lecture slides for chapter 6

8/10/2019 Lecture Slides for Chapter 6

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Chapter 6

The Schrdinger Equation

PHYS 2410

Stephens

Fall 2014

PHYS 2410 Stephens Chapter 6 Fall 2014 1 / 67
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Following the success of de Broglies matter wave hypothesis, Erwin Schrdingerdeveloped an differential equation that describes the time evolution of such amatter wave in 1926.

This model is calledwavemechanics.

Heisenberg had already presented another theory involving infinite matrices to

explain atomic phenomena.

This model is called matrixmechanics.

Although completely different on the surface, Schrdinger later proved that they

are mathematically equivalent.

Since Heisenbergs theory appears from applying Schrdingers wave mechanics,

we will concentrate on Schrdingers model (plus it is less abstract).

The theory is now called quantummechanics.

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6.1 The Schrdinger Equation in One Dimension

The Schrdinger Equation in One Dimension

Although Schrdinger developed his now-famous equation, it can not bederived.

Much like Newtons laws of motion cant be derived.

Rather, the theory is presented and left to experimenters to validate it.

As such, we will provide a heuristic argument.

Recall that a wave equation can be derived for the electric field from Maxwellsequations:

2E

t2= 1

c22E

t2 . (6.1)

Taking the solution as E(x, t)

=E0 cos (kx

t), we can show that

k2 =2

c2 = kc. (6.2)

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6.1 The Schrdinger Equation in One Dimension

The Schrdinger Equation in One Dimensioncontd

The total energy for a non-relativistic particle is

E= p2

2m+V. (6.4)

Using= E/ and p=kfor electromagnetic radiation, we can write=

2k2

2m+V. (6.5)

This differs from the similar expression for a photon, E== pc = kc,

because 1) of the presence of the potential Vand 2) that does not dependlinearly on k.

So we should expect the first time derivative (where comes from) to be related

to the second spatial derivative (where k2 comes from).


6 1 Th S h di E ti i O Di i Th S h di E ti
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6.1 The Schrdinger Equation in One Dimension The Schrdinger Equation


The time-dependent Schrdinger equation is

Time-ependent Schrdinger equation

2

2m

2(x, t)

x2 +V(x, t)(x, t)= i(x, t)

t. (6.6)

To begin, we assume a constant potential energy: V(x, t)= V0.Because the first derivative of an exponential function is proportional to its

second derivative, a possible solution is

(x, t)=Aei(kxt) =A[cos (kxt)+ isin (kxt)] . (6.7)

Inserting this expression into Eq. 6-6, we obtain Eq. 6-5:

2k22m

+V0 =.


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The Schrdinger Equationcontd

There are two significant differences between the Schrdinger equation and theclassical wave equation.

Schrdingers equation involves the imaginary number i=1 , implying that the wave

function is not necessarily a real function.The classical wave equation has a second-order time derivative while the Schrdingerequation has a first-order time derivative.

This implies that the Schrdinger equation is more like a diffusion equation than a wave

equation.

(x, t), along with the term wave function, is also called the probability density

amplitudeor just probability amplitude.

The probability densityis given as

Probability density

P(x, t) dx=(x, t)(x, t)dx=(x, t)2dx. (6.8)


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The Schrdinger Equationcontd

By convention, we normalize the wave function so that we have absolutecertainty that it will be measured somewhere within the problems domain.

In general, we know that we must find the electron somewhere in space, so we

normalize the wave function according to

Normalization condition dx= 1. (6.9)

This is called the normalization condition.

This condition has significant implications on our solutions to Schrdingersequation.

The solution must be convergent-enough to satisfy the normalization condition.

In other words, the wave function cannot diverge anywhere within the problem

domain.


6 1 The Schrdinger Equation in One Dimension Separation of the Time and Space Dependencies of(x t)
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6.1 The Schrdinger Equation in One Dimension Separation of the Time and Space Dependencies of(x, t)

Separation of the Time & Space Dependence of(x, t)

A common approach to solving differential equations resembling the

Schrdinger equation is called separation of variables.

This involves separating the spatial and temporal dependencies of the wave

function into different functions.

The spatial part was referred to as a stationarystate because it lackedtime-dependence.

