Slides EC574 Part I F2013

8/11/2019 Slides EC574 Part I F2013

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Part - 1Fundamentals of Quantum Mechanics

Quantum mechanics is the basic mathematical formalism usedto study the objects (electrons, holes, phonons, photons) that areimportant to understand the properties of semiconductor materi-

als and electronic/optoelectronic devices.

We will look at the reasons why classical mechanics can nolonger be used to describe electrons and holes. We will develop the basic QM theory by using an axiomaticapproach based on postulates. We will apply the theory to study the electrical and opticalproperties of semiconductors.

EC574 Physics of Semiconductor Materials, Fall 2008


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The limit of classical mechanics

The simplest way of studying the motion of rigid bodyin classical mechanics is to use Newtons equations.

d rdt =

pm

d pdt = V = F

The solution describes a trajectory in phase space px (t), py (t), pz (t),x(t), y(t), z(t), that for given initial condition is unique.

At each instant the three components of the vector position r and the mo-mentum p = mv are known with innite precision.

At the beginning of 1900 a series of experiment showed that this pictureit is not always applicable to study the motion of atomic scale particle, likeelectrons. (G.I. Taylor 1909)

We start looking at a simple classical system: The Harmonic Oscillator.

Readings: Brandsen-Joachain 1.2, 1.3, 1.6EC574 Physics of Semiconductor Materials, Fall 2008


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I - An Important Example

Consider a ball tied to a rope mov-

ing on a circumference of constantradius R and constant angular fre-quency 0 . The trajectory of theball is given by: r (t) = x(t)i +y(t) j .

In particular:

x(t) = Rcos(t + )

and thew second derivative is givenby:

d2x(t)dt2 =

20 Rcos (0 t + )

and

d2x(t)

dt2 =

2

0x(t)

This is the well know equation thatdescribes the Simple HarmonicOscillator (SHO) .



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II - Another Important Example

Consider a mass m on an idealspring described by a force constantkf . The other side of the spring in

xed by an immovable boundary.

If we assume that the gravity is neg-ligible (!) as we move the mass, aforce exerted by the spring will try

to take the system back at the equi-librium position. This force is:

F = md2x(t)

dt2 =

kf (x

xeq )

or:

d2x(t)dt2

= 20 (x xeq )where:

20 = kf mis the the natural frequency.



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III - Some Important QuantitiesThe potential energy V(x) is given by:

V (

x) =

F dx = kf

xdx

( x is the variation around xeq )V (

x) = 12

kf

x2 =

12

m20

x2

The kinetic energy T( x) is given by:T (x) = p22m

And the total energy E = T( x) + V( x)is a constant of the motion.E =

p2

2m +

1

2m20

x2 = const

V (

x) =

12

m20 (x xeq )2At equilibrium ( x = x

eq)the poten-

tial energy is zero, and the kineticenergy is maximum.The classical turning points (CTP) are

those values xctp of the coordinate forwhich the kinetic energy is zero and thepotential energy is maximum. At thesepoint the particle slows down, stops and

turns around.As a result during a period of oscilla-

tion, the particel it is more likely to be

found near the CTPs where its velocity

is the smallest.



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Classical Probability Density - I

To be able to dene a probability we have to select an ensemble of identicaland non interacting systems that we observe.

Since any number of identical classical systems in the same state will

always give the same measurement results, we select on of the systems asrepresentative.

We measure at random times the position of the particle in the represen-

tative systems and dene:

P cl (x)dx = Probability of nding the particle in a position between xand x + dx.

Since for the SHO T o = 2/ 0 the in the time (period) required tocomplete a full oscillation, then: P cl (x)dx = t/T 0 is the fraction of the period the particle spends in an

interval dx around the point x.



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Classical Probability Density - II

Classical probability density for tree energies:

a) E = 1 / 2 0 , n=0; b) E = 3 / 2 0 , n=1; c)E = 21 / 2 0 , n=10.

The classical probabilitycan be compute to be

(Homework-1):

P cl (x)dx = m20 1

2E m20 x2dx

(1)

It shows that the SHO isless likely to be found at

x = 0 where it moves morerapidly. On the contrary, itis more likely to be foundwhere it moves more slowly.



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The two-slit diffraction experiment

When the two slits are illuminateby a plane wave, the characteristicdiffraction pattern is formed. Thisis a behavior expected from awavelike excitation.

!!! Electrons behave like wave andparticle at the same time and this

depends on the measurement setup.

The same experiment is now per-formed with electrons. A very fainsource is used so that one electronat the time is counted with a detec-tor at either slit A or B (particle).If the experiment is run for a longtime, a diffraction pattern similar to

the wavelike excitation is observed.



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The limit of the Electromagnetic Theory

As we know from electromagnetic theory light propagates in waves accord-ingly to Maxwells equations and the wave equation.

2 E +

E t = 0

Note that in this formulation the energy ow per unit are is determine bythe Poynting vector: P

E

H and it can assume a continuum of values and

it is independent of the frequency.

The photoelectric effect was the rst experimental evidence that light canbehave both like a wave or a particle depending on the measurement condi-tions.

This conrmed Einsteins prediction of the existence of the quanta of light,i.e. the photon and Planks black body theory.



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The Photoelectric Effect

Classical: kinetic energy of the pho-toelectrons depend directly on thelight power density (intensity).

No frequency dependenceis observed.

Absorption of energy takesa nite time.

Quantum Mechanical:

As the intensity doubles the

number of photons doubles but theirenergy remains the same. there is axed energy absorbed by electrons.

The energy of each pho-

ton is h . The energy changes onin discrete amount. If h is notlarge enough (frequency threshold)

no electrons are emitted.

Absorption of energy is in-stantaneous.



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Atoms in a Magnetic Field - I

If we consider a pre-quantum-mechanical atomic model, such as the

Bohr model, the electron is suppose to

move on a circular orbit around the nu-

clei, with an angular momentum L.

The electron moving on a closedorbit forms a current loop.

A circulating current of magnitudeI enclosing a small plane dA produces

a magnetic dipole M = I dA .

If I is due to a circulating electronthen I = ev/ 2 r . Since A = r 2 ,and the angular momentum L = mvr ,the magnetic dipole is given by:

M = e2m

L = BL

and B = e / 2 m is the Bohr magne-ton. It value is:

B = 9 .27 10 24 JT 1



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Atoms in a Magnetic Field - II

If an atom with magnetic dipole M is placed in a magnetic eld B theinteraction energy is given by:

W = M Band a torque:

= M B

and a force:

F =

W or:

F x = M B

xF y = M

By

F z = M Bz

If the eld is constant there is nonet force on the dipole that precesses at

a constant angular frequency (Larmor):

L = B B (2)

If the eld is inhomogeneous thenthe dipole experience a force propor-

tional to the magnetic dipole.

One could measure the magneticmomentum of a given atom by observ-

ing the deection of an atomic beamcaused by an inhomogeneous magnetic

eld.

Stern and Gerlach experiment 1921circa.



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Stern-Gerlach Experiment - I (1921)

A beam of Silver ( 4 d10 ,5 s 1 )atoms is collimated and deected by an inhomoge-

neous magnetic eld and detected on a cool screen.

The applied magnetic eld is in the z-direction only. As a result because of thesymmetry of the applied magnetic eld: Bzx = 0 and Bzy = 0 . Consequently the force is only in the z-direction.

What do we expect to see on the screen?EC574 Physics of Semiconductor Materials, Fall 2008

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Stern-Gerlach Experiment - II (1921)

From the classical stand-point oneexpects that the atoms in beam have a

randomly oriented vector M .

As a results the expected patternof Silver atoms on the screen should be

an uniform region such that M < M z

< M .

Stern and Gerlach obtain a surpris-ing result !!!

The pattern is not uniform but hastwo well dened and separated regions.

They are symmetric respect to the point

of no-deection.

The experimental evidence suggests that the because of the spatial quantization of the magnetic moment the angular momentum (its z-component) of the atom is quan-

tized !!


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Summary

Several other experiments (Compton effect, tunneling, stability of the atomstructure, ..) shows that in case of atomic size particles the laws of classicalmechanics do not provide a correct description of the physics.

At the same time other experimental evidence shows that the electromag-netic radiation has both a wavelike and a particle nature (photon).

