QM for Nanoscience and Nanotechnology

7/30/2019 QM for Nanoscience and Nanotechnology

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Quantum Mechanics for Nanoscience and

Nanotechnology

Karl Hess

Beckman Institute, Department of Electrical Engineering and Departmentof Physics,University of Illinois, Urbana, Il 61801

1 Origins of the Quantum Theory

1.1 Black body radiation and Plancks hypothesis

Black body radiation was the topic of interest in the late 1800s and early 1900s. It evencaptured the interest of the lay people and during tours in the physics institute of theUniversity of Vienna the question often was asked: where is the famous absolutely blackbody?. To their embarrassment the tour guides had to answer it is this white box in thecorner. The absolutely black body is actually a tiny hole in basically any box, e.g. a cubeof side length L. The box is heated to a temperature T and the radiation coming out ofthe hole is measured to give the energy content U() at frequency in the frequency ranged. It has been demonstrated experimentally that in equilibrium U() depends only onthe temperature T of the walls and not on the material of which the walls are made. Theexperimental result for U() follows approximately Wiens law (later improved by Planck):

U() 3eh/kT (1)

where T is Boltzmanns constant. From all the classical knowledge at that time, however, itwas found that it should be

U() kT 2 (2)The first solution of this puzzle was given by Planck and we will derive below a brief

description how this solution was achieved.

We have to start by describing the electromagnetic waves in the black body.The wavesare characterized by a wave vector k with k = |k| = 2/ where is the wave length. Forall practical boundary conditions, the waves must be standing plane waves of all possiblewave lengths and amplitudes. Therefore we must have for the components of the wave vectork = (kx, ky, kz):

kx = 2l/L , ky = 2m/L , kz = 2n/L (3)

with l ,m,n being integers 0,1,2,....

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Figure 1: Energy content U() as a function of the frequency in arbitrary units for Planckslaw (green solid line), Boltzmans law (red dotted line) and Wiens law (blue dashed line)

Figure 2: Standing waves in a rectangular box with x = L/2,y = L (top figure) andx = 2L,y = 1.5L (bottom figure)

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Figure 3: Distribution of k-states on a 2 dimensional lattice

Using the wave vector as given above one can describe for example the electrical fieldpattern in the box by a Fourier series

kak(t)e

ikr + cc (4)

where ak is a suitable complex amplitude, t is the time-coordinate, i is the imaginary unit,r = (x,y,z) is the space-coordinate and cc indicates the complex conjugate. Thus theelectrical field that is a function of the space-time continuum is fully characterized by the set

of k vectors which is an infinite set but countable-infinite because of l ,m,n being integers0,1,2,.... Because of this fact, one can also describe the electromagnetic field as acountable sum of oscillators, each with finite energy content and one for each separate valueofl,m, and n. The frequencies of these oscillators are between and +d with = kc/(2)where c is the velocity of light. To find the energy content of the electromagnetic field wemust therefore find the energy and the density of these oscillators.

The number of oscillators N1 in the volume dkxdkydkz can be found by dividing k-spaceinto cubes with side-length 2/L. Because L is very large, we have

N1 = (L

2)3

dkxdkydkz (5)

To express N1 in terms of the frequencies it is convenient to use polar coordinates andintegrate over the solid angle which gives a factor of 4 . We therefore have

N1 = 4(L

2)3

k2dk (6)

and from = kc/2 we obtain

N1 = 4L32d/(c3) (7)

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If we also permit two directions of polarization of the oscillators then we have to multiplyEq.(7) by a factor of two to obtain the total number of oscillators N to be

N = 8L32d/(c3) (8)

It is now important to realize that classically the average energy of an oscillator is equalto kT. This is shown in detail in classical statistical mechanics. Therefore we have

U()d = kTN= 8kTL32d/(c3) (9)

which is in disagreement with experiment (approximately given by Eq.(1)).Plancks resolution of the puzzle proceeds as follows. As in classical mechanics Planck

assumed that the probability W for an oscillator to have an energy E is proportional toeE/kT. In addition, however, he assumed that the energy of the oscillators can only assumethe values Ej = jh where j = 0, 1, 2,..., and h = is the famous constant of Planck. Wetherefore have for the correctly normalized probability W

W(j) = ejh/kT

j=0ejh/kT (10)

The average energy E is then given by

j=0EjW(j) which can be found after a little algebra(HW1) to be:

E =h

eh/kT 1 (11)Multiplying by N we obtain Plancks famous law

U() = (8hL33/(c3))1

eh/kT

1

(12)

which also is in full agreement with the experimental results. Planck presented in the year1900 the following explanation for his law: We...consider-and this is the essential point-the energy E to be composed of a determinate number of equal finite parts, and employ intheir determination the natural constant h = 6.55x1027 (erg sec). This constant multipliedby the frequency of the resonator yields the energy element h in ergs, and dividing Eby h we obtain the number of energy elements.... Five years later, Einstein accepted thefull consequences of Plancks work and stated that radiation consists of quanta of energyh, which not only are important for emission and absorption but can have an independentexistence as particles in empty space. These particles are now called photons. Note alsothat the above treatment contained considerations of probability and statistics in particular

statistical mechanics. One can therefore say that quantum mechanics was brought forth bystatistical considerations.

The factor1

eh/kT 1 (13)that appears in Eq(12) is typical for photons and for all particles that are called bosons andis often called the occupation number that gives the probability to find a boson. Note thatwe will mostly use h = h/(2) instead of Plancks constant h.

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Figure 4: Bose Distribution for values around kTh 1

1.2 Einsteins 1917 work on stimulated emission

Einstein gave another derivation of Plancks law and, in the course of it, derived and statedmany important facts about the interaction of matter and radiation, including the possibil-ity of stimulated emission. His work shows, how far one can get in the understanding ofmolecular problems by use of the old (Einsteins, Plancks) quantum mechanics. Einsteinalways refers to molecules but everything he says applies basically to any nano structure.

We follow here Einsteins work from 1917 closely.According to quantum theory (that known to Einstein), a given molecule (or nano struc-ture) can exist only in a discrete series of states Z1, Z2,...,Zn,.... The inner energy of thesestates is denoted by E1, E2,...,En,.... The word state that

Einstein used is borrowed from thermodynamics where, for example, a gas of particleswas said to be in a certain state depending on temperature, pressure etc.. The word statewill be used throughout this text in various context. We will not give a precise definitionand only attempt to endow the word with more detailed properties as we go on. At theend we have to admit that we really do not exactly know what a quantum state is and canonly point to Goethes famous quote (in loose translation): whenever concepts are uncleara sounding word will fast appear.

If these molecules are part of a gas that is at temperature T, then the relative frequencyof occurrence Wn of the states Zn is given by the canonical distribution of states of statisticalmechanics:

Wn = pneEnkT (14)

Here pn is a constant that is determined by the specific molecule (nano structure) and thestate Zn. Note that pn, En and details of the state Zn could not be determined by thequantum theory known in 1917. It takes the Schrodinger equation and its solutions to

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Figure 5: Discrete series of states in a box

Figure 6: Transition with light emission

determine all of these constants. However, if we just assume that these are somehow given,e.g. by some experiment, then we will see that great progress in the general understandingcan be made. We therefore proceed with Einstein to consider the energy exchange betweenmatter and radiation.

Consider two states Zn and Zm of a nano structure with energies

Em > En (15)

The nano structure can make a transition from Zn to Zm by absorbing electromagneticradiation (light) of energy Em En.

The nano structure can also make a transition from Zm to Zn by emitting electromagnetic(em) radiation of the same energy Em En. The radiation has then a frequency thatcorresponds to the index combination (m, n). We compare now the nano structure with a

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resonator for the em-field and consider the resonator and its energy changes when it interactswith with the electromagnetic field:

(i) Effects proportional to the radiation density.

Under the influence of an em-field of frequency and radiation density U(), thenano structure (resonator) can make a transition that, depending on the phase of theresonator, will either increase or decrease the energy of the resonator. (An electricmotor can be accelerated by the applied ac-voltage and gain rotational energy, orgenerate ac-voltage and current by loosing rotational energy). Correspondingly wemake the hypothesis that the nano structure can make a transition Zn Zm byabsorbing the energy Em En according to the probability

dW = Bmn U()dt (16)

Analogously the nano structure can make a transition Zm

Zn by loosing energy to

the radiation field according to the probability

dW = BnmU()dt (17)

a process that is now called stimulated emission. Bmn and Bnm are constants now called

Einstein coefficients.

(ii) Effects independent of the radiation density.

The nano structure can radiate independent of the fact whether a radiation field ispresent or not. Einstein deduced this fact in analogy to the radioactive decay. Thiseffect is now called spontaneous emission and Einstein assigned it the probability

dW = Anmdt (18)

Anm is again a constant also called an Einstein coefficient.

