Quantum Mechanics Souri

7/24/2019 Quantum Mechanics Souri

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Souri BanerjeeSouri Banerjee

souri@[email protected]

LectLect 11

The passion begins


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HandoutHandout

Quantum

Mechanics

Matter wave, Concept

of wave packet,

Uncertainty Principle,Wave function,

Schrodinger equation,

Linearity and

superposition,

Operators andExpectation values,

Particle in a box,

Finite potential well,Tunnel effect.

46.1-46.7;

47.1-47.3

Beiser:

5.1-5.10


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Essence of QMEssence of QM

Classical Physics: See & Touch

Quantum Physics:

Inaccessible to our senses

Understood in a sort of abstractor imaginative fashion and not by

direct experience


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Essence of QM contd.Essence of QM contd.

Classical Mechanics:Completely Deterministic

Quantum Mechanics:

Act of measurement interfereswith system and modifies it

Probabilistic Simple quantities become

matrices and operators


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Essence of QM contd.Essence of QM contd.

Classical Physics:

Particles and Waves understoodas separate entities

Quantum Physics:Particle and Waves described by

one set of equations

Wave-particle duality


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Physics that tune up QMPhysics that tune up QM

Black body radiation (1900)

Photo-electric Effect (1905)

H2 Spectrum (1913)Compton Effect (1923)

Davisson Germer Experiment

(1927)


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Problem and SolutionsProblem and Solutions

Problem arose in explanation for

phenomena involving small particles(electrons, atoms) and their

interaction with EM fieldsAd hoc hypothesis and postulates

Planck, de Broglie

Physics need a reformationPhysics need a reformation


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Journey BeginsJourney Begins

Quantum Mechanics belongs to:Werner Heisenberg:

Uncertainty Principle;Accommodated indeterminism

Max Born:Probabilistic approach;

Wave functionErwin Schrodinger:

Wave Equation (1926-1927)


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Planck and dePlanck and de BroglieBroglie HypothesisHypothesis

Plancks :

hE=

de Broglies :

p

h=

Wavelength specified

by linear momentum

Energy specifiedby frequency

h=6.63x10-34J.s


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Some ExampleSome Example

Calculate the deCalculate the de BroglieBroglie wave lengthwave lengthof a dust particle of mass 10of a dust particle of mass 10--99 kg driftingkg drifting

withwith velvel 2 cm/s.2 cm/s.

m103.3

)s/m102)(kg10(s.J1063.6

mv

h

p

h

23

29

34

=

=


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More ExampleMore Example

Calculate the deCalculate the de BroglieBroglie wave lengthwave length

of a 46 g golf ball havingof a 46 g golf ball having velvel 30m/s and30m/s and

an electron withan electron with velvel 101077m/sm/s

m108.4mvh 34

ballgolf

m103.7mv

h 11e

Radius of the

H2 atom


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More ExampleMore Example

Calculate the deCalculate the de BroglieBroglie wave lengthwave length

of a 0.05eV neutronof a 0.05eV neutron

Kcm2

hc

Km2

h

p

h

2

00=

0

6

o

3

28.1

)eV05.0)(eV10940(2

A.eV104.12

=

Thermal Neutronor

Slow Neutron


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i f k


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Construction of Wave PacketConstruction of Wave Packet

We need a Wave-Packet Superposition of sine waves

Wave packet results fromsuperposition of many sine

waves of various k

= +

+


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TechniqueTechnique

Fourier Integral :

dk)kxcos()k(g)x(

It describes how the amplitudesof the waves that contribute to

wave packet, vary with k


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Results of FTResults of FT

For Sine

Wave 0

2

k

=

=

0

2

k

=

x

x k

k0= 2/0

g


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Property of a Wave PacketProperty of a Wave Packet

1) can interfere with itself,so that it can account for the

results of diffraction expts.

2) It is large in magnitude wherethe particle is likely to be.

3) corresponds to a singleparticle, not a statistical

distribution of number of quanta.


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Space and Time PacketSpace and Time Packet

A true particle is localized in

space and time

Space Packet : Made of superpositionof waves of various

Time Packet : Made of superposition

of waves of various


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Home StudyHome Study

Q. How does a wave-packet

propagate ?

Group Velocity (vg)= d/dk

Phase Velocity (vp)= /kImportant terms:

Wave-pkt moves with vg not with vp


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Some FactsSome Facts

1.x

Quantum Physics by Gasiorowicz; p27-29

For space packets:

For time packets:

1.


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What it follows?What it follows?

1.x

s.J100545.1

2h

34

=

hhp.x

Through de Broglie

Heisenberg Uncertainty Principle


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Uncertainty PrincipleUncertainty Principle

It is impossible to specify precisely

and simultaneously the values ofboth members of particular pairs of

physical variables that describe thebehavior of an atomic system

The Principle states:


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Pairs of VariablesPairs of Variables

hxp.x

hzJ.

h.E

(1)

(2)

(3)


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Physical MeaningPhysical Meaning

hxp.xA component ofp of a particlecannot be precisely specified

without loss of all knowledge ofthe corresponding component of

its position at that time


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Other OnesOther Ones

hz

J.

Precise measurement of the angular

position of a particle in an orbitcarries with it the loss at that timeof all knowledge of the componentof angular momentum perpendicularto the plane of the orbit


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The Last TermThe Last Term

h.EIf a system maintains a particularstate of motion not longer than

a time t, the energy of the systemin that state is uncertain by at least

the amount E~h/t


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EnergyEnergy--Time UncertaintyTime Uncertainty

1.

