Quantum Mechanics IBM Open Source

8/2/2019 Quantum Mechanics IBM Open Source

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Quantum Mechanics

by

Dr. Amit Kumar Chawla


2/87

Introduction

1. Stability of an atom

2. Spectral series of Hydrogen atom

3. Black body radiation

There are a few phenomenon which the classical mechanics

failed to explain.

Max Planck in 1900 at a meeting of German Physical Societyread his paper On the theory of the Energy distribution law of

the Normal Spectrum. This was the start of the revolution ofPhysics i.e. the start ofQuantum Mechanics.


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Quantum Mechanics

Quantum Physics extends that range to the region of smalldimensions.

It is a generalization of Classical Physics that includes classicallaws as special cases.

Just as c the velocity of light signifies universal constant, the

Planck's constant characterizes Quantum Physics.

sec.10625.6

sec.1065.634

27

Jouleh

ergh

==


4/87

Quantum Mechanics

1. Photo electric effect

2. Black body radiation

3. Compton effect

4. Emission of line spectra

It is able to explain

The most outstanding development in modern science wasthe conception of Quantum Mechanics in 1925. This newapproach was highly successful in explaining about thebehavior of atoms, molecules and nuclei.


5/87

Photo Electric Effect

The emission of electrons from a metal plate when illuminated

by light or any other radiation of any wavelength or frequency(suitable) is called photoelectric effect. The emitted electronsare called photo electrons.

V

Evacuated

Quartztube Metal

plate

Collectingplate

Light

^^^^^^^^ A

+_


6/87


Experimental findings of the photoelectric effect

1. There is no time lag between the arrival of light at the metalsurface and the emission of photoelectrons.

2. When the voltage is increased to a certain value say Vo, the

photocurrent reduces to zero.3. Increase in intensity increase the number of the

photoelectrons but the electron energy remains the same.

3I

Voltage

PhotoCurrent

2I

I

Vo


7/87


4. Increase in frequency of light increases the energy of the

electrons. At frequencies below a certain critical frequency(characteristics of each particular metal), no electron isemitted.

Voltage

Photo Current

v1

v2

v3


8/87

Einsteins Photo Electric Explanation

The energy of a incident photon is utilized in two ways

1. A part of energy is used to free the electron from the atomknown as photoelectric workfunction (Wo).

2. Other part is used in providing kinetic energy to the emitted

electron .

22

1mv

2

2

1mvWh o +=

This is called Einsteins photoelectric equation.


9/87


10/87

It is in form of . The graph with on y-axis andon x-axis will be a straight line with slope

oo hheV =

oVIf is the stopping potential, then

)(max ohKE =

e

h

e

h

Vo

o

=cmxy +=

ehoV


11/87

Photons

Einstein postulated the existence of a particle called a photon, to

explain detailed results of photoelectric experiment.

hc

hEp ==

Photon has zero rest mass, travels at speed of light

Explains instantaneous emission of electrons in photoelectriceffect, frequency dependence.


12/87

Compton Effect

When a monochromatic beam of X-rays is scattered from a

material then both the wavelength of primary radiation(unmodified radiation) and the radiation of higher wavelength(modified radiation) are found to be present in the scatteredradiation. Presence of modified radiation in scattered X-rays iscalled Compton effect.

electron

scatteredphoton

recoiled electron

hE=

c

hp

=

'' hE =

v

cosmv

sinmv

incidentphoton

cos'c

h

sin'

c

h


13/87

From Theory of Relativity, total energy of the recoiled electronwith v ~ c is

22 cmKmcE o+==

Similarly, momentum of recoiled electron is

22 cmmcK o=

2

22

2

1cm

cv

cmK o

o

=

= 1

1

1

22

2

cvcmK o

221 cv

vmmv o

=


14/87

Now from Energy Conversation

cos1cos

'

22 cv

vm

c

h

c

h o

+=

+= 1

1

1'

