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F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S

FLAP P11.1 Reflection and transmission at steps and barriersCOPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

Module P11.1 Reection and transmission at steps and barriers1 Opening items1.1 Module introduction

1.2 Fast track questions1.3 Ready to study?

2 Reection & transmission at a potential step when E > V 2.1 Classical description of the problem2.2 The time-independent Schrdinger eqn & its solutions2.3 Relations imposed by the boundary conditions2.4 The wavefunctions in each region and the physical

interpretation2.5 Dening particle ux in classical & quantum models2.6 Reection and transmission in the quantum model

3 Reection & transmission at a potential step when E < V 3.1 Classical description of the problem

3.2 The Schrdinger eqn & the solutions in each region3.3 Relationships imposed by the boundary conditions3.4 The wavefunctions in each region and the physical

interpretation

3.5 Reection and transmission in the quan3.6 Summary of Sections 2 and 3

4 Reection and transmission at a barrier when4.1 Classical description of the problem4.2 The Schrdinger eqn & the solutions in4.3 Relationships imposed by the boundary4.4 The wavefunctions in each region and t

interpretation4.5 Transmission in the quantum model

4.6 Examples of quantum tunnelling4.7 Summary of Section 45 Closing items

5.1 Module summary5.2 Achievements5.3 Exit test

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1 Opening items

1.1 Module introduction

In 1954, the Nobel Prize for physics was awarded to Cockcroft and Walton for their initiation of nuclear phtwenty years previously. They made use of quantum tunnelling to induce nuclear reactions in ci

which classical physics deemed impossible. In 1973, the Nobel Prize was awarded to three scientists forwork in developing useful devices called tunnel diodes which depend for their operation on the sThese devices have opened up whole new areas of physics and technology. In 1986, another Nobel Prizawarded to Binnig and Rohrer for their invention of the scanning tunnelling electron microscope. Allprizes have been awarded for the exploitation of one quantum effect, carried out with great ingenuitinsight.

The behaviour of particles such as electrons, photons and nucleons is determined by their quantum mechcharacteristics, and we rely on the Born probability hypothesis to relate particle position to the ampsquared of the associated wavefunction. Such particles may therefore be expected to exhibit wavcharacteristics. When light waves encounter a boundary where the refractive index changes, there may be breected and a transmitted wave. If the boundary is between a transparent and an opaque material, the wavstill penetrate into the opaque material for a distance roughly equal to the wavelength. It follows that light ctransmitted through a layer of a metal provided its thickness is less than the penetration depth of t

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A good example of this is the layer of silver a few atoms thick on a glass slide forming a semi-reecting m

In the context of quantum mechanics we might expect particles to show similar behaviour.In the quantum mechanical case, one can think of the particle potential energy as roughly analogourefractive index. If the potential energy changes suddenly with position, then there is usually both a transmand a reected quantum wave at the boundary, and the particles will be transmitted or reected with calcuprobability. If the increase in potential energy is greater than the particle kinetic energy, then the quantum will penetrate into the classically forbidden region though its amplitude will rapidly decrease. In consequenthe region of high potential energy is narrow enough, the wave will emerge with reduced amplitude on the side and there is a probability that the associated particle will tunnel through. A narrow region of high potenergy is called a potential barrier . Quantum mechanics therefore predicts two effects totally aliclassical mechanics based on Newtons laws: (i) particles may be reected by any sudden change in potenergy, and (ii) particles can tunnel through narrow potential barriers. Experiment conrms both of remarkable predictions.

Study comment Having read the introduction you may feel that you are already familiar with the material coveredmodule and that you do not need to study it. If so, try the Fast track questions given in Subsection 1.2. Idirectly to Ready to study? in Subsection 1.3.

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Question F2

A beam of electrons with kinetic energy 5 eV is incident on a region where the electron potentsuddenly increases from 0 to 2.5 eV. Calculate the transmission and reection coefcients at thSketch the electron density as a function of position. Account for the sequence of maxima and minima befostep. Assume the potential is constant before and after the step.

Question F3

Estimate the fraction of 2.5 eV electrons that pass through a potential barrier height 5 eV and width

Study comment

Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the mto proceed directly to Ready to study? in Subsection 1.3.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly Closing items .

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1.3 Ready to study?Study comment In order to study this module, you will need to be familiar with the following physics terms:wave , eigenfunction (of momentum), time-independent Schrdinger equation , wavefunction , Born probabilityof the wavefunction, the (stationary state ) probability density function P ( x) = | ( x) |2. You should be fam

solutions of the Schrdinger equation for a particle moving in one dimension, in a region where the potential eneconstant. In particular, you should know the different forms of the solutions: (a) when the total energy of the partgreater than the potential energy, and (b) when the total energy is less than the potential energy. Some knowledge of claNewtonian mechanics is assumed, especially the treatment of particle motion in terms of the total energy, kinetic energpotential energy. The module assumes some familiarity with the treatment of transverse waves on a string an

of travelling waves at a boundary. If you are uncertain of any of these terms, you can review them now by referrinGlossary which will indicate where in FLAP they are developed. We have to use complex numbers , so you with them in Cartesian form z = a + ib , polar form z = A(cos + i sin ) and exponential form z = A exp( icondent using a spatial wavefunction written as a complex number: for example ( x) = A exp( ikxmanipulation of complex numbers. We will frequently use differential calculus , including differentiation

functions such as sine, cosine and exponential. The following Ready to study questions will allow you to estyou need to review some of the topics before embarking on this module.
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Question R1

Write down the spatial part of the wavefunction of a particle travelling in the positive x-directiomomentum p . If the total energy of the particle is E and the potential energy has the constant valuerelation between p , E , and V ? What is the relation between the angular wavenumber k , wavelength, and the momentum magnitude p?

Question R2

Write down the time-independent Schrdinger equation for a particle moving in one dimension ( xspace where the total energy is less than the constant potential energy V . Show by substitution thwavefunction ( x) = A exp( x) + B exp( x), where A and B are constants, is a solution of this end in terms of E and V .
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Question R3

Dene the (stationary state) probability density function P ( x), for a particle moving in one dimdescribe its physical interpretation. A particle has the spatial wavefunction ( x) = A exp( x x 0, with A and real, and ( x) = 0 elsewhere. Write down an expression for the probability denFind a real positive value of A if the total probability of nding the particle somewhere between xis one (i.e. normalize the wavefunction).

Question R4

(a) Write z = a + i b in the form z = A exp( i ) by relating A and to a and b.

(b) Show that the complex number z = a iba + ib may be written as

z = exp( 2i ).

(c) Show that the complex number z = 2 aa + ib may be written z = 2 cos exp( i ).

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2 Reection and transmission at a potential step when E > V U ( x)

0

Figure 1 4 The potefunction U ( x) for a pby a repulsive force band x = d .