We now consider thermal equilibrium and ask what the radiation density U() has tobe in order not to change or disturb the probability distribution of states given by Eq.(14).For this it is necessary and sufficient that on average per unit time we have an equality ofall energy loss and gain as given by Eqs.(16) - (18). Therefore we must have for the indexcombination (m, n):

pne

EnkT

Bm

n U() = pme

EmkT

(Bn

mU() + An

m) (19)Einstein postulates now that the radiation density U() grows to infinity as T approachesinfinity which gives:

pnBmn = pmB

nm (20)

and thus we obtain for U():

U() =AnmBnm

1

eh/kT 1 (21)

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This, of course, can be considered to be a derivation of Plancks law. Inversely, if we takePlancks law of Eq.(12)as given, we can deduce the ratio A

nm

Bnmof spontaneous to stimulated

emission coefficients. As we now know, this gives us the basis for the theory of lasers. Quitean achievement of Einstein considering the simplicity of his derivation. This also teaches usa lesson: one can deduce results of quantum mechanics quite simply just by counting thestates and using the probability of occurrence of a given state from statistical mechanics onecan go a long way in the derivation. This is illustrated again in the section where we derivethe quantum conductance of nano structures in the same elementary fashion.

Finally, we would like to repeat that the factor 1/(eh/kT 1) is often called the occu-pation factor, meaning that it relates to the probability to encounter an oscillator with theenergy h. The 1 in the denominator arises from the fact that a difference is taken be-tween absorption and stimulated emission. We can guess that any field that is similar tothe em-field and interacts with matter by absorption, stimulated emission and spontaneousemission will lead to a similar occupation factor. Bose and Einstein have derived this factorlater in a different fashion and it therefore has also the name Bose-Einstein distribution.

The particles that follow such a distribution are called bosons.Fermions are particles with half-integer spin such as electrons. As we will discuss later,

they follow a different statistics. The probability f(E) to find an electron with energy E isgiven by the Fermi distribution:

f(E) =1

exp[(E EF)/kT] + 1 (22)

Here EF is the Fermi energy. As can easily be seen, for infinitely low absolute temperatureT, EF is the energy above which the probability to find an electron is zero. In spite of thealgebraic similarity of Fermi distribution and Bose occupation factor (or distribution) thetwo functions behave very differently as can easily be found by test.

1.3 The Waves of Louis de Broglie

Louis de Broglie asked the question whether one could assign a wave vector not only tophotons but also to electrons. Such assignment would not immediately mean that the elec-tron behaves like a wave. It could follow from a formal mathematical Fourier analysis ofthe physical functions that describe the electron behavior. One could just use Eq.(4) andassume that the limitations of k to a countable number corresponding to the integers l ,m,nwould follow from some periodicity or boundary conditions that the functions describing theelectrons must obey for purely mathematical reasons. However, the work of Louis de Broglie

and Schrodinger and much subsequent work indicates that there must be a deeper physicalsignificance in relating electron behavior to a wave vector. The electron does in many re-spects indeed behave like a wave and a particle at the same time. This wave particle dualityis, after all these years of research, still not entirely understood. The main purpose of thissection is to explain some of the major ideas that suggest wave like behavior of electronsand other entities that were previously considered as typical particles. This is, of course, theopposite of what we discussed in the first sections where we derived particle like propertiesof entities that were considered in the past to be wave like.

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Figure 7: De-Broglie frequencies for a clock (c), particle (p) and relativistic mass (m) as afunction of

De Broglie found the roots for his derivation of wave like electron properties in the theoryof relativity. We know from Einsteins famous work a number of important equations andfacts. For example, it is well known and experimentally established that a moving clock thatrotates with a frequency c is slowed down relative to the clock of a stationary observer thatrotates with a frequency c. Einstein showed that

c = c(1 2)1/2 (23)

where = v/c and v is the velocity of the moving clock relative to the stationary one.

De Broglie also considered Einsteins following other famous equations, first

E = mc2 (24)

with m being the mass of the particle, and second Plancks equation now also applied to aparticle to which he now assigned a frequency p

E = hp (25)

De Broglie realized now that the mass m of a moving system increases with velocity as

m = m/(1 2

)

1/2

(26)

and therefore deduced that from Eq.(25) it followed that the frequency p assigned to themoving particle must also increase in the moving system so that

p = p/(1 2)1/2 (27)

However, this is now opposite to the frequency of the clocks that slows down in a movingsystem. De Broglie therefore concluded the particle behavior can not be analogous to a

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clock. The only way out was then that it behaves analogous to a wave. If we introduce awave, then we have another parameter to work with, the wavelength or the wave vectork and by adjusting these one might get the correct results. Indeed this works out and deBroglie added the following derivations.

Consider a one dimensional model with movements in the x-direction. A wave can thenbe represented by a function sin() with = tk x where = 2. Using then mc2 = hwe have

d = dt k dx = 2mc2

h(1 2)1/2dt k dx (28)

It is typical for the wave particle duality or dilemma that no derivation can do with wave orparticle picture only. Therefore de Broglie needed now to relate the wave also to the particleby saying in essence the following: If a clock would move at the place of the particle thenthe clock also would show a phase 1 and the phase of the clock must be equal to the phaseof the wave i.e. = 1. This is one of the famous pidooma theorems (what pidooma meanscan only be explained orally in class), and is usually called the theorem of the harmony of

phases. From Eq.(23) we have d1 = 2(1 2)1/2dt and from = 1 we obtain

dt k dx = 2mc2

h(1 2)1/2dt k dx = 2(1 2)1/2dt (29)

which gives after a little algebra

hk/(2) = mv or = h/(mv) (30)

This is the famous de Broglie equation which relates the momentum of a particle to awavelength, the de Broglie wavelength. For any problem in nano-science and -technology

one should always calculate the appropriate de Broglie wavelength. When the feature sizes ofa nano structure are of the order of the de Broglie wavelength, then quantum considerationsare bound to be important. Note already at this point that in solid state type of problemsthe free electron mass must often be replaced by an effective mass that is typically but notalways smaller than the free electron mass.

The work of de Broglie and the importance of his wavelength for any considerations onthe atomic scale show that quantum mechanics was not only brought forth by statisticalconsiderations as mentioned above but also by considerations involving Einsteins relativity.I believe that the relation of statistics and relativity to quantum mechanics can not be justaccidental and is indeed very important if an understanding of quantum mechanics and its

applications, however incomplete, is to be achieved. There are, of course, many scientistswho believe that scientific theory should follow Ernst Machs suggestions who taught thefollowing: The world consists of colors, tones, warmth, pressures, spaces, times, etc., whichwe do not want to call sensations or phenomena, because in both names there liesa one-sided arbitrary theory (influence of our own eyes, ears etc.). We call them simplyelements. The comprehension of the flux of these elements, whether directly or indirectly,is the actual aim of science. Thus Mach believed that the aim of science was the orderingof the elements in the most economical way. Modern quantum mechanics is often taught in

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Figure 8: De-Broglie wavelength

this spirit. An axiomatic system and operator algebra are presented as the basic frameworkand the claim is made that this abstract system orders the elements most economicallyand beyond this avoids contradictions. Boltzmann, on the other hand, pursued a differentapproach to physical theory. He developed minute models that underlie a physical theory.For example, he explained thermodynamics by considering very detailed properties of atomsin gases and deriving from these properties macroscopic equations that are averages of themicroscopic properties. The derivations of Planck, Einstein and Schrodinger are based onsuch considerations. Schrodinger himself considered both the approach of Mach and thatof Boltzmann viable and valid provided that neither the principles of Machs nor those ofBoltzmanns approach are violated i.e. Schrodinger maintained a pragmatic balance betweenMachs positivism and Boltzmanns realism. We adopt this pragmatic approach herebecause it is, in our opinion, the most effective way to transmit an understanding to theapplications oriented nano structure area. We will give a more thoughtful discussion of thefoundations of quantum mechanics at the end of these lectures when talking about quantuminformation and quantum computing.

1.4 Discretization of wave vectors and conductance quantization

In this section we apply the Einstein-Planck approach to the conductance of a very thin,essentially one dimensional, wire. We start again from what is classically known. We assume

that the wire is very short and the electrons are not scattered very much by any imperfectionsof the wire material and the effect of charge accumulations (i.e. the electrostatics of the wires)can be neglected. We also assume that the wire is connected to two huge reservoirs thatare three dimensional and form ideal contacts. Then we know that the current through thewire is the thermionic emission current, well known from vacuum tubes, from both contacts.Lets assume for simplicity that there is a bias voltage Vbi that is large enough and directedsuch that current flows only from the left to the right and in zdirection corresponding to a

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Figure 9: Hess and Datta

current jLR. Assume now also that the length L of the wire is still long enough so that ourtreatment of counting wave vectors that is shown in Eq.(5) and (6) is valid. Then for thecase of one dimension we obtain:

N1 = (L

2)dkz (31)

Of course, we are not dealing with the oscillators of the electromagnetic field but withelectrons which we also have characterized by a wave vector and a corresponding state suchas Z1, Z2,...,Zn,... from above. As we know, the electron can have two spin values thatcorrespond to the two polarizations we discussed above and therefore we obtain

N = (L

)dkz (32)

To calculate the current we must multiply Eq.(32) by the elementary charge e and bythe velocity vz that the electron has in zdirection. There is one more problem to be takencare of: we must know that there actually is an electron in the given state with given kzand then we must sum over all the full states kz. Because L is very large we will replacethat sum by an integral. We denote the probability that the state with wave vector kz isactually occupied by the distribution function f. f is derived in all elementary quantumtexts and we assume it here as given. The form of f is similar to that shown in Eq.(13)with an important difference in sign (the minus one in the denominator becomes a plus one).