22

2

cv

cmhh o (i)

From Momentum Conversation

(ii)along x-axis

sin1

sin'0 22 cv

vm

ch o

=(iii) along y-axis

and


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Rearranging (ii) and squaring both sides

2

22

222

cos1

cos'

cv

vm

c

h

c

h o

=

(iv)

2

22

222

sin1

sin'

cv

vm

c

h o

=

(v)

Rearranging (iii) and squaring both sides

Adding (iv) and (v)

22

22

2

222

1cos'2'

cv

vm

c

h

c

h

c

h o

=

+

(vi)

From equation (i)

22

1

'

cv

cmcm

c

h

c

h oo

=+


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On squaring, we get

Subtracting (vi) from (vii)(vii)

22

22

2

222

22

1

)'(2'2'

cv

cmhm

c

hcm

c

h

c

h ooo

=++

+

0)'(2)cos1('2

2

2

=+

ohmc

h

)cos1('2

)'(22

2

=c

hhmo

)cos1('

)'(2

=c

hmo


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But

is the Compton Shift.

c=

)cos1(''

11

=

h

cmo

and'

'

c

= So,

)cos1(

''

'

=

hcmo

)cos1(' ==cm

h

o

It neither depends on the incident wavelength nor on thescattering material. It only on the scattering angle i.e.

is called the Compton wavelength of the electron

and its value is 0.0243 .cm

h

o


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Experimental Verification

Monochromatic

X-ray Source

photon

Graphitetarget

Braggs X-raySpectrometer

1. One peak is found at same position.This is unmodified radiation

2. Other peak is found at higher

wavelength. This is modified signal oflow energy.

3. increases with increase in .


19/87

0.0243 (1- cos) == )cos1( cmh

o

= max

So Compton effect can be observed only for radiation havingwavelength of few .

Compton effect cant observed in Visible Light

is maximum when (1- cos) is maximum i.e. 2.

0.05

For 1 ~ 1%

==

For 5000 ~ 0.001% (undetectable)


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Pair Production

When a photon (electromagnetic energy) of sufficient energy

passes near the field of nucleus, it materializes into anelectron and positron. This phenomenon is known as pairproduction.

In this process charge, energy and momentum remainsconserved prior and after the production of pair.

PhotonNucleus (+ve)

e

+e


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The rest mass energy of an electron or positron is 0.51 MeV(according to E = mc2).

The minimum energy required for pair production is 1.02

MeV.

Any additional photon energy becomes the kinetic energy ofthe electron and positron.

The corresponding maximum photon wavelength is 1.2 pm.Electromagnetic waves with such wavelengths are calledgamma rays .)(


22/87

Pair Annihilation

When an electron and positron interact with each other due

to their opposite charge, both the particle can annihilateconverting their mass into electromagnetic energy in theform of two - rays photon.

++ + ee

Charge, energy and momentum are again conversed. Two- photons are produced (each of energy 0.51

MeV plus half the K.E. of the particles) to conserve the

momentum.


23/87

From conservation of energy 22 cmh o=

Pair production cannot occur in empty space

In the direction of motion of the photon, the momentum isconserved if

cos2p

c

h=

ch

cosp

cospp

p

e

+

e

here mo is the rest mass and2211 cv=


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vmp o=

Momentum of electron and positron is

(i)

1cos 1


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But conservation of energy requires that

2

2 cmh o=Hence it is impossible for pair production to conserve both

the energy and momentum unless some other object isinvolved in the process to carry away part of the initial

photon momentum. Therefore pair production cannot occurin empty space.


26/87

Wave Particle Duality

Light can exhibit both kind of nature of waves and particles

so the light shows wave-particle dual nature.

In some cases like interference, diffraction and polarization itbehaves as wave while in other cases like photoelectric andcompton effect it behaves as particles (photon).


