Lecture 4

The Particle in a BoxThe Uncertainty Principle

The translational partition functionTunneling

NC State University

Chemistry 431

The particle in a box problemImagine that a particle is confined to a region of space.The only motion possible is translation. The particle hasonly kinetic energy. While this problem seems artificial atfirst glance it works very well to describe translationalmotion in quantum mechanics.

0 LAllowed Region

The solution to the Schrödingerequation with boundary conditions

Suppose a particle is confined to a space of length L.On either side there is a potential that is infinitely large.The particle has zero probability of being found at the boundary or outside the boundary.

0 LAllowed Region

The boundary conditions determine the values for the constants A and B

A = – B = 12i , Ψ = sin kx

sin will vanish at 0 since x = 0 and sin 0 = 0.sin will vanish at a if kL = nπ.Therefore, k = nπ/L.

Ψ = sin nπxLNot

Normalized !

The solution to the Schrödingerequation with boundary conditions

The boundary condition is that the wave function willbe zero at x = 0 and at x = L.

From this condition we see that B must be zero.This condition does not specify A or k.The second condition is:

From this condition we see that kL = nπ. The conditionsso far do not say anything about A. Thus, the solutionfor the bound state is:

Note that n is a quantum number!

Ψ(0) = Asin(k0) + Bcos(k0) = 0

Ψ(L) = Asin(kL) = 0 or kL = arcsin(0)

Ψn(x) = Asin(nπx/L)

The Schrödinger equation for a free particle

– h2

2m∂2Ψ∂q2 = EΨ

The solutions are:

Ψ = Aeikq + Be– ikq

eikq

e-ikq

The particle in a box has boundary conditions

Ψ(0)= 0 Ψ(L)= 0

L

The solutions to the particle in a box

E =h2k 2

2m

=h2 nπ

L2

2m

=h2n2

8mL2

ψ2

The uncertainty principleWhen we measure the properties of very small particles, wecannot help but affect them. The very act of measuring causes a change in the particle’s properties. Therefore, the description of the the measurement is a probability rather than a fixed value. We have seen the Born interpretation of the square of the wavefunction as a probability density.The consequence of this is that certain variables are linkedBy the uncertainty that is inherent in the measurement.Position and momentum are two such conjugate variables.Note that the units of position is the reciprocal of the momentum (if we factor out Planck’s constant).x has units of meter, k has units of meter-1

Momentum is p = hk.t has units of time, ν has units of time -1Energy is E = hν.

Where is the particle in the box?Since we are using a probability function we do not really know exactly where the particle is. We know that the highestprobability occurs for the position L/2. We can guess that this is the average position in the box. However, themore precisely we specify the location of the particle theless information we have about how fast the particle ismoving. This is a statement of the famous Uncertainty Principle.

ΔxΔp > h/2

Let’s look at the Uncertainty Principle using the particle-in-a-box example. If we know that the particle is in the lowestlevel then Uncertainty in its position is approximately equalto the width of the probability distribution.

The location of a particle in free space is not defined

Consider a superposition of a wavewith moment hk and h(1.1k)

The sum has a characteristic envelopefrequency at (ω2 - ω1)/2

Envelope

The sum has a characteristic beatfrequency at (ω2 + ω1)/2

Beats

As we add more frequencies wecan speak of a bandwidth Δk

Δk = 0.1

3 added cosines

Δk = 0.2

As the bandwidth increases theposition in x-space becomes more defined

5 added cosines

The superposition of waves in spaceleads to the description of a location

Δk = 0.7

15 added cosines

Relevance of the example

Although the function used in the example is periodic it isrelevant. Since in a given region of space (i.e. where a measurement can be made) the probability of observing the particle in a given region of space is dependent uponthe number of contributing waves. If more waves contribute then the momentum of the particle is lesscertain. Thus, the we can know that moment precisely ifwe are totally uncertain of the position. As we begin to specify the position more precisely we find that the momentum is less well known. Since p = hk, we can also express this condition as:

ΔxΔk > 1/2

Fourier transform related pairsPosition and momentum are related by a Fourier transform.

x pTime and energy are related by a Fourier transform.

t EThere is an uncertainty relationship for both of these related pairs. Thus, for time and energy we have

Δt ΔE > h/2

as well. These pairs can be related by a probability function that gives the width of the distribution in eachspace. Gaussian functions are particularly useful sincethe Fourier transform of a Gaussian is also a Gaussian.

Gaussian FunctionsA Gaussian function has the form exp{ -α(x – x0)2 }.The Gaussian indicated is centered about the point x0.The Fourier transform of a Gaussian in x-space is aGaussian in k-space. Since p = hk we also call this momentum space. The figure shows the inverse relationship.

x k

QuestionWhich of the following represents the hamiltonian?

