/ vis/stm/atomo.html)

• α particle decayThe heaviest particles for which tunnelling is significant are α-particles(4He nuclei). Very heavy atoms, such as U undergo α decay. Within thenucleus, a low potential energy is produced by the attractive stronginteraction. Outside the range of this interaction, the electrostatic

42

repulsion between two positive charges gives a highly positive potential.The potential barrier is estimated to be around 30 MeV, but the emergingα particles have an energy of only around 3 MeV, implying thattunnelling is important to the process. As both the barrier height and theparticle mass are high, tunnelling is only possible because of the veryshort distances involved.

α decay, showing how the potential energy for anα particle depends on its distance from a nucleus.The wavefunction for a particle tunnelling throughthe Coulomb barrier (V(r)=e2/4πε0r) is also shown.1 fermi = 10-15 m.From:hyperphysics.phy-astr.gsu.edu/.../ alptun.html

Similar considerations apply to tunnelling protons taking part in nuclearfusion reactions at the centre of the sun; thermonuclear fusion is in fact onlypossible because protons can tunnel through Coulomb barriers.

IV.b Finite potential wells

Unsurprisingly, if we return to the case of a particle in a 1D potential well,we find that the particle is only certain to be trapped by the potential (i.e. tobe inside the box) if the confining potential is infinite (as we considered insection III). If the potential is merely large, rather than infinite, there is afinite probability that the particle is outside the well.

Solutions to the SE may be found by the same general procedure as insection IV.a. (although matching the boundary conditions requires agraphical solution). If the depth of the well is V0, we can distinguish twodifferent possible types of solution:

|E|<|V0|, giving rise to bound solutions (finite in number) and|E|>|V0|, giving rise to unbound solutions (a continuum):

43

Bound states, E<V0

Within the well, the wavefunctions that satisfy TISE are oscillatory standingwave type functions similar to the infinite well, with the eigenvalue equationgiven by:

€

d2ψdx 2

= −k02ψ ,

(82)

with

€

E =h2k0

2

2m−V0

i.e. inside the well, the wavevector

€

k0 =2m(E +V0)

h.

The allowed values of k0 (and hence the allowed energies, E) are not thesame as for the infinite well, as the wavefunction does not have to go to zeroat the walls of the well – in fact it decays away exponentially in the regionsoutside the well (as might be anticipated from the discussion of tunnelling insection IV.a). The oscillatory part of the wavefunction, and its slope withinthe well have to match with those of the exponentially decaying part outsidethe well.

I II III

0 a

V(x)=-V0€

V (x) = 0

x

€

V (x) = 0 |E|<|V0|

|E|>|V0|

44

This leads to a finite number ofallowed bound solutions. Thenumber of allowed solutionsincreases as the depth of thewell increases. The figureshows the first three allowedsolutions. As in the case of theinfinite well, the solutions areeither ‘even’ or ‘odd’ wrt thecentre of symmetry.

[Figure adapted from Quantum Physics: andIntroduction, ed. J Manners, IoP Publishing,Bristol, 2000.]

The distance that thewavefunction penetrates intothe ‘classically forbidden’region outside the well dependson V0-E (as might be expectedfrom our consideration of thepenetration parameter, β, insection IV.a.)

The need for the wavefunctionto match at the boundary meansthat for shallower wells, thewavelength of the oscillatorypart of the wavefunction (insidethe well) is longer, and thus theenergy level lower than fordeeper wells as shown in thefigure overleaf.

0

0

0

a

a

a

45

[Figure adapted from Quantum Physics: and Introduction, ed. J Manners, IoP Publishing, Bristol, 2000.]

Unbound states, E>V0

In this case, the particle is unbound, and the solutions are travelling waves.The particle is not confined to the vicinity of the well. However, if weimagine the particle travelling (say) from the left of the well through thewell, we can see that in the region of the well, it is subject to a potential –V0that will affect its wavevector. If the wavevector inside the well is k0 andthat outside the well is k, then we can write the energy of the particle, E as:

€

E =h2k0

2

2m−V0 =

h2k 2

2m(83)

In classical mechanics, this would correspond with the particle havingmomentum

€

hk0 inside the well, and

€

hk outside the well. This means that in

0 0 aa

46

the well, the wavelength of the solution is shorter than outside the well(k0>k). As for the bound solutions, the different parts of the wavefunctionmust match at the boundaries. A snapshot of the travelling wave could lookas in the figure below.

[Figure adapted from QuantumPhysics: and Introduction, ed. JManners, IoP Publishing, Bristol,2000.]

We find in the case of unbound states that we can achieve a smooth join ofthe wavefunctions inside and outside the well for any value of E. So theparticle can have any value of E and a continuum of solutions exists.

Because the solutions are travelling waves, it is possible to apportion parts ofthe solution to waves travelling back and forth, being reflected by thepotential well. For example, in the region a<x<+

€

∞, where V=0, theeigenvalue equation has the form

€

d2ψdx 2

= −k 2ψ

(84)which has the general solution

€

ψ(x) = C sin(kx + δ)which may be recast as:

€

ψ(x) = Ae− ikx − Ae2iδeikx

A = −Ce−2iδ

2i(85)

Here, A, C and δ are constants. In fact, δ is called the phase shift. Eqn (85)shows that outside the well, we may identify incoming and outgoing waveswith wavenumber k, but with a phase difference of 2δ, representing a

V

-V0

47

particle of momentum

€

hk being reflected by the potential well. (In fact,when we examine a time-dependent solution to the problem we find that thisphase shift is related to the time delay that occurs during the scatteringprocess.) This will return to haunt you in 3rd year core nuclear and particlephysics.

48

V(x)

E total energy, E

extent of classicalmotion

classicallyforbidden region

CFR

kinetic energy=E-V

x0

V. The Harmonic Oscillator

V.a The classical picture

In classical physics, the harmonic oscillator describes the behaviour of manyimportant physical systems. The restoring force is proportional to thedisplacement x from an equilibrium position.

€

F(x) = −kx(86)

NB: k is the spring constant, not 2π/λ!!

This linearity in x holds also for anharmonic systems when x is sufficientlysmall. This means that almost every system that is weakly perturbed fromequilibrium behaves like a harmonic oscillator (or collection of harmonicoscillators).

The corresponding potential is

€

V (x) =12kx 2

(87)so that

€

F(x) = −dVdx

= −kx

49

The classical equation of motion for a particle moving in this potential is:

€

m˙ ̇ x = −kx

˙ ̇ x = − kxm

(88)which gives SHM with frequency

€

ω 2 =km

The energy is given by:

€

E =p2

2m+12mω 2x 2

(89)

i.e. the sum of the KE and the PE. The energy switches between PE and KEduring the motion; at the maximum extent of the motion, the energy is allPE, and KE =0 (at the ‘turning points’).

V.b The quantum-mechanical picture

The TISE for this system is:

€

−h2

2md2ψdx 2

+12kx 2ψ = Eψ

write

€

k = mω 2 :

€

−h2

2md2ψdx 2

+12mω 2x 2ψ = Eψ

(90)(The TISE for a 1D harmonic oscillator)

This equation has to be supplemented by the boundary conditions:

€

ψ(∞) = 0ψ(−∞) = 0

i.e.

€

ψ(x)→ 0 for

€

x→ ±∞, so that the particle cannot be found where thepotential is infinite; this is also required for normalisability.

50

The TISE and its boundary conditions defines an eigenvalue problem. Weexpect solutions only for discrete values (eigenvalues) of E.

Systematic solution is tedious (it uses series expansion methods – seePHYS20171). Instead, let us test a trial solution that satisfies the boundaryconditions, of the form:

€

ψ(x) = Ae−x 2

2a 2

dψdx

= −xa2Ae

−x 22a 2

d2ψdx 2

=x 2

a4−1a2

Ae

−x 22a 2

substitute in TISE:

€

−h2

2mx 2

a4−1a2

Ae

−x 22a 2 +

12mω 2x 2Ae

−x 22a 2 = EAe

−x 22a 2

€

−h2

2ma4+12mω 2

x 2 +

h2

2ma2− E

= 0

This must hold for all x. This means that

€

h2

2ma2− E = 0

and

€

−h2

2ma4+12mω 2 = 0

This gives:

€

a =h

mω

E =h2

2ma2=12

hω

ψ(x) = Ae−mωx 2

2h

(91)This is the ground state (lowest energy state), because it has no nodes. It isalso an even function. For the harmonic oscillator, we have V(x)=V(-x) (see(87)), so the eigenfunctions are either even or odd.

51

The ground state wavefunction for a 1-D harmonic oscillator with length parametera=1.(Adapted from A.C. Phillips, Introduction to Quantum Mechanics, John Wiley &Sons Ltd, Chichester, 2005.)

The first excited state:We anticipate that this will be an odd function, and will have one node atx=0.

Try

������������������������������������������������������������������������������������������������

€

ψ(x) = Axe−x 2

2a 2

dψdx

= Ae−x 2

2a 2 1− x2

a2

d2ψdx 2

= Ae−x 2

2a 2 −xa2−xa2

2 − x2

a2

= Ae

−x 22a 2 x 3

a4−3xa2

substitute in TISE:

€

−h2

2mx 3

a4−3xa2

Ae

−x 22a 2 +

12mω 2x 3Ae

−x 22a 2 = EAxe

−x 22a 2

€

−h2

2ma4+12mω 2

x 3 +

3h2

2ma2− E

x = 0

This must hold for all x. This means that

ψ0(x)

x

Units

€

a = h /mω =1

52

€

3h2

2ma2− E = 0

and

€

−h2

2ma4+12mω 2 = 0, (exactly as before)

This gives:

€

a =h

mω

E =3h2

2ma2=32

hω

ψ(x) = Axe−mωx 2

2h

(92)This is the first excited state.

General resultThe energy eigenvalues are given by

€

En = n +12

hω,

n = 0,1,2,3..........(93)

(n=0 corresponds to the ground state, and n=1 to the first excited state.)

The corresponding normalised eigenfunctions are:

€

ψn (x) =1

n!2n a π

12Hn

xa

e

−x2

2a 2

(94)where the prefactor is the normalisation factor, and Hn(x/a) is a ‘Hermitepolynomial’.

53

The first four wavefunctions are:

€

E0 =12

hω ψ0(x) =1

a π

12e−x 2

2a 2 i.e. H0(y) =1

E1 =32

hω ψ1(x) =1

2a π

122 xa

e

−x 22a 2 i.e. H1(y) = 2y

E2 =52

hω ψ2(x) =1

8a π

12−2 + 4 x

a

2

e

−x 22a 2 i.e. H2(y) = −2 + 4y 2

E3 =72

hω ψ3(x) =1

48a π

12−12 x

a

+ 8

xa

3

e

−x 22a 2 i.e. H3(y) = −12y + 8y 3

(95)ψn(x) is an even function for n even (0,2,4……..) and an odd function for nodd (1,3,5……):

[From www2.kutl.kyushu-u.ac.jp/ seminar/MicroWorld2_...]

54

V(x)

E

classicallyforbidden region

CFR

x

€

−xmaxcl

€

xmaxcl

The eigenfunctions are the eigenfunctions of a Hermitian operator, and aremutually orthogonal:

€

dxψn * (x)ψm−∞

∞

∫ (x) =1 n = m

= 0 n ≠ m

The smallest possible energy of the oscillator,

€

E0 =12

hω , is called the ‘zero-

point energy’.

Non-stationary states can be constructed as for the infinite square well, e.g.:

€

Ψ(x, t) =12ψ0(x)e

− iE0th +

12ψ1(x)e

− iE1th

is a (normalised) non-stationary state. A measurement of the energy wouldgive either E0 (with probability

€

1 2( )2

=1/2 ) or E1 (with probability 1/2).After the measurement, the system will be in the corresponding stationarystate.

V.c The probability distributions; QM vs. CM

Classically for a harmonic oscillator,

€

E =p2

2m+12mω 2x 2

(89)The limits of classical motion (where p=0) are given by:

€

xmaxcl =

2Emω 2

55

ψ0(x)

x

Units

€

a = h /mω =1P(x)=ψ0

2(x)

x

Units

€

a = h /mω =1

classically allowed region

For the QM ground state,

€

E =12

hω ,

€

xmaxcl =

2Emω 2 =

h

mω= a

(96)The QM probability of finding the particle at position x in the ground state is

€

P(x) =ψ02(x)∝e

−x 2a 2

and there is a finite probability of finding the particle outside the classicallyallowed region:

The ground statewavefunction andposition probabilitydensity for a 1-Dharmonic oscillator withlength parameter a=1.(Adapted from A.C.Phillips, Introduction toQuantum Mechanics,John Wiley & Sons Ltd,Chichester, 2005.)

56

In contrast, the classical probability distribution is governed by:The probability to be at

€

x ∝ 1v(x)

, where v(x) is the speed at x.

€

E =p2

2m+12mω 2x 2 =

12mv 2 +

12kx 2

v(x) =2E − kx 2

m

12

P(x)∝ 1v(x)

=1πω

12E

k − x2

(97)(where the 1/π is a normalisation factor.)