27/87

De Broglie Waves

Not only the light but every materialistic particle such as

electron, proton or even the heavier object exhibits wave-particle dual nature.

De-Broglie proposed that a moving particle, whatever itsnature, has waves associated with it. These waves arecalled matter waves.

Energy of a photon is

hE=

For a particle, say photon of mass, m

2mcE=

2


28/87

Suppose a particle of mass, m is moving with velocity, v then

the wavelength associated with it can be given by

hvmc =2

hcmc =2

mc

h=

mv

h=

p

h=or

(i) If means that waves are associated with movingmaterial particles only.== 0v

(ii) De-Broglie wave does not depend on whether the movingparticle is charged or uncharged. It means matter waves are

not electromagnetic in nature.


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Wave Velocity or Phase Velocity

When a monochromatic wave travels through a medium, its

velocity of advancement in the medium is called the wavevelocity or phase velocity (Vp).

kVp

=

where is the angular frequency

and is the wave number.

2=

2=k


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Group Velocity

So, the group velocity is the velocity with which the energy inthe group is transmitted (Vg).

dk

dVg

=

The individual waves travel inside the group with theirphase velocities.

In practice, we came across pulses rather thanmonochromatic waves. A pulse consists of a number ofwaves differing slightly from one another in frequency.

The observed velocity is, however, the velocity with whichthe maximum amplitude of the group advances in a medium.


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Relation between Phase and Group Velocity

dk

dVg

= )(

p

kVdk

d=

dk

dVkVV

p

pg +=

( )

2

2

d

dVVVp

pg +=

( ) 1

1

d

dV

VV

p

pg +=


32/87

d

dVVV

p

pg =

+=

d

dVVV

p

pg

2

1

1

d

dVpSo, is positive generally (not always).

pg VV


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d

dVVV

p

pg =

0=d

dVp

pg VV =

In a non-dispersive medium ( such as empty space)

=k

constant pV=


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Phase Velocity of De-Broglies waves

According to De-Broglies hypothesis of matter waves

mv

h=

wave number

h

mvk

22== (i)

If a particle has energy E, then corresponding wave willhave frequency

h

E=

then angular frequency will beh

E

22 ==

2


35/87

Dividing (ii) by (i)

(ii)h

mc22=

mvh

hmc

k

22

2

=

v

cVp

2

=

But v is always < c (velocity of light)

(i) Velocity of De-Broglies waves (not acceptable)cVp >

(ii) De-Broglies waves will move faster than the particlevelocity (v) and hence the waves would left the particlebehind.

)( pV


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Group Velocity of De-Broglies waves

The discrepancy is resolved by postulating that a moving

particle is associated with a wave packet or wavegroup, rather than a single wave-train.

A wave group having wavelength is composed of anumber of component waves with slightly different

wavelengths in the neighborhood of .

Suppose a particle of rest mass mo moving with velocity v

then associated matter wave will have

hmc

2

2= andhmvk 2= where 221 cv

mm o=


37/87

and

On differentiating w.r.t. velocity, v

22

2

1

2

cvh

cmo

=

221

2

cvh

vmk o

=

( )2

3

221

2

cvh

vm

dv

d o

=

(i)

( )2

322

1

2

cvh

m

dv

dk o

= (ii)


38/87

Wave group associated with a moving particle also moveswith the velocity of the particle.

o

o

m

vm

dk

dv

dv

d

2

2. =

Dividing (i) by (ii)

gVvdk

d==

Moving particle wave packet or wave group

Davisson & Germer experiment of electron


39/87

Davisson & Germer experiment of electron

diffraction

If particles have a wave nature, then under appropriateconditions, they should exhibit diffraction

Davisson & Germer measured the wavelength of electrons

This provided experimental confirmation of the matter waves

proposed by de Broglie


40/87

Davisson and Germer Experiment

0


41/87

0=

000=

Current vs accelerating voltage has a maximum (a bump orkink noticed in the graph), i.e. the highest number of electronsis scattered in a specific direction.