A synonym for a stationary state is an eigenstate (eigen- is German for particular,

characterstic, etc.).

So lets write the wave function as

(x, t)=(x)(t). (6.10)


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Separation of the Time & Space Dependence of(x, t)contd

Substituting this into the Schrdinger equation gives

2

2m

1

(x)

d2(x)

dx2 +V(x)= i 1

(t)

d(t)

dt. (6.13)

The left side depends only on xand the right side depends only on t.

The only way they can be equal for all values ofxand tis if the left and right sides

are constants.

This gives

2

2m

1

(x)

d2(x)

dx2 +V(x)=C. (6.14)

i 1(t)

d(t)

dt=C. (6.15)


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g q p p p ( , )


The time-dependent equation can be written asd(t)

(t)= C

i dt=iC

dt. (6.16)

Integrating each side give

(t)= eiCt/ = cos

2Ct

isin

2

Ct

. (6.17b)

So (t) describes oscillatory behavior at a frequencyf= C/h.From the de Broglie relation, we know thatf

=E/h.

So the time-dependent part of the wave function is

Time-dependent solution

(t)= eiEt/ . (6.17c)


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( , )


UsingC= E, the space-dependent differential equation is called thetime-independent Schrdinger equation:

Time-independent Schrdinger equation

2

2m

d2(x)

dx2 +V(x)(x)

=E(x). (6.18)

The probability density can be written as

(x, t)(x, t)=

(x)e+iEt/

(x)eiEt/

=(x)(x). (6.19)

So the normalization condition becomes+

(x)(x) dx= 1. (6.20)


6.1 The Schrdinger Equation in One Dimension Conditions for Acceptable Wave Functions
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Conditions for Acceptable Wave Functions

The exact form of the wave function (x) depends on the nature of the potential

energyV(x).

But must be a physically viable wave function.

Regions for which V(x) is constant lead to the simplest solutions so we tend to

analyze these situations.

But to keep things interesting, we might let V(x) assume different constant values

in different regions.

So V(x) might be discontinuous at the boundary between two regions.

To solve these problems, we solve Schrdingers equation in each region andthen match the solutions across the boundary.


6.1 The Schrdinger Equation in One Dimension Conditions for Acceptable Wave Functions
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Conditions for Acceptable Wave Functionscontd

It is unphysical for the probability of finding the particle to vary discontinuously

so the wave function must be a continuous function.

Since Schrdingers equation involves a second spatial derivative, the wave

function and its slope must be differentiable the wave functions slope must be

continuous.

The wave function must be normalizable the wave function and its derivativemust be finite-valued andmust tend to zero at infinity.

Here is a summary of the conditions on a wave function:

Conditionswave functionmust satisfy

(x) must exist and satisfy the Schrdinger equation.(x) & d(x)/dxmust be continuous.

(x) & d(x)/dxmust be finite and single-valued.

limx(x)= 0 so that (x) can be normalized.


6.2 The Infinite Square Well
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The Infinite Square Well

The simplest problem that illustrates the nature of quantum mechanics andSchrdingers equation is referred to as the infinite square well.

The potential is described mathematically as

V(x)=0, 0< x< L,, x< 0 and x> L. (6.21)

This problem could also be described as a bead on a finite string or

an electron caught between two electrodes in an evacuated tube.

Since the potential energy is infinite at the boundary, the particle cannot traveloutside of the box since there is an infinite force acting on it at x= 0 and x= Lrecall that F=dV/dx.


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The Infinite Square Wellcontd

Keeping in mind Borns probabilistic interpretation, this means the probability of

finding the particle must go to zero as we approach the box walls, i.e.,

(0)=(L)= 0.In line with de Broglies hypothesis, the wave function must resemble a standing

wave.

This means that an integral number of half-wavelengths must fit into the length L:

n

2= L, n= 1, 2, 3 . . . (6.22)

Because the particle is free inside the box its kinetic energy is

E= p2

2m= h

2

2m2= n2 h

2

8mL2. (6.23)


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It is customary to write

En=

n222

2mL2=n2E

1, n

=1, 2, 3, . . . (6.24)

Here, the lowest allowed energy is given by

E1 =222mL2

(6.25)


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To force our solution of Schrdingers equation to describe this problem, weneed to incorporate this feature into the wave function.