The smallness of h makes theUncertainty principle of interest

primarily to the atomic systems

h.E

Through Planck

.10~ 34h


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Lecture 2Lecture 2

Co i oComparison


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ComparisonComparison

Classical Mechanics:Completely Deterministic

Quantum Mechanics:


Probabilistic Simple quantities become

matrices and operators


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WaveWave--particle Dualityparticle Duality

Classical Physics:

Particles and Waves understoodas separate entities

Quantum Physics:Particle and Waves described by

one set of equations

Wave-particle duality


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Planck and dePlanck and de BroglieBroglie HypothesisHypothesis

Plancks :

hE=

de Broglies :

p

h=

Wavelength specified

by linear momentum

Energy specifiedby frequency

h=6.63x10-34J.s



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We need a Wave-Packet Superposition of sine waves

Wave packet results fromsuperposition of many sine

waves of various k

= +

+

F i I t l d T fF i I t l d T f


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Fourier Integral and TransformFourier Integral and Transform

Fourier Integral :

dk)kxcos()k(g

x

xk

k0= 2/0

g

F-T


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Concept of Time PacketConcept of Time Packet

Time Packet : Made of superpositionof waves of various

d)tcos()(g

S F tS F t


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Some FactsSome Facts

1.x

Quantum Physics by Gasiorowicz; p27-29

For space packets:

For time packets:

1.

From Space PacketsFrom Space Packets


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From Space PacketsFrom Space Packets

1.x

hp.x

Through de Broglie

Heisenberg Uncertainty Principle

From Time PacketsFrom Time Packets


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From Time PacketsFrom Time Packets

1.

The smallness of h makes theUncertainty principle of interest

primarily to the atomic systems

h.E

Through Planck

.10~ 34h


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Uncertainty PrincipleUncertainty Principle

It is impossible to specify precisely

and simultaneously the values ofboth members of particular pairs of

physical variables that describe thebehavior of an atomic system

The Principle states:


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Pairs of VariablesPairs of Variables

hxp.x

hzJ.

h.E

(1)

(2)

(3)

ThoughtThought ExperimentsExperiments


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ThoughtThought ExperimentsExperiments

What is a thought experiment ?Based on straight-forward logic.We know the results that would beobtained because there are manyexperiments that have been done

Why thought experiment ?Difficult to carry out for smallness

of scale

E i h B ll (S )E t ith B ll t (S t )


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Expt. with Bullets (Set up)Expt. with Bullets (Set up)

x

GunBackstop that

absorbs bulletswhen they hit

Movable Detector Might be a boxcontaining sand

Indestructible

Bullets 1

2

E i tE i t


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ExperimentExperiment

We wish to find out :

What is the probability that a

bullet which passes thro theholes will arrive at the back-stop

at a distancex from the center ?

C t i U d t diC t i U d t di


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Certain UnderstandingCertain Understanding

Why sayprobability?

Because we cannot say definitely

where any particular bullet goes

Byprobability we mean the chance

that the bullet arrives at detector

M tM t


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MeasurementMeasurement

How you measureprobability?

Assuming the gun shoots at thesame rate during measurement,

the probability is just proportionalto the number that reach detector

in some standard time interval

ObservationsObservations


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Bullets arrive in identical lumps

x P12

P1

P2

Gun

P12=P1+P2 Probabilities added

1

2

Expt. With LightExpt. With Light


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Expt. With LightExpt. With Light

xI12

I1

+I2

I1

I2 IntensitiesNOT added

I12= I1+I2 + 2I1I2cos

Interference Term

1

2

Expt. With ElectronsExpt. With Electrons


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Expt. With Electronsp ec o s

x

Electron Gun

Detector Geiger Counterconnected to a loud-speaker

1

2

Thought ExperimentThought Experiment


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Expts with bullets and light can be

done but not with the electronsWhy ?

The setup should be of impossiblysmall scale to show effects that we

are interested in




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ObservationsObse at o s

1) We hear sharp clicks butno half clicks

2) Clicks are erratic.

3) As detectors is moved around,clicks gets faster or slower but

of same loudness4) With two detectors, one or the

other clicks, never both at once

ConclusionConclusion


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Electrons arrive at the back-stop in identical lumps

What is the probability that anelectron lumparrive at various

distance fromx ?Proportional to average

rate of clicks at thatx

The ResultThe Result


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The ResultThe Result

x P12

ElectronGun

P12not like what we obtain for bullets

What is Happening?What is Happening?


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pp gpp g

Proposition :Proposition :Each electron eithergoes throughhole 1 orhole 2

Observed curve must be sum of

the effects of the electrons whichcome thro hole 1 or 2

Further ExperimentsFurther Experiments


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Block hole 1/2 and from clickingrate get P1/P2

P1

P2

x

ElectronGun

P1 + P2P12

Is the proposition false ?

What Next?What Next?


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What Next?What Next?

Try and locate the pathNext Attempt :

How ?

MysteryMystery


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x

Electron

Gun

LightSource

Every time we hear a click, we alsosee a flash either near hole 1 or 2,

never both at once

1

2

Mystery ContinuesMystery Continues


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Mystery ContinuesMystery Continues

Our proposition is necessarily true

Then, why P12 P1+P2

Let us keep a track of the electronsand find out what they are doing

How?How?


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Make two columns. When we heara click, put a count in column 1/

column 2 if we see a flash nearhole 1/hole 2

Every electron is recorded in twoclasses: one which comes from hole

1 and those from hole 2

What Else?What Else?


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Repeat such measurement formany values ofx to get P1 and P

2

x P1

P2

ElectronGun

LightSource

Total ProbabilityTotal Probability


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yy

Q. What is the probability that

an electron will arrive at thedetector by any route ?