The bump becomes most prominent for 54 V at ~ 50

IncidentBeam


42/87

According to de Broglie, the wavelength associated with anelectron accelerated through V volts is

o

AV

28.12

=

Hence the wavelength for 54 V electron

o

A67.154

28.12

==

From X-ray analysis we know that the nickel crystal acts as aplane diffraction grating with grating space d = 0.91


43/87

ooo

652

50180=

=

Here the diffraction angle, ~ 50

The angle of incidence relative to the family of Braggs plane

From the Braggs equation


44/87

From the Bragg s equation

which is equivalent to the calculated by de-Broglieshypothesis.

sin2d=o

oo

AA 65.165sin)91.0(2 ==

It confirms the wavelike nature of electrons

El t Mi I t t l A li ti


45/87

Electron Microscope: Instrumental Application

of Matter Waves


46/87

Resolving power of any optical instrument is proportional to thewavelength of whatever (radiation or particle) is used toilluminate the sample.

An optical microscope uses visible light and gives 500xmagnification/200 nm resolution.Fast electron in electron microscope, however, have muchshorter wavelength than those of visible light and hence aresolution of ~0.1 nm/magnification 1,000,000x can be achievedin an Electron Microscope.

Heisenberg Uncertainty Principle


47/87

Heisenberg Uncertainty Principle

It states that only one of the position or momentum can bemeasured accurately at a single moment within the instrumental limit.

It is impossible to measure both the position and momentumsimultaneously with unlimited accuracy.

or

uncertainty in position

uncertainty in momentumx xp

then

2

xpx2

h=

The product of & of an object is greater than or equal to2

xpx

0


48/87

If is measured accurately i.e.x 0x xp

Like, energy E and time t.

2

tE

2

0 L

The principle applies to all canonically conjugate pairs of quantities inwhich measurement of one quantity affects the capacity to measurethe other.

and angular momentum L and angular position

Determination of the position of a particle by a microscope


49/87

Determination of the position of a particle by a microscope

i

IncidentPhoton

ScatteredPhoton

Recoiled electron

Suppose we want to determine accurately the position and momentumof an electron along x-axis using an ideal microscope free from all

mechanical and optical defects.

The limit of resolution of themicroscope is

ix sin2

=here i is semi-vertex angle of the

cone of rays entering the objectivelens of the microscope.

is the order of uncertainty in thex-component of the position of theelectron.

x

We cant meas re the moment m of the electron prior to ill mination


50/87

The scattered photon can enter the microscope anywhere between theangular range +i to i.

We cant measure the momentum of the electron prior to illumination.So there is uncertainty in the measurement of momentum of theelectron.

The momentum of the scattered photon is (according to de-Broglie)

h

p =Its x-component can be given as

ih

px sin2

=

The x-component of the momentum of the recoiling electron has thesame uncertainty, (conservation of momentum)

xp

The product of the uncertainties in the x components of position and


51/87

The product of the uncertainties in the x-components of position andmomentum for the electron is

ih

ipx x sin

2

sin2.

=

This is in agreement with the uncertainty relation.

2.

0>= hpx x

Applications of Heisenberg Uncertainty Principle


52/87

Order of radius of an atom ~ 5 x10-15 m

then

If electron exist in the nucleus then

Applications of Heisenberg Uncertainty Principle

(i) Non-existence of electron in nucleus

mx 15max 105)(=

2

xpx

2)()( minmax

0= xpx

120

min ..101.1

2

)( =

= smkgx

px0

MeVpcE 20== relativistic


53/87

Thus the kinetic energy of an electron must be greater than 20MeV to be a part of nucleus

Thus we can conclude that the electrons cannot be presentwithin nuclei.

Experiments show that the electrons emitted by certain unstablenuclei dont have energy greater than 3-4 MeV.