Since we have a condition on the wave function at the domain boundary, this

condition is called a boundary condition.

Namely, we need to have (0)

=0 and (L)

=0.

So we need to find the wave function that satisfies

2

2m

d2(x)

dx2 = E(x) (x)=2mE2 (x)=k

2(x). (6.26)

Here, we substituted the square of the wavenumber:

k2 =p2= 2mE2 . (6.27)


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Eq. (6.26)has the standard solutions

(x)=Asin kx (6.28a)(x)

=Bcos kx. (6.28b)

We now enforce the boundary conditions (0)=(L)= 0.cos0= 1, so it does not satisfy our boundary conditions and is discarded.sin0= 0, this is the form for our wave function.

So we have

(L)=Asin kL= 0. (6.29)Recall that sin is zero for integral multiples of.




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So we have, where nis a positive integer (nZ+)

kL=n kn=n

L. (6.30)

Since we have k2

=2mE/

2, knowing the wavenumber

is equivalent to knowing the energy:

En=2k2n2m

=n222

2mL2= n2E1.

In order to speak of the particles probability, we need to normalize the wave

function: L0nndx=

L0

A2nsin2nx

L

dx= 1 A=

2

L. (6.31)



h fi ll
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So our solution is

n(x)=

2

Lsin

nx

L(6.32)

Here are plots of theprobability amplitude

and probability for the

first three levels

nis called a quantum

number.It specifies the energy

and wave function for

a state n.


6.2 The Infinite Square Well Comparison with Classical Results

C i i h Cl i l R l
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Comparison with Classical Results

Since we know the potential energyV(x), we can find the force Fx=dV/dxandthereby the acceleration ax= d2x/dt2 experienced by the particle and itsresulting motion x(t).

But there is no force on the particle betweenthe walls since V= 0.

So the particle must move with constant speed as it travels from one side of thebox to the other.

Classically, any speed and energy are allowed but we have limits on the energyEbecause of the quantum number n.

The quantum number entered the problem because of the boundary conditions placed

on (x).

We will see later that a boundary condition on a degree of freedom involves a unique

quantum number, i.e., the hydrogen atom has three quantum numbers since it is 3D.


6.2 The Infinite Square Well Comparison with Classical Results

C i i h Cl i l R l


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Comparison with Classical Results

The classical probability distribution to find the particle at some location xinsidethe box is proportional to the particles speed and the size of the box:

PCl(x)=1

L.

Quantum mechanically, the probability is

n(x)2 =n(x)2 =1

L.

This figure shows a comparisonbetween the classical and quantum

mechanical distributions for a particle

in the n= 10 state.


6.2 The Infinite Square Well The Complete Wave Function

Th C l t W F ti
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The Complete Wave Function

Recall that weve only considered the time-independent Schrdinger equation inthe above analysis of the infinite square well.

Incorporating the time-dependence, with En=n, the complete wave functionis

(x

,t

)=

(x

)

(t

)=

2

Lsin

(knx) eint.

Using the Euler relation 2isin z= eizeiz, we can write this as

n(x, t)= 1

2i

2

L ei(knxnt) ei(knxnt)

.

So the standing wave pattern of the particle in the box is analogous to a standing

wave on a string it is composed of oppositely moving, traveling waves.



E l 6 2
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Example 6-2

An Electron in aWire

An electron moving in a thin metal wire is a reasonable approximation of a particle

in a one-dimensional infinite well. The potential inside the wire is constan on

average but rises sharply at each end. Suppose the electron is in a wire 1.0 cm long.

(a) Compute the ground-state energy for the electron. (b) If the electrons energy is

equal to the average kinetic energy of the molecules in a gas at T= 300 K, about0.3 eV, what is the electrons quantum number n?



Example 6 2 Solution
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Example 6-2 Solution

An Electron in aWire

The ground-state energy is

E1 = 2(1.0551034 Js)2

2(9.1094

10

31 kg)(10

2 m)2 E1= 3.89 feV.

The quantum number is determined from En=n2E1to be

n=

En

E1=

0.3 eV

3.8

1015 eV

n 2.81106.

This is sufficiently large for the realm of classical mechanics.