Pretend we never looked at thelight flashes and then add together

detector clicks we had in columns

Surprising!Surprising!


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Surprising!Surprising!

x P1

P2

P12

ElectronGun

LightSource

Then, P12 = P1+P2

What it Suggests?What it Suggests?


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What it Suggests?at t Suggests

Ascertains Uncertainty Principle


Although we succeeded in watching

which hole electrons come through,we no longer get the old P12

Fun StudyFun Study


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For interested people:

The Heisenberg Microscope

Purpose is to measure the position

of an electron

Ref: Gasiorowicz; p33-35

Lecture 3Lecture 3


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ec u e 3

Plan of LecturePlan of Lecture


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Introducing wave function

in more quantitative fashion

Schrodinger Equation

Some application ofSchrodinger Equation

Classical MechanicsClassical Mechanics


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A particle of mass m moving inx direction under the action of

an external force F

Fdt

xdm2

2

=

The solution contains all information

about the trajectory of the particle

Task Cut OutTask Cut Out


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Q. What is the Quantum Mechanicaldescription of such a particle?

Development of a (differential)equation and the correspondingwave function (x,t) that would

represent the same particle

Building BlocksBuilding Blocks


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h

=hE

k

h

p h

Plancks Hypothesis :

de BroglieHypothesis :

Likely to beLikely to be


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(x,t) should include the notion

of the particle being Likely to be

It is large in magnitude where theparticle is likely to be and small

elsewhere

Ideal ConditionIdeal Condition


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Consider Free particle

No force acting on it

p is a constant of motion

From de BroglieFrom de Broglie


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kh

p h

(x,t) should correspond toa plane wave solution

What Type of solutions?What Type of solutions?


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Q. If a wave function (x,t) has

to represent a particle havingcompletely undetermined position

and traveling in +x direction withprecisely known momentump and

kinetic energyE, what should bethe form of (x,t) ?

Solution TypesSolution Types


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)kxcos(

)kxin(

)tkx(ie

)tkx(i

e

Form of Wave EquationForm of Wave Equation


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Two basic properties :

1) It must be linear, in order that

solutions of it can be made toproduce interference effect and

to permit the construction of awave packet

The Other PropertyThe Other Property


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2) The coefficients of the equation

must involve only constants ash and the mass or the charge of

the particle and notnot parametersof a particular kind of motion ofthe particle (p, E, kand )

Look for waveLook for wave equnequn..


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Traveling Wave-equation:

2

2

22

2

t

y

v

1

x

y

=

Q. Can we accept it?

Look FurtherLook Further


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Property 1 is satisfied as the

differential equation is linear

What about the property 2 ?

Just Check !Just Check !

Wave equationWave equation


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2

2

2

2

2

2

2

m4

p

p

E

kv =

Involves parameters of

motion (Eorp)

NOT ACCEPTED !

Is that Enough?Is that Enough?


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You cannot afford to ignore

equations linking wave-particleduality, the building blocks

m2

pE

2

= m2

k2h

=

Equivalent

What it Suggest?What it Suggest?


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m2

p

E

2

=

m2

k2h

=

Equivalent

1st

. derivative of time should berelated to 2nd. derivative of space

What Comes Out?What Comes Out?


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Conventional wave equation cannot be regarded as an equation

for describing a wave function

)tkx(ie)t,x(

)tkx(ie

Why not ?

Expected TypeExpected Type


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tx2

2

=

1st. derivative of time relatesthe 2nd. derivative of space

Free particle Schrodinger EquationFree particle Schrodinger Equation


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)tkx(i

0e)t,x(

=

..(1)

2

2

2

kx

..(2)

it

..(3)

DerivationDerivation


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From eq. 3

hh =

ti ..(4)

From eq. 2

m2

k

xm2

22

2

22 hh=

..(5)

RelateRelate


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hh

=

m2

k

m2

pE

222

..(6)

Eq 5 reads,

hh

=

2

22

xm2..(7)

Schrodinger EquationSchrodinger Equation


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t

)t,x(

ix

)t,x(

m2 2

22

=

h

h

1-D time dependent

Schrodinger Eq.

Combining eq. 4 and 7:

ComparisonComparison


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tx2

2

=

t

)t,x(i

x

)t,x(

m2

2

22

=

hh

m2

ih

=

It involves hand m only

Interpretation ofInterpretation of


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)wtkx(i

0e)t,x( =

1) (x,t) gives a complete quantum

mechanical description of a freeparticle of mass m and K.E.E

2) (x,t) is, in general, complex

IsIs ComplexComplexComplex?Complex?


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That the wave function is Complex

is NOT the defect of the formalism

(x,t) being complex in general,is not an observable quantity

Role of UncertaintyRole of Uncertainty

P iti d t i ti t i b


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(x,t) measures the probability

of finding a particle at a particularposition w.r.t. the origin of its region

Position determination uncertain byan amount ~ of linear dimensionof the wave function

BornBorns Interpretations Interpretation

P b bilit R l d ti


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Probability Real and non-negative

Define: Position Probability Density

2

)t,x(

)t,x()t,x()t,x(P

=

=

Max Borns contribution

Probability DensityProbability Density


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2

)t,x()t,x()t,x()t,x(P =

P(x,t)dx is the probability of finding

a particle in dx centered at x, at atime t

ExampleExample

A ti l li it d t thA ti l li it d t th i h thi h th


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A particle limited to the xA particle limited to the x--axis has the waveaxis has the wave

functionfunction =ax between x=0 and x=1.=ax between x=0 and x=1. =0=0

elsewhere. Find the probability that theelsewhere. Find the probability that theparticle can be found between x= 0.45 andparticle can be found between x= 0.45 and

x=0.55x=0.55

=

55.0

45.0

222

x

x

2 a025.0dxxadx2

1

Lecture 4Lecture 4


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Interpretation ofInterpretation of


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)tkx(i

0e)t,x(

=

1) (x,t) gives a complete quantum

mechanical description of a freeparticle of mass m and K.E.E

2) (x,t) is, in general, complex

IsIs ComplexComplexComplex?Complex?