The quantum mechanical description of the atomic world will move awayfrom a rigid description of either wave or particle, to introduce a more com-plex formalism that makes room for both.

This description is based on complex (for real objects !) wavefunctions andthe introduction of the concept of probability to describe the motion of atomicparticle.



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The Indetermination Principle (Sec. 2.5)

From the evidence of the electron diffraction experiment wehave seen that the classical law that describe the particle trajec-tory no longer explain the physics.

It is important then to understand under which conditions theconcept of trajectory can be used and what are its limitations. Is the classical assumption that all the variables of motion canbe be measured simultaneously at the same time still valid? Let us look again the experiment electron diffraction experi-ment and consider the case in which the slit has a size x x . How would the electron diffraction pattern change? Whatdoes this mean in terms of the other "variables of motion" of theelectron?.



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The Indetermination Principle

If we measure the position we cansay that the electron is "localized"within a space D, and is position ismeasure to be x = x

x .

What can we say about the com-ponent px along x of the momen-

tum? Note that if the electrons had awell dened momentum, we wouldnot see a diffraction pattern but

a single line, and consequently

px = 0 .

If we assume that the spread of the diffraction pattern can be mea-sured by using the position of therst dark spot, we can say that:

px px = px 2 D p x .



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The Indetermination Principle

From the diffraction experiment we obtain that px x p .

The more precisely we try to localize the electron in space, by dening the

x position, the larger is the diffraction pattern, that is a larger indeterminationon the momentum.

A question is on order .. what is the wavelength associated with an electron?

From the empirical relation introduced by DeBroglie we have that = h/p ,where h is the Planks constant.

In conclusion more precisely we can say:

px x That is the Heisenbergs indetermination principle.



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Fundamentals of Quantum Mechanics (Sec. 5.1)

There are several possible way of introducing quantum me-chanics. The most immediate is based on postulates. These arefacts that we assume as true and we build the rest of theory on it.

The formulation of these postulates is based on the experimen-tal evidence, some of which has been presented.


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20 Postulate - 1

To each physical system, for example an ensemble of quantum particle likeelectrons or holes, can be associated a wavefunction or state function . This isa complex quantity and contains all the information concerning the ensemble.

The wavefunction ( r, t ) has the following properties. It is square inte-grable:

|( r, t )|2d r <

|( r, t )|2d r = 1 (3)

The square value of the wavefunction has the meaning of position probabilitydensity .

P( r, t ) = |( r, t )|2


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21 Momentum Space Wavefunctions Instead of using a description of the system based on the real space wave-function ( r, t ), one could used the momentum space wavefunction ( p, t).As before:

|( p, t)|2d p = 1and :( p, t) =

|( p, t)

|2

has the meaning of position probability density in momentum space . In otherwords the probability of nding the momentum particle in a volume d p around

p. The momentum space and real space wavefunctions are related by the

Fourier transform.

( p, t) = 1

(2 )3/ 2

e

1 ( p r) ( r, t ) d r


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22 Postulate - 2There exist a superposition principle , the dynamical states of two quantumsystems are linearly superposable. If 1( r, t ) is a wavefunction associated toone of the possible states of the system and 2( r, t ) is another one, then theirlinear combination:

3( r, t ) = c11( r, t ) + c22( r, t )

is itself another state describing the system. Where c1 and c1 are complexnumbers. The relative phase of the two wavefunction is important because of the interference. It is important to remember that only to |( r, t )|2 is associ-ated a physical meaning. If we consider

|3( r, t )

|2 we have:

|3( r, t )|2 = |c11( r, t )|2 + |c22( r, t )|2 +2 Re {c1 c2 |1( r, t )||2( r, t )|e( [ 1 2 ])}

Where 1 and 2 are the phases of 1( r, t ) and 2( r, t ). (See also Sec. 2.1)EC574 Physics of Semiconductor Materials, Fall 2008

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23 Postulate - 3With every dynamical variable is associated a linear operator. If A is a linearoperator associated to the observable A, then

A ( 1( r, t ) + 2( r, t )) = A 1( r, t ) + A 2( r, t )The quantization is performed by replacing the conjugate variables ri and pjwith the respective operators ri and pj (See also Sec. 3.3 and 3.9).

Table 1: Observables - Operator correspondence

observable operatorposition coordinate x x

position vector r rx component of vector momentum px xmomentum p

kinetic energy T = p2

2 m 2

2 m2

potential energy V ( r, t ) V ( r, t )

total energy H = p2

2 m + V ( r, t ) H = 2

2 m

2+ V ( r, t )


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24 Postulate - 4The only results of precise measurement of the dynamical variable Ais oneof the eigenvalues an of the operator A associated to A. Eigenvalues areobtained from the solution of the eigenvalue problem:

An ( r, t ) = an n ( r, t )

The functions n ( r, t ) are said to be the eigenfunction of the system and

represent a complete orthonormal set (CON). The set of all the eigenvaluesof an operator is called the spectrum of the operator. The spectrum can bediscrete, continuos or a mix of the two. The outcome of a measurement on aphysical observable must produces a real number. Consequently the operator

associated to dynamical variables are Hermitian, and their eigenvalues real oran =an . In other words we can say that the operator is Hermitian if

n ( r, t )An ( r, t )d r =

(An ( r, t ))n ( r, t )d r


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25 Hermitian OperatorsAn operator A that represent a physical observable Aneed to have real eigen-values. Since the eigenvalue spectrum is the collection of all the possiblemeasurement outcomes, the expectation values of a physical observable Amust be real. Consequently we can write:

< A> = ( r, t )A( r, t )d rand

< A>= ( r, t )(A( r, t ))d rand since for a physical observable< A> = < A>

then

( r, t )A( r, t )d r =

(A( r, t ))( r, t )d r


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26 Postulate - 5

The expectation values of a series of measurement on a dynamical variable Aon an ensemble of systems described by the state function ( r, t ) is given by

< A> = ( r, t )A( r, t )d r ( r, t )( r, t )d rIt should always been remembered that is not the average of a classicalstatistical distribution of the dynamical variable Abetween the systems beingmeasured. Each system is identical and in the same state described by statefunction ( r, t ).

Note that if the wave-function is normalized then we can write:

< A> = ( r, t )A( r, t )d rEC574 Physics of Semiconductor Materials, Fall 2008

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27 The Diracs NotationDirac introduced a very compact notation to describe the operation of quan-tum mechanical quantities. It used the so called symbols. In thiscase we will indicate:

< 1|2 > = 1( r)2( r)d rthe symbol < 1| is called the bra and |2 > the ket . From the denition wecan also write:

< 1|2 > = < 2|1 >We can also dene the operation of multiplication by a constant and the dis-tribution property:

< 1|c2 > = c < 1|2 >< c 1|2 > = c < 1|2 >

< 1|c2 + d3 > = c < 1

|2 > + d < 1

|3 >


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Two state functions are said to be orthogonal if their scalar product is zero,

< 1|2 > = 0and the normalization conditions becomes:

< 1|1 > = 1The eigenvalue problem for an operator A is written as:

A

|n > = an

|n >

and the expectation value of the dynamical variable Ais: =

< |A| >< | >The denition of Hermitian operator can be written as:

< |A| > = < |(A) > = < (A) | > = < |(A) >


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Postulate - 6A wavefunction ( r, t ) representing any dynamical state can be representedas a linear combination of the CON eigenfuctions set {n ( r, t )}of the oper-ator A, that is associated with the dynamical variable

A.