Example 6 3
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Example 6-3

Calculating Probabilities

Suppose that the electron in Example 6-2 could be measured while in its ground

state. (a) What would be the probability of finding it somewhere in the region

0< x< L/4? (b) What would be the probability of finding it in a very narrow regionx= 0.01Lwide centered at x= 5L/8?



Example 6 3 Solution
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Calculating Probabilities

The wave function in the ground state is 1(x)=

2L

sin (x/L).

The probability of finding it in x [0, L/4] is

L/4

0P1(x) dx=

L/4

0

2

Lsin2

x

Ldx= 2

8 1

4

0.091.

The probability of finding it in x [ 58 Lx,58 L+x] can be approximated by(since x

1)

P= P(x)x= 2L

sin2xLx 0.017.



Example 6 4


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Example 6-4

An Electron in an Atomic Size Box

(a) Find the energy in the ground state of an electron confined to a

one-dimensional box of length L= 0.1 nm. (This box is roughly the size of an atom.)(b) Make an energy-level diagram and find the wavelengths of the photons emitted

for all transitions beginning at state n= 3 or less an ending at a lower energy state.





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An Electron in an Atomic Size Box

UsingE1 =n2h2/8mL2, we findE1 = 37.6 eV.

The energies are shown in the figure.The transition wavelengths are found

from mn= hc/Emn.

32 = 6.60 nm.31 = 4.12 nm.21 = 11.0 nm.


6.3 The Finite Square Well

The Finite Square Well
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Quantization occurs by enforcing a constraint on the system.

The infinite square well had quantized energy levels because of the constraint of

placing the particle in the box.

Here is another common

potential energy function,known as the finite square

well.

Since the potential energy is finite, the particle could be found outside the well.

IfE>V0, the particle is not confined and the energy is not quantized although theparticles motion is position dependent.

We consider E


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The Finite Square Wellcontd

Inside the well, V(x)= 0 so that Schrdingers equation reduces to

(x)=k2(x) k2 =2mE2 .

Outside the well, for 0

>x

>L, Schrdingers equation is

(x)= 2m2 (V0E)(x)=2(x) (6.33)

2 = 2m2 (V0E)> 0. (6.34)

We can find the solution for each region (eikx inside and ex outside) andenforce the matching conditions (continuity of and ) but finding the energylevels requires solving a transcendental equation.



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We still need to enforce some other necessary conditions on the DE solutions forthem to be viable wave functions.

Firstly, we need 0 when x.Examining Eq.(6.33) shows that outside the well has the same sign as thefunction .

Remember that the second derivative yields the curvature of the function.

This means that when > 0 then > 0,Figure (a), or when < 0 then < 0,Figure (b).

This means that the solutions to Schrdingers equation outside the well are not

necessarily well-behaved since they diverge at infinity and cannot be wave

functions.



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But inside the well, we have , so the function always bends toward the xaxis.

When the function is positive (negative), the curvature is negative (positive) like sine

or cosine.

So the function is oscillatory.

Here is a plot of a satisfactory wave function with wavelength 1and two other

solutions to Schrdingers equation with similar wavelengths.

For>1, + as xincreases.For

0.


6.5 The Simple Harmonic Oscillator

The Simple Harmonic Oscillatorcontd


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cont d

As for the finite square well, consider the curvature to a certain energyE:

(x)=k2(x) where k2 =2m2 [EV(x)] .

For x

0.

Classically, the answer is simple: ifE< V0then each particle reflects, ifE>V0then each particle passes the step with a reduced speed.

The quantum mechanical result is more interesting.

For E


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Schrdingers equation in each region is:

Region I For x< 0,d2(x)

dx2 =k21(x), k21 =2mE

2 . (6.61)Region II For x> 0,

d2(x)

dx2 =k22(x), k22 =

2m(EV0)2 . (6.62)


6.6 Reflection and Transmission of Waves Step Potential

Step Potentialcontd
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The solutions to these DEs are

Region I For x< 0,

1(x)=Aeik1x+Beikx. (6.63)

Region II For x> 0,

2(x)=Ceik2x+Deik2x (6.64)Recall the conditions on the solution to be a viable wave function:

and d/dxmust be continuous across x=

0.