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That the wave function is Complex

is NOT the defect of the formalism

(x,t) being complex in general, is

not an observable quantity andshould includelikely to be concept

BornBornss InterpretationInterpretation

Probability Real and non negative


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Probability Real and non-negative

Define: Position Probability Density

2

)t,x(

)t,x()t,x()t,x(P

=

=

Max Borns contribution

Probability DensityProbability Density


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2

)t,x()t,x()t,x()t,x(P =

P(x,t)dx is the probability of finding

a particle in dx centered at x, at atime t

Example 1Example 1


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Evaluate the probability density forEvaluate the probability density for

a simple harmonic oscillator lowesta simple harmonic oscillator lowestenergy state wave function:energy state wave function:

t

m

C

2

ix

2

Cm 21

2

eAe)t,x(

=

h

SolutionSolution

Cm


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2xCm

2

eAP

=

h

P independent of time but is not

Consequence of the fact thatthe particle associated with

is in a single energy state

Plot ofPlot of P(x)P(x) vs.vs. xx

C


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2xCm

2

eP

=

h

x

P(x)

0

P is max atx=0

This is where theparticle is mostlikely to be found

Example 2 (Home Assignment)Example 2 (Home Assignment)

Calculate the normalization constantCalculate the normalization constant


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Calculate the normalization constantCalculate the normalization constantand the probability density for a waveand the probability density for a wave

function given by (at t=0):function given by (at t=0): (compre.(compre. 04)04)

ikx2

x

eAe)x(

22

=

Ans:2

1

A

=

22xeP

=

GaussianWave Packet

Change of NotionChange of Notion


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QM deals with where the particle

is, without the particle beingthought of as what it made of

Probabilistic

interpretationNotion of

Wave-packet

NormalizationNormalization

l


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1dV

2

=

Particle must

be somewherein space

If is a wave packet, the aboveintegral must converge and the

coefficient of adjusted to getvalue of the integral equal to 1

ProblemProblem

Normalize:


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tmC

2ix

2Cm 2

1

2

eAe)t,x(

=

h

Normalize:

1dxeAdxP

2

xCm

2=

h

SolutionSolution

21

xCm

)(2

h


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Use:

4

1

2

0

x

)Cm(2

)(dxe

hh =

4

1

8

1

)(

)Cm(A

h

=

Final ResultFinal Result

1 1


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tm

C

2

ix2

Cm

4

1

8

12

1

2

ee)(

)Cm()t,x(

=

h

h

Normalized wave function

Why Normalization?Why Normalization?

Before normalization amplitude of


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Before normalization, amplitude ofthe wave-function was arbitrary

The linearity of Sch. Eq. allows awave function to be multiplied by anarbitrary constant and still remain asolution to the equation

Effect of NormalizationEffect of Normalization


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What has normalization done?

Normalization has the effect offixing the amplitude by fixing the

value of the multiplicative constant,such as A

Box NormalizationBox Normalization

Normalize a free particle wave-function


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Normalize a free particle wave function

= 1dxdx 20

0 must be zero Tackle it!

Approximation of Free ParticleApproximation of Free Particle


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Free particle

Ideal Case

Can we have a physical situation

very close to the ideal one ?

A Physical ScenarioA Physical Scenario


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How about this?

A proton moving in a highly

mono-energetic beam emergingfrom a cyclotron & hitting a target

nucleus inserted in the beam

How?How?

From point of view of target


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From point of view of targetnucleus and in terms of distances

of the order of nuclear radius r,thex position of a proton in the

beam, for all practical purposes,completely unknown, i.e. x>>r`

proton

near nucleus, plane wave

Solve the ProblemSolve the Problem

The proton beam is limited ond b l t d th


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pone end by cyclotron and other

by laboratory wall (distance, L)

Normalize by restricting between0 and L, taking =0 outside

1dxL

0

=

BoxNormalization

Expectation ValueExpectation Value

We know what information we have


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Souri BanerjeeSouri Banerjeesouri@[email protected]

from probability density, regarding

Expectation value gives informationnot only about position but also of

momentum, energy and all otherquantities characterizing its behavior

What isWhat is ExpectationExpectation??

Imagine you make a measurement


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g yof the position of a particle at aninstant t, then, the probability offinding it betweenx andx+dx is :

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

=

Repeat Expt.Repeat Expt.

Repeat the same measurement ab f d l


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pnumber of times on identical system

(same wave function) at the samevalue of tand record the observed

value ofx where we find the particle

Some sort of Average position

of the particle

Mathematical ExpressionMathematical Expression

Average value is the ExpectationV l f di t f th ti l


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pValue ofx coordinate of the particle

at the instant t

In Mathematical Notation:

=

dx)t,x(Pxx

ExplanationExplanation

dx)t,x(Pxx


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=

dx)t,x(Pxx

Integrand is the value ofx weightedby probability of observing that value

Integration gives the average of

the observed value

Final FormFinal Form

The expectation value of position


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dxxx

p pcoordinate is given by:

From Other WayFrom Other Way

iixNx


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=

i

Nx

Average position of the number ofidentical particles distributed along

Xaxis in such a way, that there areN1particles at x1, N2particles at x2,

N3particles at x3and so on.