( r, t ) =n

cn n ( r, t )

The expansion coefcient cn can be found using the orthonormality relation

m ( r, t )( r, t )d r = n cn m ( r, t )n ( r, t )d r = n cn n,m = cmThe physical meaning of the expansion coefcient:

P n = |cn |2 = n ( r, t )( r, t )d r2

(2)

is the probability amplitude , i.e. the probability of nding at time t the system

in the state n ( r, t ) as a result of a measurement.EC574 Physics of Semiconductor Materials, Fall 2008

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Matrix Representation of operatorsConsider again the eigenvalue equation,

An ( r, t ) = an n ( r, t )

where A is an Hermitian operator, or given two state function and wehave, < |(A) > = < (A)| > . Then the eigenvalues an are real number.Let us know consider a CON set of wavefunctions {n }where n can be niteof innite. Accordingly to postulate 5 any wavefunction can be expanded interms of a orthonormal set, then ( r, t ) = n cn n ( r, t ). We can say thecn are a representation of ( r, t ) in the basis {n }. Consider now a linearand Hermitian operator A, and let A operate on ( r, t ) producing anotherwavefucntion (r, t ), then (r, t ) = A( r, t ). We can expand (r, t ) interms of {n }and we have:

( r, t ) =m

dm m ( r, t ) (3)


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and the expansion coefcients can be written as

dm = < m | > = < m |A| > = < m |A|n

cn n > (4)

follows that

dm =n

< m |A|n > cn =n

Am,n cn (5)

then the matrix elements Am,n transform the

{n

} of ( r, t ) into the one

of (r, t ). Since, the coefcients dm that uniquely dene (r, t ) are relatedthrough the matrix elements to cm that uniquely dene ( r, t ), then we cansay that the operator A is completely specied by the set of matrix elementsAm,n ,

Am,n = < m |A|n > (6)Clearly the matrix representation of an operator depends on the chosen basisset.


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Postulate - 7

The time evolution of the wavefunction describing a quantum system is deter-mined by the time dependent Schrdinger equation:

t

( r, t ) = H( r, t )

where H is the total hamiltonian of the system.

The hamiltonian H represent the physical observable total energy. In generalwe write:

t

( r, t ) = 2

2 m2 + V ( r, t ) ( r, t )

In its general form the potential energy operator V ( r, t ) will depend on time.For example in the case of an oscillating electromagnetic eld coupled to thesystem.


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The Schrdinger Equation

We have seen that the time evolution of the state of the quantum mechanicalsystem is described by the Schrdinger equation:

t

( r, t ) = 2

2 m2 + V ( r, t ) ( r, t )

This is in general a PDE, and a solution is a function of four coordinates, three

spatial and time. The hamiltonian H represent the physical observable totalenergy.

Consider rst the case where the potential energy operator only depends onthe spatial coordinates but not on time:

t

( r, t ) = 2

2 m2 + V ( r) ( r, t )

We can try to solve is by using separation of variables.


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We can write the solution as:

( r, t ) = ( r)(t)

Substituting we obtain:

( r) t

(t) = (t) 2

2 m2 + V ( r) ( r)

Dividing RHS and LHS by ( r)(t) we have:

1(t)

t

(t) = 1( r)

2

2 m2 + V ( r) ( r)

this means that the LHS is a function only of time and the RHS only of space.

Or:g(t) f ( r)

this means that the two sides have to be equal to a constant


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1

(t)

t(t) = E

1( r)

2

2 m2 + V ( r) ( r) = E

Now we haver that the solution of the rst equation:

t

(t) E (t) = 0is:

(t) = (t0) e E ( tt 0 )

and describe the time evolution of the state function.

The solution of the time independent Schrdinger equation provides with the

stationary states of the system, described by the eigenvalues spectrum En anda set of CON functions {n ( r)}.

Hn ( r) = E n n ( r)


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Or:

2

2 m

2

( r) + ( V ( r) E ) ( r) = 0Accordingly to postulate 4 every time we perform a measurement of energy ona quantum mechanical a system described by the state function ( r) the resultwe obtain is a value among the spectrum E

n and because we have measured

it the system has collapsed on one of the states of the CON set {n ( r)}.Example I - Free Particle

Consider a free particle (V ( r) = 0) moving in one dimension:

2

2 m 2

x2 (x) E (x) = 0

the solution is:

(x) = A e k x + B e k x (7)where:

k =

2 m E

2


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Let us assume that because of the boundary conditions in (7) A = 1 andB = 0 , then

(x) = e k x (8)

This means that if an electron (or a quantum mechanical particle) in not sub- jected to an external force, then it is described as a plane wave. Note that the

eigenvalue spectrum is a continuum as if it would be for a classical particle.

Furthermore we have |(x)|2 = 1 x and this means the probability to ndthe particle is 1 in all the points of the space. This means the uncertainty on

the position is innite.What are the possible values of the momentum? To answer the question weneed to solve the eigenvalue problem:

P x (x) = px (x) x

(x) = px (x)

or px = k


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Note that the time dependent solution for this problem is readily written as:

(x, t ) = e [ px x E t ] (9)

Consider again equation (8), is this wavefunction normalized? How wouldyou compute the normalization coefcient?

Example II - Wavepacket

The wavefunction in equation (8) describe a particle that has denite value of momentum and energy but is completely delocalized in space.

To describe a particle partially localized in space we can use a superpositionof free particle states.

(x, t ) = 1 2 e

[ px x E t ] ( px ) dpx (10)


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The (envelope) function ( px ) is in general complex (magnitude and phase)

and give the amplitude of the component with momentum px .

The center of the wavepacket is determined by:

d[ px x E t ]dpx px = p0

= 0 (11)

And moves with the group velocity:

vg =dE ( px )

dpx px = p0(12)


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Conservation of the Probability Density (Sec. 3.2)From classical electrodynamics we have a continuity equation for charge:written as:

J + t

= 0

What can we say about the probability density (or charge density distribu-tion)?

P( r, t ) = ( r, t )( r, t ) = |( r, t )|2

Let us start from the Schrdinger and multiply the left and right terms by( r, t ):

2

2 m ( r, t )

2( r, t ) + ( r, t ) V ( r, t ) ( r, t ) = ( r, t )

t ( r, t )

take the complex conjugate

2

2 m ( r, t )

2 ( r, t )+( r, t ) V ( r, t ) ( r, t ) =

( r, t )

t ( r, t )


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and subtract equation the second equation from the rst:

2

2 m( r, t )

2 ( r, t ) ( r, t )2 ( r, t ) =

( r, t ) t ( r, t ) + ( r, t )

t ( r, t )

since:

( r, t ) t ( r, t ) + ( r, t )

t ( r, t ) =

t (( r, t ) ( r, t ))

we have:

2

2 m( r, t )

2 ( r, t )

( r, t )

2 ( r, t ) +

t

(( r, t ) ( r, t )) = 0


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Using the vector identity: ( A) = A + A

with A = , we have:( r, t )

2 ( r, t )

( r, t )

2 ( r, t ) =

[ ( r, t )( r, t ) ( r, t )( r, t ) ]consequently we can write:

t (( r, t ) ( r, t )) +

2 m [ ( r, t )( r, t ) ( r, t )( r, t ) ] = 0Notice that if we dene the probability current density as:

J =

2 m [ ( r, t )

( r, t )

( r, t )

( r, t ) ]

we obtain the continuity equation for the probability density:

t

P( r, t ) + J = 0


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Operators, Commutation Relations and Uncertainty

Consider the operator x px , and compute the expectation value of the observ-able product of position and momentum < x p x > (page 93, Sec. 3.3)

< x p x > = +

(x, t ) x x (x, t )dx

Integrating by parts we have:

< x p x > = [ x (x, t )( x, t )]+

+ +

(x, t ) x (x (x, t ))d x

since |(x, t )| 0 as x , then:

< x p x > = +

(x, t ) x

x (x, t )dx +

+

(x, t )( x, t )dx

or:

< x p x > = < x p x > +


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The operator x px is evidently not hermitian since its expectation value is notreal. Similarly we can nd that:

 = 

Consequently the order in which the operators are applied counts and x px hasdifferent expectation values than px x, or as it is said they do not commute .

Given two operators A and B, the commutator of A and B is dened as:

[A, B ] = AB BAif

[A, B ] = AB

BA = 0

AB = BA

then the two operators are said to commute.

In general:

[x, px ] = [y, py ] = [z, pz ] =


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Commutating Operators and EigenfunctionsLet us assume that Aand Bare two observables represented by the operatorsA and B, and assume there exist a complete set {n }of eigenfunctions thatis simultaneously eigenfunction of A and B. Then A and B are said to becompatible , and their associated eigenvalues an and bn are given by:

A n = an n B n = bn n

Also we have:

A B n = A bn n = bn A n = bn an n = B A n

Consequently if we have a state function (x, t ) this can be expanded usingthe basis set {n }, then:

(A B B A) ( x, t ) = n cn (A B B A) n = 0and:

(A B

B A) = [A, B ] = 0


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the two operators commute. The converse is also true: two operators thatcommute have a set of CON eigenfunctions {n }in common.