This prediction is very significantly different from the QM prediction for thelowest energy levels. For example, for E0, the QM prediction is that theparticle will spend most time around the middle of the well (for a vibratingmolecule, this would correspond to the equilibrium internuclear separation –see next section.) In contrast, the classical prediction indicates that theparticle would spend most time at the turning points of the vibration (whereit is moving most slowly). For

€

E >> hω 2 , the QM probability distributionapproaches the classical form:

[From :hyperphysics.phy-astr.gsu.edu/.../ hosc6.html, vertical red lines indicate

€

xmaxcl ]

57

V.d Diatomic molecules

To a good approximation, the force between atoms in a diatomic molecule(H2, O2, HCl……) is well-described by a simple harmonic oscillator modelfor small displacements from the equilibrium separation, i.e. the potentialV(r) is quadratic near the minimum.

(For large displacements, this approximation breaks down – so a realmolecule eventually dissociates into 2 atoms as the internuclear separation isincreased, and

€

V (r)→ 0 as r→∞. As r decreases, the repulsion betweenatoms increases more rapidly than the parabolic behaviour of the harmonicoscillator. This more realistic behaviour is known as an anharmonicoscillator, represented by the solid potential curve above.)

Writing

€

r − r0 = x for the separation of the atoms, we have classically:

€

E = −V0 +12kx 2 +

p12

2m1+

p22

2m2

(98)where p1, p2 are the momenta and m1, m2 are the masses of the atoms at theends of the bond.

V(r)

-V0

0

r0

equilibriuminternuclearseparation

r

interatomic potential

n=

parabolicapproximation

58

In the centre of mass frame, the total momentum is zero, so

€

p1 = −p2 = p (say)

From (98), the total energy in the centre of mass frame is:

€

E = −V0 +12kx 2 +

p2

21m1

+1m2

= −V0 +12kx 2 +

p2

2µ(99)

where

€

µ =m1m2

(m1 + m2), known as the ‘reduced mass’.

The term –V0 represents a constant shift in the overall energy (it is the well-depth). It is often ignored, i.e. the energies are measured relative to thebottom of the potential well.) The energy then becomes:

€

E =12kx 2 +

p2

2µ(100)

which is equivalent to a particle of mass µ moving in a one-dimensionalSHO potential with

€

V =12kx 2.

We therefore expect the SHO solutions to apply to the vibrational states of adiatomic molecule, i.e.:

€

En = n +12

hω,

ω =kµ

(101)where k is the ‘spring constant’ of the molecule.

(For small displacements, the force between the atoms can be modelled by

using a ‘spring’ of spring constant

€

k =∂ 2V∂r2 r= r0

using a Taylor expansion:

59

€

V (r) =V (r0) + (r − r0)∂V∂r r= r0

+12(r − r0)

2∂2V∂r2 r= r0

=V (r0) +12kx 2

(as the second term is zero, and (r-r0)=x), hence

€

k =∂ 2V∂r2 r= r0

,

and a model where the diatomic consists of two masses at the end of a springis a good model for vibrational states.)

Example:

The observed frequency of EM radiation from transitions between adjacentvibrational levels in carbon monoxide is 6.43 x 1013 Hz. Estimate the springconstant, k.

€

ω =kµ,

k = µω 2

For CO,

€

µ =m1m2

(m1 + m2)=

12 ×1612 +16

amu = 6.86 amu =1.14 ×10-26 kg

The spring constant is:

k=1.14 x 10-26 x (2π x 6.43 x 1013)2 Nm-1 = 1860 Nm-1.

The spacing of the vibrational levels (see (101)) is

€

ΔE = hω .

So the frequency of the emitted photon, ω, is the frequency of the classicaloscillator. The energy, ΔE=6.63 x 10-34 x 6.43 x 1013 Hz = 4.26 x 10-20 J =0.266 eV, which is in the infrared.

Note: In practice, diatomic molecules also have rotational energy, and it isnot trivial to deduce vibrational energy levels directly from the spectrum ofthe emitted radiation.

m1 m2

k

60

VI. Quantum Mechanics in Two and Three Dimensions

VI.a The 2D Simple Harmonic Oscillator

For a particle moving in the xy plane in a 2D parabolic potential,

€

V (x, y) =12k(x 2 + y 2)

=12mω 2(x 2 + y 2).

(102)

The paraboloidal potential well of a 2D SHO [from AP French and E F Taylor, An introduction to quantum physics, Van NostrandReinhold (International), London, 1989.]

The kinetic energy is

€

T =px2 + py

2

2m

The corresponding QM operator is:

€

ˆ T =ˆ p x

2 + ˆ p y2

2m=−h2

2m∂ 2

∂x 2 +∂ 2

∂y2

(103)and the TISE reads:

€

−h2

2m∂ 2ψ(x,y)∂x2

+∂ 2ψ(x,y)∂y2

+12mω 2(x 2 + y 2)ψ(x,y) = Eψ(x, y)

(104)

|ψ(x,y)|2dxdy is the probability to find the particle in an area element dxdyaround the point (x,y).

61

The TISE can be written as

€

ˆ H ψ = ( ˆ H x + ˆ H y )ψ = Eψwhere

ˆ H x =−h2

2m∂ 2

∂x 2 +12

mω 2x 2 (which depends only on x)

ˆ H y =−h2

2m∂ 2

∂y 2 +12

mω 2y 2 (which depends only on y)

(105)The eigenfunctions of

€

ˆ H x are

€

ψnx(x) , where

€

ˆ H xψn x(x) = nx +

12

hωψn x

(as in the 1D case)

ˆ H yψn y(y) = ny +

12

hωψn y

(106)It follows that the product functions

€

ψnx ,ny(x, y) =ψnx

(x)ψny(y)

are eigenfunctions of

€

ˆ H = ˆ H x + ˆ H y:

€

( ˆ H x + ˆ H y )ψnx(x)ψny

(y) = ˆ H xψnx(x)ψny

(y) +ψnx(x) ˆ H yψny

(y)

= nx +12

hωψnx

(x)ψny(y) +ψnx

(x) ny +12

hωψny

(y)

= (nx + ny +1)hωψnx(x)ψny

(y)

(107)It follows that the eigenvalues of

€

ˆ H are

€

Enx ,ny= (nx + ny +1)hω,

nx = 0,1,2.....ny = 0,1,2.....

(108)and the corresponding eigenfunctions are:

€

ψnx ,ny(x, y) =

1nx!2

nx a π1

ny!2ny a π

12

Hnx

xa

Hny

ya

e

− x 2 +y 2( )2a 2

(109)

62

Degeneracy

Notice that we can now have more than one state with the same energy.

€

nx ny Enx ,ny

0 0 hω

0 11 0

2hω

0 21 12 0

3hω

Two or more states with the same energy are said to be ‘degenerate’, so forexample the energy level

€

3hω has degeneracy 3.

If two (or more) states have the same energy, then any linear superpositionof them is also an eigenstate of the same energy (since TISE is linear).(Note that a linear combination of states with different energy does notsatisfy TISE, however.) So there is more than one way to write the solution.

Example:

€

ψ10(x, y) and ψ01(x, y) have the same energy E = 2hω

⇒ ˆ H aψ10 + bψ01( ) = 2hω aψ10 + bψ01( )where a and b are arbitrary constants

because ˆ H aψ10 = 2hωaψ10 and ˆ H bψ01 = 2hωbψ01

Any linear superposition of degenerate eigenfunctions is also aneigenfunction with the same energy.

63

VI.b The z-component of angular momentum

For a particle moving in a ‘central potential’, i.e. V(r,θ,φ)=V(r), with nocomponent in the direction perpendicular to the radius vector, the angularmomentum of the system, L, is conserved:

(110)i.e. L (=rxp) is a constant.

Such potentials are abundant in physics, e.g. the 2D or the 3D SHO, where

€

V (x, y,z) =12mω 2 x 2 + y 2 + z2( ) =

12mω 2r2, i.e. independent of θ and φ, and the H-

atom, where

€

V (r) =−e2

4πε0r (r is the electron-proton separation.) In

QM,therefore, the eigenfunctions of the angular momentum operator play akey role for any central potential.

Angular Momentum Operators

As angular momentum is the vector product of r and p, it lies at right anglesto the plane of rotation in a right-handed screw sense:

L

L

Classically,

€

L = r × p

=

i j kx y zpx py pz

€

r × F =dLdt

= 0

64

and we may write the components of angular momentum as follows:

€

Lz = xpy − ypxLx = ypz − zpyLy = zpx − xpz

We can write down the operator for these components, by replacing theclassical variables by the QM operators for position and momentum:

(111)

Considering first two dimensions, we can change to plane polar coordinates,(r,φ), where

€

x = rcosφy = rsinφ

€

ˆ L x = ˆ y ̂ p z − ˆ z ̂ p y = −ih y ∂∂z− z ∂

∂y

€

ˆ L z = ˆ x ̂ p y − ˆ y ̂ p x = −ih x ∂∂y

− y ∂∂x

€

ˆ L y = ˆ z ̂ p x − ˆ x ̂ p z = −ih z ∂∂x

− x ∂∂z

(r,φ)

x

y

r

φ

65

€

⇒∂∂φ

=∂x∂φ

∂∂x

+∂y∂φ

∂∂y

= −rsinφ ∂∂x

+ rcosφ ∂∂y

= x ∂∂y

− y ∂∂x

⇒ ˆ L z = −ih ∂∂φ

(112)[In 3D, we can use cylindrical polar coordinates,

€

x = rcosφy = rsinφz = zto get the same result.]

In plane polar coordinates, we write the wavefunction as ψ(r,φ) instead ofψ(x,y). Since the 2D SHO potential (or any central potential) depends onlyon r and not on φ, we seek solutions of the separable form

€

ψ(r,φ) = R(r)Φ(φ)

If ψ(r,φ) is an eigenfunction of

€

ˆ L z (which is expected since then it woulddescribe a system with definite angular momentum, which is suggested bythe classical result that angular momentum is conserved for a centralpotential), then ψ(r,φ) would satisfy

€

−ih ∂∂φ

R(r)Φ(φ)( ) = LzR(r)Φ(φ)

⇒−ih∂Φ∂φ

= LzΦ

⇒Φ(φ) = eiLzφ / h

(113)where Lz is the eigenvalue of

€

ˆ L z , and we have ignored an arbitrary constantprefactor.

66

Boundary ConditionΦ(φ) must be a single-valued function everywhere in space.

€

Φ(φ + 2π ) =Φ(φ)

⇒ eiLz (φ +2π ) / h = eiLzφ / h

⇒ e2πiLz / h =1

so:

€

Lz = mh m = 0,±1,±2,±3.........(114)

and:

€

ψ(r,φ) = R(r)eimφ

(115)

Here we have obtained the very important result that angular momentum isquantised in units of

€

h. (Remember that this was assumed by Bohr in hismodel for the H-atom, in order to explain experimental results, but arisesnaturally from wave mechanics.)

This result is borne out by experiment – every measurement of onecomponent of orbital angular momentum (conventionally taken to be the z-component) in atomic and sub-atomic systems always gives an integermultiple of

€

h.

€

h is the fundamental quantum of orbital angular momentum

(We use the term ‘orbital angular momentum’ here to distinguish it fromspin angular momentum – see later.)

(r,φ)

rφ

x

y

67

VI.c Angular momentum eigenfunctions for 2D SHO

Can we find linear functions of the degenerate energy eigenfunctions of 2DSHO which are also eigenfunctions of

€

ˆ L z , i.e. which have the form deducedin (115), i.e.:

€

ψ(r,φ) = R(r)eimφ ?We know that:

€

ψnx ,ny(x, y) = AHnx

xa

Hny

ya

e

− x 2 +y 2( )2a 2

where A is a normalisation constant, from (109).

For the ground state (nx=0, ny=0):

€

ψ00(x, y) = Ae−(x2 +y 2 ) / 2a 2

= Ae−r2 / 2a 2 (r2 = x 2 + y 2)

(116)This has the desired form, with

€

R(r) = Ae−r2 / 2a 2

m = 0

This state has circular symmetry and Lz=0.

For the first excited states (nx=1, ny=0 and nx=0, ny=1):

€

ψ10(x, y) = Cxe−x2 / 2a 2e−y

2 / 2a 2

= Cxe−r2 / 2a 2

(note in the first line, the first term is the 1st excited state of a 1D oscillator,and the 2nd is the ground state of a 1D oscillator.)

€

ψ01(x, y) = Ce−x2 / 2a 2 ye−y

2 / 2a 2

= Cye−r2 / 2a 2

68

We can write these in polar form using:

€

x = rcosφy = rsinφ

€

⇒ψ10(x, y) = Ccosφ re−r2 / 2a 2

€

ψ01(x, y) = C sinφ re−r2 / 2a 2

Neither of these has the required form. Is it possible to find a combinationthat has the form

€

ψ(r,φ) = R(r)eimφ ?

We note by inspection that:

€

ψ10 + iψ01 = C(cosφ + isinφ)re−r2 / 2a 2

= Cre−r2 / 2a 2eiφ

ψ10 − iψ01 = C(cosφ − isinφ)re−r2 / 2a 2

= Cre−r2 / 2a 2e−iφ

(117)

Hence

€

ψ10 + iψ01 has the form R(r)eimφ with m =1ψ10 − iψ01 has the form R(r)eimφ with m = −1

These two states have the same total energy

€

2hω , and

€

Lz = +h (m =1), Lz = −h (m = −1).

So we have found alternative eigenfunctions that correspond to definitevalues of energy and of Lz. This can be done for any of the allowed valuesof the energy.