This form is typical for particles like electrons that are called Fermions. f contains also thequantity EF which is called the Fermi energy. Each state can only hold two electrons withopposite spin and this gives:

f =1

e(EEF)/kT + 1(33)

E is the energy of an electron with wave vector kz. Using the above, the absolute value ofthe current becomes:

|jLR| = e

0

vzf dkz (34)

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Figure 10: Quantum conductance

We now use the relation of velocity and energy to the wave vector as given by de Broglie i.e.Eq.(30) to obtain:

|jLR| = ehm

0

kz1

e(h2kz

2

2mEF)/kT + 1

dkz (35)

Here h = h/(2). The integral is easily performed by introducing a new integration variableequal to k2z and one obtains:

|jLR| = 2eh

kTln(1 + eEF/kT) (36)

Because the Fermi energy is raised by the applied bias voltage we use a suitable energy scaleand put EF eVbi >> kT to obtain:

|jLR| = 2e2

hVbi (37)

We have up to now only considered the counting of states in zdirection. Of course we havealso standing waves perpendicular to the zdirection in the wire. For each such additionalstanding wave or mode we obtain a current equal to that in Eq.(38) so that we have:

|jLR| = n2e2

hVbi (38)

where n is an integer.

The quantity 2e2

h is called the quantum conductance. The corresponding resistance wouldbe about 13k. The existence of a quantum conductance was experimentally proven after1980, a very long time after the work of Einstein and de Broglie. Nevertheless, an under-standing of the quantum conductance could have been achieved by them, using their methodsof the old quantum mechanics. What they could not achieve was a calculation of the energiesof the electrons in the various quantized modes or states. This was only possible after thework of Schrodinger which is discussed next.

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2 Schrodingers Quantum Mechanics

2.1 Hamiltons Classical Mechanics and Action Function

Hamilton formulated classical mechanics in terms of his function H that represents the total

energy as function of spatial and momentum coordinates of a particle. For example, if wehave a particle with number i = 1,...,N and its x-coordinate is xi while its momentumcoordinate in the x-direction is pxi. Then we have H = H(xi,pxi) and first

dpxidt

= Hxi

(39)

and seconddxi

dt= +

H

pxi(40)

and similar equations for the y, z space components. A trivial example would be a would be

one free particle (we give it no number) moving in x-direction. Then H = p2x/(2m) and thefirst equation gives px = constant and the second gives v = px/m. Other examples are givenas HW. We can see that in this simple case the function H does not depend on the spacecoordinate x and therefore the corresponding momentum is constant. For general problems,however, H does depend on the space coordinate which complicates things.

Hamilton-Jacobi had now an interesting idea. They suggested that one should transformthe coordinate system by a function S in such a way that the function H of the new systemthat they denoted by H would not depend on the space coordinates but only on new mo-mentum coordinates that we denote by pxi. As in the trivial example above, these pxi arethen constants and the two Hamilton equations are immediately solved. Naturally it is not

always easy to find such a function S. If it exists, then S is a function of the old spatialcoordinates and the new momentum coordinates i.e. in one dimension S = S(xi, pxi). Alittle algebra shows that we have

pxi =S(xi, pxi)

xiand xi =

S(xi, pxi)

pxi(41)

We consider now a system with time independent constant total energy E. Then we haveH = H(xi,pxi) = E and using Eq.(41 obtain the following differential equation for S interms of the original space coordinates, called the Hamilton-Jacobi equation:

H(xi,S(xi, pxi)

xi ) = E (42)

The function S is now a function of the space coordinates and of the constant momentumcoordinates of the transformed system. I am adding now a few speculations in order toexplain Schrodingers procedure to derive his famous equation that is presented below. Letsassume that we have a system of particles that we want to describe. Lets also assume that weknow from the origins of quantum mechanics that there are constants of the particle systemthat are essentially described by the quantum numbers l ,m,n or related quantities. Lets

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speculate then that these constants are also related in a simple way (e.g. are proportional)to the constant momenta that appear in the function S. This gives us already a lot ofinformation about S and how it must look like for a particle system that we want to treatquantum mechanically. Before we discuss this further, though, lets back up and discuss thesituation with which Schrodinger was dealing before he derived his equation.

2.2 Particles, waves and Hamiltons equations, and what Schrodinger

made of it

Schrodinger had carefully studied relativity and not only immediately understood the sig-nificance of de Broglies work but had already entertained considerations that were basedon writings of Hermann Weyl about relativity and came to similar conclusions as de Brogliedid. From that time on, Schrodinger was trying to explain electronic properties by wave-typethought experiments and gave seminars on the topic. During one of these seminars PeterDebye made the remark that in order to deal properly with waves one needs a wave equation.

And, of course as we know now, Schrodinger came up with one.It is well known from electro-magnetics that the space dependence in three dimensions

of an amplitude of a wave motion, independent of time, is given by the following partialsecond order differential equation (often called Helmholtz equation):

(2

x2+

2

y2+

2

z2) + k2 = 0 (43)

were k is the absolute value of the wave vector. Using de Broglies relation from Eq.(30) andthe fact that total energy E minus potential energy V equal E V = mv2/2 one obtainsfrom Eq.(43:

(2

x2+

2

y2+

2

z2) +

2m

K2(E V) = 0 (44)

where K = h/(2) h.Schrodinger noticed now a great similarity of this wave equation with the Hamilton-

Jacoby equation. He renamed Hamiltons function S, the action function, in terms of

S = Kln (45)

Now, everyone who has been at Ludwig Boltzmanns grave knows that Eq.(45) is one ofBoltzmanns greatest achievements only Boltzmann wrote S = klnW which is the famous

law connecting entropy S with thermodynamic probability W. Schrodinger did not use thisform just accidentally. He is telling us something and must certainly have had some thoughtsconnecting with probability. Using Eq.(45) together with the Hamilton-Jacobi equationEq.(45) and replacing we get:

K2

2mi[(

xi)2

+ (

yi)2

+ (

z i)2

] (E V(xi,yi,zi))2 = 0 (46)

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Here we have considered only one particle that numbered by i and have assumed that it hasthe mass mi.

Anyone who is good in the calculus of variations (and Schr odinger certainly was) knowsnow that Eq.(44) is the Euler-Lagrange equation for Eq.(46) if the following variationalproblem is considered: We seek those functions that fulfill Eq.(46) which are finite andcontinuously twice differentiable and make the integral of Eq.(46) over all space an ex-tremum. The last sentence in quotation marks is called the Schrodinger condition. Thus,applying the Schrodinger condition to Eq.(46) we obtain:

K2

2mi(

2

(xi)2+

2

(yi)2+

2

(zi)2) + (E V(xi,yi,zi)) = 0 (47)

which is identical to the wave equation derived from the Helmholtz equation except thatnow x,y,z are replaced by the particle coordinates xi,yi,zi. The remarkable point of thislatter derivation is that we have used only the Hamilton-Jacobi equation and the Schrodinger

condition. We have made no reference to waves or to the de Broglie wavelength except thatwe later will use K = h.Eq.(47) is for one particle (the is). Had we derived it for two particles, we would have

obtained:

K2

2m1(

2

(x1)2+

2

(y1)2+

2

(z1)2) +

K2

2m2(

2

(x2)2+

2

(y2)2+

2

(z2)2) + (EV(r1, r2)) = 0

(48)where r1 = (x1, y1, z1) and r2 = (x2, y2, z2). Of course, a generalization for a large numberof particles is easy only the solution becomes more complicated. With each particle oneadds three more variables that enter the differential equation. Even the most powerful

computers of the day take a month of cpu time to solve the Schrodinger equation exactly for10-20 particles. A small nano structure that contains about 50 atoms still typically containshundreds of electrons in addition to the 50 nuclei. It is therefore clear that approximatemethods are needed to understand nano structures.