Commutator AlgebraWe also have the simple rules that are easy to prove:

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

[A, [B, C ]] + [B, [C, A]] + [C, [A, B ]] = 0

These are useful to manipulate sequences of operators.


47

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More on the Uncertainty RelationsConsider two dynamical variables A and B and < A > = < |A| > , = < |B | > the respective expectation values for a given statefunction . We dene the uncertainty A on the observable Aand Bas:

A = < (A< A > )2

>, B = < (B )2

>it can be shown (homework) that:

A B

1

2 |< [A, B ] >

|Two non-commuting observables that are canonically conjugated then wehave [A, B ] = i and:

A B

2As we have seen, for x and px we have [x, px ] = i , consequently:

x px

2


48

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Connection to Classical Mechanics:The Ehrenfest Theorems (Sec. 3.4)

This theorem states that Newtons classical dynamics equations:

d r

dt =

p

m (13)

andd pdt

= V (14)are exactly satised by the expectation (average) values of the correspondingoperators in quantum mechanics.

Note that in quantum mechanics we can not talk of trajectory in the classical

sense and consequently we can not compute dr/ d t or d p/ d t but rather wecan compute the rate of change of the expectation values of position < r >and momentum .

If we start by computing the time derivative of the expectation values of theEC574 Physics of Semiconductor Materials, Fall 2008

49

i i di h

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position coordinate x, we have:ddt < x > = dd t ( r, t )x( r, t )d rafter some straightforward manipulation of the right hand side we can write:

ddt < x > =

m ( r, t ) ( r, t )

x dror

ddt

< x > = 1m

( r, t )

x

( r, t )d r = 1m



which is the equivalent of equation 13:

ddt

< x > = 1m

 dr

dt =

pm

(15)

In the same way by considering the time derivative of the momentum (forexample x component of P ) we have:

d

dt =

d

dt ( r, t )

x( r, t )d r


50

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With some straightforward manipulation of the right hand side we can write:

ddt

< px > = ( r, t ) V ( r, t ) x ( r, t )d r = V ( r, t ) xthat is analogous to the classical equation 14.

ddt

< px > = V ( r, t )

x d p

dt = V (16)


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51 Solvable Problems in Quantum Mechanics


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Solvable Problems in Quantum MechanicsThe solution of the the Schrdinger equation in a closed form it is possibleonly for a limited set of problems. In particular this depends on the form of the Hamiltonian operator. In general we can write:

t

( r, t ) = H( r, t ) = [ 2

2m2 + V( r)]( r, t )

if the potential energy does not depend on the time coordinate time, the so-lution of equation 17 can be separated in a time dependent part and a spatialdependent part. The solution of the spatial dependent part provides the sta-tionary states of the system. The situations where a solution can be found ina closed form are:

Free particlePotential stepPotential barrier

Innite potential square well

Square wellHarmonic oscillatorPeriodic Potential

Central Potential, Hydrogen Atom.


52 Innite Potential Well (Sec. 4.5)


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( )

V (x) = 0 |x| a |x| a

Consequently:

(x) = 0 for

|x

| > a

2

2 m 2

x2 (x) E (x) = 0

the solution is:

(x) = A e k x + B e k x

also:

(x) = A cos(k x) + B sin (k x)

where:

k = 2 m E

2

the boundary condition is:

(x) = 0 f or x =

a


53


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Acos(k x) = 0

and

B sin (k x) = 0

then

kn = n 2 a

= n

Lwhere L = 2 a . Then the solutions

are:

n (x) = An cos(kn x) for

n = 1 , 3, 5, 7,...

m (x) = Bm sin (km x) form = 2 , 4, 6, 8,...

using the normalization condition:

n (x)n (x) dx = 1

The nal wavefunctions are:

n (x) = 1 a cos(n 2 a x) for

n = 1 , 3, 5, 7,...m (x) = 1 a sin (

n 2 a x) for

m = 2 , 4, 6, 8,...

Consequently the possible energythat the particle can assume in thewell are quantized .

En = 2

2

n2

8 m a 2 =

2

2n

2

2 m L 2

Note that there is only one eigen-function for each energy level.


54


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Innite Potential Well Wavefunctions


55


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Note:

There is a non-zero minimum energy En = 2 2 / (8 m a 2), this energyincreases as the well becomes narrower (reaction to connement).

The zero point energy is a direct consequence of the indetermination prin-ciple. In fact the uncertainty on the position is x = a consequently theuncertainty on the momentum is

px = /a and a minimum energy of the

order 2 / (m a 2).

The nth eigenfunction has (n-1) nodes. The ground state never has a node. Wavefunctions corresponding to two different energy eigenvalues E n andEm are orthogonal:

n (x)m (x) dx = 0 n = mEC574 Physics of Semiconductor Materials, Fall 2008

56 Parity


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ParityNote that for the solution of the previous example all the wavefunctions havea dened parity, they are either:

n (

x) = n (x) even parity

m (x) = m (x) odd parityThe wavefunctions posses a denite parity since the potential is symmetric

respect the origin x = 0 : V (x) = V (x). The Schedinger equation doesnot change if x x .We can also dene a parity operator such that the observable parity P repre-sented by the operator P:

P n (x) = p n (x)

when applied to the eigenfunction give the eigenvalues p = 1.EC574 Physics of Semiconductor Materials, Fall 2008

57 Finite Square Well (Sec. 4.6)


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V (x) = V0 |x| a0 |x| > aNote that now the wavefunction for

|x| a will no longer be 0.

We need to use a suitable set of con-tinuity condition for the wavefunc-tion at the interface of the differentregions. There are two region inter-

faces: x = a and x = a , thecontinuity boundary conditions be-tween two regions I and II are:

I (x0) = II (x0)1

m I I (x )

dx x = x 0 = 1m II

II (x )dx x = x 0

The second boundary condition is

equivalent to use the continuity of the probability current density at theinterface:

J I = J II EC574 Physics of Semiconductor Materials, Fall 2008

58 We consider two different situations for the solution of this problem:

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a) V 0 < E < 0 and b) E 0Case a) For region IId2

dx2 (x) +

2 (x) = 0 =

2 m

2 (V0

|E

|)

and for regions I and III

d2

dx2 (x)

2 (x) = 0 =

2 m

2

|E

|Since the potential is symmetric respect the origin then the wavefunctionsneed to have a denite parity. Then we can consider the solution only forx > 0. The even solutions are given by:

(x) = Acos(x ) 0 < x < a

(x) = C ex x a


59 Requiring that (x) and its rst derivative be continuous at the interface lo-

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cated at x = a, we have two transcendental equations:Acos(a ) = C ea Asin (a ) = C ea

or

tan (a ) = (17)

The odd solutions are given by:

(x) = B sin (x ) 0 < x < a

(x) = C ex x aRequiring that (x) and its rst derivative be continuous at the interface lo-cated at

x = a, we have the transcendental equation:

cot (a ) = (18)The energy level are computed solving equations 17 and 18.


60 Setting = a and = a we have

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tan = cot = (19)Since and have to be positive and:

2 + 2 = 2 = 2 m V 0 a

2

2

We can nd graphically the solutions by nding the intersection between thecircles dened by 2 + 2 = 2 of radius and 19 for even and odd wave-functions.

Note:

The bound states are non-degenerate and their number is nite.

Even and odd states alternates in the energy scale.

The gure in the next slide shows the graphical solutions of the problem. Thesame equations can be easily solved numerically.


61 G hi l S l i f V < E < 0

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Graphical Solutions for

V 0 < E < 0


62 Finite Square Well - wavefunctions


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q


63 Case b) E 0

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For region I and III the solutions for the external regions x < a and x > aare given by:

(x) = A e k x + B e k x

C e k x

Where as usual k = 2 m E 2 .For region II the solutions is written as:

(x) = F e x + G e x

where =

2 m (V 0 + E )

2

The ve constants A,B,C,F,G are computed by imposing the continuity of thewavefunction and its derivative at x = a and x = a.Note that the spectrum of energies is continuous.


64 Solving for A,B, and C we can dene a Transmission Coefcient T and aReection Coefcient R:


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Reection Coefcient R:

R = |B |2|A|2

T = |C |2|A|2

and clearly R+T=1.