If E and Lz can both have definite values simultaneously, this implies that thecorresponding operators commute, i.e. that

€

ˆ H ̂ L z = ˆ L z ˆ H .

(r,φ)

x

y

r

φ

69

We can check this for the 2D SHO (using plane polar coordinates):

€

ˆ H = −h2

2m∇2 +

12

mω 2r2

=−h2

2m∂ 2

∂r2 +2r∂∂r

+1r2

∂ 2

∂φ 2

+

12

mω 2r2

ˆ L z = −ih ∂∂φ

We can see that these commute, because

€

∂∂φ

commutes with

€

∂∂r

and also

with any function of r, leaving only

€

∂∂φ, ∂

2

∂φ 2

=

∂ 3

∂φ 3−∂ 3

∂φ 3= 0

The fact that

€

ˆ H , ˆ L z[ ] = 0 implies that one can find common eigenfunctions ofboth operators, i.e. E and Lz can both have definite values for a given state.However, this is not true for the original states

€

ψnx ,ny(x, y). Only the

appropriate linear combinations of degenerate states are eigenfunctions of

€

ˆ L z .

70

VI. Quantum Mechanics in Two and ThreeDimensions…..contd.

VI.d Angular momentum in 3D

Classically, L=r x p

€

⇒ ˆ L = ˆ r × ˆ p

=

i j kˆ x ˆ y ˆ z ˆ p x ˆ p y ˆ p z

and so, as we have seen,

(111)A particle can only be simultaneously in an eigenfunction of any two ofthese operators (e.g.

€

ˆ L x and

€

ˆ L z ) if they commute.

Consider the commutator

€

ˆ L z, ˆ L x[ ] acting on some wavefunction ψ:

€

ˆ L z, ˆ L x[ ]ψ = ( ˆ L z ˆ L x − ˆ L x ˆ L z )ψ

= −h2 x ∂∂y

− y ∂∂x

y ∂ψ

∂z− z ∂ψ

∂y

+ h2 y ∂

∂z− z ∂

∂y

x ∂ψ

∂y− y ∂ψ

∂x

= −h2 x ∂ψ∂z

+ xy ∂2ψ

∂y∂z− y 2 ∂

2ψ∂x∂z

− xz∂2ψ∂y 2 + yz ∂

2ψ∂x∂y

+ h2 xy ∂

2ψ∂y∂z

− y 2 ∂2ψ

∂x∂z− zx ∂

2ψ∂y 2 + z ∂ψ

∂x+ zy ∂ 2ψ

∂x∂y

= h2 z ∂∂x

− x ∂∂z

ψ = ih ˆ L yψ

€

ˆ L x = ˆ y ̂ p z − ˆ z ̂ p y = −ih y ∂∂z− z ∂

∂y

€

ˆ L z = ˆ x ̂ p y − ˆ y ̂ p x = −ih x ∂∂y

− y ∂∂x

€

ˆ L y = ˆ z ̂ p x − ˆ x ̂ p z = −ih z ∂∂x

− x ∂∂z

71

This is true for all ψ, so we can write:

(118)(Note the cyclic symmetry.)

All three commutators are non-zero. This is important, as this means thatthere is a fundamental restriction on the simultaneous measurement of theangular momentum components.

• We cannot know simultaneously the value of two components ofangular momentum;

• We may determine precisely the value of any one component of theangular momentum, but we cannot simultaneously determine another.Lz is usually arbitrarily chosen as the component that is determinedprecisely.

However, it transpires that it is possible also to know the magnitude, |L| ofthe angular momentum, at the same time as the value of one component,through the operator for L2:

€

ˆ L 2 = ˆ L x2 + ˆ L y

2 + ˆ L z2

(119)We need to work out the commutator with

€

ˆ L z in order to determine whetherwe can determine the values of L2 and Lz simultaneously:

€

ˆ L 2, ˆ L z[ ] = ˆ L x2 , ˆ L z[ ] + ˆ L y

2 , ˆ L z[ ] + ˆ L z2, ˆ L z[ ]

= ˆ L x2 , ˆ L z[ ] + ˆ L y

2 , ˆ L z[ ]

We can use the identity:

€

ˆ A ̂ B , ˆ C [ ] = ˆ A ̂ B ˆ C − ˆ C ̂ A ̂ B − ˆ A ˆ C ̂ B − ˆ A ˆ C ̂ B ( )= ˆ A ˆ B ˆ C − ˆ C ̂ B ( ) + ˆ A ˆ C − ˆ C ̂ A ( ) ˆ B

= ˆ A ˆ B , ˆ C [ ] + ˆ A , ˆ C [ ] ˆ B

€

[ ˆ L x, ˆ L y ] = ih ˆ L z

€

[ ˆ L y, ˆ L z ] = ih ˆ L x

€

[ ˆ L z, ˆ L x ] = ih ˆ L y

72

to write:

€

ˆ L x2 , ˆ L z[ ] = ˆ L x ˆ L x, ˆ L z[ ] = ˆ L x ˆ L x, ˆ L z[ ] + ˆ L x, ˆ L z[ ] ˆ L x

= −ih ˆ L x ˆ L y + ˆ L y ˆ L x( )and

€

ˆ L y2 , ˆ L z[ ] = ˆ L y ˆ L y, ˆ L z[ ] = ˆ L y ˆ L y, ˆ L z[ ] + ˆ L y, ˆ L z[ ] ˆ L y

= ih ˆ L y ˆ L x + ˆ L x ˆ L y( )

€

⇒ ˆ L x2 , ˆ L z[ ] + ˆ L y

2 , ˆ L z[ ] = 0

⇒ ˆ L 2, ˆ L z[ ] = 0

similarly also, ˆ L 2, ˆ L x[ ] = 0 and ˆ L

2, ˆ L y[ ] = 0

(120)Thus we can know L2 and Lz at the same time, i.e. a system can have definitevalues of L2 and Lz.

VI.e The Operator

€

ˆ L 2

Not surprisingly, in general when we deal with central potentials,transforming from Cartesian to spherical polar coordinates leads tomathematical simplifications, as V(r,θ,φ)=V(r), which depends on only the rcoordinate. We therefore need to obtain the form of

€

ˆ L 2 in spherical polarcoordinates.

€

ˆ L 2 in spherical polar coordinates

z=rcosθy=rsinθ sinφx=rsinθ cosφ

Figure from M Alonso and E J Finn, ‘Fundamental UniversityPhysics’, volume III, ‘Quantum and Statistical Physics’, 1968

73

In Cartesian coordinates (using (111)):

In a central potential, simplifications are generally achieved by transformingto spherical polar coordinates. This is non-trivial (see e.g. French andTaylor, box 11.1), but we obtain:

(as found previously for 2 dimensions)

(121)and we find :

(122)Clearly,

€

ˆ L 2 and ˆ L z commute because

€

∂∂φ

commutes with

€

∂∂θ

and with any

function of θ.

€

⇒ ˆ L 2, ˆ L z[ ] = 0 (2nd proof).

A classical analogy to

€

ˆ L 2

Consider the TISE in 3D with a central potential, V(r,θ,φ)=V(r):

(123)

€

ˆ L 2 = −h2 ∂ 2

∂θ 2 + cotθ ∂∂θ

+1

sin2θ∂ 2

∂φ 2

€

ˆ L z = −ih ∂∂φ

€

ˆ L x = −ih −sinφ ∂∂θ

− cotθ cosφ ∂∂φ

€

ˆ L y = −ih cosφ ∂∂θ

− cotθ sinφ ∂∂φ

€

ˆ L 2 = −h2 y ∂∂z− z ∂

∂y

y ∂

∂z− z ∂

∂y

+ z ∂

∂x− x ∂

∂z

z ∂

∂x− x ∂

∂z

+ x ∂

∂y− y ∂

∂x

x ∂

∂y− y ∂

∂x

€

−h2

2m∇2 +V (r)

ψ = Eψ

74

where

€

−h2

2m∇2 =

ˆ p x2 + ˆ p y

2 + ˆ p y2

2m=−h2

2m∂ 2

∂x2 +∂ 2

∂y2 +∂ 2

∂z 2

is the kinetic energy operator

in three dimensions.

In spherical polar coordinates

(124)Substituting in the TISE (123), taking the nucleus as the origin of thecoordinates gives:

(125)

By comparing with (122) we can see that this contains the operator for L2,i.e.:

(126)The first term contains the r-dependence of the KE operator (the ‘radial part’of the KE operator), whereas the second term contains the dependence onangles (θ,φ) (the ‘angular part’ of the KE operator). (We can also see thatbecause V(r) is simply a central force field, dependent only on r, it appearsthat the problem will be very much simplified by separation of variables,separating the radial parts from the bits dependent on angles – see nextsection).

∇2 =∂ 2

∂x2+∂ 2

∂y2+∂ 2

∂z2

€

−h2

2m∂ 2

∂r2 +2r∂∂r

ψ +

ˆ L 2

2mr2ψ + V (r)ψ = Eψ

€

=∂ 2

∂r2+2r∂∂r

+1r2

∂ 2

∂θ 2+ cotθ ∂

∂θ+

1sin2θ

∂ 2

∂φ 2

€

−h2

2m∂ 2

∂r2+2r∂∂r

+1r2

∂ 2

∂θ 2+ cotθ ∂

∂θ+

1sin2θ

∂ 2

∂φ 2

ψ +V (r)ψ = Eψ

75

Now consider classical motion in a central potential:

Angular momentum,

€

L = mrvφ⇒ L = mr2 ˙ φ

total KE = 12m vr

2 + r2 ˙ φ 2( )

⇒'rotational part' of KE = 12mr2 ˙ φ 2 =

L2

2mr2

(127)

In QM, the angular or rotational part of the KE operator is given by

€

ˆ L 2

2mr 2

VI.f Eigenfunctions and Eigenvalues of

€

ˆ L 2

We seek wavefunctions ψ(r,θ,φ) which are simultaneously eigenfunctions ofenergy,

€

ˆ L z and ˆ L 2. Since

€

ˆ L z and ˆ L 2 do not depend on r , and we are working ina central force field, where V(r) is independent of θ and φ, we try separatingthe variables. We write:

(128)i.e. a radial part and an angular part. The angular part is sometimes calledthe ‘spherical harmonics’ (see later).

The eigenvalue equation for

€

ˆ L 2 is:

€

ˆ L 2ψ = L2ψ ,

where L2 is the eigenvalue of

€

ˆ L 2.

Hence

€

ˆ L 2R(r)Y(θ,φ) = L2R(r)Y(θ,φ)

€

R(r) ˆ L 2Y(θ,φ) = R(r)L2Y(θ,φ)

(since

€

ˆ L 2 does not involve r, but operates only on the angular parts of ψ.)

€

ψ(r,θ,φ) = R(r)Y (θ,φ)

y

x

rm

φ

€

vφ = r ˙ φ

€

vr = ˙ r

76

(129)

Now eigenfunctions of

€

ˆ L 2 that are also eigenfunctions of

€

ˆ L z must have the φ-dependence that we found in (115) for

€

ˆ L z , i.e.

€

eimφ for some integer m. Thisimplies:

€

Y(θ,φ) = P(θ)eimφ (m is an integer)

⇒∂ 2Y∂φ 2 = −m2Y

Substituting this in (129), and cancelling

€

eimφ gives

Now write

€

L2 = λh2, where λ is a positive dimensionless constant.

(130)What are the boundary conditions? We seek solutions to (130) which arefinite (rather than infinite) at θ=0 and π (i.e. the ‘poles’). We can writedown some of these solutions by inspection:

a) P(θ) = constant

This works if

€

λ =m2

sin2θ. But this must hold for all θ

€

⇒ λ = 0, m = 0

b) P(θ) = cosθ

€

⇒dPdθ

= − sinθ , d2Pdθ 2 = − cosθ

€

−h2∂ 2

∂θ 2+ cotθ ∂

∂θ+

1sin2θ

∂ 2

∂φ 2

Y = L2Y

€

−h2d2

dθ 2+ cotθ d

dθ−

m2

sin2θ

P = L2P

€

d2Pdθ 2

+cosθsinθ

dPdθ

+ λ −m2

sin2θ

P = 0

77

substituting in the eigenvalue equation (130) gives:

€

− cosθ +cosθsinθ

(− sinθ )+ λ−m2

sin2θ

cosθ = 0

⇒ λ−m2

sin2θ− 2

cosθ = 0

This must hold for all θ

€

⇒ λ = 2, m = 0

c) P(θ) = sinθ

€

⇒dPdθ

= cosθ , d2Pdθ 2 = − sinθ

substituting in the eigenvalue equation (130) gives:

€

− sinθ +cos2θsinθ

+ λ sinθ − m2

sinθ

= 0

⇒1− m2

sinθ+ (λ− 2)sinθ = 0 (using cos2θ = 1− sin2θ )

This must hold for all θ

€

⇒ λ = 2, m = ±1

d) P(θ) = 3cos2θ-1

€

⇒dPdθ

= −6cosθ sinθ, d2Pdθ 2

= 6(sin2θ − cos2θ)

substituting in the eigenvalue equation (130) gives:

78

€

6(sin2θ − cos2θ) − 6cos2θ + λ −m2

sin2θ

(3cos2θ −1) = 0

⇒ 6 −18cos2θ + λ(3cos2θ −1) − m2

sin2θ(3cos2θ −1) = 0 (using cos2θ − sin2θ =1− 2cos2θ)

This must hold for all θ

€

λ = 6, m = 0

e) P(θ) = sinθcosθ

€

⇒dPdθ

= cos2θ − sin2θ, d2Pdθ 2

= −4 sinθ cosθ

substituting in the eigenvalue equation (130) gives:

€

−4cosθ sinθ +cosθsinθ

(cos2θ − sin2θ) + λ −m2

sin2θ

cosθ sinθ = 0

1−m2

sinθ+ (λ − 6)sinθ = 0 (using cos2θ − sin2θ =1− 2sin2θ and cancelling cosθ throughout)

This must hold for all θ

€

λ = 6, m = ±1

f) P(θ) = sin2θ

€

⇒dPdθ

= 2sinθ cosθ, d2Pdθ 2

= 2cos2θ − 2sin2θ

substituting in the eigenvalue equation (130) gives:

€

2(cos2θ − sin2θ) +cosθsinθ

2sinθ cosθ + λ −m2

sin2θ

sin2θ = 0

⇒ 4cos2θ − 2sin2θ + λ sin2θ −m2 = 0⇒ 4 − 6sin2θ + λ sin2θ −m2 = 0 (using cos2θ =1− sin2θ)

79

This must hold for all θ

€

λ = 6, m = ±2

etc………….