2.1 Classical description of the problemThe physical situation we are modelling is quite simple. A particle is movingwith constant velocity in the positive x-direction and encounters at some

point a strong force directed in the negative x-direction. The force is constantin time and acts over a short distance d . It is usually best to represent thesituation in classical (Newtonian) mechanics with a potential energy functionU ( x). The derivative of the potential energy function with respect to x givesthe force component in the negative x-direction:

F x ( x) = dU ( x)

dx

The potential energy function looks like that drawn in Figure 1. Notice thatfor convenience we have taken U ( x) = 0 for x < 0 and U ( x) = V for x d . It is the change in the pthat is physically signicant, and the force acts in the region between 0 and d . For obvious reasonsis referred to as a potential step , and we wish to know what happens to a particle encountering it.
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Classical mechanics gives a denite answer, and all we need to know is the initial speed of the particle, its

m and the potential energy function. Classically, the total energy E of the particle, kinetic plus potenconstant everywhere. So, if we let the speed of the particle be u as it approaches the step, and if wespeed beyond the step, then

E = 12 mu2 + 0 = 12 mv 2 + V

Hence

v = 2( E V )m

This equation has a real solution only if E V . The particle then passes across the step and cont

reduced speed. If the potential energy has a value greater than E beyond the step, then the solution foimaginary and does not represent a physical situation. In fact, the particle is reected from the step andseen in the region x d .

In the classical sense, reected means that the repulsive force is so strong that the particle comes to rest and

moves back in the direction from which it came.
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U ( x)

0

Figure 1 4 The potefunction U ( x) for a pby a repulsive force band x = d .

Question T1

Suppose the potential energy function in Figure 1 is U ( x) = 0 for x < 0 andU ( x) = 6 J for x d. A particle of mass 6 kg approaches from the left withspeed 2 m s1. Find its speed after the step. 4

If the force on the particle is in the positive x -direction, then the potentialenergy decreases and we have a step down rather than a step up! It is easy toshow that the particle will always gain speed at this step and classicalphysics predicts that it will never be reected.
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2.2 The time-independent Schrdinger equation and its solutions

U ( x)

0

E

E V

region (I) regio

Figure 2 4 A schemarepresentation of a poteThe distance over whicacts is assumed to be nThe total energy of the

In order to nd out what quantum mechanics can tell us about a physicalsituation, we have to make a simple model and then solve theSchrdinger equation to nd the appropriate wavefunctions . As in classicalmechanics, we model the potential function as a step and make the

approximation that d is small, but small compared with what?The relevant dimension is now the de Broglie wavelength of the incidentparticle = h / p , where h is Plancks constant and p is the magnitude of the

particle momentum initially. The potential energy diagram is the same as

Figure 2, and it is convenient to call the space x < 0 region (I) and the space x 0 region (II). The time-independent Schrdinger equation in onedimension is:

22 m

d 2 ( x)dx 2 +U ( x) ( x) = E ( x) (1)

where E is the total energy of the particle, U ( x) the potential energyfunction and ( x) the spatial part of the wavefunction.
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Study comment The full wavefunction ( x, t ) is time-dependent and satisfies the time-dependent SchrdiFor a stationary state of definite energy E the wavefunction takes the form

( x, t ) = ( x)exp i E

t

Since this module is entirely concerned with states of given energy E , we need only determine ( x), which wconveniently refer to as the wavefunction. The full wavefunction ( x, t ) follows immediately. Similarly, weEquation 1 as the Schrdinger equation. 4

22 m

d 2 ( x)dx 2

+U ( x) ( x) = E ( x) (Eqn 1)

In region (I), U ( x) = 0 and ( x) = 1( x) so the Schrdinger equation is 22 m

d 2 1 ( x)dx 2

= E 1 ( x)

The solution of this equation takes the form: 1( x) = A exp( ik 1 x) + B exp(ik 1 x) (2)

with A and B arbitrary complex constants and k 1 = 2 mE .

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In region (II), U ( x) = V and the Schrdinger equation is:

22 m

d 2 2 ( x)dx 2

= ( E V ) 2 ( x)

The solution then has the form:

2( x) = C exp( ik 2 x) + D exp( ik 2 x) (3)with C and D arbitrary complex constants and k 2 = 2 m( E V ) .Notice that k 2 is a real number because E > V .

The rst term on the right-hand side of Equation 2

1( x) = A exp( ik 1 x) + B exp(ik 1 x) (Eqn 2)

represents a particle moving with momentum component k 1 in the x-direction, and threpresents a reected particle moving with momentum component k 1. The rst term in Equation 3

particle transmitted across the step moving with momentum component k 2, and the second termparticle moving in the negative x-direction. Clearly, we must set coefcient D = 0, since this phyinvolves a particle approaching the barrier from the left, not from the right and here it cannot returninfinity.
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U ( x)

0

E

E

V

region (I) regio

Figure 2 4 A schemarepresentation of a poteThe distance over whicacts is assumed to be n

The total energy of the

In summary, the solutions to the Schrdinger equation in the two regionsare:

1( x) = A exp( ik 1 x) + B exp(ik 1 x) in region (I) (4a)

2( x) = C exp( ik 2 x) in region (II) (4b)

The important question of normalization of the wavefunctions must bepostponed until we have discussed fully the constraints imposed on theconstants A , B and C by the conditions at the boundary between the tworegions.

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2.3 Relations imposed by the boundary conditionsGeneral solutions of differential equations always contain arbitrary constants, and we can determine thesethe conditions assumed to apply at certain positions or times. For example, the general solution of the equation for transverse waves on a string stretched between two xed points has two arbitrary constantWe can determine these constants from the initial configuration of the string including the requirement thdisplacement is zero at the ends.

We can apply two boundary conditions , at x = 0, to the solutions of the Schrdinger equatapplication. We will state them and then make some justication.

boundary condition (1) 1( x) = 2( x) at x = 0

boundary condition (2) d 1( x)

dx= d 2 ( x)

dxat x = 0

The rst condition ensures that the wavefunction has a single value at x = 0. The Born probabilisays that the probability density is given by | ( x) |2 , so that the solution in region (I) must match the region (II) at the boundary between the regions.
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The second condition

boundary condition (2) d 1( x)dx

= d 2 ( x)dx

at x = 0

is clear when we examine the Schrdinger equation itself (Equation 1).

2

2 md 2 ( x)

dx 2 +U ( x) ( x) = E ( x) (Eqn 1)

If both E and U ( x) are nite quantities everywhere, then the second derivative d 2 1( x)/ dx2 musThis means that the rst derivative of the wavefunction d 1( x)/ dx cannot suddenly change

including the boundary at x = 0. Therefore the slope of the wavefunction in region (I) at x = 0slope of the wavefunction in region (II) also evaluated at x = 0.

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0

0

0(a)

(b)

(c)

The possibilities are illustrated in Figure 3:

(a) shows a discontinuity in at x = 0,(b) has continuous but d / dx discontinuous at x = 0,

and in

(c) both and d / dx are continuous.

Only in case (c) can we say that varies smoothly across the boundary,and this is what we require .

Figure 3 4 Boundary conditions at x = 0. Only (c) is allowed. (a) 1 2 ,(b) 1 = 2 but d 1 / dx d 2 / dx, (c) 1 = 2 and d 1 / dx = d 2 / dx.