Eq.(48) shows the basic problem of Schrodingers quantum mechanics. At the one side wewish to describe particles such as electrons as waves, on the other side we can not abandonthe individualization of particles and the distinction between their coordinates. In orderto proceed we need to just say that the electron is particle and wave at the same time or,perhaps, that some properties of an electron can be described by assigning a wave to theelectron while other properties, for example the mass, are best described by imagining the

electron as a point-like particle. Of course, even such a picture has its limitations. A crystallattice, for example, has the consequence that we need to assign an effective mass to theelectron. This point-like effective mass is, however, a consequence of the spatially extendedcrystal. Such logical difficulties may persuade the purist to not consider any pictures andjust work with operators in an abstract Hilbert space. Quantum mechanics can be donethat way. However, I believe when it comes to nanoscience and technology pictures areextremely helpful and fun and no real mastership can be achieved without them. Thereforewe just accept the particle-wave schizophrenia (or nicer said duality). Again, a deeper

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Figure 11: One dimensional potential well

discussions of the foundations of quantum mechanics is given at the end of the course inconnection with quantum information.

3 Important special solutions of the one particle Schrodinger

equation

In this section we do some practising by solving problems that are important for nanotechnology. What we want to obtain from the Schrodinger equation are results for thequantities that the old quantum mechanics had to assume as given. Most important in theold quantum mechanics are the energy levels E1, E2,...,En,... of a nano structure. We haveencountered these when we discussed Einsteins approach of interaction of radiation with

nano structures. We start with the calculation of the energy levels of a quantum well andthen proceed to a crystal lattice and finally the two-dimensional exciton which is equivalentto the exciton in a narrow quantum well. All of these cases we will first treat for one particleonly and therefore can drop the indices that number the particles, i.e. we will just use xinstead of xi etc..

3.1 The infinite potential well

Consider a one dimensional (xdirection) potential energy profile that is infinite everywhereexcept for d x d.

The Schrodinger equation for an electron in the well then becomes:

h2

2m

2

x2 = E (49)

Solving this equation, it is important to note that both the energy E of the particleand its wave function are unknown. This is typical for equations of the type of Eq.(49).Such equations are called eigenvalue equations and are characterized by the fact that theresult of a (differential) operator acting on the function () is proportional to the function

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(equals E). As a consequence of the structure of this type of equation, the unknown E andthe unknown are determined by completely different methods or requirements as we willillustrate below. Usually it is much simpler to determine E and often that is all one needsanyway.

If we had no potential well, the electron would be free and as can be seen by inspectionthe following equation for solves Eq.(49):

= A+eikx (50)

Inserting Eq.(50) into Eq.(49) we obtain E = h2k2

2mwhich is the kinetic energy of an electron

with de Broglie wavelength and corresponding wave vector k (k = |k|). We see here alreadythat we have obtained the energy E without the necessity to know A+ and the completewave function . The full is determined by normalization conditions that we will discusslater.

We now return to the quantum well. Because of the boundaries the solutions that we

now can guess will be standing waves. Therefore in addition to the wave from Eq.(49) whichpropagates in one direction we will need at least a second reflected wave that propagates inthe opposite direction and therefore:

= A+eikx + Aeikx (51)

Because the potential boundaries are infinite, must be zero at the points d,d which givesfor d:

A+eikd + Aeikd = 0 (52)

and for d:A+e

ikd + A

eikd = 0 (53)

The last two equations contain the unknowns A+, A and are homogeneous. This meansthat they have only then a solution if the determinant of the coefficients vanishes, that is

eikd eikd

eikd eikd

= 0. (54)

This leads to the requirement that:

sin(2kd) = 0 (55)

which is equivalent to:k =

n

2dfor n = 1, 2,... (56)

Using Eq.(49) one obtains

En =(nh)2

8md2for n = 1, 2,... (57)

As one can see, we did not need to know the wave function and we obtained the resultof discrete energies En that the electron can have. These discrete energies correspond to the

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discrete photon energies and to the standing waves of the photons in the box that representedthe source of the black body radiation. Only now we defined standing waves for the electronby using the Schrodinger equation. The real miracle is, of course, that the discrete energies,the energy levels of the well, are in complete agreement with experiments. We note at thispoint that this astonishing agreement with experiments is limited to the discrete energies.Other results of Schrodingers quantum theory are more difficult to compare with appropriateexperiments. As we will see, the wave function does not correspond to a measurable quantityand the square of the wave function that will be interpreted as the probability to find theelectron is, naturally, much more difficult to measure than En.

The wave functions for the quantum well can be determined in the following way. UsingEq.(52) together with Eq.(56) we have:

A+A

= ein = (1)n+1 (58)

which givesn(x) = A+cos(nx/(2d)) for n = 2, 4, 6... (59)

andn(x) = A+sin(nx/(2d)) for n = 1, 3, 5... (60)

The coefficient A+ is determined through the interpretation that the square of the wavefunction is the probability to find the electron. Therefore, the integral of the square of thewave function from d to +d must be 1. One can easily see that therefore A+ = 1/

d. We

will discuss the probability interpretation later in some detail.The solutions of Eqs.(59) and (60) can be combined into one by offsetting the position

to obtain:

n(x) = A+sin[n(x + d)/(2d)] for n = 1, 2, 3... (61)

Any wave function can be written as a linear combination of the functions given in Eq.(61)i.e.

(x) =n

cnd

sin[n(x + d)/(2d)] (62)

with n

|cn|2 = 1 (63)

There is a general principle behind it. The solutions of the Schrodinger equation form acomplete orthogonal system (orthonormal if normalized). The system is called completebecause any can be written as a linear combination of the elements of the system. Itis called orthonormal because

j ldr = j,l. Here denotes the complex conjugate,

r = (x,y,z) and j,l is the Kronecker symbol.We finish this example by the following remarks. Actual quantum wells are not sand-

wiched between infinite but only between finite potentials. Furthermore the sandwiching isdone usually between two different materials. This means that the well and the boundariesare having different atomic structure. It is then actually surprising that the solutions still

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Figure 12: Two dimensional square well

have anything to do with the idealized case. However, we will see that different materialsand different atomic structure often necessitate only the use of an effective mass for the

electron instead of its free electron mass and, whenever appropriate, of the dielectric con-stant of the material instead of that of the free space. The finite height of the potential ormore complicated forms of the potential necessitate either the use of perturbation theory ornumerical methods to arrive a result. If, in addition, the atomic structure plays a role, thenwe need to use more basic methods (ab initio methods) to arrive at valid solutions of theSchrodinger equation. We will discuss all of these methods.

3.2 Quantum wires and dots

Rather than considering a well, we might construct a system with vanishing boundary condi-

tions for in two dimensions say x, y so that the electrostatic potential is zero for 0 x Lx,0 y Ly and z + which defines a slab of infinite extension. Outside the slabthe potential is infinite. For small Lx, Ly one calls such a slab a quantum wire. The energylevels of such a wire can be derived similarly to those of a well and are:

Ekz =(n1h)

2

2mL2x+

(n2h)2

2mL2y+

(hkz)2

2mfor n1, n2 = 1, 2,... (64)

Such a quantum wire, connected to two three dimensional contacts, would show thequantum conductance that we had discussed before. The integers n1, n2 describe the modesperpendicular to the direction of conduction and must be inserted into Eq.(38) instead of

the generic integer n that we have used in this equation.If we confine the vanishing electrostatic potential also to a small distance in the zdirection

i.e. we have zero potential for 0 x Lx, 0 y Ly and 0 z Lz and infinite potentialotherwise, then we have a quantum dot. The name derives, of course, from the fact that wenow have quantization, called in these cases size quantization, in all direction. Free motionof the electron is therefore not possible in any direction and we have, in this sense, a zero-dimensional structure, a dot. Of course, if the de Broglie wavelength is very large as couldbe the case in a crystal where the electron has a very small effective mass, then the quantum

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Figure 13: A quantum wire

Figure 14: A quantum dot

dot can also be rather large. In the semiconductor GaAs, for example, we have an effectivemass that is 0.067 of the free electron mass and a quantum dot with Lx = Ly = Lz = 10nmshows still significant size quantum effects i.e. even the lowest energy level is still of theorder of 150 meV. The equation for the energy levels En of the quantum dot described aboveis easy to derive and one obtains:

En =(n1h)

2

2mL2x+

(n2h)2

2mL2y+

(n3h)2

2mL2zfor n1, n2, n3 = 1, 2,... (65)

As in the case of quantum wells, wires and dots are made in nano technology by using

different materials that lead to the confinement. Again, the different materials can often becharacterized just by use of an effective mass. When the structures become extremely small,however, then the nature of the atoms that constitute the material may become importantand also the approximations that lead to the possibility to use an effective mass becomeinvalid. Then one needs to use more advanced methods, e.g. ab initio methods to calculatethe energy levels.

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Figure 15: One dimensional quantum well with different materials: AlAs-GaAs-AlAs

Figure 16: Two dimensional quantum wells and wires

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-10 -5 0 5 10

-10

-5

0

5

10

Figure 17: An exciton in two dimensions

3.3 Energy levels of an exciton in a quantum well

In the following and approximate the quantum well simply by taking the two dimensionallimit, i.e. we calculate everything in a plane instead of in three dimensions. Therefore thespace coordinate are r1 = (x1, y1) and r2 = (x2, y2).