Concept of Bohr Orbit violates Uncertainty Principle


54/87

But

Concept of Bohr Orbit violates Uncertainty Principle

m

pE

2

2

=2

.0

px

mppE =

mpmv= p

tx

=

pxtE = ..

2.

0 tE

A di t th t f B h bit f l t i


55/87

According to the concept of Bohr orbit, energy of an electron in aorbit is constant i.e. E = 0.

2.

0

tE t

All energy states of the atom must have an infinite life-time.

But the excited states of the atom have lifetime ~ 10-8 sec.The finite life-time t gives a finite width (uncertainty) to the energylevels.

Two slit Interference Experiment


56/87

Two-slit Interference Experiment

Laser

Source

Slit

Slit Detector

Rate of photon arrival = 2 x 106/sec

Rate of photon detection = 105

/secTime lag = 0.5 x 10-6 sec

Spatial separation between photons = 0.5 x 10-6 c = 150 m

1 meter

T l i t (1908) d bl lit i t ith di


57/87

Taylors experiment (1908): double slit experiment with very dimlight: interference pattern emerged after waiting for few weeks

interference cannot be due to interaction between photons, i.e.

cannot be outcome of destructive or constructive combination ofphotons

interference pattern is due to some inherent property of eachphoton - it interferes with itself while passing from source toscreen

photons dont split light detectors always show signals of same intensity

slits open alternatingly: get two overlapping single-slit diffractionpatterns no two-slit interference

add detector to determine through which slit photon goes:no interference

interference pattern only appears when experiment providesno means of determining through which slit photon passes

Double slit experiment QM interpretation


58/87

Double slit experiment QM interpretation

patterns on screen are result of distribution of photons

no way of anticipating where particular photon will strike impossible to tell which path photon took cannot assign specifictrajectory to photon

cannot suppose that half went through one slit and half throughother

can only predict how photons will be distributed on screen (orover detector(s))

interference and diffraction are statistical phenomena associatedwith probability that, in a given experimental setup, a photon will

strike a certain point high probabilitybright fringes

low probabilitydark fringes

Double slit expt wave vs quantum


59/87

Double slit expt. -- wave vs quantum

pattern of fringes:

Intensity bands due tovariations in square ofamplitude, A2, of resultant

wave on each point onscreen

role of the slits:

to provide two coherent

sources of the secondarywaves that interfere on thescreen

pattern of fringes:

Intensity bands due tovariations in probability, P, ofa photon striking points on

screen

role of the slits:

to present two potential

routes by which photon canpass from source to screen

wave theory quantum theory

Wave function


60/87

Wave function

*|| 2 =

The quantity with which Quantum Mechanics is concerned is thewave function of a body.

||2 is proportional to the probability of finding a particle at aparticular point at a particular time. It is the probability density.

Wave function, is a quantity associated with a moving particle. It isa complex quantity.

Thus if iBA += iBA =*222222

*|| BABiA +===

then

is the probability amplitude.

Normalization


61/87

Normalization

d

||2 is the probability density.

The probability of finding the particle within an element of volume d2||

Since the particle is definitely be somewhere, so

1|| 2 =

d

A wave function that obeys this equation is said to be normalized.

Normalization


62/87

Schrodingers time independent wave equation


63/87

Schrodinger s time independent wave equation

xA 2sin=

One dimensional wave equation for the waves associated with amoving particle is

From (i)

xA

x

2sin

42

2

2

2

=

is the wave amplitude for a given x.where

A is the maximum amplitude. is the wavelength

(i)

2

2

2

2 4=

x(ii)

h


64/87

vm

h

o

=

2

22

2

1

h

vmo= 2

2

2

12

h

vmm oo =

22

21

h

Kmo

=where K is the K.E. for the non-relativistic case

(iii)

Suppose E is the total energy of the particle

and V is the potential energy of the particle

)(21

22VE

h

mo =

Equation (ii) now becomes


65/87

This is the time independent (steady state) Schrodingers wave

equation for a particle of mass mo, total energy E, potential energyV, moving along the x-axis.