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Applying the boundary conditions gives us two equations linking A, B and C :

Using Equations 4,

1( x) = A exp( ik 1 x) + B exp(ik 1 x) in region (I) (Eqn 4a)

2( x) = C exp( ik 2 x) in region (II) (Eqn 4b)

boundary condition (1) gives:

1(0) = 2(0)

so that A exp

0 + B

exp

0 = C

exp

0

i.e. A + B = C (5)

Applying boundary condition (2):

A(ik 1) exp 0 + B(ik 1) exp 0 = C (ik 2) exp 0

so that Ak 1 Bk 1 = Ck 2 (6)
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Equations 5 and 6

A + B = C (Eqn 5)

Ak 1 Bk 1 = Ck 2 (Eqn 6)

can be used to determine the two ratios B / A and C / A:

B A

= k 1 k 2k 1 + k 2

C A =

2 k 1k 1 + k 2

(7)

Question T2

Consider a string lying along the x-axis under tension F . The density per unit length of the string isand 2 for x 0. Considering transverse waves on the string, describe how this problem in classical mecmight be similar to the quantum-mechanical problem of particles incident on a potential step.(The speed of transverse waves is F .)4

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2.4 The wavefunctions in each region and the physical interpretation; normalizationIn all our discussions of quantum mechanics so far, we have assumed that the wavefunction ( xto unity and represents a single particle , so | ( x) |2 x gives the probability of nding the particle in to x + x. Now it is convenient to modify the prescription and say that the wavefunction can representone particle . In particular, the wavefunction ( x) = A exp( ikx) can represent a set of particles, m

x-direction, all with the same momentum k .

The position of any one particle is not specied (to do so violates the Heisenberg uncertainty principsay the average number of particles per unit length is given by | ( x)|2 . In this context, averagaverage of a large number of observations made on particles with this wavefunction.

The wavefunction ( x ) = A exp( ikx) represents a stream of particles moving in the x-directionShow that, on average, there are | A |2 particles per unit length.
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Now we must examine the wavefunctions (Equations 4)

1( x) = A exp( ik 1 x) + B

exp(ik 1 x) in region (I) (Eqn 4a)


in the potential step problem in the light of the new prescription.

In region (I), we have the wavefunction: 1( x) = A exp( ik 1 x) + B exp(ik 1 x) with k 1 = 2 mE

The rst term of this single wavefunction now represents a stream of particles moving in the positivetowards the step; these are the incident particles. The second term represents a stream of particles monegative x-direction away from the step, the reected particles . The average number of incident punit length is | A |2 , and the average number of reected particles is | B |2 . Notice that the single wavrepresent both the incident and reected particles.

In region (II), the wavefunction is:

2( x) = C exp( ik 2 x) with k 2 = 2 m( E V ) This simply represents a stream of particles moving in the positive x-direction away from transmitted particles . The average number of transmitted particles per unit length is | C |2 .

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A further simplication of the equations is made, with no loss of generality, if the wavefunctions arein a special way. If the coefcient A is set equal to one, then the average number of incident particlelength is equal to one.

Question T3

Electrons with kinetic energy 5 eV are moving in the positive x-direction in a region of constaThere are on average 5 106 electrons per millimetre. Find a suitable wavefunction to descnd A and k .4

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2.5 Dening particle ux in the classical and quantum modelsIn classical mechanics, it is straightforward to dene the concept of particle ux . We will restrict tto motion in one dimension, but the extension to three dimensions is not difcult. If there are N palength and each one has speed u in the positive x-direction, then all the particles in a length u t wpoint in time interval t . The number passing a xed point per unit time is the particle ux F :

F = Nu t t = Nu

In quantum mechanics, the denition of particle ux is equally simple, provided we are dealingwavefunctions of particles with denite momentum (i.e. momentum eigenfunctions). Consider a streparticles represented by the wavefunction ( x) = A exp( ikx). The average number of particles per unthe constant | A |2 , and their speed u is obtained from the momentum magnitude:

p = k 4 and 4 u = p / m

so that: u = k / m

Since k > 0, so is u.

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The ux is then given by:

F = | A |2 k m

(8)

A word of warning is necessary here! Equation 8 can only be used when the wavefunction is a momeigenfunction; the momentum is then the same for all the particles, and the average number of particles pelength is a constant, independent of position.

Question T4

What is the ux of electrons in Question T3?

[Electrons with kinetic energy 5 eV are moving in the positive x-direction in a region of constaThere are on average 5 106 electrons per millimetre.]

What is the corresponding current in amps?

( Hint : The current is the total charge in coulombs passing a point per second.) 4

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2.6 Reection and transmission in the quantum model

U ( x)

0

E

E V

region (I) regio


We are now in a position to complete the quantum description of particlereection and transmission at potential steps. Two coefcients are denedwhich characterize the behaviour of particles when they encounter apotential step such as that illustrated in Figure 2.

The reection coefcient is dened as follows:

R = ux of reected particles ux of incident particles

The wavefunction in region (I) is: 1( x) = A exp( ik 1 x) + B exp(ik 1 x)

The rst term represents the particles incident on the step, and, usingEquation 8,

F = | A |2 k m

(Eqn 8)

the incident ux is | A | 2 k 1 / m.

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The second term

1( x) = A exp( ik 1 x) + B exp(ik 1 x)

represents the reected particles which have average density | B | 2 per unit length and the re| B |2 k 1 / m. The reection coefcient is therefore:

R = | B|2 k 1 m

| A |2 k 1 m=

| B|2

| A |2Using Equations 7,

B A

= k 1 k 2k 1 +k 2

(Eqn 7)

we get the result:

R = k 1 k 2k

1 + k

2

2

(9)

The reection coefcient at a potential step with E > V

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The transmission coefcient is dened as:

T = ux of transmitted particles

ux of incident particles

The wavefunction in region (II) is 2( x) = C exp( ik 2 x), and this represents the particles transmittedstep. The transmitted ux is given by | C | 2 k 2 / m and hence the transmission coefcient is:

T = | C |2 k 2 m| A |2 k 1 m

= | C |2 k 2

| A |2 k 1

Using Equations 7 again:

C A = 2 k

1k 1 +k 2 (Eqn 7)

T = 4 k 1k 2

(k 1 + k 2 )2

(10)

The transmission coefcient at a potential step with E > V
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In order to check that our equations for R and T are self-consistent, we must show that particles are nthe step. The number of particles crossing the step per second (the transmitted ux) added to the nureected per second (the reected ux) must be equal to the incident ux. From the denitions ofmeans that R + T = 1. This is indeed the case!

Question T5

Use Equation 9 and Equation 10

R = k 1 k 2k 1 + k 2

2

(Eqn 9)

T = 4 k 1k 2(k 1 + k 2 )2

(Eqn 10)

to show that R + T = 1. 4

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Remember that R and T are both ratios of uxes ; this means that the formulae for R and T whatever the value of A. It is also legitimate to regard R and T as reection and transmission probare dealing with the behaviour of a single particle at the step.