This leads to the matching conditions:1(0)=2(0) and 1(0)=2(0).



Step Potentialcontd
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Enforcing continuity of the wave function, we have

1(0)=2(0) A+B=C+D. (6.65a)

Continuity of the wave functions slope gives

1(0)=2(0) ik1 (AB)= ik2 (CD) . (6.65b)

We can set D= 0 since we expect no incoming particles from +.eik2x represents a plane wave moving in the xdirection.

SinceArepresents the probability amplitude of the incoming particles, we need

to determine Band Cin terms ofA.

eik1x represents a plane wave moving in the +xdirection.



Step Potentialcontd
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The result for these probability amplitudes is

B

=

k1k2k1+k2

A

=

EEV0

E+EV0A. (6.66)

C= 2k1k1+k2

A= 2

EE+EV0

A. (6.67)

Next, we form the ratios B/Aand C/Ato get the relative probability amplitude.



Step Potentialcontd
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Taking the norm-squared of these ratios gives the relative probabilities:

R=BA

2

=

k1k2k1+k2

2(6.68)

T=k2k1

C

A

2

= 4k2k1(k1

+k2)

2 (6.69)

The factor ofk2/k1for Tis necessary to normalize Tsince there will be different

currents in the two regions because of the differing potentials.

Ris called the reflection coefficient (or probability of reflection) and Tis calledthe transmission coefficient (or probability of transmission).

Rtells us what the probability is that a particle of energyEwould be reflected back into

x< 0.Ttells us what the probability is that a particle of energy Ewould be transmitted into

x> 0.

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Step Potentialcontd


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Here is a sequence of images showing a wave packet encountering a steppotential.

Notice the increased amplitude as the

particle strikes the step this is

related to the longer wavelength.Notice the ripples this is

interference between the incident

components and reflected

components comprising the packet.

The dot represents the motion of a classical particle.

Note the reflected wave packet.



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6.6 Reflection and Transmission of Waves Barrier Potential

Barrier Potential


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The next step is to consider a barrier that is only of finite width, not infinitely

wide as for the step.

V(x)=

V0, 0< x< a,0, 0> x> a. (6.73)

The wave function must now be divided into three regions:

I(x)=Aeik1x+Beik1x x< 0 II(x)=Cex+De+x 0< x< a (6.74)III(x)= Eeik1x+Geik1x x> a

Where k21 = 2mE/2 and 2 = 2m(V0E) /2.



Barrier Potentialcontd
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This problems requires a more involved solution, but the steps are the same.

I(0)=II(0) and II(a)=III(a).

I(0)=

II(0) and

II(a)=

III(a).

The solution for the wave function (with the correctcoefficients) is show to the right.

Note the decay in amplitude through the barrier.

Classically, we should have R= 1 for E


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The transmission coefficient from Region I to Region III is

T= FA

2

=1+ sinh2z

4 EV0

1 E

V0

1

. (6.75)

For thick barriers, a 1, this simplifies to

T 16 EV0

1 E

V0

e2a. (6.76)

If

, we recover the step barrier and T

=0.

Tunneling is important is many technical applications as well as chemical

bonding, to mention two.



Scanning Tunneling Microscope


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Optical microscopes cannot image atoms since visible light involves wavelengths

orders of magnitude larger than atoms.

But electrons can image atoms.

One device to do so, and hinges of quantum mechanics, is the scanning

tunneling microscope(STM).

It involves an atomically-sharp tip that scans across asurface.

A potential difference is applied between the tip and

sample (so an STM is restricted to conductive samples).

Depending on the distance between the tip and sample, electrons can tunnel

across the separation.

Since T e2x, monitoring the potential difference or resulting current resultsin very accurate topographical maps of the electrons in the samples surface.



Scanning Tunneling Microscopecontd
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Here are some sample STM images taken at room temperature.

Gold nanoparticles on

a TiC surface.



NH3Atomic Clock
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Ammonia-based atomics clocks also use tunneling.

Here is the NH3molecule.

The potential energy of the N atom is shown.

The nitrogen atom oscillates between the two potential wells, a process that is

classically forbidden, but occurs only because of tunneling.

The oscillation frequency is 2.37861010 Hz, a rather modest molecularfrequency.

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lecture slides for chapter 6

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