Systematic exploration shows that solutions exist forλ=0,2,6,12,20,30,42……i.e. λ=l(l+1), for l=0,1,2,3………

For each value of l, there are (2l+1) solutions (i.e. the degeneracy of l is(2l+1)), corresponding to integer values of m in the range

m=-l,……….,+l

l is called the ‘orbital angular momentum quantum number’ (or just ‘orbitalquantum number’).

m (sometimes written ‘ml’) is the ‘azimuthal angular momentum quantumnumber’ (or just ‘azimuthal quantum number’).

Simultaneous eigenfunctions of

€

ˆ L z and ˆ L 2 have eigenvalues

€

L2 = l(l +1)h2

Lz = mh

where l = 0,1,2......and, for a given l :m = −l,−(l −1),......,l −1,li.e. m ≤ l; m is an integer

(131)Solving the TISE in 2D (θ,φ) has led to 2 quantum numbers, l and m.

Note:

€

Lz2 ≤ L2

⇒ m2 ≤ l(l +1)hmaximum value of m is l

80

m cannot be larger than l, because Lz would then be larger than L, which isimpossible.

The vector model for orbital angular momentum (use with caution!)

It is possible to visualise the orbital angularmomentum as a vector L of length

€

l(l +1)hwhose projection on the z-axis has avalue

€

mh between

€

−lh and lh.

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’,volume III, ‘Quantum and Statistical Physics’, 1968

The angle the vector L makes with the z-direction is not arbitrary, but isrestricted in direction (called space quantisation). In the presence of a field(normally regarded as defined by the z-direction), the vector precessesaround the field direction.

As an illustration, thecases where l=1 and l=2are shown; m may take 3and 5 possible values,respectively.

Figure from M Alonso and E J Finn,‘Fundamental University Physics’, volumeIII, ‘Quantum and Statistical Physics’,1968

81

[Figure from P A Cox, Introduction to QuantumTheory and Atomic Structure, Oxford UniversityPress, Oxford, 1996.]

The diagram above shows the possible orientations of the angularmomentum vector for l=2. Each cone shows the range of directions possiblefor a given m value. The z-component is well-defined, but the x- and y-components are undefined.

The Spherical Harmonics

The eigenfunctions corresponding to the solutions to (130) (the θ-dependence of the wavefunctions) are the ‘associated Legendrepolynomials’, Pl,m(cosθ).

The combination

€

Yl,m (θ,φ) = Pl ,m (cosθ)eimφ

(132)

is called a ‘spherical harmonic’. It provides a description of the shape of thewavefunction. The spherical harmonics represent the allowedwavefunctions for a particle on a sphere, where the distance to the origin (r)is fixed, but otherwise the particle is allowed free movement.

The complete eigenfunctions have the form

€

ψ(r,θ,φ) = R(r)Yl ,m (θ,φ)(133)

82

SUMMARY OF ANGULAR MOMENTUM EIGENFUNCTIONS

The simultaneous eigenfunctions of

€

ˆ L z and ˆ L 2 are the spherical harmonicsYl,m(θ,φ) (where the φ-dependence is just eimφ).

€

ˆ L 2Yl,m (θ,φ) = l(l +1)h2Yl ,m (θ,φ)ˆ L zYl,m (θ,φ) = mhYl ,m (θ,φ)

where l = 0,1,2,3.......and m ≤ l for a given l

(134)

Normalisation of Ym.l(θ,φ)

The volume element in 3D spherical polar coordinates is

€

dV = r2drsinθdθdφ ,where sinθdθdφ is the angular part. The spherical harmonics are normalisedwith the angular part of the volume element.

€

sinθdθ dφYl,m (θ,φ)2

0

2π

∫0

π

∫ =1

(135)The lowest few spherical harmonics are shown below:

[Table from M Alonsoand E J Finn,‘FundamentalUniversity Physics’,volume III, ‘Quantumand Statistical Physics’,1968]

83

Shapes of Ym.l(θ,φ)

For l=0, the wavefunction is spherically symmetric (and of constant phase).

Figure from M Alonso and E J Finn, ‘Fundamental UniversityPhysics’, volume III, ‘Quantum and Statistical Physics’, 1968

For l=1, m can take 3 values, 0, ±1. The m=0 state has a cosθ dependencewhich leads to a dipole-type elongation along the z-axis. This also meansthat the phase of the wavefunction changes at the origin:

+

-

(We have cheated a bit here, as r is not constant; we have combined thespherical harmonic with the radial part of the wavefunction, R(r), seeSection VIII, later. For m=±1, the shape is less obvious, but twosymmetrised combinations can be arrived at, which have dipoles lying alongthe x- and y-axes. This will be particularly important when we consider theH atom – see later. In the H atom, l=0 is called an s-orbital, and l=1 is a p-orbital. We will return to these, but just to whet your appetite, have a look atthe news item on the next page – which shows us that we can now measurewhat they look like…….)

84

From Chemistryworld, vol. 6, no. 10, October 2009:

News in briefElectron clouds unveiledFor years, undergraduate chemists have been shown pictures depicting the atomic orbitalsof atoms as described by the Schrödinger equation. But now, researchers from theKharkov Institute for Physics and Technology, Ukraine, have gone one better andmanaged to directly image the electron density surrounding a carbon atom - unveiling theshapes of both s- and p-like orbitals.

FEEM has allowed us to 'see' s- and p-like orbitals

© PHYSICAL REVIEW B

Using a technique known as cryogenic field-emission electron microscopy (FEEM), theresearchers studied the carbon atoms at the end of a chain of single carbon atoms formedby unravelling a sheet of graphene. By improving the resolution and sensitivity of thetechnique the team observed both spherical and dumbbell-shaped areas of electrondensity. They found that these electron density maps correspond closely to the electron densitypatterns predicted by the Schrödinger equation for 2s and 2p orbitals.

85

Spherical Harmonics in Helioseismology

Helioseismology is the study of the resonant modes of the Sun. Acousticwaves in the interior of the Sun perturb the solar material, so gentleoscillations of the surface occur with periods of minutes. A set of standingwaves (made up of regions of high and low pressure) are allowed –essentially a set of spherical harmonics labelled by l and m, and a thirdnumber, n, which describes how the mode varies radially outwards from thecentre of the Sun:

Figure courtesy of Prof Y Elsworth,BiSON website, University ofBirmingham.

(The blue and redregions are movingin oppositedirections relative tothe observer. Blackindicates nodalregions)

m is related to the number of nodal planes found around the equator, and l isrelated to the number of nodal planes found around the azimuth:

Figure from www.oca.eu.

The allowedwavefunctions fora particle on thesurface of a spheremay berepresented in thesame way.

86

Orthogonality of Ym.l(θ,φ)

For the spherical harmonics, the eignvalues are real, the operators for

€

ˆ L 2 and ˆ L z are Hermitian, and the spherical harmonics are orthogonal:

€

sinθdθ dφY *l ,m (θ,φ)0

2π

∫0

π

∫ Yl ',m ' (θ,φ) = 0, unless l = l' and m = m'

(136)For

€

m ≠ m' the orthogonality can easily be checked, since

€

dφei(m '−m)φ0

2π

∫ =1

i(m'−m)[e2πi(m '−m) −1]

= 0, m ≠ m'

For

€

m ≠ m' the orthogonality is enforced by the φ- dependent factors.For

€

m = m', l ≠ l' it is enforced by the θ-dependent factors.

Example: Y00 and Y10

€

€

sinθdθ dφ ⋅ 14π0

2π

∫0

π

∫ ⋅34π

cosθ

€

= constant sinθ cosθdθ = constant 12sin2θ

0

π

∫0

π

= 0

Therefore Y00 and Y10 are orthogonal.

87

VII. Rotational States of Diatomic Molecules

VII.a Energy eigenvalues for a rigid rotor

We can treat diatomic molecules as rigid bodies – so a rotating diatomic isknown as a ‘rigid rotor’. (This neglects vibrational modes – in fact theenergy (and time) scales of electronic, vibrational and rotational transitionsare quite different, allowing us in general to treat each form of energyquantisation independently and add the results together – an approximationknown as the Born-Oppenheimer approximation.)

[Figure adapted from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968]

A rigid rotor or dumbbell (in our case a diatomic molecule) possesses kineticenergy of rotation by virtue of its rotation about its centre of mass.

€

Erot =12Iω 2

(137)

Here, I is the moment of inertia about an axis through the centre of mass andω2 is the angular frequency of rotation about an axis through the centre ofmass.

m1 m2

€

m2

m1 + m2

r0

€

m1m1 + m2

r0

88

Angular momentum about this axis is

€

L = Iω

⇒ Erot =L2

2I

[c.f. translational KE, Etrans =p2

2m]

(138)In QM,

€

L2 → ˆ L 2, with eigenvalues

€

l(l +1)h2 . So we expect the rotationalenergies to be quantised, with allowed values

€

El =l(l +1)h2

2I(139)

For a given value of l, the degeneracy is (2l+1), corresponding to the (2l+1)different values of the quantum number m.

If the atoms making up the diatomic molecule have masses m1, m2 withseparation r0, then

€

I = m1m2

m1 + m2

2

r02 + m2

m1

m1 + m2

2

r02

=m1m2

m1 + m2

r02

= µr02

where µ is the reduced mass, m1m2

m1 + m2

(140)The moment created by the motion of the electrons on rotation along the z-axis is negligible and the moment of inertia along z can be taken as zero.

Hence

€

El =l(l +1)h2

2µr02

(141)

Historically, the quantum number l associated with the rotation has beencalled J! So often the rotational energy is given as:

89

(142)As a gauge of the size of this energy, for the hydrogen molecule,

€

h2 2I = 0.8 ×10−3 eV, giving λ=0.15 mm, in between the µ-wave and IR partsof the spectrum. As the rotational energy is inversely proportional to I, forheavier molecules, the rotational energy is smaller, with values in the µ-wave region. If we compare this with thermal energy (kT = 2.5 x 10-2 eV atroom temperature), this means that the rotational energy levels are well-populated at room temperature, with a distribution of excited statesoccupied.

In the µ-wave region of the EM spectrum, spectra are commonly discussedin terms of wavenumber (cm-1, E/hc). In these units the rotational energy isgiven by:

(143)where B is the rotational constant given by:

(144)(care here – if B is in cm-1, c also needs to be in cm s-1).

The energy level spacing (from (141)) is:

€

El − El−1 =h2

2µr02 l(l +1) − (l −1)l[ ]

=lh2

µr02 = 2Bl,

where B is the rotational constant, B = h2 2I (in energy units)(145)

€

Erot =h2

2IJ(J +1)

J = 0,1,2,3......

εr =Erhc

=h

8π 2IcJ(J +1) = BJ(J +1)

J = 0,1,2,3......

B =h

8π 2 Ic

90

(Or, in spectroscopic notation, the wavenumber frequency of a transitionfrom the state J to the state (J+1) is given by

(146))The rotational energy-level spacings thus increase with l, in contrast to theequal spacings in the vibrational spectra:

[Figure adapted from C N Banwell and E M McCash, ‘Fundamentals of Molecular Spectroscopy’, 4th edition, McGraw-Hill,London,1994] B is the rotational constant.

Because the gaps between levels are successively 2B(

€

= h2 /µr02 in energy

terms), 4B, 6B, etc, we find that in the microwave absorption spectrum (oremission spectrum), lines appear at intervals of 2B, as shown in the diagram.As B involves I, microwave spectroscopy thus provides a powerful methodof determining I, and hence r0 if µ is known.

ν J→ J+1 = B(J +1)(J + 2) − BJ(J +1) = 2B(J +1)

l=

€

3h2

µr02

€

2h2

µr02

€

h2

µr02

91

Example:

HCl has

€

r0 ≈1Å and

€

µ ≈1.62 ×10−27 kg

Substituting in (141) gives energies of the first few rotational levels as 0(l=0); approx. 0.0043 eV (l=1); 0.013 eV (l=2), 0.026 eV (l=3).The rotational gas phase absorption spectrum of HCl is shown below; theenergies are in the microwave/far-IR part of the spectrum. The lines areseparated by 2B.