R = k 1 k 2k 1 + k 2

2

(Eqn 9)

T = 4 k 1k 2(k 1 + k 2 )2

(Eqn 10)

Notice that the reection coefcient approaches the limit R = 1 and the transmission coefcient gwhen k

2 0. This happens when V is equal to the incident particle energy E .

The function P ( x) = | ( x) |2 represents the probability density when is the wavefunction for a sin

However, when represents a set of particles, the function P ( x) = | ( x) |2 gives the average parti

the position x. P ( x) is then the particle density function and it is interesting to plot a graph ofpotential step.
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P ( x)

Region (I)

Figure 4 4 The particle

function P ( x) in the regpotential step ( E > V ). P ( x) shows a pattern of amaxima and minima cauinterference of the incide

reected waves.

This is shown in Figure 4 for the case when A , B and C are real.In region (I), we have to use the full expression for 1( x) (Equation 4):



P1 ( x) = 1* ( x) 1 ( x) = A2 + B2 + 2 ABcos(2 k 1 x) (11)In region (II)

P2 ( x) = 2* ( x) 2 ( x) = C 2 (12)P 1( x) and P 2( x) join smoothly at x = 0 because of the boundary conditions(Equations 5 and 6).

A + B = C (Eqn 5)

Ak 1 Bk 1 = Ck 2 (Eqn 6)

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P ( x)

Region (I)



reected waves.

In region (I), P ( x) has a cosine component due to the interference betweenthe incident and reected waves.

This example of quantum-mechanical interference with material particlesis analogous to the interference of light beams reected from boundariesbetween materials with different optical properties. The interferenceminima are not at zero because the amplitude of the reected wave is less

than the amplitude of the incident wave.

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Question T6

Starting from Equations 4,



verify Equations 11 and 12.P1 ( x) = 1* ( x) 1 ( x) = A2 + B2 + 2 ABcos(2 k 1 x) (Eqn 11)

In region (II)

P2 ( x) = 2* ( x) 2 ( x) = C 2 (Eqn 12)Show that P 1(0) = P 2(0). (Remember Equations 11 and 12 assume A, B and C are real.) 4

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P ( x)

Region (I)



reected waves.

Question T7

Figure 4 shows clearly that the probability per unit length of nding aparticle in region (II) is greater, on average, than the probability in region(I). How can that be when a fraction of the incident particles isreected? 4
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3 Reection and transmission at a potential step when E < V

3.1 Classical description of the problemU ( x)

0

P

Figure 5 4 The functiopotential step. The heigV is greater than the en

incident particle. Accorclassical mechanics, thereach the point P and arreected back.

A particle travelling in the positive x-direction encounters a step where thepotential energy U ( x) increases by an amount V greater than the particleenergy. The physical situation is modelled by the illustration in Figure 5and the total energy of a particle is represented by the horizontal dashedline. Applying the law of energy conservation to solve for the speed of theparticle, we get:

12 mv

2 = E U ( x)

v = 2 E U ( x)( )m

P marks the point where the total energy and the potential energy are equal,

E = U ( x) and v = 0. There is no real-number solution to the speed equationanywhere to the right of P. According to classical mechanics, each particlewill reach the point P and then be reected back. The classical reectioncoefcient is one.

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3.2 The Schrdinger equation and the solutions in each regionU ( x)

0

E V

Figure 6 4 A schemrepresentation of a powith E < V . Particlesfrom the left.

For the quantum mechanical problem we again make the assumption that thedistance over which the potential is increasing is small compared with the deBroglie wavelength of the incident particles. The situation is illustratedschematically in Figure 6 with the region x < 0 designated (I) and x 0

designated (II). The solution of the Schrdinger equation follows exactly asin Subsection 2.2, and we can write the solutions by referring to Equations4:



The angular wavenumbers k 1and k 2 are given by:

k 1 = 2 mE

4 and 4 k 2 = 2 m( E V )

In this case it is clear that k 2 is a purely imaginary number since E < V , but we cannot abandon this s

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Remember that in quantum mechanics, the solutions make sense if the predictions for the valuobservable quantities are real. We will continue the analysis and examine the form of the solution in regionIf we replace k 2 by i , then is a real number and we nd:

2( x) = C exp[ i(i ) x]

= C exp( x)4 since i2 = 1

The real constant may be written in terms of E and V as follows:

= k 2 / i = ik 2

However k 2 = 2 m( E V ) = i 2 m(V E )

so that = 2 m(V E )

In fact, there is a second part of the general solution in region (II) corresponding to a rising expthan a falling exponential: 2( x) = C exp( x) + D exp(+ x). The second term is ruled out ahere because it predicts that the probability of observing the particles increases without limit as x

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In summary, we have:

1( x) = A exp( ik 1 x) + B exp(ik 1 x) in region (I)

2( x) = C exp( x) in region (II) (13)

with 4 k 1 = 2 mE

4 and 4 = 2 m(V E )
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3.3 Relationships imposed by the boundary conditions

The general requirements discussed in Subsection 2.3 that the wavefunction and its rst derivative mcontinuous everywhere apply in this example. The relations between the arbitrary constants Aexactly the same as in Equations 7:

B

A= k 1 k 2

k 1 + k 24 and 4

C

A= 2 k 1

k 1 + k 2However now k 2 is a purely imaginary number, and we have equated it to i so:

B

A

= k 1 i

k 1 +i 4 and 4

C

A

= 2 k 1

k 1 +i Show that the complex number B / A can be written in the form: B / A = exp( 2 i ) where

Show also that | B | = | A |.

Question T8

Show that the complex number C / A can be written in the form: C / A = 2 cos exp( i ) and | C | =

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In summary, we have the following relations for the ratios of the constants:

B A

= k 1 i k 1 +i

= exp( 2 i ) so that 4 | B|| A |

= 1

C A = 2 k 1

k 1 +i = 2 cos exp( i ) so that4 | C || A | = 2 cos (14)

with =

arctan

k 1

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3.4 The wavefunctions in each region and the physical interpretation

The wavefunctions for particles in regions (I) and (II) are given by Equations 13,

1( x) = A exp( ik 1 x) + B exp( ik 1 x) in region (I)

2( x) = C exp( x) in region (II) (Eqn 13)

and the ratios of the constants B / A and C / A by Equations 14. B A

= k 1 i k 1 +i

= exp( 2 i ) so that 4 | B || A |

= 1

C A

= 2 k 1k 1 +i

= 2 cos exp( i ) so that 4 | C || A |

= 2 cos (Eqn 14)

The incident beam A exp( ik 1 x) has an average density of | A | 2 particles per unit length and the re

B exp( ik 1 x) has an average density |

B

| 2

. Our analysis showed that | B

| 2

= | A

| 2

(Equation 14), anof the reected and incident particles are equal.

