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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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Souri BanerjeeSouri Banerjee
souri@[email protected]
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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Souri BanerjeeSouri Banerjee
souri@[email protected]
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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souri@[email protected]
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:
Act of measurement interfereswith system and modifies it
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
Construction of Wave PacketConstruction of Wave Packet
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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
= +
+
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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souri@[email protected]
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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souri@[email protected]
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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ObservationsObservations
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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Thought ExperimentThought Experiment
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
Thought ExperimentThought Experiment
ObservationsObservations
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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
Act of measurement interfereswith system and modifies it
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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souri@[email protected]
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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souri@[email protected]
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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souri@[email protected]
)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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souri@[email protected]
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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souri@[email protected]
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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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
)
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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
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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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souri@[email protected]
21
22)xx(x
2
12
x
2
xx )pp(p
Final ResultFinal Result
20
2x
2
12
=
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souri@[email protected]
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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souri@[email protected]
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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souri@[email protected]
)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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souri@[email protected]
)x(E)x(H=
)x(E)x()x(Vdxm2 2 =
Eigen Function EquationEigen Function Equation
)x(E)x(H =
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souri@[email protected]
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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souri@[email protected]
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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souri@[email protected]
3) 1st. Derivative of (x) in spacemust befinite, continuous andsingle-valued
2) (x) must be single valued
Lecture 6Lecture 6
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souri@[email protected]
Eigen Function EquationEigen Function Equation
)x(E)x(H =
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Souri BanerjeeSouri Banerjee
souri@[email protected]
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 =
ConclusionConclusion
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
ConclusionConclusion
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 ???
Final ResultFinal Result
)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
Final ResultFinal Result
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
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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
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Souri BanerjeeSouri Banerjee
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Le)DC(LsinB2e)DC(LcosB2
Set-I