A. hk

B. h2k2

2mC. – h2

2m∂2

∂x2

D. – ih ∂∂x

QuestionWhich of the following represents the hamiltonian?

A. hk

B. h2k2

2mC. – h2

2m∂2

∂x2

D. – ih ∂∂x

QuestionWhat is the hamiltonian?

A. it is the energy

B. it is the momentum

C. it is the energy operator

D. it is the momentum operator

QuestionWhat is the hamiltonian?

A. it is the energy

B. it is the momentum

C. it is the energy operator

D. it is the momentum operator

Statistical averaging over translational energy levels

Quantum mechanics must agree with classicalphysics (mechanics) at high temperature or when the average quantum number becomes verylarge. This is the case for translational energylevels since the spacing of those levels is verysmall compared to thermal energy, kT.

Here, we consider how to average over theenergy levels given by the particle-in-a-box solutions.

The translational partition functionThe translational partition function consists ofa sum over a very large number of states

q = e–(n2–1)βε = e–(n2–1)βεdn1

∞

Σn = 1

∞

This sum can be expressed as an integral overthe states n. The integral can be expressed asa Gaussian.

Energies from particle in the box can be used to calculate energy level spacingThe difference in energy levels n in the particle-in-the-box solutions has the general form.

where ε n gives the energy of a large number oftranslational energy levels derived from the particle in the box solutions. We can write ε = h2/8mX2 to use a factor in the Boltzmanndistribution, so that εn = ε(n2 - 1)

εn = h2

8mX 2 n2 – 1

The translational partition function is a Gaussian

The Gaussian in n integrates to π1/2/2.

Substitute for ε to obtain

e–n2βεndn0

∞

= 1βε

1 / 2

e–x 2d0

∞

= 12

πβε

1 / 2

qx = 2πmh2β

1 / 2

X

The translational partition function in three dimensions X,Y, and Z

The volume is V = XYZ

This expression for the translational partitionfunction derived from the particle in the boxis the same as that derived classically from the integral over all velocities.

q = 2πmh2β

3/2

XYZ= 2πmh2β

3/2

V

The translational partition function can be expressed in terms of a

thermal wavelength Λ

where the thermal wavelength is defined as:

q = 2πmkTh2

3/2

V = VΛ3

Λ = 2πmkTh2

Tunneling

Tunneling of electrons, protons or other small particles is not possible according to classicalmechanics. However, in quantum mechanicsa particle can penetrate a barrier even if its kinetic energy,E is less than the potential energybarrier height, V.

TunnelingE

V

TunnelingIf the walls of the box are not infinitely high, the The wavefunction of the particle does not decayTo zero. Instead it decreases exponentially.Ultimately it has some probability for tunnelingThrough a barrier even when E < V.

– h2

2m∂2

∂x2Ψ + VΨ = EΨ

TunnelingE

V

The evanescent waveSince V is greater than E the solutions in the Barrier are not oscillatory functions, but rather areExponentially decaying functions.

Note that the order of E-V is reversed and the i isalso absorbed so the the wavefunction is nolonger oscillatory. The function Cekx increases without bound and is not a practical solution.

Ψ = Ceκx + De–κx

κ = 2m(V – E)h

Boundary conditionsThe oscillatory part and the evanescent part of thewavefunction must match up at the boundary.If we think about it the wave as a propagating wave and the exponential decay as a transmissionacross a barrier, then there will also be a reflectedwave.

We can use the property that the wave functionand its first derivative must be continuous to find

conditions at the boundary.

Ψ = Aeikx + Be–ikx (oscillatory part left side)Ψ = Ceκx + De–κx (exponential part)Ψ = A′eikx + B′e–ikx (oscillatory part right side)

Boundary conditionsThe boundaries are at x = 0 and x = L. Thus, theconditions are:

The first derivatives are:

There are four equations and six unknowns. If we assume that the particle is coming from theleft then we can set B = 0. If we calculate the transmission coefficient then we need the ratio A’/A.

A + B = C + D at x = 0A′eikL + B′e–ikL = CeκL + De–κL at x = L

ikA – ikB = κC – κD at x = 0ikA′eikL – ikB′e–ikL = κCeκL – κDe–κL at x = L

Transmission coefficientWith these constraints we can solve for the transmission coefficient, T.

where ε = E/V.

The tunneling phenomenon is important inelectron transfer theory. Electron transfer isa key aspect of energy transduction in biology(e.g. photosynthesis and respiration, among others).Electron tunneling is also the effect used in thescanning tunneling microscope (STM).

T = 1 + eκL – e–κL

16ε(1 – ε)

– 1