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The transmitted particles are represented by the wavefunction in region (II):

2( x) = C exp( x)

The density of transmitted particles is given by

| 2( x) |2

= | C

|2

exp( 2 x)

This is a most important result: quantum mechanics makes the clear prediction that particles can be obsinside region (II) . The particle density decreases exponentially with distance from the step. The scalepenetration is set by the constant = 2 m(V E ) . In interesting cases, this is usually of the sammagnitude as the angular wavenumber k 1 = 2 mE in region (I).It is helpful to visualize the situation by plotting a graph of the density function P ( x) as we didSubsection 2.6 .

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P ( x)In region (II) we have with the help of Question T8 :

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P ( x)

0

4

rregion (I)

Figure 7 4 The density functioregion of a potential step at xincident beam is from the left. Tof each particle is less than the step V . In region (I), there is a maxima and minima due to the between incident and reected wequal amplitude. In region (II), decreases exponentially.

In region (II), we have with the help of Question T8 :

2( x) = C exp(

x)

P2 ( x) = 2* ( x) 2 ( x) = | C |2 exp( 2 x)

and since C = 2 cos exp( i )

P 2( x) = 4 cos 2 exp( 2 x) (16)

P 1( x) = 2 + 2 cos(2 k 1 x + 2 ) (Eqn 15)

The density function is drawn in Figure 7 using Equations 15 and16. You can see a perfect interference pattern in region (I) with theregular array of maxima and minima. The amplitudes of the incidentand reected waves are equal so the maximum value of P 1( x) is 4

and the minimum 0. At the boundary, P 1(0) = P 2(0) = 4 cos

2 , andthe density function passes smoothly to the decreasing exponential

in region (II).

Question T9

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Question T9

Why does the density function have a maximum of four particles per unit length in region (I) when the incparticle density is one per unit length? 4

We can characterize the penetration of particles into region (II) by a distance D = 1/(2 ). At thi

density, given by Equation 16,

P 2( x) = 4 cos 2 exp( 2 x) (Eqn 16)

falls to 1/e of its value at x = 0:

P 2( D ) = P 2(0) exp( 2 D ) = P 2(0) exp( 1) = P 2(0)/e

h i d h i i i f E d V i i 13

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The penetration depth is given in terms of E and V using Equations 13:


D = 12

=

2 2 m(V E ) (17)

The penetration depth for particles incident on a potential step with E < V

Question T10

A beam of electrons of energy 5 eV approaches a potential step of height 10 eV. Assume there is onincident particle per unit length. Calculate the following: (a) the average particle density at x = 0;(b) the point nearest the step where P 1( x) is minimum; (c) the penetration depth D .4

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In this subsection, we have obtained two remarkable predictions of quantum mechanics. The predictions sedefy common sense but they have been veried experimentally in many situations.o Particles can penetrate into classically forbidden regions where V > E .o Quantum-mechanical interference arises from a potential step and produces a regular set of points whe

particle density is zero.

3 5 R i d i i i h d l

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3.5 Reection and transmission in the quantum model

The reection coefcient R was dened in Subsection 2.6 :

R = ux of reflected particleslux of incident particles

The wavefunction in region (I) is: 1( x) = A exp( ik 1 x) + B exp(ik 1 x)

Equation 8 gives the incident ux | A | 2 k 1 / m and the reected ux | B |2 k 1 / m. We have already that when E < V , | B | 2 = | A | 2 (Equation 14), so that the reected ux is equal to the incident uxreection coefcient is one.

R = 1 (18)

The reection coefcient at a potential step with E < V

Thi l i h h di i f l i l h i b h b h

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This result agrees with the prediction of classical mechanics, but what can we say about the transmcoefcient? Particles can be observed across the step in the classically forbidden region, but is there anyIf particles are not to be created or destroyed at the boundary, then R + T = 1, and if R =Even though particles are observed in region (II), there is no ux. Remember that the wavefu 2( x) = C exp( x) is not an eigenfunction of momentum, and so does not represent a travelling wave associated momentum and an associated ux. Equation 8

F = | A |2 k / m (Eqn 8)

cannot be used to calculate ux since k is not dened.

T = 0 (19)

The transmission coefcient at a potential step with E < V

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3.6 Summary of Sections 2 and 3

At a step where the potential energy increases by an amount V , classical mechanics makes denitefor the transmission and reection of particles. If the total energy E of the incident particles is grethen the particles are always transmitted across the step and are never reected. If E is less thparticles are always reected and never transmitted.

When the distance over which the potential is changing is of the same order as, or smaller than, the de Bwavelength of the incident particles, then it is inappropriate to use classical mechanics. A quantum-mechatreatment is required. Quantum mechanics makes some unexpected predictions which reveal the wave natuparticles in an interesting way. In the case E > V , there is always a nite probability that a parti

reected and the theory allows a calculation of the reection and transmission coefcients, R and(Equations 9 and 10).

R = k 1 k 2k 1 + k 2

2

(Eqn 9)

T = 4 k 1k 2(k 1 + k 2 )2

(Eqn 10)

I th E < V th q t di ti f R d T g ith th l i l di ti b

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In the case E < V , the quantum predictions for R and T agree with the classical predictions buimportant new features. There is always a nite probability of particles being found in the classicregion beyond the step where the wavefunction has an exponential shape and the particle has no dmomentum. The theory allows a calculation of the penetration depth D in terms of E and V (Equat

D = 12

=

2 2 m(V E ) (Eqn 17)

The penetration depth for particles incident on a potential step with E < V

Critical elements in the theory are the selection of appropriate solutions of the Schrdinger equation regions before and after the potential step and the matching of the solutions at the boundary with both

d ( x)/ dx being continuous at the boundary.

P ( x) P ( x)The wave nature of material

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0 x

Region (I) Region (II)

Figure 4 4 The particle densityfunction P ( x) in the region of apotential step ( E > V ). In region (I),P ( x) shows a pattern of alternatingmaxima and minima caused by theinterference of the incident andreected waves.

0

4

rregion (I)

Figure 7 4 The density functioregion of a potential step at xincident beam is from the left. Tof each particle is less than the step V . In region (I), there is a maxima and minima due to the between incident and reected wequal amplitude. In region (II), decreases exponentially.

particles is made very clear

by the appearance of fringes caused by theinterference of the incidentand reected waves. There isa regular pattern of pointswhere the particle density isa minimum separated bypoints where the density ismaximum (Figure 4 andFigure 7).

4 Reection and transmission at a barrier when E < V

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4 Reection and transmission at a barrier when E V U

( x

)

0

E V

P

B C

A

Figure 8 4 A potential barrier with Particles with energy E are incidenand, according to classical theory, arthe point P.

4.1 Classical description of the problemWe can now model a slightly more complicated physicalsituation by forming a potential barrier from two closelyspaced steps. This is shown in Figure 8. The particles approachfrom the left and encounter rst a negative force from point A topoint B; from B to C there is no force, and from C to D there is apositive force. Particles with energy E < V will reach point P,according to classical mechanics, before being reected back.The classical reection coefcient for a barrier of this nature isone.

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4.2 The Schrdinger equation and the solutions in each region

U ( x)

0 x

E V

P

B C

DA

Figure 8 4 A potential barrier with height V .Particles with energy E are incident from the leftand, according to classical theory, are reected atthe point P.