[Figure adapted from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968]

92

VIII. The Hydrogen Atom

VIII.a Central Force Problems and the Hydrogen Atom

In order to obtain the allowed eigenfunctions for the H atom, we need tosolve the Schrödinger equation in three spatial dimensions:

The TDSE is (from lecture 1):

€

−h2

2m∂ 2Ψ∂x 2

+∂ 2Ψ∂y 2

+∂ 2Ψ∂z2

+V (x, y,z,t)Ψ = ih∂Ψ

∂t

which is often expressed:

€

−h2

2m∇2Ψ+V (r,t)Ψ = ih∂Ψ

∂t(time evolution of the wavefunction)

or

€

−h2

2m∇2 +V (r,t)

Ψ = ih∂Ψ

∂t(6)

i.e.

€

ˆ H Ψ = ih∂Ψ∂t

, where Ψ = Ψ(r ,t),

in other words,

€

ˆ H governs the time evolution of the wavefunction.

For the TISE:If we separate the space and time variables, exactly as in 1D, we find, for thecase where

€

V(r ,t) = V(r ) (a central potential):

€

Ψ(r,t) =ψ(r)e−iEt / h

where −h2

2m∇2ψ(r) +V (r)ψ(r) = Eψ(r)

(147)for stationary states (states with definite E).

In polar coordinates:

93

(124)but

(122)

so

€

∇2 =∂ 2

∂r 2 +2r∂∂r−

ˆ L 2

h2r 2

This is sometimes written

€

∇2 =1r 2

∂∂r

r 2 ∂∂r

−

ˆ L 2

h2r 2

So the TISE reads, for a central force (

€

V (r,t) =V (r)):

(126)or

(148)This is the 3D TISE for a central potential.

VIII.b The Coulomb Potential and the Hydrogen Atom

The hydrogen atom has one electron circulating thenucleus (one proton), and atomic number, z=1. Thepotential V(r) is simply the electrostatic attractionbetween the electron and the nucleus, the Coulombpotential, which is spherically symmetric (a centralpotential):

€

∇2 =∂ 2

∂r2+2r∂∂r

+1r2

∂ 2

∂θ 2+ cotθ ∂

∂θ+

1sin2θ

∂ 2

∂φ 2

€

ˆ L 2 = −h2 ∂ 2

∂θ 2 + cotθ ∂∂θ

+1

sin2θ∂ 2

∂φ 2

€

−h2

2m∂ 2

∂r2 +2r∂∂r

ψ +

ˆ L 2

2mr2ψ + V (r)ψ = Eψ

€

−h2

2mr2∂∂r

r2 ∂ψ∂r

+

ˆ L 2

2mr2ψ + V (r)ψ = Eψ

-e

+e

r

94

€

V(r ) =−e2

4πε0r(149)

Hence we seek solutions to (148) in the separable form

(133)In (148), the only term operating on the angular part of the wavefunction isthe term in

€

ˆ L 2. We already know that the spherical harmonics Yl,m(θ,φ) arethe simultaneous eigenfunctions of

€

ˆ L z and ˆ L 2. Then, using

€

ˆ L 2Yl,m (θ,φ) = l(l +1)h2Yl ,m (θ,φ)(134)

equation (148) becomes:

€

−h2

2mer2ddr

r2 dR(r)dr

Y (θ,φ) +

l(l +1)h2

2mer2 R(r)Y (θ,φ) − e2

4πε0rR(r)Y (θ,φ) = ER(r)Y (θ,φ)

or

€

−h2

2mer2ddr

r2 dR(r)dr

+

l(l +1)h2

2mer2 R(r) − e2

4πε0rR(r) = ER(r)

(150)(Essentially, we have separated the variables, so that we have two TISE’s,one for the angular part (spherical harmonics) which we solved earlier in(129), and one for the radial part, (150).)

It is convenient to write

€

R(r) =U(r)r

(151)Then

€

r2 dRdr

= r2 1rdUdr

−Ur2

= r

dUdr

−U

and

€

ddr

r2 dRdr

= r

d2Udr2

+dUdr

−dUdr

Then substituting in (150) and multiplying by r gives:

€

ψ(r,θ,φ) = R(r)Ylm (θ,φ)

95

€

−h2

2me

d2Udr2 +

l(l +1)h2

2mer2 U(r) − e2

4πε0rU(r) = EU(r),

where l(l +1)h2

2mer2 is the 'centrifugal potential'.

(152)This is effectively a one-dimensional equation in the coordinate r.

VIII.c Dimensionless variables

The radial functions are often expressed in so-called ‘dimensionless’ units.We define distance relative to the radius of the H-atom in its ground state(the Bohr radius, a0, see below), and energy again relative to the magnitudeof the energy of the H atom in its ground state, ER.

We note that the quantities

€

h2

mer2 and e2

4πε0r have the same dimensions, so

have the same dimensions.

We remember two important results from the Bohr model of the H-atom (seePHYS10121):

1. The radius of the H-atom, according to Bohr is given by:

For n = 1, r= 0.53 Å = a0, known as the ‘Bohr radius’:

(153)This is the characteristic length scale of the system – the ‘size’ of thehydrogen atom in its ground state (n=1).

€

Note also that a0 = 1α

h

mec, where α =

e2

4πε0hc=

1137

, the 'fine structure constant'

€

r =4πε0n2h2

mee2

€

a0 =4πε0h2

mee2

96

2. The energy of the hydrogen atom in the Bohr model is given by:

(154)and more generally, for any one electron system (He+, Li2+, etc), the sametreatment gives

(155)where ze is the nuclear charge.

We note that the quantities

€

e2

2 4πε0( )r and E have the same dimensions.

Substituting n=1 in (154), and noting that when n=1, r=a0, and using (153),we find that the magnitude of the energy of the H atom in its ground state,ER, is given by:

€

ER =e2

2 4πε0( )a0

=mee

4

2(4πε0)2 h2 =13.6 eV, the 'Rydberg energy'

(156)(This is so-called because it was introduced by Rydberg to explain thespacing of the spectral lines of atomic hydrogen before the advent ofquantum mechanics.)

Thus the H atom has a characteristic length scale defined by a0, and acharacteristic energy scale, defined by ER.

We define a ‘dimensionless’ distance and energy,

(157)

€

˜ E = EER

ρ =ra0

€

E = −mee

4

2 4πε0( )2h2

1n2

€

E = −mee

4

2 4πε0( )2h2

z2

n2

97

(Expressed in this way, the energy of an orbital in the Bohr model is simplygiven by

(158))

The equation for U (152) then becomes:

€

−d2Udρ2 +

l(l +1)ρ2 U −

2ρ

U = ˜ E U

(159)(The radial equation), or:

€

−d2Udρ2 + Veff (ρ)U = ˜ E U

where Veff (ρ) =l(l +1)ρ2 −

2ρ

Veff(ρ) is the effective potential (see next section).

VIII.d Radial Wavefunctions for Hydrogen

The radial equation (159) (or (152)) contains an ‘effective potential’composed of the Coulomb potential, and the centrifugal potential, arisingfrom the angular momentum of the electron:

€

Veff (r) =l(l +1)h2

2mer2 −

e2

4πε0ror in dimensionless units:

€

Veff (ρ) =l(l +1)ρ2

−2ρ

(160)As V(r) is dependent on 1/r, this has an asymptotic form for l=0, but isincreasingly dominated by the centrifugal potential for higher l, as shown inthe figure. (This has some interesting consequences for the forms of theradial functions for different l values – see later. It means that there is no‘centrifugal barrier’ preventing an electron with l=0 (known as an s electron)from being at the nucleus (r=0) – whereas there is such a potential barrier forany other value of l.)

€

˜ E n= −1n2

98

[Figure adapted from F Yang and J H Hamilton, ‘Modern Atomic and Nuclear Physics’, McGraw-Hill, 1996]

Let us now examine the behaviour of the radial equation (159) in thelimiting cases of large distance,

€

ρ →∞ and small distance,

€

ρ → 0.

1. For

€

ρ →∞, the terms proportional to 1/ρ2 and 1/ρ are negligible(compared to the term

€

˜ E U ):

€

⇒d2Udρ2 ≈ − ˜ E U, for ρ →∞

Note that bound states correspond to negative total energy; theCoulomb potential is attractive and

€

V (r)→ 0 for r→∞; the zero ofenergy is taken as when the proton and the electron are at an infinitedistance apart – bound state energies are negative w.r.t. this, and workhas to be done to separate the proton and the electron held in a boundstate.

€

⇒ ˜ E < 0 (− ˜ E > 0) for bound states. (For

€

˜ E > 0, a continuum of allowedsolutions exists – c.f. the particle in a finite potential well, section IV.These are known as ‘scattering states’.)If we set

€

− ˜ E = b2 , we obtain

€

d2Udρ2

= b2U

with independent solutions

€

U ≈ ebρ and U ≈ e−bρ . But

€

U ≈ ebρ is notacceptable, as it is not normalisable (

€

ebρ →∞ for ρ →∞), and we knowthat the wavefunction (and hence the probability of finding the

99

electron) must become zero at an infinite distance from the atom.Thus we infer that:

€

U(ρ) ≈ e−bρ for ρ →∞where

€

˜ E = −b2

2. For

€

ρ → 0, the term

€

l(l +1)Uρ2

is much larger than

€

−2Uρ

and

€

˜ E U

(provided

€

l ≠ 0).The leading order terms give

€

−d2Udρ2

+l(l +1)ρ2

U = 0

If we attempt a power law solution

€

U ≈ Aρa

in the above equation we find

€

A[−a(a −1)ρa−2 + l(l +1)ρa−2] = 0,which works with a(a-1)=l(l+1)

€

⇒ a = l +1 or a = −lSo the independent solutions for

€

ρ → 0 have the forms

€

U(ρ) ≈ ρ l+1 and U(ρ) ≈ ρ− l

But

€

U(r) = rR(r)⇒U(ρ)→ 0 for ρ → 0, so the

€

ρ−l solution is not allowed.Therefore, we assume

€

U(ρ) ≈ ρ l+1 for ρ → 0

Combining the large-ρ and small-ρ limits, we can write:

€

U(ρ) = ρ l+1e−bρ f (ρ)(161)

Let us substitute this form into the full radial equation

€

−d2Udρ2 +

l(l +1)ρ2 U −

2ρ

U = ˜ E U

(159)or, using

€

˜ E = −b2 ,

€

−d2Udρ2

+l(l +1)ρ2

U −2ρU + b2U = 0

(162)

We need to determine

€

dUdρ

and d2Udρ2 .

100

€

dUdρ

= e−bρ (l +1)ρ l f (ρ) − bρ l+1 f (ρ) + ρ l+1 dfdρ

and

€

= e−bρ [l(l +1)ρ l−1 − 2b(l +1)ρ l + b2ρ l+1] f (ρ) + [2(l +1)ρ l − 2bρ l+1] dfdρ

+ ρ l+1 d2 fdρ2

Inserting all of this into (162) and cancelling

€

e−bρ throughout gives:

€

⇒ ρ l +1 ′ ′ f + 2ρ l (l +1− bρ) ′ f + [l(l +1)ρ l−1 − 2b(l +1)ρ l + b2ρ l +1 − l(l +1)ρ l−1 + 2ρ l − b2ρ l +1] f = 0

Dividing through by ρl gives

€

ρd2 fdρ2

+ 2(l +1− bρ) dfdρ

+ 2(1− b(l +1)) f = 0

(163)Physical solutions only exist for certain values of b (i.e. certain values of

€

˜ E = −b2 = E / ER ).

We need to find the forms of the function f that satisfy (163) – for insertioninto (161), in order to find the allowed forms of the radial part of thewavefunction U(ρ). In order to avoid U(ρ) having unphysical behaviour(e.g. becoming very large at large distances from the nucleus), f(ρ) is likelyto be some finite polynomial.

The simplest solution is f(ρ)=constant.

This requires (see (163))

€

b =1l +1

,

with energy E = −b2ER = −ER

(l +1)2

l = 0,1,2,3,.......

€

d2Udρ2

= e−bρ[l(l +1)ρ l−1 f (ρ) + (l +1)ρ l dfdρ

− b(l +1)ρ l f (ρ)

−bρ l+1 dfdρ

+ (l +1)ρ l dfdρ

+ ρ l+1 d2 fdρ2

− b(l +1)ρ l f (ρ) + b2ρ l+1 f (ρ) − bρ l+1 dfdρ]

101

The next simplest solution has the form

€

f (ρ) = c0 − c1ρ

€

dfdρ

= −c1,d2 fdρ2

= 0

Substituting in (163) gives

€

−2(l +1− bρ)c1 + 2(1− b(l +1))(c0 − c1ρ) = 0

Equating constant terms:

€

⇒−(l +1)c1 + (1− b(l +1))c0 = 0(gives c1 in terms of c0)

Equating terms in ρ:

€

c1b − [1− b(l +1)]c1 = 0(fixes b)

€

⇒ b =1l + 2

⇒ E = −b2ER = −ER

(l + 2)2

[ and

€

c1c2

=1− l +1

l + 2l +1

=1l +1

−1l + 2

=1

(l +1)(l + 2)]

The next simplest solution has the form

€

f (ρ) = c0 − c1ρ − c2ρ2

with

€

b =1l + 3

E = −ER

(l + 3)2

etc.