R = 1 + 4E (V 0 + E )V 20 sin 2( 2a)

1T = 1 +

V 20 sin 2( 2a)4 E (V 0 + E )

1

and clearly R+T=1.


65


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Note - I: The transmission coefcient is less than unity, as opposed towhat happens in the case of a classical particle.

Note - II: The transmission peaks for which T = 1 occur at L = n.The minima are reached when L = (2 n 1)/ 2. Where as dened =

2mE/ 2 .

Note - III: Perfect transmission occurs when the dimension of the well L isequal to a integral of half-integral number of de Broglie electron wavelengths2/ (Resonance).

Note - IV: If E V then T becomes asimptotically equal to unity.


66 Potential Step (Sec. 4.3)

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In this case we have again two distinct cases a) V 0 < E < 0 and b) E V 0Case a) We have 0 < E < V

0 then

d 2dx 2 (x) + k

2 (x) = 0 k = 2 m2 E x < 0d 2dx 2 (x) +

2 (x) = 0 = 2 m2 (V0 E) x > 0


67 with solutions

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(x) = A e k x + B e k x x < 0(x) = D e x x > 0

By using the boundary conditions on the wavefunction and its derivative we

obtain:(x) = 2 A e / 2 cos (k x / 2) x < 0(x) = 2 A e / 2 cos(/ 2) e x x > 0

where:

= 2 tan 1 V 0E 1

1/ 2

As in the case of the nite square well we can dene a reection coefcientfor the potential step.

R = |B |2|A|2


68


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Potential Step - Wavefunctions


69 Notice that the probability density on the left of the barrier:


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P( x) = 4 |A|2 cos2 (k x / 2)has an oscillatory behavior due to the interference of incident and reectedwaves.

Under the barrier it becomes:

P( x) = |D |2 e2 x

Note! There is a nite probability of nding the particle under the barrier, thisis clearly impossible from the classical standpoint. Why do we have this?

x

1

px

x

=

2m(V 0

E )

E =

2 px2m (V 0 E )

The particle energy can no longer be said to be lower than V 0


70 Case b) We have E > V 0 thend 2 2 2 m

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dx 2 (x) + k (x) = 0 k = 2 E x < 0

d 2dx 2 (x) + k2 (x) = 0 k = 2 m2 (E V0) x > 0

with solutions

(x) = A e k x + B e k x x < 0(x) = C e k

x + D e k x x > 0

since there is no reected wave once the particle has move to the region x > 0

then D=0 As before we can dene a transmission and reection coefcients:

R = |B |2|A|2

= (k k)2(k + k)2

=1 1 V 0 /E 21 +

1

V 0 /E

2

T = |C |2|A|2

= 4kk (k + k)2

= 4 1 V 0 /E 1 + 1

V 0 /E

2


71 Potential Barrier (Sec. 4.4)

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V (x) =

0 x < 0

V0 0 x a0 x > a

We are going to have as usual three solu-

tions, on the left of the barrier, under thebarrier and on the right of the barrier, andwe will consider the case in which E > V0or E

V0 for

(x) =A e k x + B e k x x < 0C e k x x > a

As before we will dene:

R = |B |2|A|2

T = |C |2|A|2


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73 Clearly there is a nite probability for the particle to tunnel under the barrier


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and appear on the other side. This effect is extremely important for a lot of practical applications in electronic devices, for example ohmic contact.

For E > V0

then the solution for 0

x

a is:

(x) = F e k x + G e k

x

and as before we have can compute the relations between the coefcients and

determine the transmission and reection coefcients:

R = |B |2|A|2

= 1 + 4E (E V 0)V 20 sin 2(k a)

1

and

T = |C |2|A|2

= 1 + V 20 sin 2(k a)4 E (E V 0)

1


74


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Note: The transmission peaks for which T = 1 occur at ka = n.The minima are reached when L = (2 n 1)/ 2. Where as dened k =

2m(E V 0)/ 2 . Note - III: Perfect transmission occurs when the barrier thickness a isequal to a integral of half-integral number of de Broglie electron wavelengths

2/k (Resonance).


75 The Linear Harmonic Oscillator (Sec. 4.7)


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The potential in this case has the form:

V (x) = 12

k x2

The potential is character-ized by the spring constant kin analogy with the classicalharmonic oscillator. We de-ne:

0 = km

and the quantity 0 is theequivalent harmonic oscilla-

tor energy. We also dene:

= m 0 EC574 Physics of Semiconductor Materials, Fall 2008

76 The Schrdinger equation becomes: 2 2 1


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2 m x2 (x) + 2 k x2 (x) = E (x)This type of equation is not easily solved. Many techniques can be used, theusual one it to nd a solution by series expansion. In this case the eigenvalues

that one obtains are:

E n = n + 12

0 n = 0 , 1, 2, 3, . . .

The wavefunctions are given by:

n (x) = N n e2 x 2 / 2 H n ( x)

where N n is a normalization constant and and H n ( x ) is the Hermite poly-

nomial of order n. These polynomial are a CON set of functions, denedas:

H n ( ) = ( 1)n e2 dn e

2

d n



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78

The normalization constant is obtained by requiring that the square of theabsolute value of the wavefunction be 1. It is:


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N n = 2n n!Consequently:

n (x) = 2n n! e 2 x 2 / 2 H n ( x )Example: H-O MoleculeConsider the H-O molecule. The smaller Hydrogen atom is bounded to alarger Oxygen atom. The force between them can be expressed as:

F = U ( r)Where U ( r) is the potential energy. Since the molecule is stable U ( r) has aminimum corresponding to the bond length r0 .


79

Around r0 the potential U ( r) can be expanded in Taylors series:( ) ( 0) ( 0) ( ) 1( 0)2

2

( )


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U (r ) = U (r 0) + ( r r 0) U (r )r r = r 0 + 12(r r 0)2 U (r )r 2 r = r 0 + ...

Since r = r0 is a minimum then

U (r )r r = r 0= 0

then around the equilibrium positionthe potential can be represented by (as-suming U (r 0) = 0 )

U (r ) = 12

(r r 0)2 2U (r )

r 2 r = r 0

That is an harmonic oscillator with

k = 2U (r )

r 2 r = r 0


80

The Linear Harmonic Oscillator Revisited(M h R i ODE)


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(Math Review: ODE)In this section we look in detail at the solution of the equation:

2

2 m

2

x2 (x) +

1

2 k x2 (x) = E (x)

or:

T (x) + V (x) = 2

2 m 2

x2 (x) +

12

m20 x2 (x) = E (x)

that can be rewritten as:

d2

d x2 (x)

m 20 2

x2 (x) + 2mE

2 (x) = 0 (20)

we dene:

= m0 and = xEC574 Physics of Semiconductor Materials, Fall 2008

81

then

d d d d (21)


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ddx dd d dx = dd (21)and

d

dx2 =

d

dx

d

dx =

d

d

d

d

d

dx = 2

d2

d 2 (22)

Substituting (21) and (22) in (20) we obtain:

dd 2

2 + ( ) = 0 (23)

and:

= 2E

0

Equation (23) is the Webers equation, that has a well known solution. We willderive the solution rst studying the asymptotic behavior of (23) and solvingit by using a power series expansion.


82

Solution of the Webers EquationLet us consider: d


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Let us consider: dd 2

2 + ( ) = 0 (24)

we want to determine a solution. The rst thing it is normally done in the

presence of a unknown ODE is to study the asymptotic behavior.

Asymptotic behavior: In this case we have that 2 and we can write:

dd 2

2 ( ) = 0 (25)

And we have also assumed that as then ( ) ( ). The solutionof equation (25) has the form:

( ) = A e2 / 2 + B e

2 / 2 (26)

We set B = 0 since (x) has to be bounded as

.


83

Consequently we have that: ( ) = A e2 / 2 , and the solution of (24) is:( ) = A H ( ) e2 / 2 .


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We need now to factor out the asymptotic behavior and determine an ODEfor H ( ). We can immediately see that:

dd ( ) = A

dH ( )d e

2 / 2

A H ( ) e2 / 2

(27)

and:

d2

d 2 ( ) = A d2H ( )

d 2 e2 / 2

2A dH ( )

d e2 / 2

A H ( ) e

2 / 2 + A 2H ( ) e2 / 2

(28)

Substituting (27) and (28) in (23) we have:

d2H ( )d 2 2

dH ( )d

+ (1)H ( ) = 0 (29)That is the Hermite DE.