U ( x)

0

E V

region (I)

region

Figure 9 4 A schematrepresentation of a potewidth d . The energy Eis less than the height o

A quantum treatment of this barrier problem isrequired if any of thedistances AB, BC, CD in

Figure 8 are of the sameorder as, or less than, thede Broglie wavelength of the incident particles.We will assume that thedistances AB and CD arethe small ones and thebarrier is then drawnschematically as in

Figure 9.In most physical situations, the width of the barrier, d , is greater than the de Broglie wavelength.

x < 0 is designated (I), the region 0 x < d designated (II) and the region x d designated (III).

As usual particles are incident on the barrier from the left with energy E The solutions of the

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As usual, particles are incident on the barrier from the left with energy E. The solutions of theequation in regions (I) and (III), where the potential is zero, have the same form. You can refer baSubsection 2.2 and Equation 2:

1( x) = A exp( ikx) + B exp( ikx)4 in region (I) (Eqn 2)

3( x) = G exp( ikx) + H exp( ikx)4 in region (III)

with k = 2 mE

We can immediately put H = 0 because particles cannot return from innity. The general solutio

Schrdinger equation in region (II), where E < V , was derived in Subsection 3.2 : 2( x) = C exp( x) + D exp( x)

with = 2 m(V E )

Notice that we have included the rising exponential term here because region (II) does not extend to innity

In summary, we have the general forms of the wavefunctions in the three regions:

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In summary, we have the general forms of the wavefunctions in the three regions:

1( x) = A exp( ikx) + B exp( i kx) in region (I)

2( x) = C exp( x) + D exp( x) in region (II) (20)

3( x) = G exp( ikx) in region (III)

with k = 2 mE

4 and 4 = 2 m(V E )

4 3 R l ti hi i d b th b d diti

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4.3 Relationships imposed by the boundary conditions

The situation now seems to be getting out of hand with ve arbitrary complex constants and the bouconditions to be applied at x = 0 and x = d . Matching the wavefunctions and their derivatives at the bgives us four independent simultaneous equations which can, in principle, be solved for the four ratios

D / A , and G / A . This piece of algebraic manipulation is extremely tedious, and we will not bore you w

details. We are interested mainly in the constant G because this tells us what the transmission coWe will quote the result and not even ask you to conrm it!

G A

= 4 k exp( d )(k +i )2 (k i )2 exp( 2 d )

Now look closely at the denominator of this expression. We have already stated that the width of the barrieusually much greater than the de Broglie wavelength of the incident particles. This in turn makes the arguof the exponential large and the second term in the denominator negligibly small compared with the rst To a very good approximation, we have the following expression:

G A =

4 k exp( d )(k +i )2 (21)

This can be rewritten more conveniently as:

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G A =

4exp( d )exp( 2 i )k + k (22)

Question T11

Show that Equation 21G A

= 4 k exp( d )(k +i )2 (Eqn 21)

can be written in the form of Equation 22.

What is the modulus of G / A when = k ?4

The presence of the exponential factor exp( d ) in the numerator of Equation 22 ensures that |

small and very sensitive to the value of d .

We can now make similar approximations in region (II), and we nd that the wavefunction is dominated bf lli i l ( ) C ( ) d h i i i l k li ibl

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falling exponential term, 2( x) = C exp( x), and the rising exponential makes a negligible co

The ratios B / A and C / A are then to a good approximation the same as for the potential step ( ESubsection 3.3 ). In summary, we have the following results:

B A

k i k + i = exp( 2

i ) and| B || A |

1

C A

2 k k + i = 2 cos exp(

i ) and| C || A |

2 cos

D A 0 (23)G A

4exp( d )k + k exp( 2

i ) and| G || A |

4exp( d )k + k

with tan = k


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The wavefunctions in the three regions are given by Equations 20 , with the ratios of the constaEquations 23 . Remember that we have made the assumption that the width of the barrier is much greater thde Broglie wavelength. In region (I), the particles are represented by travelling waves, the incident particl

A exp( ikx) and the reected particles by B exp( ikx). The density of the reected particles, | B | 2 , to the density of the incident particles, | A |2 , since | B / A | 1. In the classically forbidden regwavefunction is dominated by the falling exponential so that the density of particles is given approximate| 2( x) | 2 = | C |2 exp( 2 x). Finally, in region (III), the transmitted particles are represented by the twave G exp( ikx), where | G | is much smaller than | A |. The density of the transmitted particles is covery much smaller than the density of the incident particles.

Quantum mechanics unambiguously predicts that particles can tunnel through a potential barrier and canobserved with reduced density but with their original momentum. This process is known astunnelling .

Question T12

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Show that the density of particles in region (III) is constant and is given by 16 | A|2 exp 2 d (

k + k (

P ( x)

0

region (II)region (I) regio

d

Figure 10 4 The particle density function P ( xpotential barrier of height V . Particles are incidleft, each with energy E less than V .

All these facts can be neatly represented by the graphof the density function P ( x) = | ( x) | 2 shown inFigure 10. The form of P ( x) in regions (I) and (II) isvery similar to that shown in Figure 7 for thepotential step, because of the realistic approximationsdescribed in Subsection 4.3 , so there is no need torepeat the calculations. In region (I), there is analmost perfect pattern due to the interference betweenincident and reected waves of almost equalamplitude. At the boundary, x = d , P ( x) has not quite

fallen to zero and joins smoothly to its constant valuein region (III).

4.5 Transmission in the quantum model

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U ( x)

0

E V

region (I)

region


The method for nding the particle uxes and hence the reection andtransmission coefcients follows from Equation 8

F = | A |2 k m

(Eqn 8)

and the pattern set in Subsections 2.6 and 3.5 . The transmission coefcientfor the potential barrier illustrated in Figure 9 is given by:

T = ux of transmitted particlesux of incident particles

The incident ux in region (I) is given by | A |2 k / m and the transmittedux in region (III) by | G | 2 k / m. Consequently:

T = |G |2

| A|2

Equation 21

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G A =

4 k exp( d )(k +i )2 (Eqn 21)

gives the constant G in terms of A , and from this we can get the square modulus of G (seeQuestion T12 ):

|G |2 | A|2 16exp( 2 d )k + k ( )2

It follows immediately that:

T 16 exp( 2 d )k + k ( )2

(24)

The transmission coefcient for a potential barrier

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We can show that the phenomenon of barrier penetration is consistent with the Heisenberg uncertainty prinE t

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E t .

The Heisenberg uncertainty principle tells us that a particle can borrow energy E sufcient tobarrier provided it repays the debt in a time t of order / E . The value of E is given by V ELet us assume a typical barrier of width d = 2/ so that T 4 exp( 4) = 0.07 (using Equation 25).

T 4 exp( 2 d ) (Eqn 25)

where = 2 m(V E )

and d is the barrier width.