If the particle is moving in 3-dimensional space then

)(24

2

2

2

2

VEmhx

o =

0)(2

22

2

=+

VEm

x

o

0

0)(222

2

2

2

2

2 =++

+

VEm

zyx

o

0


66/87

For a free particle V = 0, so the Schrodinger equation for afree particle

02

2

2 =+ Emo0

0)(2

2

2 =+ VEmo

0

This is the time independent (steady state) Schrodingers waveequation for a particle in 3-dimensional space.

Schrodingers time dependent wave equation


67/87

Schrodinger s time dependent wave equation

)( pxEt

i

Ae

=

Wave equation for a free particle moving in +x direction is

(iii)

2

2

2

2

p

x=

where E is the total energy and p is the momentum of the particle

Differentiating (i) twice w.r.t. x

(i)

2

222

xp

=

0 (ii)

Differentiating (i) w.r.t. t

0

iE

t=

tiE

= 0

For non-relativistic case


68/87

Using (ii) and (iii) in (iv)

(iv)

Vxmt

i +

=

2

22

2

00

E = K.E. + Potential Energy

txVm

pE ,

2

2 +=

V

m

pE +=

2

2

This is the time dependent Schrodingers wave equation for aparticle in one dimension.

Linearity and Superposition


69/87

Linearity and Superposition

2211 aa +=

If 1 and 2 are two solutions of any Schrodinger equation of a

system, then linear combination of 1 and 2 will also be a solution

of the equation..

Here are constants

Above equation suggests:

21 & aa

is also a solution

(i) The linear property of Schrodinger equation

(ii) 1 and 2 follow the superposition principle

If P1 is the probability density corresponding to 1 and P2 is the


70/87

21 +ThenTotal probability will be

2

21

2 |||| +==P

due to superposition principle

)()( 21*

21 ++=

))(( 21*2*1 ++=1

*

22

*

12

*

21

*

1 +++=

1

*

22

*

121 +++= PPP

21 PPP +Probability density cant be added linearly

1 p y y p g 1 2probability density corresponding to 2

Expectation values


71/87

Expectation values

dxxf

= 2||)(

Expectation value of any quantity which is a function of x ,say f(x) isgiven by

for normalized

Thus expectation value for position x is

>< )(xf

dxx

= 2||>< x

Expectation value is the value of x we would obtain if we measuredthe positions of a large number of particles described by the samefunction at some instant t and then averaged the results.

Q. Find the expectation value of position of a particle having wave


72/87

dxx=1

0

2||

Solution

>< x

1

0

42

4

=

x

a

p p p gfunction = ax between x = 0 & 1, = 0 elsewhere.

dxxa =1

0

32

>< x4

2a=

Operators


73/87

p

pix 0

=

(Another way of finding the expectation value)

For a free particle

An operator is a rule by means of which, from a given functionwe can find another function.

)( pxEti

Ae

= Then

Here

xip

=0^

is called the momentum operator

(i)

iSimilarly


74/87

Ei

t 0=

Similarly

Here

tiE

= 0

^

is called the Total Energy operator

(ii)

Equation (i) and (ii) are general results and their validity is thesame as that of the Schrodinger equation.

If a particle is not free then^


75/87

Uximt

i +

= 2

2

1 00

This is the time dependent Schrodinger equation

^^^

.. UEKE +=^

^2^

2U

m

pE

o

+=

UU =^

Uxmt

i += 222

200

U

xmt

i +

=

2

22

2

00

If Operator is Hamiltonian


76/87

If Operator is Hamiltonian

Then time dependent Schrodinger equation can be written as

U

xm

H +

=

2

22^

2

0

EH =^

This is time dependent Schrodinger equation in Hamiltonianform.