An exciton is a combination of an electron and a hole in a solid.We will discuss electrons and holes in connection with crystal lattices. For now we just

can assume that an electron is a negatively charged particle with mass m1 while a hole ispositively charged with mass m2. The electron and hole attract each other and form anentity that is similar to an atom. As one does for atoms we assume for the moment that thehole mass is very large, basically infinite. We then need only consider one moving particleand therefore can again drop for simplicity the indices (1, 2) and only consider the electronwhos mass we denote by m where the asterisk reminds us of the fact that we are in a crystaland the mass is an effective mass not the free electron mass. The hole just forms the zeroof the coordinate system. If it moves i.e. performs a translation in space, the electron justtranslates with it. Therefore the Schrodinger equation is given by:

h2

2m(

2

(x)2+

2

(y)2+) + (E+

e2

40r) = 0 (66)

where e is the absolute value of the elementary charge, r = |r|, 0 is the dielectric constantof the crystal in question and we have used Coulombs law to express the potential energy.Now we introduce cylinder coordinates r = (r, ) and write for the wave function:

(r) = R(r)eil (67)

where i is the imaginary unit and l is an integer. We therefore have in cylinder coordinates:

[ h2

2m{1

r

d

dr(r

d

dr) l

2

r2} e

2

40r]R(r) = ER(r) (68)

A differential equation, a so called eigen-value equation, such as Eq.(68) is, of course difficultto solve. One therefore has only two possibilities: either one finds a differential equation

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2 4 6 8 10x

-1

1

2

3

4

5

x S t at e Z3, wi th n2, l1

2 4 6 8 10x

0.5

1

1.5

2

2.5

3


2 4 6 8 10x

0.5

1

1.5

2


Figure 18: The radial wavefunction Rn,l for (n,l) = (1,0),(2,0) and (2,1)

in the mathematical literature that is equivalent after some substitution or one solves theequation numerically. We will discuss numerical solutions as special projects. Fortunately,

however, Eq.(68) is equivalent to Laguerres differential equation after substituting R =bearr|l|L. One can see then that L represents a Laguerre polynomial (HW problem).

It is now important to note that we are often not interested in the functions L. These canbe identified as describing the states that were denoted by Einstein as Z1, Z2,...,Zn,.... How-ever, as stated at the beginning what we want is mostly the discrete energies E1, E2,...,En,....Characteristically for all these type of problems a solution of the differential equation existsonly for certain discrete energies when En < 0 which means that the particle is confined orbound by the form of the electrostatic potential of the other particle or the quantum welletc. These discrete energies for which solutions exist are usually given by simple expressions.In our case of Laguerres differential equation we have:

En = me4

8h22220(n + 1/2)2

(69)

It is interesting to note that this solution is distinctly different to the solution for thehydrogen atom in three dimensions that can be found in all quantum mechanics texts. Thereis a 1/2-number in the denominator, and the Ens are almost a factor of two larger thanthose of the hydrogen atom in 3 dimensions which are given by:

En = me4

32h22220n2

(70)

The formula for hydrogen atoms in free space does, of course, not contain the dielectricconstant . However, a donor atom in a semiconductor such as arsenic in silicon would bedescribed exactly by Eq.(70)

This larger value of En is indeed observed for excitons in quantum wells. Why is thisinteresting? Well, the classical problem of planets going around the sun has actually solutionsin a plane and one obtains the same solutions whether or not one starts the calculation in twoor three dimensions. Not so in quantum mechanics! The Schrodinger condition apparently

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makes a difference for the cases when solutions are sought in spaces of different dimensionalityand experiments on nano structures, e.g. the exciton in a quantum well, have confirmed this!Here we have one important example why nano structure research is even important for veryfundamental questions in quantum mechanics. It gives us the unique possibility to investigatequantum solutions as a function of dimensionality. One can sandwich an atom in a quantumwell and make it almost two dimensional. One can go further and create quantum wiresthat are almost one dimensional and finally one can create entities that are very small in allthree dimensions, quantum dots.

3.4 Crystals, energy band structures and effective masses

We have repeatedly referred to the fact that in semiconductor crystals the mass of the electronmust be replaced by an effective mass. In this section, we present justification for this andbeyond that show how an electron propagates in a crystal. In general, this propagationcan not be calculated from a one electron treatment of the Schrodinger equation because

a crystal contains, of course, many electrons that interact with each other. However, thereexists a very powerful approximation that applies for most of the interesting problems ofnano science and technology. Within this approximation one can consider one electron, theelectron whose propagation we want to calculate, and describe all the other electrons and thenuclei of the crystal by a potential V(r) that is periodic with respect to the space coordinater. Of course, it is extremely complicated to actually compute this potential. To do this,one would have to solve the Schrodinger equation for the many electron system. It turnsout, however, that one can deduce the most important properties of the crystal electron inquestion, if one only knows a few Fourier-components of V(r). We will discuss below howone can obtain these Fourier components. For now we just assume that somehow we know

them and we derive a solution of the Schrodinger equation from this knowledge.

3.4.1 Energy bands derived by Fourier analysis

The crystal is a set of atoms or molecules whose location is described by points Rl that aregenerated by three vectors a1, a2, a3 from the relation:

Rl = l1a1 + l2a2 + l3a3 (71)

In the simplest case, a1, a2, a3 are the orthogonal vectors of three dimensional Euclideanspace (a, 0, 0), (0, a, 0), (0, 0, a).

The corresponding crystal lattice is called the cubic simple lattice with distance a betweenthe lattice points. Because the crystal is periodic, the crystal potential is also periodic i.e.

V(r + Rl) = V(r) (72)

It is natural then to assume that the physical phenomena in the crystal are also of thesame periodicity as shown in Eq.(72) and that therefore Fourier analysis is the appropriate

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Figure 19: Cubic simple crystal in 2 dimensions

mathematical tool to solve the Schrodinger equation for this case. Fourier analysis teachesus that if we have a function f that obeys:

f(r + Rl) = f(r) (73)

then we can expand f into the Fourier series

f(r) = h

Ah

eiKhr (74)

with

Ah =1

f(r)eiKhrdr (75)

Here h = 0,1,2,... is an integer (not to be mixed up with Plancks constant), is thevolume of the cell that when repeated over and over fills the whole crystal which is for thecubic simple crystal = a3. The vectors Kh are called reciprocal lattice vectors and, for thecubic simple crystal, are given by:

Kh = h1b1 + h2b2 + h3b3 (76)

where b1 = (2a

, 0, 0), b2 = (0,2a

, 0) and b3 = (0, 0,2a

).We can now guess the wave function from the following considerations. For a free electron

is proportional to eikr. Because of the periodicity of the crystal the wave function willalso be modified to reflect the periodicity so we might have:

(k, r) = eikruk(r) (77)

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where uk has the periodicity of the crystal i.e. uk(r + Rl) = uk(r). This guess of thewave function can in fact be proven mathematically and Eq.(77) is called Blochs theorem.Because uk is periodic, we can write according to Eq.(74):

uk(r) = h

AheiKhr (78)

Inserting Eq.(78) into the Schrodinger equation we obtain:

h2

2m

h

|k + Kh|2Bh eV(r)h

Bh = Eh

Bh (79)

where Bh = Ahei(k+Kh)r.

The termsEh(k) = |k + Kh|2 (80)

appearing in Eq.(79)are called the empty lattice energy terms or empty lattice energy bandsbecause they form the solution for the energy when V(r) = 0. Although these terms arejust stroboscopic replicas of the free electron parabola energy E vs k relation, they areactually very useful for the following reasons.

Eq.(79) contains all the information on the crystal structure through the vectors Kh. Allthe rest of the information on the crystal material and atomic properties is buried in thecrystal potential V(r). Assume for a moment that the crystal potential is very smooth andpresents only a small perturbation. Then one can deduce from perturbation theory, thatwe will treat later, that significant changes of the solution due to the crystal potential areonly introduced around certain values of k which correspond to the so called Brillouin zone

boundary. The solution of the Schrodinger equation is therefore identical to the empty latticebands except around these boundary points. This means that one can get a pretty goodidea about the solution just by considering the empty lattice bands. To gain this insight isparticularly important when the crystal lattice is more complicated than just cubic simple.Then also the Kh are more complicated and it becomes very valuable to obtain first a plotof the empty lattice bands. Examples for simple one dimensional lattices are shown in class(Figures).