102

VIII.e Energy Eigenvalues for the hydrogen atom

The allowed energies are:

€

E =

−ER

(l +1)2, l = 0,1,2,3.....

−ER

(l + 2)2, l = 0,1,2,3.....

−ER

(l + 3)2, l = 0,1,2,3.....

We can simplify the description by introducing the principal quantumnumber, n:

€

En = −ERn2

(164)where n=l+1, or l+2, or l+3 etc.i.e. n=1,2,3,4…………(where higher values can be obtained in many ways).

(In dimensionless units, the energy of an orbital is simply given by

(158))For a given value of n (i.e. a given energy), the maximum value of l is (n-1),corresponding to n=l+1, so, for a given value of n, l takes the values

l = 0,1,2,…………,n-1

For a given value of l, the quantum number m (often written ml todistinguish it from the spin angular momentum quantum number for the zcomponent, ms- see later) takes the values

ml = -l, -(l-1)…….-1,0,1,……..,(l-1),li.e. (2l+1) values.

€

˜ E n = −1n2

103

These quantum numbers specify the values of E, Lz and L2:

€

E = −ER

n2, L2 = l(l +1)h2, Lz = mlh

(165)Note that the energy of the eigenfunction is dependent only on n. (NB: Thisin fact is true only for one-electron systems such as H, He+, Li2+. Forsystems with more than one electron, the interactions between the electronscauses the energy of the orbitals to depend on l also. You will meet manyelectron atoms in PHYS20502 next semester.)

The result for energy is exactly the result obtained from the Bohr model(which was given in (154) and (158)). This was a major success for wavemechanics.

VIII.f The radial wavefunctions and radial probabilitydistributions

The function U(ρ) has the general form

€

U(ρ) = ρ l+1e−bρ (c0 − c1ρ + c2ρ2 .......+ (−1)n− l−1cn− l−1ρ

n− l−1)(166)

(

€

c0 − c1ρ + c2ρ2.......+ (−1)n− l−1cn− l−1ρ

n− l−1 is the associated Laguerre polynomial),

where

€

b2 = − ˜ E = − EER

= +1n2

⇒ b =1n

(taking the positive root)Hence the function U(ρ) depends on n and l:

€

Un,l (ρ) = ρ l+1e−ρ n (−1)kk= 0

n− l−1

∑ ckρk

(The alternation of sign for sequential powers of ρ is important. The numberof changes in sign for such a sequence gives the number of nodes forpositive ρ.)

104

Recall that

€

R(r) =U(r)r

and ρ = ra0

€

Rn.l (r) = constant × ra0

l

e−r na0 (−1)kk= 0

n− l−1

∑ ckra0

k

(167)

The first few radial wavefunctions Rn.l(r) (NOT NORMALISED) are shownbelow:

n l=0 l=1 l=21

€

e−r / a0 No solution No solution2

€

1− r2a0

e−r / 2a0

€

ra0e−r / 2a0 No solution

3

€

1− 2r3a0

+2r2

27a02

e−r / 3a0

€

ra01− r6a0

e−r / 3a0

€

ra0

2

e−r / 3a0

Notes:• The exponential is

€

e−r / na0 (allowing the value of n to be determined byinspection).

• The behaviour at small r is

€

r a0( )l (allowing the value of l to bedetermined by inspection).

• The polynomial has degree (n-l-1), which corresponds to the numberof nodes in the radial part of the wavefunction.

• The functions are dependent upon two quantum numbers, n and l.• Orbitals with l=0 have historically been called ‘s orbitals’, l=1 gives p

orbitals, and l=2, d orbitals (after this, l=3 gives f orbitals).

The form of these solutions is shown overleaf.

105

Radial Probability Distributions

The square of R(r) gives a measure of the radial probability density forcharge (and hence the electron density) along a given radius vector. A moreuseful quantity is the probability of finding the electron between radii r and(r+dr). As the surface area of a sphere is 4πr2, the probability of finding theelectron in the shell is 4πr2R(r)2dr. 4πr2R(r)2 is the radial electron density atradius r, known as the radial distribution function (or radial probabilitydensity):

€

P(r) = 4πr2R(r)2 ,(168)

where P(r)dr is the probability to find the electron at a distance

€

r→ r + drfrom the proton.

106

(Note that as there is no angular part of R, this result is essentially obtainedfrom integrating over all angles, θ and φ:

€

P(r)dv = sinθdθ dφ0

2π

∫0

π

∫ r2R2dr = 2 × 2π × r2R(r)2dr = 4πr2R2dr

See more on normalisation of the complete wavefunction later.)

The effect of this is seen in the diagram: it forces a node at the nucleus, evenwhen l=0, and also weights the electron density to a region beyond the lastnode. The function is largest at the radius with the greatest probability offinding the electron.

For example, for a 1s orbital (the ground state of the H atom, with n=1, l=0):

Here the maximum radial density occurs at a0, the Bohr radius. Thisillustrates nicely the difference between the results generated through theBohr model (‘old’ quantum mechanics) and the Schrödinger equation (wavemechanics). In the Bohr model, a0 was the only radius at which the electroncould be found; in wave mechanics, a0 is the most probable radius.

VIII.g Complete wavefunctions for the hydrogen atom

The complete wavefunction, including the angular dependence, is given by

107

€

ψn,l,ml(r,θ,φ) = Rn,l (r)Yl ,ml

(θ,φ)

= Rn,l (r)Pl ,ml(θ)eimlφ

where Pl,ml(θ) are the associated Legendre polynomials

(169)Note that we have solved the TISE in three spatial dimensions, andgenerated three quantum numbers, n, l, and ml. One can deduce the valuesof these quantum numbers for a given wavefunction (and check that it is apossible wavefunction) by inspecting the wavefunction:

The full wavefunction has the following form:

€

ψn,l,ml(r,θ,φ) ≈ r

a0

l

e−r na0 × (polynomial in r a0) ×Pl ,ml(θ) ×eimlφ

where Pl,ml(θ) are the associated Legendre polynomials

and the polynomials in r a0 are known as the associated Laguerre polynomials(170)

n = division of r/a0 in the exponentiall = power of the leading (r/a0) term (also corresponds to the number ofnodes in the angular part of the wavefunction)ml = integer in

€

eimlφ factor(n-1) =total number of nodes in the complete wavefunction (radial plusangular parts)(n-l-1) = number of nodes in the associated Laguerre polynomial = order ofthe polynomial (and the number of nodes in the radial part R(r).)

Normalisation

The usual normalisation condition is

€

ψn,l,ml(r,θ,φ)∫∫∫

2dv =1

i.e.

€

r= 0

∞

∫θ = 0

π

∫ ψn,l ,ml(r,θ,φ)

2

φ= 0

2π

∫ r2drsinθdθ dφ =1

(171)

As

€

ψn,l,ml(r,θ,φ) = Rn,l (r)Yl ,m(θ,φ) , and the usual normalisation of the spherical

harmonics (from (135)) is

108

€

sinθdθ dφ0

2π∫0

π

∫ Yl ,ml(θ,φ)

2=1

then we have (as a normalisation condition for Rn,l(r)):

€

r2

0

∞

∫ Rn,l2 (r)dr =1

or Un,l2 (r)

0

∞

∫ dr =1

where Rn,l (r) =Un,l (r)r

(172)VIII.h Representation of atomic orbitals

The angular parts of the wavefunctions are described by the sphericalharmonics that we determined earlier in part VI.

For l=0 (s states), the wavefunction is spherically symmetric(|Y00|2=constant), as we saw in part VI.

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volumeIII, ‘Quantum and Statistical Physics’, 1968

For l=1, ml can take 3 values, 0, ±1. The ml=0 state has a cosθ dependencewhich leads to a dipole-type elongation along the z axis. This is known as apz orbital. For ml=±1, the spherical harmonics that we arrived at in part VIdo not naturally point along the two Cartesian axes, x and y. This makes itdifficult to consider (for example) how the atomic orbitals on one atom mayinteract with those on another in chemical bonding. However, because theY1±1 are degenerate eigenstates, it is possible to make 2 symmetrised linearsuperpositions of Y1±1 which have dipoles lying along the x and y axes. It isin fact these that you may be familiar with from chemistry as px and pyorbitals:

109

For ml=±1

(173)The form of these symmetrised functions for l=1 and l=2 (d orbitals) isshown below:

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968

Y1,±1 =38πsinθe±iϕ

px =12(Y1,1 + Y1,−1) =

34πsinθ cosφ

py =12(Y1,1 − Y1,−1 ) =

34πsinθ sinφ

110

Any pictorial representation of the complete orbitals must recombine theradial and angular parts to produce a distribution of the electrons in space.In addition we should note that our schematic representations of the orbitalsindicate probabilities, rather than certainties of finding electrons. As theorbitals do not have definite radii, we can only define surfaces whichrepresent a 90% (say) probability of finding the electron within it.

An example is shown below for a 1s orbital:

[Figure adapted from Chemical Structure and Bonding, R L DeKock and H B Gray, Benjamin Cummings, Menlo Park, 1980.]

As examples of the combination of the angular and radial parts in therepresentation of atomic orbitals we take the 2s and 2p orbitals.

For the 2s orbital, l=0, so there are no nodes in the angular part of thewavefunction – it is spherically symmetric - so the orbital must be ‘round’.The node must be in the radial part of the wavefunction, as illustratedoverleaf. The presence of a node means that the phase of the electrondensity changes sign on either side of it, leading to the graphicalrepresentation also shown overleaf.

111

[Figures from Chemical Structure and Bonding, R L DeKock and H B Gray, Benjamin Cummings, Menlo Park, 1980 and Structureand Spectra of Atoms, W G Richards and P R Scott, John Wiley and Sons, Chichester, 1986.]

In the case of the 2p orbital, the node is now in the angular part (as l=1, sothere are no nodes in the radial part (this excludes any at r=0, and in fact theonly radial functions which do not have a node here are s orbitals – this is adirect consequence of the centrifugal part of the effective potential in (160),and becomes very important in considering the energy ordering of orbitalsin many electron atoms). The angular part thus has the dipolar cosθ shapewhich we saw earlier for 2p orbitals. The radial part has the overall shapeshown on pg. 104. The combination of the two functions must be taken inarriving at an overall representation; the overall resulting representation isnormally something akin to the diagram below for 2pz:

[Figure from Structure and Spectra of Atoms, W G Richards and P R Scott, John Wiley and Sons,Chichester, 1986.]

112

VIII.i Other single electron atoms

The analysis for H (nuclear charge z=1) is easily extended to other singleelectron atoms such as He+ (z=2) and Li2+ (z=3) by replacing e by ze for thenuclear charge in the expressions for a0 and ER.

Then the Coulomb potential

€

V (r) = −ze

4πε0r

The new Bohr radius is

€

a'0 =a0z

and the energies are

€

En = −z2ER

n2.

So for He+, the electron is more tightly bound to the nucleus (the boundlevels are more negative by a factor of 4) and the ‘orbits’ are a factor of 2closer to the nucleus.

Reduced mass correction

The expression (155) for the energy of the radial functions contains the massof an electron. In fact, it should contain the reduced mass of the system ofelectron and nucleus, µ, giving:

(174)where

(175)where Mn is the mass of the nucleus.Because the nucleus is much heavier than the electron, this has a barelynoticeable effect. The difference in the energy levels for the hydrogenisotopes H, D and T is approximately:(ΔE/E)H= 0.05%, (ΔE/E)D=0.027%, (ΔE/E)T=0.018%

En = −

z2µe4

2h2 (4πε0 )2n2

€

µ =meMn

me + Mn

=me

1+me

Mn

113

VIII.j The spectrum of hydrogen

A transition from a state with principal quantum number n, to one withprincipal quantum number m (<n), creates a photon with energy

€

hν = ER1m2 −

1n2

(ER =13.6 eV)

(176)The resulting spectral lines in the emission spectrum can be organised intoseries according to the value of m:

Lyman series (1906)

€

hν = ER112−1n2

n = 2,3,4,......

(ultraviolet, λ=91-122 nm)

Balmer series (1885)

€

hν = ER122−1n2

n = 3,4,5,......

(optical/ultraviolet, λ=365-656 nm)

Paschen series (1908)

€

hν = ER132−1n2

n = 4,5,6,......

(infrared, λ=820-1875 nm)

The lines above are observed in the emission spectrum (photon given out).The series may also be observed in the absorption spectrum, where a photonis absorbed in exciting an electron from a state with principal quantumnumber m to n (m<n).

The ionisation limit/ionisation energyThe energy required to ionise the electron from a state m (remove it to thecontinuum) is obtained by inserting

€

n =∞ as the end limit of the transition,for example for the ground state of H (m=1):

€

EIP = ER112 −

1∞2

=13.6 eV =1307 kJ mol-1

The experimental value is found to be 1317 kJmol-1.

114

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968

115

SUMMARY:Solution of the TISE (3 dimensions) for a one electron species generates 3quantum numbers, which together specify orbital shape, size and direction:

The principal quantum number, n

Specifies the orbital energy En, and the size of an orbital.

The orbital angular momentum (or orbital) quantum number, l

Specifies the magnitude of the angular momentum and the shape of anorbital.