Associate Probability DensityAssociate Probability Density

For single particle,Ni atxi replacedb b b l h h l b


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dxxdxx

dx

dxxx

2

2

2

by Probability Pi that the particle be

Found in an interval dx atxi

For other functionsFor other functions

d22


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dxxx 22

dx)x(f)x(f

Point to NotePoint to Note

If, ),x(


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,

Can we still use our definition?

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

OK, as all measurements made to

evaluate V(x,t) are made at same t

Looking DeepLooking Deep

Q. Can we write the following?


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dxtxptxp ),(),(

=

Or,

dxtxEtxE ),(),(

=

Uncertainty Plays the RoleUncertainty Plays the Role

In QM, it is NOT possible to writep


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Q , s O poss b e o e p

as a function ofx, sincep andx cannot be simultaneously known with

complete precision

What is the way out?

Operators OperateOperators Operate

For Free-particlewave function:

)wtkx(i0e


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wave function: 0=

xi

p

pi

x

=

=

h

hIt associates thedynamical variable

p with a differentialoperator

Energy OperatorEnergy Operator

)wtkx(ie)tx( Again start with:


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0e)t,x( =

iE

E

i

t

=

h

hE is written interms of adifferentialoperator

OperatorsOperators

xx


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xx

xi

p

h

tiE

h

)


130/343

Words of UncertaintyWords of Uncertainty

2

0

Independent of x


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0=

The particle is equally likely tobe found anywhere, i.e. x=

p=0; momentum precisely knownp h Single value k

Steady State SolutionSteady State Solution

)wtkx(i

0e)t,x(


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0e)t,x(

=

xip

tiE

0 ee

=

hh

t

iE

0 e)x(

h

.(1)

Lecture 5Lecture 5


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Mathematical ExpressionMathematical Expression

Average value is the ExpectationValue of x coordinate of the particle


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Value ofx coordinate of the particle

at the instant t

In Mathematical Notation:

=

dx)t,x(Pxx

Final FormFinal Form

The expectation value of positioncoordinate is given by:


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dxxx

coordinate is given by:

=

dx)t,x(Pxx

For other functionsFor other functions

dxxx 22


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dxxx

dx)x(f)x(f

Looking DeepLooking Deep

Q. Can we write the following?


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

=

Or,

dx)t,x(E)t,x(E *

=

In Inherent ProblemIn Inherent Problem

Take for example:


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dx)t,x(p)t,x(p *

=

To get the integral, the integrandmust be expressed asf(x,t)

Uncertainty Plays the RoleUncertainty Plays the Role

In QM, it is NOT possible to writep


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as a function ofx, sincep andx cannot be simultaneously known with

complete precision

What is the way out?

Operators OperateOperators Operate

For Free-particlewave function:

)tkx(i0e

=


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xi

p

kxi

=

=

h

hh

It associates thedynamical variable

p with a differentialoperator

The StatementThe Statement

The effect of multiplying thefunction (x t) by a dynamical


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function (x,t) by a dynamical

quantityp is same as theeffect of operating it with theDifferential operator:

xi

h

Energy OperatorEnergy Operator

)tkx(i

0e)t,x(

Again start with:


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0e)t,x(

=

iE

ti

=

=

h

hhE is written interms of a

differentialoperator

OperatorsOperators

xx


143/343


xi

p

h

tiE

h

)

Final ExpressionsFinal Expressions

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

=


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),(p),(p

dx)t,x(E)t,x(E *

=

Note: The order of * and

and the operator

Order ofOrder of ,, ** and coordinateand coordinate

Alternatives:


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

=

or

dxp)t,x()t,x(p

=

See What HappensSee What Happens

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


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dx)(xi

h

[ ]

i

h

Move AheadMove Ahead

[ ]

h

= ?


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iImpose a mathematical demand on that it must be normalizable

,y,xfor,0

?

NonNon--sensesense

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


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0

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

=

=

Makes no sense

Take the other oneTake the other one

dxp)t,x()t,x(p

=


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p),(),(p

dxx

)(i

=

h

No Meaning

Matching ClassicallyMatching Classically

For Free-particle :

)tkx(i0e

=


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Evaluate:

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

=

mE2

Extending for GeneralityExtending for Generality

Are the operator relations onlyfor free particles ?


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Take a particle in a potential V

EVm2

p2

=

TrickTrick

2

Replace dynamical quantitiespandEby differential operator


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tiV

m2p

2

=

h

tiVxm2 2

22

=

h

h

Operator Equation

Give Life to the EquationGive Life to the Equation

The operator equation has thesignificance when applied to a


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wave function (x,t)

t)t,x(i

)t,x(Vx

)t,x(

m2 2

22

=

h

h

Sch. Equation ina general form

Hamiltonian OperatorHamiltonian Operator

22


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V

xm2

H2

22

h

Hamiltonian Operator

ExampleExample

FindFind , andand forforthe Gaussian wave packet:the Gaussian wave packet:


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xik2

x2

1

2

1

02

2

ee1)x(

=

StepStep--II

1 22x


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0

dxex1

x

=

=

Integrand is an odd function ofx

Step IIStep II

1 22

x

22


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dxex2

dxex1

x

2

2x

0

2

22

=

=

StepStep--IIIIII

Use:


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aandxexnn

axn 1

0

2

2)12.....(5.3.12

+ =

Putting ValuesPutting Values

2 22


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2

.4.x2

=

=

Step IVStep IV

dx)x(i)x(px

h


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x

dxx

)x()x(i

h

Step ContinuesStep Continues

x2


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=

xik2expxx 02

xik

2

xexpik

x02

2

02

2

Expectation ofExpectation ofppxx

dx)x(

)x(ipx

h


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x

2

2

02

xexpik

xi

h

0x kp h

Last EvaluationLast Evaluation

2

2 dx)x(i)x(p

h


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2

2

2

0

2

x

2k

)(x

)(p

hh

Work out yourself!