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85

( j + 2)( j + 1) cj +2 j

2

jc j j

+ ( 1)

cj j

= 0 (34)


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j =0 ( j + 2)( j + 1) cj 2 2 j =1 jc j + ( 1) j =0 cj = 0 (34)

and, collecting j :

c2 + (1)c0 + j =0

[( j + 2)( j + 1) cj +2 2 jc j + (1)cj ] j = 0(35)

For this expression to be zero it is needed that:

c2 + (1)c0 = 0 (36)and

( j + 2)( j + 1) cj +2 2 jc j + (1)cj = 0 (37)or:c2 = (1 )c0 j = 0 (38)


86

andcj +2 =

2 j + (1 )( j + 2)( j + 1) cj j = 1 2 3 (39)


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cj = ( j + 2)( j + 1) cj j = 1 , 2, 3,... (39)nally:

cj +2

cj=

2i + (1 )( j + 2)( j + 1)

(40)

Note - I: Successive coefcients cj +2 and cj are related by the recurrencerelation (40).

Note - II: The energy appears in the recurrence relation (40). Note - III: Does the series converge? Using the dAlembert ratio test we

have:

cj +2 j +1

cj j = 2 j + (1 )( j + 2)( j + 1) 2 j 2

2

j 0 (41)Consequently the series converges.


87

Note - IV: What does the series converge to? To begin with we canassume c0 and c1 arbitrarily, and use the recurrence relation to determine the

other coefcients. In this case we can write two expressions for H (), one for


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other coefcients. In this case we can write two expressions for H ( ), one foreven:

H e ( ) = c0 1 + c2c0

2 + c4c2

c2c0

4 + .... (42)

and one for odd functions:

H o( ) = c1 1 + c3c1

+ c5c3

c3c1

5 + .... (43)

Remembering that

ex =

n =0

xn

n! ex

2=

n =0

x2n

n! (44)

we have that as | | , then H ( ) e2. Consequently:

( ) = A H ( ) e2 / 2 = A e

2e

2 / 2 = A e2 / 2 (45)


88

Consequently, as | | , ( ) diverges and can not be normalized and as aresult it is physically unacceptable. To obtain a viable solution we can enforceall the coefcients above a certain order let us say j to be zero


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all the coefcients above a certain order, let us say jmax to be zero.

If cj = 0 f o r j > j max then cj max +2 = 0 , and using the recurrencerelation we have: = 1 + 2 jmax

Note - V: We have force the Hermite Series to be nite, we have HermitePolynamials :

H ( ) =j max

j =0

cj j = 2 jmax

1 (46)

Note - VI: Chosen a jmax we have:

jmax = 0 = 1 H ( ) = c0 jmax = 1 = 3 H ( ) = c1


89

and jmax = 2

= 5 H ( ) = c0 + c1 2

j = 3 = 7 H () = c1 + c33


99/175

jmax = 3 = 7 H ( ) = c1 + c3 Furthermore

c2 = 1

2 c0 = 1

5

2 c0 = 2c0c2 =

2 + (1 )2x3

c1 = 2 + 1 7

6 c1 =

23

c1

More in general if we use n to indicate the order of the Hermite polynomialwe have:

n = 2n + 1 = E n

0or:

E n = n + 12 0

Note - VII: The requirement of square integrability (or boundary conditionsif you want) has led to the energy quantization.


90

More on the Linear Harmonic OscillatorClassical Turning Points In correspondence to the classical turning points


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g p g pwe have: E n = V (xn ), that is:n +

1

2

0 = 1

2m20 x

2n

x2n = (2n + 1) 0

m20

xn = 2n + 1 2 Probability Density P n (x) a) The claossical oscillator does not penetratein the forbidden regions. b) The QM oscillator can be found in the found in theclassical forbidden regions. c) The QM oscillator is more likely to be foundat x = 0 , d) As n increases the probability density approached the classicalone (correct classical limit or Principle of Correspondence)



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fOQ_ I Po t ~ t o . P ~l C(.l. S I) J t7 o ~ I 0 oiL... :rf--l~ r y C:YP ~ @ N ~f - i t S } 1 - I i l ~

111/175

{

,}

~ t c ~ = o\ ) ( \ ~ ~e a c ~ J:; oX 1 ? < 9 0

f a_+ ~ )~ = ) tt< Cc:wrn(+-hi. + w ~ - t ) ) ~c. 9 0 ~

Cl-t- := L _ + t f + k V X

-\Q:~ t ~ ~t ~ d

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-{a ct

=~ C t

t~ a -c

tW

@ -----------------{ ..fcc

c t - * c ~ : ; - l } ) ~j c < t ~ ~ ~ } d x


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X X == fUt;;; ~ + a )a (bi 0- q JJ

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X X == fUt;;; + a . _ )a (bi 0 q_JJ11 (q+ a ) a..++ ~ -zwrn

/Jere fk : * < r n . c - s S / o 0 S t : )F r )< { XL A:fV 1-. c--11 ..v e>r ~ 4 = A t t us c. r u 1'"'-='

6.:lrt e l T ~

I

tiw [ a , ~ +:r.fj = e tfn d=- Hlf. -= Cny.,

En== 1ftv t1~ )

@

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i \ ~~ +{_ ~ .. - ~ l J ~ ~ V \= 1fwH +ft vv)~ ~

Y ~ ~ -~ . ~ . , = Y fn }

etra._~ V t= ~ Y 1

a_ a + ~ . , =

~ a _ u ~ ~- ~ ~ ~ Y 1n f ~ ,~ ~ ~ ,

1 J o f 2 f t ~t ~ Iev ~ 6 FP t e t t.JJ l:S ( \ f \ Z I @j \ ~ t == i.

~ +~ ' l LR < h . VJore \ j f ~ ~ d . =iS l ~ J ~ ~~ -

116/175

et ~ = d. ~ y , - t

. ~ F O I H

l1 C~ ~ V l ~ O V

cPtPF tc . - t t l JU

_1

+oo

J ~ )'*l A r ~n dAc= t4,f ) \ Vl iteJ M>;r l {90too CO .:> s- tov\ L a . - ~ + ~ ~ ~ -qtl < x = l ~ c j LR +'f2~

- . ) \ 0 0

:l Qo

{fttd ~ / Q ' ~ ~2


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2 m( ) ( ) ( ) ( )

Depending on the particular form of V ( r) we will be able to solve Eq.(47) ina closed form. We will consider two cases that are quite important. The rstis the quantum box, which is the simple example of a structure where all thecoordinate of motion are quantized. In this case we have:

V ( r) = V x (x) + V y (y) + V z (z)

The second case is the central potential, which represents the starting pointfor the study of atoms.

V ( r) = V r (|r|) V , ( , )and where V ( ) = 1 . In both cases we are able to solve Eq.(47) by separa-tion of variables.


94

Quantum BoxConsider a box with sides of lengtha b and c Lets assume the poten


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a,b and c. Let s assume the poten-tial V (x,y,z ) = 0 inside the box

and V (x,y,z ) = outside thebox. Consequently the wavefunc-tion (x,y,z ) = 0 outside the box.

In this case we can solve by separation and we write:

n ( r) = X (x) Y (y) Z (z)

It is easy to show that we can write:

2

2 md2

dx2 + V x (x) X (x) = E n x X (x)



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96

If we consider a cubic box, a=b=c than we have :E n =

2 2

2 m a 2n2

1 + n2

2 + n2

3

different combinations of the values of n1 , n 2 , n 3 give the same energy. No-


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tice that for each of these combinations the wavefunctions will be different.We refer to these wavefunctions as degenerate and their level of degeneracy.


97


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Additional topics to be considered for a good performance in the exams:

- You should solve the problem for the free particle in 3D. Once you solvethis problem consider the issue of the normalization of the wavefunction youobtain.

- Solve the problem for the 3D harmonic oscillator. Determine the degeneracyfor the case of the isotropic and anisotropic system.

Note - The solutions are in Sec. 7.1 of the textbook.