The azimuthal angular momentum (or azimuthal) quantum number, ml

Specifies the component of angular mometum in a specific direction (oftendefined by an external field). Related to the direction of an orbital and itsbehaviour in a magnetic field (where the (2l+1) degeneracy is lifted – theZeeman effect).

n = 1,2,....∞

l = 0,1,2.....(n −1)

€

ml = 0,±1,±2....± l

116

IX. Electron spin, the Pauli Exclusion Principle and the He atom

IX.a Electron Spin

The Stern-Gerlach experiment (1924), showed that electrons have intrinsic(‘spin’) angular momentum. In this experiment, a beam of metal atoms isobtained by heating a sample of the metal in a furnace which has a hole topermit a jet of atoms to escape. This beam is then collimated by slits and isthen passed into an inhomogeneous magnetic field:

[Figure from Structure and Spectra of Atoms, W G Richards and P R Scott, John Wiley and Sons, Chichester, 1986.]

A charged particle with angular momentum possesses a magnetic moment.In an inhomogeneous field, a body with a magnetic moment experiences adeflecting force due to the difference in field strength at the two poles of thebody. The body is therefore forced in the direction of increasing ordecreasing field strength depending on the orientation of its moment:

(177)where M is the magnetic moment, and B is the external field.

If the beam is made up of atoms with one outer s electron, e.g. Na (outer 3selectron) or Ag (outer 5s electron), then it is found that the beam is split intotwo by the inhomogeneous field. This suggests that the atoms possess someangular momentum. However, the outer s electron has l=0, i.e. no resultantangular momentum from this source. Thus the atoms must have some othersource of angular momentum. This is characterised as electron spin. The

€

F z = M.dBdz

117

fact that the beam is split into two shows that the electron spin may have onlytwo orientations relative to the magnetic field, either parallel or antiparallel.We may think of the electron spinning on its axis in one of two directions,one of which produces a moment in line with the applied field, and one ofwhich opposes it.

We have already seen that in the case of orbital momentum, the angularmomentum vector may be oriented relative to a fixed axis in 2l+1 ways (i.e.any l value has a degeneracy, g=(2l+1)). Extending this to spin, we haveg=(2s+1)=2, giving s=1/2 (where we have denoted the spin angularmomentum by a quantum number s). Drawing analogies with orbital angularmomentum (l and ml), we may characterise spin quantisation by twonumbers, s=1/2 (analogous to l) and ms=±1/2, (the spin angular momentumquantum number for the z component, analogous to ml) representing the twopermitted orientations of the spin angular momentum. Then the magnitude ofthe spin angular momentum is given by:

(178)where Sz is z- component of the spin angular momentum.

The two possible orientations are generallycalled ‘spin up’ and ‘spin down’, and areshown here:

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volumeIII, ‘Quantum and Statistical Physics’, 1968

€

S2 = s(s+1)h2 =3h2

4

s =12

ms = ±12

Sz = msh = ±12

h

118

In fact, although it is convenient to think of ‘electron spin’ in this way, it is inreality a relativistic correction to the TISE. The time-dependent treatment ofwas developed by Dirac in 1928. As this treatment includes 4 dimensions (3spatial dimensions and time), we find that four quantum numbers appearnaturally from this treatment.

So, to completely specify the quantum state of an electron in an atom requires4 quantum numbers:

n, l, ml and msFor each possible (n,l,ml), one can have ms=±1/2.

IX.b How important are relativistic effects?

Its actually reasonably straightforward to make an estimate of how importantDirac’s relativistic correction to the energy levels in the H atom is.

The energy of the electron in the ground state of the H-atom from TISEputting n=1 is:

(178)We would like to know how this might be affected when the electron velocityis close to c.

In order to obtain an estimate, we note that the kinetic energy of the electronwill be of similar magnitude to E (an application of the virial theorem). Thisgives:

(179)allowing us to extract an estimate of velocity:

(180)

€

E =−z2mee

4

2h2(4πε0)2

€

EK =mev

2

2≈ E =

z2mee4

2h2(4πε0)2

€

v ≈ ze2

h(4πε0)

119

The relativistic factor, v/c is therefore:

(181)α is known as the fine structure constant.

€

Note also that a0 = 1α

h

mec, where a0 is the Bohr radius

It is clear therefore that relativistic effects will become important for atomswith z>>1.

IX.c Symmetry properties of 2-particle states

Consider 2 non-interacting electrons in the same potential. The wavefunctionfor the combined state can be written as a product of the individualwavefunctions:

€

Ψ(r1,r2,t) =ψA (r1)ψB (r2)e− iEt / h

(182)Here,

€

ψA (r1) is the wavefunction for electron 1 (with position r1) which is instate A, and electron 2, at r2 is in state B. The energy of the total system isE=EA + EB, where EA,EB are the energy eigenvalues corresponding toeigenfunctions ψA and ψB.

(The wavefunction of the whole system is the product of the wavefunctionsof the individual electrons because if the electrons are independent, theprobability distributions of each electron are uncorrelated, so the probabilityof finding one electron at one given position and of finding the other atanother given position simultaneously is the product of the probabilities offinding the individual electrons at these positions.)

IndistinguishabilityIt is a fundamental property of nature that electrons (and other fundamentalparticles) are ‘indistinguishable’ (or ‘identical’).

This suggests that we must have

€

vc≈

ze2

hc(4πε0)= zα =

z137

α =e2

hc(4πε0)=1137

120

€

Ψ(r1,r2,t) = Ψ(r2,r1,t)(183)

and the square of the wavefunction (relating to the probability of finding theelectrons) must be symmetric with respect to exchange of particles. This isnot satisfied by (182) unless ψA = ψB.

However, there are two possible (normalised) superpositions of the ‘spatialparts’ of wavefunctions (one symmetric and one antisymmetric) that have|Ψ|2 which is symmetric with respect to exchange of electrons:

€

Ψs =12ψA (r1)ψB (r2) +ψB (r1)ψA (r2)[ ]e− iEt / h

Ψa =12ψA (r1)ψB (r2) −ψB (r1)ψA (r2)[ ]e− iEt / h

(184)Both are eigenstates of the system (since

€

ψA (r1)ψB (r2) and

€

ψB (r1)ψA (r2) havethe same energy, E=EA + EB). Both have |Ψ|2 which is symmetric withrespect to both electrons. However the two wavefunctions (184) differ in animportant respect. The wavefunction with the plus sign is symmetric withrespect to exchange of electrons, i.e.

(185)However, the wavefunction with the minus sign is antisymmetric with respectto exchange of electrons, i.e. it changes sign when

€

r1↔ r2:

(186)The energy of the atom associated with Ψa cannot be the same as thatassociated with Ψs. If the two electrons are very close together, then the twoterms making up the wavefunction are very similar, and Ψa becomes verysmall. In fact if the two electrons are in the same place at the same instant,Ψa vanishes. Thus the antisymmetric wavefunction represents one in whichthe two electrons cannot be in the same region of space at the same time.This lowers their mutual repulsion by an amount known as the exchangeenergy. This is a purely quantum-mechanical contribution to electron-electron repulsion.

€

Ψs(1,2) = Ψs(2,1)

€

Ψa (1,2) = −Ψa (2,1)

121

Thus, a two-electron atom (such as He) has two sets of stationary states andenergy levels, one described by symmetric orbital wavefunctions, and theother by antisymmetric wavefunctions. This is true for all cases, except whenA=B. In this case, Ψa=0, and only the symmetric orbital is possible.

IX.d The helium atom (a two-electron species)

Unfortunately, for any system which contains more than one electron, wefind that an exact analytical solution of the SWE is impossible. This isbecause the potential is now influenced not only by the Coulombic attractionbetween the electrons and the nucleus, but also by the repulsion between theelectrons. This term provides a coupling between the electrons which cannotbe calculated exactly.

For He (and other 2 electron species like H- and Li+) we have the situationbelow:

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968

For a many-electron system, the potential energy of the whole atom is givenby:

(187)Because of the last term, the motion of one electron cannot be consideredindependently of the others, and the energy of an individual electron in anatom cannot be calculated. Approximate methods must therefore be used toobtain energies for the wavefunctions. The potential energy in this case is:

€

V = −ze2

4πε0riall e −∑ +

e2

4πε0rijall pairs∑

€

V = −ze2

4πε0r1−

ze2

4πε0r2+

e2

4πε0r12

122

(188)

The first two terms correspond to the attraction between the nucleus and eachof the electrons, while the last term represents the repulsion between the twoelectrons. This term cannot be evaluated exactly.

Possible Approximations:

1. Independent particle model

This approximates the repulsion by ignoring it! Crude but simple. The twoelectrons therefore move independently of each other. The electrons feel nomutual repulsion, and therefore each can be assigned hydrogen-likewavefunctions. The possible ‘space’ wavefunctions are therefore as in (184):

€

Ψs =12ψA (r1)ψB (r2) +ψB (r1)ψA (r2)[ ]e− iEt / h

Ψa =12ψA (r1)ψB (r2) −ψB (r1)ψA (r2)[ ]e− iEt / h

(184)where ψA and ψB are now wavefunctions for a hydrogen-like atom with z=2,characterised by quantum numbers n,l,ml. (Since

€

ψA (r1)ψB (r2) and

€

ψB (r1)ψA (r2)have the same energy, E=EA + EB, the energy of these linear combinations isalso E=EA + EB.)

The energy of the He atom in its ground state (na=nb=1, parahelium, see later)is then given by:

(189)The actual ground state energy is around –79 eV. The difference is large andis caused by the neglect of the electron-electron repulsion, which is a largepositive contribution to the energy, tending to destabilise the atom.

2. Self-consistent field approach

A better approach is to consider the electrons as independent, but moving inan averaged potential provided by the central field of the nucleus and anaverage central field produced by the other electron. The net effect of eachelectron on the motion of the other is to screen the charge of the nucleus by acertain amount.

€

Eg = −2ERz2 = −2× (13.6 eV) × 4 = −108.8 eV

123

Overall, this approach gives the energy of the He atom (parahelium, na=nb=1)in its ground state as:

(190)where s is the screening constant. It must have a value of about 0.32 to agreewith the observed ground state energy of He. In fact, s has different valuesfor differing values of l, leading to a lifting of the degeneracy of the differentl states which we saw in H.

Each electron may be assigned a hydrogen-like wavefunction, which has anangular part and a radial part:

(191)As we saw for the H-atom, the angular part of the wavefunction isindependent of the central potential field, and is therefore not affected by theaverage central field introduced to describe the electron-electron repulsion.However, the radial part is dependent on the form of the potential, and theconsequence is that the radial part Rn.l(r) differs from that for a hydrogenatom. In the presence of the electron screening effect, the radial parts areaffected differently, depending on the form of the function. For example, sorbitals have a non-zero amplitude at the nucleus, and penetrate through anyshielding outer electrons better than p orbitals. For fixed n, they thusexperience a higher effective nuclear charge, (z-s), than say, p orbitals, i.e. thescreening constant, s is smaller. This means that s orbitals have a lowerenergy than p orbitals, and in general, the (2l+1) degeneracy of the l levels forfixed n is lifted.

IX.e The exchange term

The origin of the exchange term in the electron-electron repulsion can be seenby considering the electron repulsion term as a small perturbation of theHamiltonian of the system:

(192)where

€

ˆ H 1 and ˆ H 2 are hydrogen-like Hamiltonians for the two electrons, and

€

ˆ H 12 is the Hamiltonian for their interaction:

€

Eg = −2ER (z − s)2

€

ψ(r,θ,φ) = Rn,l (r)Yl .ml(θ,φ)

€

ˆ H = ˆ H 1 + ˆ H 2 + ˆ H 12

124

(193)

(194)

Individually, the hydrogen functions ψA and ψB are eigenfunctions of

€

ˆ H 1 and ˆ H 2 , with energies EA and EB. Thus:

(195)and the energy of the He atom, E, is given by:

(196)

The last term contains all the electron repulsion, and has different values forsymmetric and antisymmetric wavefunctions. It can be divided into twoterms:

(197)The third term represents the purely classical Coulomb repulsion between theelectron clouds, called the ‘Coulomb integral’ and given by:

(198)

The last term is the exchange term, which has no classical counterpart:

(199)

€

ˆ H 1 = −h2

2m∇1

2 −2e2

4πε0r1

€

ˆ H 2 = −h2

2m∇2

2 −2e2

4πε0r2

€

ˆ H 12 =e2

4πε0r12

€

( ˆ H 1 + ˆ H 2)ψatom = (EA + EB )ψatom

€

E = ψ *atom∫ ˆ H ψatomdv1dv2 = EA + EB + ψ *atomˆ H ∫∫ 12ψatomdv1dv2

€

E = EA + EB + E12c ± E12

ex

€

E12c = dv1∫∫ dv2ψA (1)

2 e2

4πε0r12ψB (2)

2

€

E12ex = dv1∫∫ dv2ψ *A (1)ψB (1)

e2

4πε0r12ψ *B (2)ψA (2)

125

In equation (197), the minus sign corresponds to the antisymmetricwavefunction, while the positive sign corresponds to the symmetricwavefunction. We find that E12

ex>0, so the antisymmetric orbital state ψa hasa lower energy than ψs.