Of course, for any precise calculation, the crystal potential can not be neglected. In-cluding the potential, one can solve the Schrodinger equation in the well known way that isprescribed by Fourier analysis. We multiply Eq.(79) by the factor ei(k+Kl)r and integratethe resulting equation over all the crystal volume V

oland normalize (i.e. divide) by the

volume. Using this procedure, we encounter the following integrals:

1

Vol

ei(k+Kh)rei(k+Kl)rdr = h,l (81)

The simple equality to the Kronecker follows from the fact that the integrand equals 1if h = l. For h = l the integrand oscillates rapidly over the large volume and the integral

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Figure 20: Empty lattice bands for one dimensional lattice

Figure 21: Empty lattice band for two dimensional cubic simple crystal

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Figure 22: Empty lattice band for two dimensional cubic face centered crystal

therefore equals 0. For the term that contains the crystal potential, however, things becomemore complicated. Then we have integrals of the form

< Kl|V(r)|Kh >= 1Vol

ei(k+Kl)rV(r)ei(k+Kh)rdr (82)

Here we have used the customary shorthand notation < Kl|V(r)|Kh > for the integral.There is a deeper mathematical sense in this shorthand and we refer the interested reader tomore detailed descriptions of the Dirac notation in standard texts on quantum mechanics.We also will give a brief outline of the general features of this notation below. For now weonly need to know that Eq.(82) represents nothing but the Fourier transform of the crystalpotential with respect to Kh Kl i.e.

< Kl|V(r)|Kh >= 1Vol

ei(KhKl)rV(r)dr (83)

Using Eqs.(82) and (83) we obtain then from (79) for each given index < l < +the following algebraic equation:

[El(k) E(k)]Al +h

Ah < Kl|V(r)|Kh >= 0 (84)

If we still set aside the fact that we do not know the crystal potential, then Eq.(84) containsthe unknown E(k) and Al, Ah while El(k) is given by Eq.(80). Thus we have an infinitesystem of homogeneous linear algebraic equations. We can, of course, not solve such aninfinite system. One therefore usually attempts a numerical solution by limiting l, h and bythen obtaining a finite number of equations. A computer can easily deal with about one

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Figure 23: Empty lattice bands with a gap

hundred values of l and h. Note, however, that the situation is exactly the same as it wasin the case of the quantum well. We have a system of homogeneous equations that has onlythen a solution if its determinant vanishes. This condition gives us then a result for E(k)even if we do not know the coefficients Al, Ah and, in fact, E(k) is all we want to know. Toshow the procedure explicitly we seek now a solution for the values h, l = 0,1. To makethings simpler we choose an energy scale so that < 0|V(r)|0 >=< 1|V(r)| 1 >= 0 andthen obtain the following determinant.

E0(k) E(k) < 0|V(r)| 1 >

< 1|V(r)|0 > E1(k) E(k)

= 0. (85)

With the notation S(k) = E0(k) + E1(k) and using the fact that < 1|V(r)|0 > is thecomplex conjugate of < 0|V(r)| 1 > one obtains from Eq.(85):

E(k) =1

2S(k) 1

2

S(k)2 + 4| < 0|V(r)| 1 > |2 (86)

The curve corresponding to this result looks exactly like the corresponding empty lattice

curves except that for k = a , which is the Brillouin zone boundary in one dimension, weobtain a splitting of the curves with an energy gap EG = 2| < 0|V(r)| 1 > |.

This is the well known energy gap of band structure theory that separates, for exam-ple, valence and conduction bands in semiconductors. Such a band gap can also easily bemeasured by investigating the optical properties of the semiconductor. Roughly speaking,the semiconductor is transparent for light energies h that are below the gap energy i.e.h EG. From the experimentally known energy gap we can therefore deduce the Fouriertransform | < 0|V(r)| 1 > | of the unknown crystal potential. In practice it turns out that

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Figure 24: Illustration of tight binding Hamiltonian

even if we use a much larger number of integers h, l, one needs only a relatively small numberof Fourier components of the crystal potential which can be determined from a number ofexperimental results. For various historical reasons, this method of calculation of E(k) iscalled the empirical pseudo potential method.

3.4.2 Energy bands by nearest neighbor interactions

Energy bands have been calculated by a variety of methods that, on the surface, look com-pletely different. Accordingly, these methods have also received very different names. Inprinciple, however, the methods are often mathematically very similar if not identical. Toillustrate this we describe here a so called tight binding nearest neighbor method and takethe opportunity to explain the Dirac notation a little better.

We can formally write the Schrodinger equation as

H0n = Enn (87)

where H0 is called the Hamiltonian operator, which can be a differential operator as givenin the above examples for the Schrodinger equation. We know then that the solutions n

form a complete orthogonal system and if they are properly normalized we have:j ldr = j,l (88)

It is known from mathematics that such a system of functions also forms a vector space. Wecan therefore denote the functions as vectors e.g.

l |l > (89)

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and can define a scalar product of vectors such that

< j|l >=

j ldr = j,l (90)

Because this basic system of vectors (functions) |n > is complete, any vector | > can bewritten as a sum of these vectors with coefficients that are the projections of | > on |n >i.e.

| >=

n|n >< n| > (91)

From this equation we can see that we can take the symbol |n > < n| as a projectionoperation (or projection operator) that projects any vector | > onto |n >. One can alsosee that

n|n >< n| is the identity operator because it turns any | > into itself. It is also

very useful and practical to represent other operators in terms of such projection operators.For example, the Hamiltonian H0 can be written as:

H0 =

n|n > En < n| (92)because then we have:

H0|n >=

n|n > En < n|n >= En|n > (93)

We turn now to describe a simple way to write the Hamiltonian of a crystal. If oneconsiders a crystal and likes to solve band structure problems, one could think of H0 asbeing the Hamiltonian describing totality of all the single separated atoms. The atoms areat this point assumed to be independent. We can attach to each atom a wave function |i >.In contrast to the examples above, the index i labels now simply the different atoms and

not the different eigenvalues of the energy for a given atom. We assume, for simplicity, thateach atom has a single Eigenvalue E0 and a single corresponding atomic wave function |i >.Addition of more energy eigenvalues and wave functions for each single atom complicatesthe notation but needs to be done for realistic calculations. If we now think of any wavefunction | > defined by a certain amplitude ci of finding an electron at a given site, we canrepresent this wave function by:

| >=

ici|i > (94)

or| >=

i|i >< i| > (95)

where we have used ci =< i| > as above. Thus the localized atomic wave functions |i >form the complete ortho normal system generated by H0 exactly as we had it for the caseof e.g. one quantum dot and its energy eigenvalues as well as corresponding wave functions.This illustrates the great flexibility of the Dirac notation and illustrates the fact that thegeneral mathematical formalism of eigenvalue equations is very flexible and can be adjustedto and adopted for many different physical situations.

To make a crystal out of these unconnected atoms and to permit propagation of elec-trons between them i.e. to be delocalized, one needs to define the contributions to the

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Figure 25: Nearest neighbour interaction

Hamiltonian due to the overlap of the atomic potentials of the different sites. In general,this overlap can be described by a term

lm

Vlm|l >< m| (96)

which then gives in total the so called tight binding Hamiltonian

H =

i|i > Ei < i| +lmVlm|l >< m| (97)Vlm represents, of course, a potential energy not a potential. We are sometimes using the sym-bol V for potential and at other instances for potential energy because it is often convenientnot to include the symbol e for the elementary charge as it can conflict with the exponent e.g.ex. The physical meaning of these two terms can be further explained as follows. The totalHamiltonian of an electron in a crystal is the sum of kinetic energy, T, and potential energyV(r). The potential energy is to first approximation given by the sum of the potential v(riof the single isolated atoms. Therefore we have H0 = T +

i v(ri =

i|i > Ei < i|. The

contributions of all atoms to the potential energy at site l are Vl =

m=l v(rm and thesegive rise to the off diagonal terms in Eq.(97. The contributions of these non diagonal termslead to a connection of the originally separated states of H0 and to a nonzero probability foran electron to escape from one distinct state to another.

A good approximation to this Hamiltonian is often the nearest neighbor interaction thatis we have

Vlm = V if l, m are nearest neighbors (98)

andVlm = 0 if l, m are not nearest neighbors (99)

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We assume now a one dimensional chain ofN atoms with lattice spacing a and use the trialwave function:

|k >= 1N

j

eikaj |j > (100)

Inserting this |k > into the nearest neighbor tight binding Hamiltonian and using Ei = E0for all i = 1, 2,... we obtain for the energy Eigenvalues E(k):

E(k) = E0 + V(eika + eika) = E0 + 2V cos(ka) (101)

3.5 The effective mass

See reference [1].

3.6 Counting and filling states

See reference [2].

3.7 Tunnelling

We know that electromagnetic waves are not necessarily propagating anywhere but undercertain circumstance decay exponentially within fractions of a wavelength and die out. Onecalls this decaying portion of the wave an evanescent wave. A well known case is that ofan em-wave propagating in a medium of high dielectric constant with a neighboring mediumof lower dielectric constant. When a light ray falls at a shallow angle on the interface it isreflected. However, this does not mean that the electromagnetic field is zero in the medium

of smaller dielectric constant. There exists still the evanescent wave. One could check thisin the following way, If one sandwiches a medium of small dielectric constant between twolayers with large dielectric constant then there will still be propagation of the light beamdepending how thin the medium with the small dielectric constant is.