Final ResultsFinal Results

x2

2

=


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0x=

2

0x kp h 2

2

2

0

22

x2

kp

hh

Other DefinitionsOther Definitions

Standard Deviation:


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21

22)xx(x

2

12

x

2

xx )pp(p


20

2x

2

12

=


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2k

2kp

2

1

2

0

2

2

2

2

0

2

x

hh

hh =

22.

2p.x

xhh

=

Uncertaintyverified

Steady State SolutionSteady State Solution

)tkx(i

0e)t,x(

=


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xip

tiE

0 ee

=

hh

tiE

0 e)x(

h

.(1)

E= Total

energy ofthe particle

Time independentTime independent SchSch.. EqEq..

Putting (1) in Sch. Eq.

d22

h


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)x(E)x()x(Vdxm2 2 =

1) Not a partial diff. equation

2) (x) is called the stationary stateof the particle as does not

depend on time

2

With OperatorWith Operator

d22

h


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)x(E)x(H=

)x(E)x()x(Vdxm2 2 =

Eigen Function EquationEigen Function Equation

)x(E)x(H =


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1) Eigen value equation. is theeigen-function of the operator (H).The multiplying constant E is thecorresponding eigen-value .

Mathematical Demand onMathematical Demand on

1) must be normalizable.

h f


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2) (x,t) = C11(x,t) + C22(x,t)

vanishes at infinity

Linearity and superposition

On Eigen FunctionOn Eigen Function

1) (x) must befinite and continuous


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3) 1st. Derivative of (x) in spacemust befinite, continuous andsingle-valued

2) (x) must be single valued

Lecture 6Lecture 6


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Eigen Function EquationEigen Function Equation

)x(E)x(H =


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Eigen value equation. is theeigen-state of the operator (H).The multiplying constant E is thecorresponding eigen-value .

Hamiltonian OperatorHamiltonian Operator

VH22

h


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V

xm2

H2

h

Hamiltonian Operator

Mathematical Demand onMathematical Demand on

1) must be normalizable.

i h t i fi it


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vanishes at infinity

You know why

Condition 2Condition 2

2) (x,t) = C11(x,t) + C22(x,t)


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Linearity and superposition

Check !Check !

StatementStatement

t)t,x(i)t,x(V

x)t,x(

m2 2

22

=

hh

If ( ) d ( ) th t


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If 1(x,t) and 2(x,t) are the twosolutions of the above equationfor a particular V, then,

(x,t) = C11(x,t) + C22(x,t)

is also a solution to that equation

Go aheadGo ahead

Substitute the value of (x,t) in theSchroedinger Equation:


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0)

t

C

t

C(i

)CC(V)

x

C

x

C(

m22

2

1

1

22112

2

2

22

1

2

1

2

=

h

h

RearrangeRearrange

0)iV(C

)tiVxm2(C

2

22

2

22

2

1

12

1

22

1

h

h

hh


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0)iVxm2(C222

=

h

If linear combination is also asolution, this equality should besatisfied, which is so for all valuesof C

1and C

2

QuizQuiz

Convince yourself:

This essential result would not


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This essential result would notbeen obtained if the Schroedinger

equation contained terms thatare NOTNOT proportional to the first

power of (x,t) .

Concepts on Eigen StatesConcepts on Eigen States

Consider :

ikx

11 C)(


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11 eC)x( =

andikx

22 eC)x(

=

Insert Momentum OperatorInsert Momentum Operator

)eC(x

i)x(p ikx11

h


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)x(k1

h=

1(x,t) is an eigen state of themomentum operatorand eigenvalue is h

Insert AgainInsert Again

)eC(x

i)x(p ikx22

h


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)x(k2

h

2(x,t) is also an eigen state of

the momentum operatorand eigenvalue is h

Use LinearityUse Linearity

ikx

2

ikx

1 eCeC)x(


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Q. Is (x) also an eigen state of

the momentum operator?

CheckCheck

)x()k()x()k(

)eCeC(x

i)x(p ikx2ikx

1

hh

h


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)x()k()x()k( 21 hh

(x) is a mixed state and it is notan eigen state of momentum operator

Mixed StateMixed State

Q. What you expect with such

mixed states for


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mixedstates for,

dxpp

=

???

Use HUse H

eCk

)eC(xm2)x(H

ikx

22

ikx

12

22

1

h

h


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)x(m2

k

eCm2k

1

22

1

h

h

=

=

1(x,t):An eigen state of H-operator

On the Other StateOn the Other State

eCk

)eC(xm2)x(H

ikx

22

ikx

22

22

2

h

h


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)x(

m2

k

eCm2k

2

22

2

h

h

=

=

2(x,t):Also an eigen state of H-operator

On Mixed StateOn Mixed State

ikx

2

ikx

1 eCeC)x(


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?)x(H =


k

)eCeC(xm2

)x(H

22

ikx

2

ikx

12

22

h

h


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)x(

m2

k

h=

(x) is an eigen state of the

Hamiltonian operatorbut it is not aneigen state of momentum operator

Energy Eigen StateEnergy Eigen StateWhen a particle is in a state that

a measurement total energy leadsto single eigen valueE,

iEt


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h

iEt

e)x()t,x(

=

independent of time(Stationary States)

Two Energy StateTwo Energy State

hh

tiE

22

tiE

11

21

e)x(Ce)x(C)t,x(

time dependent


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time dependent

An example: An electron thatis in the process of making atransition from excited state tothe ground state

(Non-stationary States)

On Eigen FunctionOn Eigen Function

(x) and its first derivative inspace must befinite, continuous

and single valued.