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100

of dimensionality d as:D (E ) = 1Ld N E

D(E) i th b f t t t


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Consequently for a particle in a verylarge box, in the limit case innitelylarge, that is a free particle, the densityof states is proportional to the squareroot of the energy.More generally we can dene the thedensity of states for a system

or D(E) is the number of states atthe energy E in the volume Ld . Itcan also be written in terms of k as:

D (E ) = 1Ld

N E

= 1Ld

N k

kE

and N k

= 2 Ld

(2)d

Using the previous expression the

density of states of systems of dif-ferent dimensionality can be com-puted.


101

Angular Momentum and Central Potential Problems (Sec. 6.1)We consider in this situation the case in which the potential depends only on

the radial coordinate and does not have any angular dependence. We haveV ( r) = V (r ). Note that in this case the force is a central force , i.e. directed


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along the radius, since F = V .Note that in this situation the net torque on the system is zero. In fact, sincethe torque is given by: = F r and the momentum arm and the force arecollinear, as a consequence = 0 .

Furthermore if no external torque is applied then the angular momentum is

conserved. In classical mechanics the use of conservation law, allows to writethe equation of motion of a system.

A theorem exits that states that quantities that the conservation principles thatcome from particular symmetries are the same in classical and quantum me-chanics.

This means that the observable angular momentum is conserved in a quantummechanical problem involving a central potential.


102

Angular Momentum Operator(s) (Sec. 6.1)In classical mechanics the angular momentum is given by:

L = r p =

i j k

x y z


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L = r p = x y z px py pz

For the principle of correspondence, the quantum mechanical operator for themomentum is given by:

L = r = i j k

x y z

x

y

z

it is easy to show that

L = Lx i + Ly j + Lz kEC574 Physics of Semiconductor Materials, Fall 2008

103

whereLx =

y z

z y

Ly = z x x zL = x y

with the operator that associated to

the magnitude of the angular momen-tum.

[L L ] L


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Lz = x y y xor

Lx = ypz zpyLy = zpx xpzLz = xpy

ypx

and in vector form

L = ( r )It is quite straightforward to show thatthe operators that represent the an-gular momentum components do notcommute, but each one commutes

[Lx , Ly ] = Lz

[Ly , L z ] = Lx

[Lz , Lx ] = Ly

Notice that the operator associated to

the magnitude of the angular momen-tum is:

L2 = L2x + L2y + L

2z

and

[L2 , Lx ] = 0 [L2 , L] = 0


104

Note that:

- Only one component of the angular momentum can be measured with inn-ity precision.


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- In addition to one component the magnitude of the angular momentum can

be measured.- The eigenfunction of an operator representing the one of the angular mo-mentum components are not eigenfunctions of the others.

- Note that in the case of a central potential, as we discussed the angularmomentum is a constant of motion. If the operator L commutes with theHamiltonian then the angular momentum and the total energy can be knownat all the times.

- In this case the eigenvalues of the two quantities are said to be good quantumnumbers .


105

Angular Momentum Operator in Polar Coordinates (Sec. 6.3)Since we want to study problems involving a central potential, is is moreconvenient to use polar coordinates {r,, }. This are related to the cartesiancoordinates as follows:

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coordinates as follows:

x = r sin cos y = r sin sin z = r coswith the limits:

0 r < 0 0 2it is simple to show that

Lx = sin cot cos

Ly = cos cot sin


106

andLz =

nally the operator for L2 in polar coordinates is:

1 1 2


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L2 = 2 1sin

sin

+ 1sin 2

2

Note that L x , Ly , Lz and L2 do not operate on r .

Eigenfunctions and eigenvalues of L z

We need to solve the eigenvalue equation:

Lz m () = m m ()

m () = m m ()

that has solutions:

m () = 12 em m (0) = m (2) m = 0 , 1, 2,...EC574 Physics of Semiconductor Materials, Fall 2008

107

Schdinger Equation in Polar Coordinates (Sec. 6.3 and 7.2)In order to write the the Schdinger equation in polar coordinates for a centralpotential, we have rst:

H 2 + V( )


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H = 2 m p2 + V (r )

The momentum operator is then given by:

p = = r r

+ 1

r sin

+

1r

and the unit vectors r ,, are given by:

r = sin cos i + sin sin j + cos k

=

sin i + cos j

= cos cos i + cos sin j sin kThe spherical representation of L is found as:


108

L = r p = + 1sin and in from the product L L we nd:

L2 = L L = 2 1i

sin

+ 1i 2

2

2


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sin sin 2 2As seen before L

2commutes with any function or derivative of r. Conse-

quently L2 commutes with V(r) in the central potential case. Since L ia alsoindependent of r then it commutes with V(r) as well.

To build the hamiltonian H we need to compute p2 , now we note that:

L2 = ( r p) ( r p) = r2 p2 r ( r p) p + 2 r pand

r

p =

r

rconsequently

p2 = L2

r 2 2

r 2

rr 2

r



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110

Since L2 depends only on and we can think of nding a solution bywriting:(r,, ) = R(r ) Y (, )

and Y (, ) is the solution of the eigenvalue equation for L2 . Notice also that:

2 2 2


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L2 (r,, ) = L2 R(r ) Y (, ) = R(r ) L2 Y (, ) = ..

R(r ) 2 Y (, ) = 2

the we can substitute to L2 the term 2 .

To solve We are going to solve Schdinger equation in polar coordinates we

need then to solve:

2

2 m r 2 2

2 m r 2

rr 2

r

+ V (r ) = E

We proceed by using separation of variables and we write (r,, ) = ..= R(r ) Y (, ). Using the ususal procedure we obtain two equations: onefor the radial part of the and one for the angular part of the wavefunction.


111

Solution of the Schdinger Equation by Separation (Sec. 7.2)Using the usual procedure we obtain:

2

2 m r 2

rr 2

r

+ 2

2 m r 2 + V (r ) R = E R


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2 m r r r 2 m r

for the radial part and:

2 1sin

sin

+ 1sin 2

2

2 Y = 2 Y

This second equation can be further separated by using:Y (, ) = ( ) ()

it is easily shown that for the equation is:

d2

d2() + m2() = 0


112

which results inm () = C ei m

Where C =1 for normalization and since m () has to be single-valued then

ei m = ei m ( + 2 )


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the follows:m = 0, 1, 2, ....

As it can be seen easily these are eigenfunctions of Lz as well.

Lz m () = ei m + 2 = m ei m + 2 = m m ()

The solutions for the second equation with unknown () is more complex.

sin 2

1sin

dd

sindd

+ = m2


113

We consider two separate cases: when m=0 and m=0 .If m=0 the equation becomes:

1sin

dd

sindd

+ = 0


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sin d d

This is the Legendre equation and has solutions only if: = l(l + 1)

with l = 0 , 1, 2, .... Consequently

L2 () = 2 () = l(l + 1) 2 ()

Consequently the the functions () are eigenfunctions of L2 with eigenvaluel(l + 1) 2 . The solutions of the Legendre equation is once again a set of com-

plete ortho-normal function called the Legendre Polynomials. If we considerthe change of variables:

= cos d = sin d F ( ) = ( )EC574 Physics of Semiconductor Materials, Fall 2008

114

the Legendre equation can be written as:d

d 1

2

dF

d + F = 0

the solution of the Legendre equation can be written as:


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P l( ) =

1

2l l!

dl

d l( 2

1)l

and the polynomials are:

P 0( ) = 1 P 0(cos) = 1

P 1( ) = P 1(cos) = cos

P 2( ) = 12 (3 2 1) P 2(cos) = 12 (3cos2 1)

the legendre polynomials are a CON set of functions:

0 P l (cos)P q(cos) sin d = 0 l = q EC574 Physics of Semiconductor Materials, Fall 2008

115

the normalization coefcient is found as:

0 |P l (cos)

|2sin d =

2

2l + 1

If m=0 the solution depends both on l and m, in this case the general solutioni f d i f h i d d l i l h i


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is found in terms of the associated Legendre polynomials. They are written

as:

P ml ( ) = (1 2)m/ 2 dm P l ( )

d m =

12l l!