A very similar phenomenon is known for the Schrodinger waves. Consider instead of aquantum well a thin quantum barrier that is designed as follows.

The static electrical potential is zero everywhere except for a region d x +dwhere it assumes a (negative) value V0 so that we have a positive potential energy eV0. ASchrodinger-De Broglie wave impinging upon this barrier will decay and become evanescent.However, if the barrier is thin enough, it will leak through and propagate on the other

side. With the probability interpretation for the wave function, this means that an electronwill have a certain non zero probability of going through the barrier. W can now representthe propagating free electron that has a kinetic energy E by a wave vector k whose absolutevalue is as usually given by k = (

2mE)/h. The exponentially decaying wave in the barrier

can be characterized by a wave vector = i that is imaginary ( is real), just as we knowit from the Helmholtz equation and from electromagnetic waves. We have

= [

2m(V0 E)]/h (102)

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Figure 26: Schematic picture of Tunneling through a rectangular barrier

The wave function has then the general form:

(x) = Aeikx + Beikx for x < d(x) = Cex + Dex for |x| < +d

and(x) = Eeikx + F eikx for x > +d (103)

We must now match the wave functions and their derivatives at the boundaries x = d.This is a laborious algebraic exercise and we perform the procedure therefore with an ap-proximation. We assume that left of the barrier we have an incoming and a reflected waveand therefore both A, B = 0. At the left barrier we assume we have only a wave decayingfrom left to right and therefore C = 0. At the right barrier we need to admit a reflection andtherefore both C, D = 0. Finally we have only one wave outgoing to the right and thereforeF = 0. This approximation is obviously good for thicker barriers. It certainly simplifies thealgebra. The condition of continuous wave function and derivative gives for x = d:

Aeikd + Be+ikd = Ded

andik[Aeikd Be+ikd] = Ded (104)

For the barrier at x = d we obtain:

Ced + Ded = Eeikd

and[Ced Ded] = ikEeikd (105)

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Figure 27: Exact transmission coefficient for tunneling

Eliminating C from the last two equations gives us D in terms ofE and eliminating B fromEqs.(104) gives us therefore A in terms of E i.e. the transmitted amplitude in terms of theincoming amplitude. We know that the transmission probability T is equal to the square of

the amplitude ration T = (EA

)2

which is:

T

42

k2 + 2e4d (106)

which reflects the exponential decay of the evanescent mode. For thinner barriers this ap-proximation does not hold and one obtains T 1/[1 + (kd)2]. The transmission coefficientT above the barrier also can only be calculated by the full matching procedure without theabove approximation. It shows an oscillating behavior as can be seen in the figure.

3.7.1 Arbitrary forms of potentials, WKB

The wave function matching method outlined above, works only for simple rectangular po-tential shapes. If one deals with arbitrary potentials, a numerical approach is necessary.

There also exists an approximation due to Wentzel, Kramers and Brillouin (WKB) that isoften very good and that we will describe here.

Assume that we have a one dimensional barrier of potential energy V(x) > 0 for d x d and zero potential energy outside this region exactly as we had it above except thatV is now not constant. The Schrodinger equation reads then:

(d2

dx2 2) = 0 (107)

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Choosing an arbitrary point x = b left ofd we try now the solution

=1

ex

b(x)dx (108)

Inserting Eq.(108) into Eq.(107)we see that Eq.(108) indeed solves Eq.(107) provided thatone can neglect the term proportional to

d2

dx2(

1

) (109)

One can indeed neglect this term if (i) the potential energy varies relatively smoothly and if(ii)

= 0 (the latter because is in the denominator of Eq.(109)). These two conditions

are unfortunately a lot to ask for because of the following reason.

vanishes exactly forE = V. If a particle approaches the potential barrier from the left having a kinetic energyE, and if the barrier rises smoothly, then we have E = V just at the point where the particle

hits the barrier. If, on the other hand, the barrier rises abruptly at d, then the potentialenergy does not change smoothly and we need to suspect that this part of the approximationfails. Therefore it appears that we have no possibility for any practical case that Eq.(109) isa great approximation. Lets therefore just apply the WKB approximation to the rectangularbarrier from above by calculating the transmission coefficient T exactly as we did above. Wefind T by dividing the wave function slightly to the right of d by the wave function slightlyto the left ofd and squaring:

T = e4d (110)

which reproduces the important exponential factor of Eq.(106). It is the fact, that WKBgets the exponential factor correctly that makes it often a reasonable approximation forrelatively thick barriers. One can then fit most experimental data by just slightly modifyingthe thickness 2d of the barrier that is often not even well known. Because of this fact,the WKB approximation takes an honorable place among the approximations to tunnellingproblems.

3.7.2 Coulomb blockade

Tunnelling as described above is based on the solution of the Schrodinger equation for asingle electron and does not consider any effects of the electrons charge. For a long timethis deficiency was not noticed and experiments could be satisfactorily explained. These

experiments typically involved a thin insulator sandwiched between two metal plates usuallyhaving a rather large area. The insulator represents a tunnel barrier because the conductionband of the insulator is above the energy of the conduction electrons in the metals, the Fermisea, as we have pointed out in the section on band filling. As the technology related to thetunnelling experiments was developed further, a strange phenomenon was observed. Thetunnelling current should start flowing as soon as a voltage, no matter how small, is appliedto the tunnelling barrier because the probability of tunnelling is nonzero and can be largefor thin barriers (with a thickness of the order of a few Angstrom). However, the tunnelling

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Figure 28: Coulomb Blockade

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current was observed to be zero as long as the applied voltage was below a certain thresholdvoltage Vth.

This lack of current for small (close to zero) voltages was called the zero point anomalyand the literature of the time was full of speculations where this anomaly could arise from.The explanation turned out to be simple. The two metal contacts with an insulator inbetween form a capacitor with capacity C. This capacity can be calculated from the parallelplate formula to be approximately C = Ar0/d where A is the area of the metal plates andd the thickness of the insulator. Elementary courses in electrical engineering tell us thatthe energy Eth that is necessary to transfer a single electron from one capacitor plate to theother is given by

Eth =e2

2C(111)

which requires a voltage Vth =eC

. This means that no tunnelling current can flow unless theapplied bias is above this voltage threshold.

It becomes now clear why for large contact areas and a correspondingly large capacitance

this threshold was not noticed. In that case and for relatively large temperatures T thethreshold voltage is smaller than the always available thermal voltage kT/e. Only if thecapacitance is very small, of the order of Ato-Farad, becomes the threshold noticeable atroom temperature because then we have:

|eVth| >> kT (112)The existence of such a threshold, or the visibility of the zero point anomaly, is now calledthe coulomb blockade effect. This effect permits the control of tunnelling down to singleelectrons and so called single electron transistors have been proposed.

These devices would have the advantage to deal with the absolute smallest amount of

charge that any electron device can deal with. It has the disadvantage that any imperfectionthat carries charge will significantly disturb its operation. In addition, the feature sizes ofsuch devices must not exceed the length of about one nanometer if one wants to work at roomtemperature because otherwise Eq.(112) is not fulfilled. As a consequence it is generally seenas very difficult to develop integrated circuits based on single electron transistors. One needsa whole circuit of nano structures with very few charge carrying imperfections that woulduncontrollably shift the threshold of the single devices.

3.7.3 Resonant tunnelling

3.7.4 Inter band tunnelling

4 Stationary perturbation theory

We have seen above that the various explicit method of solving the Schrodinger equationdepend sensitively on the details of the problem and that an explicit solution can usually befound only for the simplest geometries. For more complicated structures there are mainly twomethods available: (i) a numerical solution of the problem and (ii) a solution by perturbationtheory. We will first describe perturbative methods and deal with numerical approaches later.

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Figure 29: IEEE spectrum paper

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4.0.5 The perturbation series

The idea of perturbation theory is the following. We have given a Hamiltonian H in theSchrodinger equation that defies any attempt of explicit solution. However, there is a portionofH that we denote by H0 and for which we know how to obtain the complete orthogonal and

normalized system of Eigenfunctions |n > and the eigenvalues En. The remaining portion ofH we denote by H1. The factor epsilon is thought to be a small number, indicating that thisterm is not the main term and represents just a correction. The idea is then to constructseries expansions in powers of that are smaller and smaller as the power of increases.Thus we have

H = H0 + H1 (113)

andH0|n >= En|n > n = 0, 1, 2, ... (114)

and we wish to find eigenvalue E and wave function | > of the equation

H| >= E| > (115)To achieve this goal we consider expansion around the lowest zero order solution |0 >, E0using the following power series expansions:

| >= |0 > +|1 > +2|2 > +... (116)and

E = E0 + E1 + 2E2 + ... (117)

As usual with such power series expansions, we now insert Eqs.(116) and (117) int Eq.(115)and group all terms with the same coefficient n together since all these terms need to vanish

separately. Thus we obtain for the three lowest equations:

0 : H0|0 >= E0|0 > (118)1 : H0|1 > +H1|0 >= E0|1 > +E1|0 > (119)

2 : H0|2 > +H1|1 >= E0|2 > +E1|1 > +E1|0 > (120)Eq.(118) is just the zero order equation. Eq.(119) is the basis for first order perturbationtheory Expanding 1 in terms of the complete ortonormal system of H0 we have

|1 >=n

an|n > (121)

We put a0 = 0 because |0 > appears separately in Eq.(116). Then multiplying Eq.(119) by< 0| gives immediately:

E1 =< 0|H1|0 > (122)This is the important result for the perturbed energy level. Multiplying Eq.(119) with< i| =< 0| gives

ai =< i|H1|0 >

E0 Ei (123)

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Figure 30: Scattering of an electron by a small object

from which we get the perturbed wave function:

| >= |0 > +1

< n|H1|0 >E0 En (124)

Of course, we would have been able to expand around |j > instead of |0 > and would havethen obtained similar results just with 0 > replaced by |j > in the last three equations.