We estimate t from the speed of the particle:

t d

v

4 and 4

v = pm

= k m

Hence t 2

k m

= 2 m

k

However = 2 m( E V ) 4 and 4 k = 2 mE

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However =

4 and 4 k =

Substituting these gives t

E (V E )

Both E and V E are of the same order of magnitude, so that:

t

V E

E

which is the required relation.4

potential energy4.6 Examples of quantumtunnelling

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total energyof the

- particle

oscillating

r n

r c

smooth changesin shape

oscilla

smoothlyfalling

wavefunction of the - particle

Figure 11 4 The potential barrierseen by an -particle in aradioactive nucleus.

g

The idea of tunnelling was rst usedto explain -decay in radioactivenuclei. -particles may be formedwithin a nucleus. In heavy nuclei theymay be formed with enough energyto escape. They are held withinthe nucleus by a potential barrierwhich consists of an attractive part(due to nuclear forces), and a

repulsive part (due to the electrostaticrepulsion between the -particle andthe residual nucleus), as shownschematically in Figure 11.

potential energyThis combination produces apotential well in which in which the
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total energyof the

- particle

oscillating

r n

r c

smooth changesin shape

oscilla

smoothlyfalling

wavefunction of the - particle

Figure 11 4 The potential barrierseen by an -particle in aradioactive nucleus.

-particle is trapped. Notice that theenergy is positive and states of thissame energy exist outside thepotential well of the nucleus, so the -particle can tunnel out of its

potential well and escape from thenucleus. This is -decay .

1412

The shape of the barrier is more complex than we have envisaged so far,but it should be clear that the higher the energy of the -particle, the

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l o g

( /

s

1 )

1086420

2468

1012

141618

0.40.3

Figure 12 4 The variationwith -particle energy for radioactive nuclei.

more likely it will be to penetrate the barrier, both directly because of the increased energy, and indirectly because of the decreased width of the barrier. These two effects lead to an extremely wide variation indecay rates for a relatively small change in energy. Figure 12 shows avariation of a factor 10 24 in decay rates for a change of a factor 2 in

energy. This same idea, used in reverse, suggested to Cockcroft and Walton thatit would be possible to induce nuclear reactions using low-energyprotons (about 0.5 MeV) without sufcient energy to overcome the

electrostatic repulsion. This began the whole study of nuclear physicsusing particle accelerators.

The same process is involved in the nuclear reactions that supply theSuns energy. The principal reaction here involves the eventual

formation of helium by the fusion of hydrogen nuclei (protons). It isessential to take tunnelling into account to be able to explain the rate of energy production. In laboratory fusion experiments the same principleallows nuclear fusion at attainable temperatures.

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P ( x)


4.7 Summary of Section 4This section describes one of the most powerful

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0


d

Figure 10 4 The particle density function P ( xpotential barrier of height V . Particles are incidleft, each with energy E less than V .

Finally, good approximations are derived for the transmission coefcient T . In our approxdominated by the exponential factor exp( 2 d ).

p

predictions of quantum mechanics. We have seen thatmaterial particles can penetrate into classicallyforbidden regions due to their wave nature. Here wego further and show that particles can penetratethrough classically forbidden barriers ( Figures 8 and9). In most realistic situations, the width of the barrieris such that the transmission coefcient is small andconsiderable simplications can be made in thetheory. The nature of the particle wavefunctions is

described in the regions before the barrier, inside thebarrier and after the barrier. Matching thewavefunctions at the two boundaries allows the particle density ratios to be calculated. A graph of the padensity function near the barrier is given in Figure 10. Three features are important:o The interference pattern due to the incident and reected waves;o The exponential decay within the barrier;o The constant density beyond the barrier.

U ( x)region (I) regio5 Closing items

5 1 Mod le s mmar

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0

E

E V


5.1 Module summary1 A particle with kinetic energy E approaches a step where the potential

increases by V . Classical mechanics predicts that if E > V then theparticle will always pass over the step.

2 If the dimensions of the step are of the same order or less than the deBroglie wavelength of the incident particle, then quantum mechanicsrather than classical mechanics must be used.

3 A potential step where E > V is shown in Figure 2. The solutions of theSchrdinger equation are:

1( x) = A exp( ik 1 x) + B exp( ik 1 x) in region (I) (Eqn 4a)


with k 1 = 2 mE

4 and 4 k 2 =

2 m( E V )

The ratios of the coefcients are determined by boundary conditions . The wavefunction must both be continuous at x = 0. This gives:

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B A

= k 1 k 2k 1 + k 2

4 and 4 C A

= 2 k 1k 1 + k 2

(Eqn 7)

4 The wavefunction 1( x) represents the incident particles, with density | A | 2 per unit length,momentum k 1 in the positive x-direction, and the reected particles, with density | B |

momentum k 1. The wavefunction 2( x) represents the transmitted particles, of density | C momentum k 2.

5 For a wavefunction ( x) = A exp( ikx), which is a momentum eigenfunction, the ux of particl

F = | A |2 k / m (Eqn 8)

6 At a potential step, the reection coefcient R and the transmission coefcient T are dened:

R = ux of reected particles ux of incident particles

4 and 4 T = ux of transmitted particlesux of incident particles

P ( x)

Region (I)

If E > V , quantum mechanics predicts:

k1 k22

4 k1k2

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g ( )

Figure 4 4 The particle function P ( x) in the regpotential step ( E > V ). P ( x) shows a pattern of amaxima and minima cauinterference of the incidereected waves.

R = k 1 k 2

k 1 + k 2

4 and 4 T =

4 k 1k 2(k 1 + k 2 )2 (Eqns 9 and 10)

As expected R + T = 1. The prediction that R > 0 when E > V is adirect contradiction of classical mechanics.

7 The wave nature of particles is demonstrated by the particle density function P ( x) in region (I), which shows theinterference between the incident and reected waves (Figure 4).

U ( x)8 A potential step where E < V is shown in Figure 5. The solutions of the

Schrdinger equation are:

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0

P

Figure 5 4 The functiopotential step. The heigV is greater than the en

incident particle. Accorclassical mechanics, thereach the point P and arreected back.

1( x) = A exp( ik 1 x) + B exp( ik 1 x) in region (I)


with k 1 =

2 mE

4 and 4 =

2 m(V E )

The ratios of the constants are given by the boundary conditions at x = 0:

B A =

k 1 i k 1 +i = exp( 2 i ) so that 4

| B || A | = 1

C A

= 2 k 1k 1 +i

= 2 cos exp( i ) so that 4 | C || A |

= 2 cos (Eqn 14)

with = arctan k 1

P ( x)rregion (I)

9 In region (I), the wavefunction 1( x) represents the incident andreected particles with the same density, | B |2 = | A |2 .

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0

4

Figure 7 4 The density functioregion of a potential step at x

incident beam is from the left. Tof each particle is less than the step V . In region (I), there is a maxima and minima due to the between incident and reected w

equal amplitude. In region (II), decreases exponentially.