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and single valued.

To ensure that the eigen functionbe a mathematically well-behavedfunction so that measurable quantities

evaluated from eigen functions arealso well behaved.

Why the Condition?Why the Condition?

Q. Why (x) and its first derivativein space must befinite, continuous

and single valued ?


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and s ng va d ?

Probability density cannothave more than one valueat a particularx and t.

ClarificationClarification

dx

)x(dand)x(

If notfinite


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dx

dx

)x(de

x

)t,x(

)x(e)t,x(iEt

iEt

h

h

=

=

Also not

finite


If (x) and its first derivative inspace are notfinite, we wont be

able to arrive at finite and definitel f bl i i


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able to arrive at finite and definitevalues of measurable quantities

Expectation values ofx, p, Eetc.involve and its first derivatives

ContinuityContinuity

In order that 1st. derivative of (x)in space must befinite, it is necessary

that (x) be continuous


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( )

Any function always has an

infinite first derivative whenit has a discontinuity

Other Way RoundOther Way Round

Look at time-independent Sch. Eq.

[ ]

)x(E)x(Vm2

dx

)x(d

22

2

h


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dx h

For finite V(x),Eand (x),2

2

dx

)x(d must be

finite

dx

)x(d

must be continuous

Geometrical InterpretationGeometrical Interpretation

f(x)Not Finite


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xx0

Single ValuedSingle Valued

f(x) Not Single

valued


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xx0

ContinuityContinuity

f(x) Not Continuous


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x0 x

Comments on ContinuityComments on Continuity

Is the continuity of the wavefunction well and truly satisfied

every time?


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y

Normalizing a particle in a box,we do not have a continuous

derivative at the walls as (x)is zero outside

Real LifeReal Life

Discontinuity arises from the factthat we assume, walls are rigid,that is V= at the walls


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In real life, walls are never rigidand there is no sharp change in

at the walls and the derivativesare continuous

Lecture 7Lecture 7


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Particle in a boxParticle in a box

Find the quantum mechanicaldescription of a particle free tomove inside a one-dimensionalbox of length L having rigid walls


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g g g

0 Lx

V=V=

ConditionsConditions

1) Walls are rigid and the particleis free within the boundaries of

the wall


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V at the walls

0V=

inside the box

Schroedinger EquationSchroedinger Equation

[ ]

)x(E)x(Vm2

dx

)x(d22

2

h


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0)x(mE2

dx

)x(d

22

2

=

h

SolutionSolution

kxcosBkxin)x(

..(1)

2

2 mE2k =


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Next task is to evaluate theconstants A and B using theboundary conditions

2k

h

Boundary ConditionsBoundary Conditions

Condition 1:

0xat0


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0a0 =

Wave-function must vanish atthe rigid walls

0B=

22ndnd. Condition. Condition

Lxat0 =


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kLin0=

InferenceInference

kLin0=


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0But,

Why ???


)nin(0kLin

Mathematically,


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)(

n = 0, 1,2,3

Look DeepLook Deep

nkL

)nin(0kLin


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L

nk

nkL

=

= n=0 meansk=0 which isunacceptable

Point to be NotedPoint to be Noted

)nin(0kLin

n = 1,2,3


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n 1,2,3

n =0 excluded!

The WaveThe Wave--functionfunction

)L

xnsin(A)x(

nn

=


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Normalize it!

Final Form of WaveFinal Form of Wave--FunctionFunction

)L

xnsin(

L

2)x(

n

=


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Why (x) is not complex ?

Geometric InterpretationGeometric Interpretation

)L

xnsin(L2)x(n

=

3 | |2


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1

2

3

0 L0 L

|1|2

|2|2

|3|2

Reading FiguresReading Figures

|2|2 dx)x(

2x

x

n

2

1

Check:

Between, sayx1=0.45L


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0L

|1|2

x1 x2

y 1

andx2=0.55L and n=1;n=2

2n1n PP

=

>

EnergyEnergy

2

22222

nmL2

n

m2

k

E

hh

=


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

NoteNote

2

222

nmL2nE

h

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

1) Energy levels are discrete


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2) Lowest (ground)energy state

2

22

1 mL2E

h=

(As n = 0 excluded)

Zero Point EnergyZero Point Energy

2

22

1

mL2

0E h=

Zero PointEnergy


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Consequence of Heisenberg

Uncertainty Principle

ReasoningReasoning

If,E=0 thenp=0

x =


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But x cannot be greater thanLwhich prohibitsEto become zero

Energy SpacingEnergy Spacing

The spacing between the successiveenergy levels:

n)1n(E 2222

h


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[ ]

[ ]

1n2mL2

n)1n(

mL2

E

2

22

2

h

Interesting PointInteresting Point

[ ]

1n2mL2

E2

22

h

As L , E 0


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,

Energy forms a continuum!

Discreteness of energy is an inherentproperty of quantum mechanics

VisualizeVisualize

Origin of the band gap (Eg) andformation of energy bands are

results of quantum mechanicalcalculation of electron transport


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through a crystal

Periodic Boundary Condition

A Subtle PointA Subtle Point

x

V= V=


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-L/2 +L/2

x

0

Where does the calculation differ?