IX.f The Spin Wavefunction

So far, this discussion is incomplete in that we have only considered the‘space’ wavefunctions of the two electrons, and have taken no account oftheir spin. If the atom contains two electrons, each with spin s=1/2, thenthese may be oriented either parallel or antiparallel, giving a resultant spin,S‡, of 1 or 0 respectively:

(200)States with S=0 are called singlets, because they can be obtained in only oneway. However, when S=1, the allowed values of the resultant spin angularmomentum quantum number for the two electrons, MS, are 0,±1, as shownoverleaf. The spin degeneracy (2S+1)=3, and these states are known astriplets.

The total spin wavefunction of the singlet state is antisymmetric with respectto exchange of electrons, while the triplet state wavefunctions are allsymmetric. If we label the one-electron spin wavefunctions for spin up anddown electrons as χ±, the four possible spin wavefunctions are as givenbelow:

€

χa =12[χ+(1)χ−(2) − χ+(2)χ−(1)], MS = 0

χ s =

χ+(1)χ+(2) MS = +112[χ+(1)χ−(2) + χ+(2)χ−(1)] MS = 0

χ−(1)χ−(2) MS = −1

(201)

‡ When we consider only one electron, quantities such as the orbital angular momentumand spin angular momentum etc. are given lower case symbols, l, s etc. For a many-electron system, the allowed resultants of these vector quantities for the whole system ofelectrons are normally written with capital letters, L, S etc.

€

s1 = s2 =12

S = s1 ± s2 = 0,1

126

Figure from M Alonso and E J Finn, ‘Fundamental University Physics’, volume III, ‘Quantum and Statistical Physics’, 1968

We obtain the total wavefunction of the atom by multiplying the ‘space’ ororbital wavefunction by the spin wavefunction:

Ψtotal = (orbital wavefunction)x(spin wave function)(202)

As there are two possible symmetries of the orbital wavefunction, ψs and ψa,and two symmetries for the spin wavefunction, χs and χa, there are inprinciple four possible combinations of the wavefunctions.

In practice, we find that only two of the possible combinations are everobserved. States which have antisymmetric spin wavefunctions, χa, S=0 arealways combined with symmetric orbital functions, ψs, while the tripletstates, S=1, which have a symmetric spin function, χs are combined with anantisymmetric space function, ψa.

IX.g The Pauli exclusion principle

More generally we find that for fermions (half-integer spin particles such aselectrons, protons, neutrons…), the total wavefunction must be antisymmetricwith respect to exchange of electrons. This is a statement of the PauliExclusion Principle.

If the orbital wavefunctions refer to the same state, ψA = ψB, then n, l and mlmust be the same, and ψa=0 (see (184)). The possible spin functions are:

127

• χa, S=0 (ms=+1/2 for one electron and ms=-1/2 for the other), whichcombines with ψs and

• χs, S=1 (electrons have the same (parallel) spin ms=+1/2 or ms=-1/2 forboth electrons), which combines with ψa (and ψa=0, so this state doesnot exist).

It is thus not possible for the two particles to be in the same quantum state.This leads to the widely used statement of the Pauli exclusion principle:

‘No two electrons in the same atom can have all four quantum numbers (n, l,ml, ms) the same’

So if a pair of electrons have n, l, ml the same, they must have opposite spin,ms=±1/2. This has very far-reaching consequences for the electronconfigurations of all atoms. In the case of helium, it essentially means that itsatoms exist in two separate states, S=0 and S=1, which may essentially beregarded as two separate gasesfl, as there are two possible ways of obtainingthe total wavefunction:

(203)For parahelium, the electrons have opposite spin, ms=+1/2 for one electronand ms=-1/2 for the other. The lowest energy state has both these electronspaired in the state with n=1 (called the 1s state). The electron configurationis written 1s2, and the electrons have quantum numbers n=1, l=0, ml=0,ms=±1/2.

For orthohelium, the electrons both have the same value of ms (parallel spin).Hence the 1s2 configuration cannot exist, because otherwise all electronswould have the same set of all four quantum numbers. The ground state istherefore 1s2s – where the second electron has n=2. This is the lowestenergy configuration that can be achieved that satisfies the Pauli exclusionprinciple.

Note that the 1s2s configuration has a lower energy in orthohelium than the1s2s configuration in parahelium due to the lower exchange repulsion whenthe electron spins are parallel.

fl Transitions which change the electron spin, ΔS≠0 are forbidden by the selection rules(see later lecture courses), so it is very difficult to transform from parahelium toorthohelium and vice versa.

€

Ψtotal =ψsχa ⇒ singlets (parahelium)Ψtotal =ψaχ s ⇒ triplets (orthohelium)

1

PHYS20101Introduction to Quantum Mechanics

SUMMARY OF IMPORTANT CONCEPTS

The following is a summary originally prepared by A CPhillips, adapted by G D Lafferty, A J Bray and W R Flavell

QUANTUM WAVEFUNCTIONS

• Particle and wave properties are described by a wavefunction Ψ whichebbs and flows in accordance with the time-dependent Schrödingerequation (TDSE),

€

ˆ H Ψ = ih∂Ψ∂t

where

€

ˆ H is the ‘energy operator’, usually called the Hamiltonianoperator.

•

€

Ψ2 is a probability density for position.

In one dimension,

€

P(x, t)dx = Ψ(x,t) 2dx = the probability of finding the particle at time t between x and x + dx

• If you look everywhere, you will be certain to find the particle. Ψ isnormalised at all times t such that, integrated over all space, theprobability of finding the particle is unity.

In one dimension:

€

Ψ(x, t) 2−∞

+∞

∫ dx =1

2

QUANTUM STATES

Wavefunctions represent the possible states of motion of real particles.They bear only a passing resemblance to the well-defined particletrajectories encountered in classical physics, and are called quantum states(QS).

• A QS provides precise predictions for the probabilities of the resultsof measurements.

• In the absence of measurements, a QS evolves deterministically inaccordance with the time-dependent Schrödinger equation.

• Any linear superposition of solutions of the TDSE is also a solution.(This is not true of superpositions of solutions of the time-independentSchrödinger equation (TISE) unless they have the same energy, i.e.they are degenerate.)

• A QS is fragile. A measurement destroys it and replaces it by a newquantum state which is compatible with the outcome of themeasurement.

• However, we still do not understand the link between the statisticalnature of some of the predictions of quantum mechanics and thecertainties we measure in the macroscopic world.

QUANTUM EVOLUTION

• Time evolution is governed by the TDSE

€

ˆ H Ψ = ih∂Ψ∂t

.

• If the QS is a state of certain energy E, then

€

Ψ =ψ e−iEt / h ,where ψ satisfies the TISE

€

ˆ H ψ = Eψ ,

3

and all observable properties are constant in time. Such a state iscalled a stationary state.

• If the QS is a state of uncertain energy with normalised wavefunctiongiven by

€

Ψ = c1ψ1e− iE1t / h + c2ψ2 e

− iE2t / h ,

an energy measurement results in E1 with probability |c1|2 or E2 withprobability |c2|2, and observable properties oscillate with a period

€

2πh /E2 − E1 .Thus if the QS is a state of uncertain energy ΔE, the timescale (δt) forchange of observable properties is of the order

€

δtΔE ≈ h .

QUANTUM MECHANICAL TUNNELLING

• If a quantum particle is subject to a confining potential V, there is afinite probability of finding the particle in classically forbiddenregions (where E<V) unless the confining potential is infinite.

• A particle may thus ‘tunnel’ through a thin barrier of thickness a witha tunnelling probability that depends upon

€

e−2βa , where β =2m(V − E)

h.

Hence the wavefunction decays exponentially in the classicallyforbidden barrier region.

QUANTUM OBSERVABLES

A measurable quantity or observable, A, is represented in quantummechanics by an operator

€

ˆ A .

In general, the outcome of a measurement of A is uncertain:

• For a system in the state Ψ(x,t), the expectation value of A is

€

A = Ψ* (x,t) ˆ A −∞

∞

∫ Ψ(x, t)dx

4

• The expectation value of A2 is

€

A2 = Ψ* (x, t) ˆ A 2−∞

∞

∫ Ψ(x,t)dx

• The uncertainty in the outcome is

€

ΔA = A2 − A 2

Sometimes the outcome is certain:

• If the quantum state is an eigenstate of

€

ˆ A ,

€

Ψ =ψn where ˆ A ψn = Anψn ,

the outcome is equal to the eigenvalue An.

QUANTUM COMPATIBILITY

When are observables A and B compatible?

• Physically, if we can know both precisely at the same time.

• Mathematically, if the commutator

€

ˆ A , ˆ B [ ] is zero, so that there exists acomplete set of QS’s with certain values for both A and B.

Examples

• Position and momentum are incompatible because

€

ˆ x , ˆ p [ ] = ih

• The x and y components of angular momentum are incompatiblebecause

€

ˆ L x, ˆ L y[ ] = ih ˆ L z

5

• The z component of angular momentum and its magnitude arecompatible because

€

ˆ L 2, ˆ L z[ ] = 0

ANGULAR MOMENTUM IN QUANTUM MECHANICS

• Orbital angular momentum has uncertain direction. At best, themagnitude and only one component can be determined with certainty.

• For a particle moving in a central potential V(r,θ,φ)=V(r), angularmomentum is conserved.

• In such systems, the separable wavefunction

€

ψ(r,θ,φ) = R(r)Yl ,ml(θ,φ) is

an eigenfunction of certain L2 and Lz, but uncertain Lx and Ly.

• The eigenvalues L2 and Lz are quantised:

€

L2 = l(l +1)h2

Lz = mlh

where l = 0,1,2......and, for a given l :ml = −l,−(l −1),...,0,...,l −1,li.e. ml ≤ l; ml is an integer

l is called the ‘orbital angular momentum quantum number’ (or just‘orbital quantum number’).

ml is the ‘azimuthal angular momentum quantum number’ (or just‘azimuthal quantum number’).

• The eigenfunctions have specific angular shape. For example,

€

Y1,+1 = −38πsinθ e+ iφ , Y1,0 =

34π

cosθ , Y1,−1 =38πsinθ e− iφ

6

QUANTUM STATES IN A CENTRAL POTENTIAL

For a particle in a central potential:

• There exist eigenfunctions with certain E, L2 and Lz of the form

€

ψ(r,θ,φ) =U(r)r

Yl,ml(θ,φ)

where the radial wavefunction R(r) is given by U(r)/r.

• For a system such as an atom, where the central potential is aCoulomb potential, the possible energies, E, for each value of theorbital angular momentum quantum number, l, are found by solvingthe radial TISE,

€

−h2

2me

d2

dr2 +l(l +1)h2

2mer2 −

e2

4πε0r

U(r) = EU(r),

where l(l +1)h2

2mer2 is the 'centrifugal potential',

subject to the boundary conditions U(r)=0 at r=0 and at r=

€

∞.

• The allowed energies of bound state solutions depend only on thevalue of the principal quantum number, n, as En

€

∝ -1/n2. n has values(l+1), (l+2), (l+3)…….., hence n=1,2,3……..

€

∞. The maximumvalue of l is thus (n-1).

• The complete wavefunctions for the hydrogen atom have the form

€

ψn.l ,ml(r,θ,φ) = constant × r

a0

l

e−r na0 (−1)kk= 0

n− l−1

∑ ckra0

k

× Pl ,ml(θ) × eimlφ

€

where Pl,ml(θ) are the associated Legendre polynomials

and the polynomials in r a0 are known as the associated Laguerre polynomials.

The index n-l-1 corresponds to the number of nodes in the radial partof the wavefunction.

7

ELECTRON SPIN ANGULAR MOMENTUM AND THEPAULI EXCLUSION PRINCIPLE

• The electron in the hydrogen atom has angular momentum arisingfrom its orbital motion, and also from another source, characterised asthe ‘spin angular momentum’. In reality this is a relativisticcorrection to the TISE.

• By analogy with orbital angular momentum, the magnitude of the spinangular momentum is given by:

where Sz is z- component of the spin angular momentum and ms is thespin angular momentum quantum number.

• It is a fundamental property of nature that electrons (and otherfundamental particles) are ‘indistinguishable’ (or ‘identical’).

• Thus, the probability of finding an electron must not change whenelectrons are exchanged. Possible ‘space’ or ‘orbital’ wavefunctionsfor a two-electron system are thus:

€

Ψs =12ψA (r1)ψB (r2) +ψB (r1)ψA (r2)[ ]e− iEt / h

Ψa =12ψA (r1)ψB (r2) −ψB (r1)ψA (r2)[ ]e− iEt / h

• Ψa represents a wavefunction in which the two electrons cannot be inthe same region of space at the same time. This lowers their mutualrepulsion by an amount known as the exchange energy. This is apurely quantum-mechanical contribution to electron repulsion.

€

S2 = s(s+1)h2 =3h2

4

s =12

ms = ±12

Sz = msh = ±12

h

8

• For fermions (half-integer spin particles such as electrons, protons,neutrons…), the total wavefunction (‘space’x‘spin’ parts) must beantisymmetric with respect to exchange of electrons. The more widelyused statement of the Pauli exclusion principle is:

‘No two electrons in the same atom can have all four quantumnumbers (n, l, ml, ms) the same’

• For atoms containing more than one electron (such as He, which hastwo), the repulsion term between the electrons cannot be calculatedexactly. This means that the TISE for the atom cannot be solvedexactly, and approximate methods must be used.