In region (II), the wavefunction 2( x) is not a momentumeigenfunction and there is no ux of particles. The particledensity function P ( x) (Figure 7) shows a perfect interferencepattern in region (I) and joins smoothly to the fallingexponential in region (II). There is a nite probability of

particles being observed in the classically forbidden region:P 2( x) = 4 cos 2 exp( 2 x). At the penetration depth D = 1/(2 ),the particle density is 1/e of its value at x = 0.

10 The reection coefcient at the step where E < V is given by:

R = | B |2| A |2

= 1 (Eqn 18)

This agrees with the classical prediction.

U ( x)region (I)

region 11 A potential barrier with width d and V > E is shown schematically in

Figure 9. The distances over which the potential is increasing ordecreasing are assumed to be smaller than the de Broglie wavelength

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0

E V


decreasing are assumed to be smaller than the de Broglie wavelengthof the incident particles. The solutions of the Schrdinger equation are

1( x) = A exp( ikx) + B exp( ikx) in region (I)

2( x) = C exp( x) + D exp( x) in region (II) (Eqn 20)

3( x) = G exp( ikx) in region (III)

with k = 2 mE

and = 2 m(V E )

The ratios of the constants are given by matching the wavefunctionsand their slopes at the two boundaries.

With the assumption that d >> 1/ , the following approximations are valid:

B k i | B |

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B A

k i k + i = exp( 2 i ) and

| B || A | 1

C A

2 k k + i = 2 cos exp(

i ) and| C || A |

2 cos

D A

0 (Eqn 23)

G

A 4exp( d )

k + k exp( 2 i ) and | G |

| A | 4exp( d )

k + k with tan =

k

P ( x)


12 In region (I), the wavefunction 1( x) representsboth the incident and reected particles withapproximately the same density. In region (II),

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0 d

Figure 10 4 The particle density function P ( x

potential barrier of height V . Particles are incidleft, each with energy E less than V .

pp y y g ( ),the wavefunction 2( x) is not a momentumeigenfunction. In region (III), the wavefunction 3( x) represents the transmitted particles.The density function P ( x) (Figure 10) shows a

typical interference pattern in region (I), fallsexponentially in region (II) and joins smoothly tothe constant value | G |2 in region (III).

13 The transmission coefcient at a barrier whereV > E is given by T = | G |2 / | A |2

so that:

T 16 exp( 2 d )k + k ( )2

(Eqn 24)

When and k are of the same order of magnitude, a good approximation is T The transmission coefcient is normally much less than one so that the reection coefcient R

14 Quantum tunnelling also allows particles to tunnel into or out of a potential well . Suchresponsible for nuclear fusion and also allow many technological developments.

5.2 Achievements

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Having completed this module, you should be able to:A1 Dene the terms that are emboldened and agged in the margins of the module.A2 Describe the motion of a particle at a potential step using classical mechanics. Identify the circums

when quantum mechanics should be used rather than classical mechanics.

A3 Recall the form of the solutions of the Schrdinger equation for particles of energy E eidealized potential step when E > V . Interpret the solutions in terms of the density and ux of thereected and transmitted particles.

A4 Recall conditions on the wavefunctions at a boundary, and use the conditions to determine the rel

between the constants appearing in the wavefunctions.A5 Derive expressions for the transmission and reection coefcients at a potential step when E

and contrast the predictions of quantum and classical mechanics.A6 Recall the form of the solutions of the Schrdinger equation for particles of energy E e

idealized potential step when E < V . Interpret the solutions in terms of the density and ux of thand reected particles. Explain how these solutions predict that particles can penetrate into the classforbidden region.

A7 Use the theory to conrm that the reection coefcient for a step is unity when E < V , in aclassical theory.

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A8 Recall the form of the solutions of the Schrdinger equation for particles of energy E < V eidealized potential barrier of width d . Interpret the solutions in terms of the density and ux of thereected and transmitted particles.

A9 Use the approximate formula for the transmission coefcient through a potential barrier.

A10 Use the quantum theory results to calculate uxes of particles transmitted by, and reected from, spesteps and barriers.A11 Outline the use of barrier penetration, or tunnelling, in different physical situations.

Study comment You may now wish to take the Exit test for this module which tests these AIf you prefer to study the module further before taking this test then return to the Module contents to revtopics.

5.3 Exit test

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Study comment Having completed this module, you should be able to answer the following questions, each of whione or more of the Achievements.

N o t e We have used electron beams with energies in the eV region to illustrate the general principles of quantum mechapplied to steps and barriers. This is an attempt to keep to a minimum the amount of repetitive arithmetic in the ques

You should be fully aware that similar examples can be generated using nuclear particles (e.g. protons orenergies in the MeV region and lengths of the order of 10 15 m.

Question E1

( A2 and A10 )4 A beam of electrons, each with kinetic energy 5 eV, is incident on a step where tenergy decreases by 2.5 eV. Calculate the transmission and reection coefcients in the quantumWhat does classical mechanics predict? How can you decide which theory of mechanics to use?

Question E2

( A4 , A5 and A7 ) Explain qualitatively how the boundary conditions on wavefunctions lead to Equations 5

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A + B = C (Eqn 5)

Ak 1 Bk 1 = Ck 2 (Eqn 6)

Verify Equations 7 B A

= k 1 k 2k 1 +k 2

and C

A= 2 k 1

k 1 +k 2(Eqn 7)

from Equations 5 and 6. Derive expressions for the transmission and reection coefcients when

Show that for a potential step with V > E the ratio

B A

= k 1 i k 1 +i

4 where 4 = 2 m(V E )

Hence show that | B | = | A | and the reection coefcient is one.

Question E3

( A3)4 Find a suitable wavefunction to describe the incident electrons in Question E1 if the current

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[A beam of electrons, each with kinetic energy 5 eV, is incident on a step where the potential energyby 2.5 eV.]

What is the current after the step?

Question E4

( A6 )4 At the junction between copper and an insulator, the potential energy of a conduction electron incby 11.1 eV . Find the penetration depth, into the insulator, of a typical conduction electron with kinetic 7.0 eV.

Sketch the graph of the density function and account for the series of maxima and minima inside the copper

Question E5

( A8 , A9 and A10 )4 A beam of electrons with kinetic energy 5 eV approaches a potential barrier of h

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and width 1 nm. Sketch the graph of the density function near the barrier.Calculate the transmission coefcient, T , according to Equation 24.

T 16 exp( 2 d )

k + k ( )2 (Eqn 24)

Are the approximations made to derive the formula valid in this case?

What is the value of T when the barrier potential is reduced to 6 eV?

Is Equation 25 also a good approximation?

T 4 exp( 2 d ) (Eqn 25)

Question E6

( A9 , A10 and A11 )4 In a modern scanning tunnelling microscope, the tip of a ne needle is 1 nm

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surface. The needle to metal gap behaves like a potential barrier of height 5 eV. Calculate thecoefcient T for electrons of energy 2.5 eV.

By what factor does T change when the gap increases to 1.1 nm? What is the signicance of this resul

Study comment This is the final Exit test question. When you have completed the Exit test go back to Subtry the Fast track questions if you have not already done so.

If you have completed both the Fast track questions and the Exit test , then you have finished the module andhere.

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