Higher order perturbation terms require a little more algebra and we just list the per-turbed energy arising from E0 to second order:

E = E0+ < 0|H1|0 > +1

< 0|H1|n >< n|H1|0 >E0 En (125)

Perturbation theory is extremely useful for a host of problems including scattering ofparticles by small obstacles, corrections to well known problems due to small additions ofarbitrary potentials etc...

We will return to several applications of perturbation theory in the course of these lec-tures, here we give only two examples, that of the effects of an electric field (called starkeffect) in a quantum well and scattering by a periodic potential.

Consider the infinite quantum well problem but add a constant electric field F for d x +d. Note that the addition

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Figure 31: Quantum well with an additional constant electric field

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Figure 32: Electron impinging on periodic crystal

of the field is not enough to specify the problem because the Schrodinger equation dependson the potential. We could, for example specify a potential energy that is symmetric with

respect to the middle x = 0 of the quantum well, negative to the left and positive to theright e.g. V = eF x. Then we obtain the first order energy correction from Eq.(122) andEq.(61):

E1 =< 0|eF x|0 >= eFd

+dd

x{sin[(x + d)/(2d)]}2dx = 0 (126)

which tells us that for such a symmetric potential there is no first order Stark effect. Hadwe chosen a potential energy eF d + eF x then we would have obtained the non-zero answereF d.

For the second example we consider a free electron with a wave function proportional tieikr impinging on a periodic crystal with periodic potential energy

V(r) =h

BheiKhr (127)

where the Kh are reciprocal lattice vectors as discussed in connection with band structurecalculations. We use now the fact that the perturbation series involves matrix elements ofthe form

< k|V(r)|k >=

eikr

h

BheiKhreikrdr (128)

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The integrand in this equation varies rapidly and the integral therefore vanishes except if wehave

k + Kh = k (129)

Only when this condition is fulfilled do we have a non zero contribution to the perturbation

series. The collision of a single electron with the huge crystal must be elastic and therefore

|k| = |k| (130)

Eq.(129) and (130) together represent the condition for Bragg diffraction, in other words theelectron is scattered by the crystal if this condition is fulfilled. In one dimension the Braggconditions are equivalent to k = Kh

2. This is exactly the Brillouin zone boundary where

we had the opening of the band gap in the band structure calculation i.e. the point wherethe empty lattice bands crossed. The same is true in two or three dimensions.

4.0.6 Greens functions and perturbation theory

Given a Schrodinger equation of the usual form (HE) = 0 with a Hamiltonian differentialoperator H one defines the Greens function as the solution of the equation

(H E)G(r, r, E) = (r r) (131)

If one has the Greens function, what is the use of it? Most commonly, it is used to ob-tain the solution of the inhomogeneous differential equation if we have the solution of thehomogeneous one. For, if we wish to solve the equation

(H

E) = f(r) (132)

then

(r) =

f(r)G(r, r, E)dr (133)

represents one solution as can be immediately found out by inserting Eq.(133) into (132).This fact can now also be used in a special way to actually obtain a Greens function G byperturbation theory if the Greens function G0 is known for the unperturbed problem. Letsassume that H = H0 + V(r) where V(r) is just any given small perturbation of the potentialenergy and lets further assume that we know the solution G0 of

(H

E)G0(r, r, E) =

(r

r) (134)

We now like to obtain the solution G of

(H+ V(r) E)G(r, r, E) = (r r) (135)

To obtain this solution, we first observe that the perturbed problem can also be writtenas:

(H0 E) = V(r)(r) (136)

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which is just the inhomogeneous equation corresponding to the homogeneous equation of theunperturbed problem. Therefore substituting V(r)(r) for f(r in Eq.(133) we obtain

(r) =

V(r)(r)G0(r, r

, E)dr (137)

Adding to this solution of the inhomogeneous equation the homogeneous solution we havethe general solution of the perturbed problem:

(r) = 0(r) +

V(r)(r)G0(r, r, E)dr (138)

The only problem is now that Eq.(138) is actually an integral equation because the unknown appears on both sides. However, we can now try to get a perturbation series by iteration.We start by approximating by 0 on the right hand side of Eq.(138). This gives

(r) = 0(r) +

V(r)0(r)G0(r, r, E)dr (139)

Inserting now this into Eq.(138) gives

(r) = 0(r)+

V(r)0(r)G0(r, r, E)dr+

G0(r, r

, E)V(r)G0(r, r, E)V(r)0(r)drdr

(140)This process can be continued with more and more elaborate integrals. Of course, theformalism can also be put into matrix and vector form by the following discretization. Wewrite the potential energy as a vector V with

V = (V(r1), V(r2), V(r1),...) (141)

The Greens function as a function of two variables becomes then a matrix which we denote byG and G0. For the integral we can then substitute a Riemann-sum which is just a product ofmatrix and vector multiplied by a constant characteristic for the distance between dicretizedpoints. We then have

V(r)G(r, r, E)dr = GV (142)

With these changes, we can write the whole series expansion as

| >= |0 > +

G0V|0 > +

G0V

G0V|0 > +... (143)It is easy to see that this is equivalent to

| >= |0 > +G0V| > (144)

which replaces the original integral equation Eq.(137). | > and |0 > are here, of course,also vectors obtained by the same discretization procedure. Remember now that the Greensfunction is by its definition closely related to the wave function. We therefore can go through

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the whole procedure that we just went through using the Greens function instead of thewave function. This gives

G = G0 + G0VG (145)

This equation is usually called a Dyson equation. The Dyson equation is useful in numerous

ways. Here we just show its power by calculating the transmission coefficient for a T-shapednanostructure as shown in the Figure.We treat the T-shaped nanostructure in the tight binding formulation for a one dimen-

sional infinite chain of atoms numbered 0,1,2,.... At the atom with number 0 we attach asidearm of finite length with atom numbers 1, 2, 3,.... as shown in the figure. The tight bind-ing wave functions we denote then by |0 >, | 1 >,... and for the sidearm by |1 >, |2 >, ....As in all such problems, it is now important to choose the unperturbed problem in the bestpossible way which mostly means as close to the complete perturbed problem but still simpleenough so that we can either solve it explicitly ourselves or at least find a solution in theliterature. There exists a solution in the literature for the infinite chain and also one fora finite chain (sidearm) [3]. Therefore we take these two still unconnected chains as theunperturbed problem. The perturbed problem just includes the coupling of the chains. Thiscoupling can be described in the tight binding method by the nearest neighbor interactionthrough:

V01 = V(|0 >< 1| + |1 >< 0| (146)Now we use the Dyson equation

G = G0 + G0V01G (147)

We do not need to solve this equation because all we want is the transmission coefficientTs of an electron propagating from the left of the sidearm to the right. Therefore we need

only certain matrix elements of the Greens functions G (we know those of G0). As in thecase of tunnelling, the transmission coefficient Ts is given by the square of the transmissionamplitude ts which is

ts =< d|G| + d >< d|G0| + d >

(148)

Here < +d| denotes a given site to the left of the sidearm and |d > a corresponding one tothe right of the sidearm. We can create now equations for the unknown matrix elements ofG0 the following way. We multiplying the Dyson equation from the left by < d| and fromthe right by | + d > as well as remembering that G0 and therefore all its matrix elementsare known from reference [3], we obtain:

< d|G| + d > = < d|G0| + d > + < d|G0|0 >V < 1|G| + d > (149)Here we have also used the fact that < d|G0|1 > = 0 because the unperturbed Greensfunction G0 can not lead to any coupling to the sidearm. We now have an equation for thematrix element < d|G| + d > that we are seeking but we have introduced another unknownmatrix element < 1|G| + d > because the Dyson equation contains G twice. The procedureis therefore the following. We use the Dyson equation again and multiply it now by the two

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appropriate vectors (one from the left one from the right) to obtain the new unknown matrixelement. We continue the procedure until we have as many equations as we have unknownsand then solve these equations e.g. numerically. For the particular case of the T-shapednanostructure we need to repeat this procedure

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