(1 2)m/ 2 dl+ m

d l+ m( 2 1)l

and they are a CON set as well. Notice that now we have:

+11 |P ml ( )|2d = 22l + 1 (l + m)!(l m)!Note:

l can assume integer values from 0,1,2,...

m can assume integer values ranging from l to + lEC574 Physics of Semiconductor Materials, Fall 2008

116

Spherical Harmonics (Sec. 6.3)The complete solution of the angular part of the Schrdinger equation is givenby:

Ylm ( ) =2l + 1 (l m)! ( 1)m e m P ml (cos)


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Y lm (, ) 4 (l + m)! (1) e P l (cos)These functions are called spherical harmonics, and are simultaneous eigen-functions of the operator L2 and Lz . They are orthogonal:

2

0

0Y

lm(, )Y

qn(, )d =

lq

mn

where d = sin dd.

We also have:

Y 00 (, ) = 1 4 Y 10 (, ) = 34 cos Y 11(, ) = 38 e sin


117


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118


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120

If we write: u(r )El = r R El (r )the we can write the radial part as:

2

2 md2

d r 2 + V eff (r ) uEl (r ) = E u El (r )


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then

0 |REl (r ) |2 r 2 dr = 1

Free Particle

Remember that in the case of the solution of the free particle problem in 1Dthe solutions we obtained where simultaneous eigenfunctions of the Hamil-tonian and the linear momentum operator. If we set V(r)=0 for a problem

described in terms of spherical coordinates we obtain the solutions that are si-multaneous eigenfunctions of the Hamiltonian, and the L2 and LZ operators.We write the radial equation for this case as follows:



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123

Spherical Potential Well (Sec. 7.4)

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In this case the central potential is not zero but

V (r ) =0 r r0

V 0 0 < r < r 0EC574 Physics of Semiconductor Materials, Fall 2008


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125

The solution is:REl (r ) = B [ j l ( r ) + n l ( r )] = B h

(1)l ( r )

Where = (2 m/ 2)E and nl (x) is the spherical Neumann function.We also have that:(1) 1 (1) 1 1


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h(1)0

( r ) =

1

re r h(1)

1 ( r ) =

1

r +

1

2 r 2e r

The energy level are then computed by solving the transcendental equation:

d j l (Kr )d r

j l (Kr )r = a

=

d h (1)l ( r )d r

h(1)l ( r ) r = ain case of l = 0 this is written as:

K cot (K a ) = This equation can be solved in the same way the similar problem for the one-dimensional square well was approached.


126


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127

The Hydrogen Atom (Sec 7.5)For an ion with charge Z the central potential has the form:

V (r ) = Ze2

(40)r

where Ze is he nucleus charge and r the distance between the nucleus and

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where Ze is he nucleus charge and r the distance between the nucleus andthe electron. If we dene, as before, the radial equation solution as: R(r ) =u(r )/r then the radial equation becomes:

2

2

d2

d r 2 +

2 l(l + 1)

2 r 2 + V (r ) uEl (r ) = E u El (r )

The effective potential is:

V eff (r ) =

Ze2

(40)r +

2 l(l + 1)

2 r 2Where is the reduced mass. The connement potential depends on the or-bital angular momentum and for l = 0 then V eff (r ) < 0r .


128

Effective Conning Potential for the Hydrogenic Atom


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129

With a suitable change of variables (see Brandsen-Joachain pp.332) to dimen-sion less quantities this equation becomes:

12

d2d 2

+ l(l + 1)22

1

u = u

The solution of this equation can be obtained by using a power series expan-


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sion and requires that: =

12n2

and

E n = 12n2

Z 2 e2 m 2

for the hydrogen atom Z = 1 and we have:

E n =

13.06

n2 eV (49)

The ground state (when n = 1 ) of the hydrogen atom has a binding energy of -13.602eV.


130

Comments:- The energy eigenvalues provided by Eq.(49) depend only on the principalquantum number n. This is always the case for a potential that has a depen-dence of r1 .- For the 3D square well (in spherical coordinate) seen before the energy


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q ( p ) gy

eigenvalue depends both on n and l.- In the case of alkali atoms (one valence electrons) the potential is not com-pletely r1 because of the presence of the core electrons.

- To have better agreement with the experimental measurement several cor-rections have to be included (relativistic and nuclear effects).

- As we have seen solving the central potential problem, the value of n sets a

limit on the maximum value of l, in other words:0 l n 1


131

More in general since the solution of the Schrdinger equation for the hy-drogen atom is a three dimensional problem, three different coordinates of motion have to be known. These three coordinates are represented by thethree quantum numbers n (for r ), l (for ) and m (for ).

Table 2: Allowed set of Quantum Numbers for the Hydrogen Atom


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n l m1 0 02 0 0

1 -1, 0, +13 0 0

1 -1, 0, +12 -2, -1, 0, +1, +2

4 0 01 -1, 0, +12 -3, -2, -1, 0, +1, +2, +3


132

The corresponding radial wavefunctions are computed by a series expansion.

3/ 2


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R1(r ) = 1 Z a 03/ 2

eZr/a 0

R2,0,0(r ) = 14 2 Z a 0

3/ 21 Z ra 0 eZr/ 2a 0

R2,1,0(r ) = 14 2 Z a 0

3/ 2Z ra 0 e

Zr/ 2a 0 cos

R2,1,1(r ) = 18 2

Z a 0

3/ 2 Z ra 0 e

Zr/ 2a 0 sin e


133

Plot of the Radial Wavefunctions


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Note: r2R2nl (r ) is the probability of nding the electron between r and r+dr .


134

Plot of the Radial Wavefunctions


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Note: r2R2nl (r ) is the probability of nding the electron between r and r+dr .


135

Approximation MethodsIn many situations of practical importance the Schrdinger can not be solved

in closed form. In these cases even an approximation of the solution has to bedetermined. It is then important to be able to have the possibility of comput-ing, an approximate solution and the degree of accuracy of the approximationitself. Several approximation methods have been developed and they can be


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pp p ydivided in two groups:

Methods for stationary problems. Methods for time-dependent problems.

In particular we will look at stationary and time dependent perturbation the-ory.

Time-Independent Perturbation Theory (Sec. 8.1)

Let us assume that the hamiltonian of a system can be expressed as:

H = H 0 + H (50)


136

where H 0 is the hamiltonian of the unperturbed system and H is the per-turbation. The parameter 0 1 allows to change the strength of theperturbation. We suppose that the solution of for the unperturbed problem isknown:

H 0(0)n = E (0)n

(0)n (51)

and we want to nd an approximation of the real solution of the problem:


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and we want to nd an approximation of the real solution of the problem:

H n = E n n (52)

Non-degenerate perturbationLet us consider rst the case where the eigenvalue is not degenerate. In thiscase to each eigenvalue corresponds only one eigenfuction. Suppose now wecan expand both the eigenvalues and eigenfunctions in power series of the

parameter . E n =j

j E ( j )n n =j

j ( j )n (53)


137

with the condition that:lim

0

E n = E (0)n lim

0n = (0)n (54)

Substituting into equation (52) we obtain:

(H 0 + H )((0)n + (1)n + 2(2)n + ) =


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(E (0)n + E (1)n + 2E (2)n + )((0)n + (1)n + 2(2)n + ) (55)we then obtain,

H 0(0)

n + H

0 (1)

n + H

02(2)

n +

+ H (0)n + 2H (1)n + 3H (2)n + =E (0)n

(0)n + E

(0)n

(1)n +

2E (0)n (2)n +

+ E (1)n (0)n + 2E (1)n (1)n + 3E (1)n (2)n + + 2E (2)n

(0)n +

3E (2)n (1)n +

4E (2)n (2)n +

(56)


138

now we need to equate the coefcients of equal power on both sides of equa-tion 56 for different powers of . For 0 we have the unperturbed system:H 0

(0)n = E

0n

(0)n , (57)

for we have:

H 0(1)n + H (0)n = E (0)n (1)n + E (1)n (0)n (58)


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and for 2 we have:

H 0(2)n + H (1)n = E (0)n (2)n + E (1)n (1)n + E (2)n (0)n (59)

The expressions for all the n coefcients can be computed accordingly. Therst order correction can be computed multiplying equation (66) by (0)n andintegrating over the space. We have:

(0)n H 0

(1)n dr +

(0)n H (0)n dr =

Slides EC574 Part I F2013

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Transcript of Slides EC574 Part I F2013