Work OutWork Out

kxcosBkxin)x(

Boundary Condition:


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y

2

Lxat0)x(

ContinueContinue

2

LkcosB

2

LksinA0

And (1)


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2

LkcosB

2

LksinA0

(2)

ContinueContinue(1) + (2)

02

LkcosB2 =

(3)


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(1) - (2)

02LksinA2

=

(4)

What We getWhat We get

02LkcosB2

=

02LksinA2

=


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A and B cannot be zerosimultaneously as the (x)

becomes zero everywhere

PossibilitiesPossibilities

1) 0

2

kLcos,0A =


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2) 0

2

kLsin,0B =

What Then?What Then?

From 1):

2

n

cos02

kL

cos

=


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odd...,3,1n=

L

nk

=

Wave FunctionsWave Functions

)L

xncos(L2)x(

n

=

xn2

oddn


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)L

xnsin(L

2)x(n

=

evenn

Choose proper (x) dependingthe particles state

An ExampleAn Example

Ground-state wave function foran electron trapped inside a box

bounded by L/2 and +L/2 is:


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souri@bitssouri@bits

--pilani.ac.inpilani.ac.in

)Lxcos(

L2)x(

n

=

Calculate Expectation ValueCalculate Expectation Value

dx)t,x(x

)i)(t,x(p2L

2L

h

Expectation value of p for the 1st

excited state:


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souri@bitssouri@bits

--pilani.ac.in

pilani.ac.in

2L

dx)L

xsin()

L

xcos()

L

2)(

L)(i(

2L

2L=

h


dx)Lxsin()

Lxcos(

2L

2L

Integrand is an odd function as it


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is product of an even function andan odd function

0p=

????

RecollectRecollect

)x(k)x(p 2/12/1 h

For,ikx)/(

2/12/1 eC)x( =


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, 2/12/1 )(

Unhindered Plane waves

The Situation HereThe Situation Here

Here you put the rigid walls sothat particle is confined between

-L/2 and +L/2


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It must bounce back and forth and

constantly reversing momentumssign and giving standing wave

Solution We ExpectSolution We Expect

ikxikx ee~)x(


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)L

xnsin(L2)x(

n

=

Standing wave solution

ExpectationExpectation

)x()k(

)x()k()x(p

2

1

h

h

It is equally probable that the


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0dxpp =

sign ofp is either + or -

CheckCheck

Lxncos

Ln

L2)i(

)L

xnsin(L

2

x)i()x(pn

h

h


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)x(pnn

n(x) is not an eigen state of themomentum operator as expected

Home AssignmentHome Assignment

Take:

h

iEt

eLxcos

L2)t,x(

=

2Lx

2L


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FindFind , andand forforthe above wave function and verifythe above wave function and verify

uncertainty principleuncertainty principle

Lecture 8Lecture 8


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Finite Potential wellFinite Potential well

x

V=V=

x

V(x)

V=V0 V=V0

E


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-L +L0 -L +L0

Square Wellwhere increase in pot.energy at the walls is abruptbutfinite

Defining WellDefining Well WellWell

x

V(x)

V=V0 V=V0

E

Conditions:

LxL0

Lx;LxV)x(V 0

1)


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-L +L0

2) The particle with total energyE


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-L +L0 -L +L0

The motion of a classical particlewith E


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One should consider an existence

of (x) outside the well as well

Quantum MechanicsQuantum Mechanics

Why we should have solutionsoutside the well?


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We did not take it for theparticle trapped inside a box

of length L with rigid walls

Ideal vs. RealIdeal vs. RealRigid walls demand V=at the

walls but now we have finite V

The finiteness, along with continuity


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conditions on and d/dx for allvalues ofx, are strictly obeyed andtaken into consideration

In Real LifeIn Real Life

1) Condition,=0 at the walls is lifted

2) For the finiteness of at theboundary and imposing thecontinuity conditions, one would


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y ,expect (x) to exist outside the

well and (x), outside and insidemust join smoothly at the walls

RegionsRegionsV(x)

V=V0 V=V0

EI II III


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-L +Lx

0

All three regions to be studiedand matched at the walls

RegionRegion--IIII

0)x(mE2)x(d22

2

=

h

Things that do NOTNOT change:

1) Wave eqn. inside the region


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dx2) The general solution

kxcosBkxin)x(

RegionRegion--IIII

Things that changechange:

1) The values of the constants Aand B as boundary conditions


254/343



are different2) The eigen function will take

a different look.

Task in HandTask in Hand

When Vis finite, since Sch. Eq. isunaltered inside the well, we have to

supplement the general solution forL


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Remember:Remember:

The condition thatE


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Where,

xx DeCe)x(

h

)EV(m2 0

=

MatchingMatching

cosBin)(

Region-II

1)

2

2 mE2h

=

Region-I and III


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xx DeCe)x(

2)

h

)EV(m2 02

=

Some More ConditionsSome More Conditions

xx DeCe)x(

Impose: vanishes at infinity

Rewrite Equation 1:


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D=0, for eq.1) to be solution forx>L

andandC=0, for eq. 1) to be solution forxL

xDe)x(

=

Forx


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xcosBxin)x(

ForL>x>L

Plot of Wave FunctionsPlot of Wave Functions

2

3


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1

0 L

Evaluating ConstantsEvaluating Constants

Impose other requirements oneq. 1 and 2 that and d/dxmust be continuous atx = L


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What We GetWhat We Get

L

L

CeLsinBLcosA

CeLcosBLsinA

=

Set-I

LDLBLiA


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Set-II

L

L

DeLsinBLcosA

DeLcosBLsinA

=

=

Some ManipulationSome Manipulation

L

L

e)DC(LcosA2e)DC(LsinA2

Set-III

L

)DC(LB2

S t IV


263/343



Le)DC(LsinB2e)DC(LcosB2

Set-I

Quantum Mechanics Souri

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Transcript of Quantum Mechanics Souri