Eigen values and Eigen function


77/87

g g

Schrodinger equation can be solved for some specific values ofenergy i.e. Energy Quantization.

a=^

Suppose a wave function () is operated by an operator such that

the result is the product of a constant say a and the wave functionitself i.e.

The energy values for which Schrodinger equation can be solved arecalled Eigen values and the corresponding wave function are calledEigen function.

then

is the eigen function of

a is the eigen value of

^

^

Q Suppose is eigen function of operator then findxe2=

2d


78/87

Q. Suppose is eigen function of operator then findthe eigen value.

The eigen value is 4.

Solution.

e=2dx

2

2^

dxdG =

2

2^

dx

dG

= )( 2

2

2xe

dx

d=

xeG 2^

4=

4^

=G

Particle in a Box


79/87

Consider a particle of rest mass mo enclosed in a one-dimensional

box (infinite potential well).

Thus for a particle inside the box Schrodinger equation is

Boundary conditions for Potential

V(x)=

0 for 0 < x < L

{ for 0 > x > LBoundary conditions for

=

0 for x = 0

{0 for x = L

02

22

2

=+

Em

x

o

0

x = 0 x = L

=V =V

particle

0=V

0=V inside(i)

h 2


80/87

Equation (i) becomes

p

h=

k

2=

0pk =

0

Emo2=

2

2 2

0

Emk o=

(k is the propagation constant)

(ii)

022

2

=+

kx

(iii)

General solution of equation (iii) is

kxBkxAx cossin)( += (iv)

Boundary condition says = 0 when x = 0


81/87

Equation (iv) reduces to

Boundary condition says 0 when x 0

(v)

0.cos0.sin)0( kBkA +=

1.00 B+= 0= B

kxAx sin)( =

Boundary condition says = 0 when x = L

LkAL .sin)( =

LkA .sin0 =0A 0.sin = Lk

nLk sin.sin =

nkL =


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Put this in Equation (v)

(vi)

L

nk

=

L

xnAx

sin)( =

When n # 0 i.e. n = 1, 2, 3., this gives = 0 everywhere.

nkL =

Put value of k from (vi) in (ii)

2

2 2

0

Emk o=

2

22

0

Em

L

n o=

k 22022h


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Where n = 1, 2, 3.

Equation (vii) concludes

om

kE

2

220

=2

22

8 Lm

hn

o

= (vii)

1. Energy of the particle inside the box cant be equal to zero.

The minimum energy of the particle is obtained for n = 1

2

2

18 Lm

hE

o

= (Zero Point Energy)

If momentum i.e.01 E 0 0p xBut since the particle is confined in the box of dimensionL.

Lx = max

Thus zero value of zero point energy violates the Heisenbergs


84/87

uncertainty principle and hence zero value is not acceptable.

2. All the energy values are not possible for a particle inpotential well.

Energy is Quantized

3. En are the eigen values and n is the quantum number.

4. Energy levels (En) are not equally spaced.

n = 1

n = 3

n = 2

3E

1E

2E

xnAx

sin)( =


85/87

Using Normalization condition

LAxn sin)( =

1sin0

22 =

dx

L

xnA

L

1|)(| 2 =

dxxn

12

2 =

L

AL

A2

=

The normalized eigen function of the particle are

L

nx

Lxn

sin

2)( =

Probability density figure suggest that:


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1. There are some positions (nodes) in the box that will never beoccupied by the particle.

2. For different energy levels the points of maximum probability arefound at different positions in the box.

|1|2 is maximum at L/2 (middle of the box)

|2

|2 is zero L/2.

Particle in a Three Dimensional Box


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Eigen functionzyx =

L

zn

L

yn

L

xnAAA z

yxzyx

sinsinsin=

L

zn

L

yn

L

xn

L

zyx sinsinsin2 3

=

2

2222

8)(

mL

hnnnE zyx ++=

zyx EEEE ++=Eigen energy

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Quantum Mechanics IBM Open Source