Quantum Mechanics

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K.S. MAHESH LOHITHAssistant Professor, Center for emerging technologies,SBM Jain College of Engineering,BANGALORE-562 112.

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PHOTOELECTRIC EFFECT

• OBSERVED BY HENRICH HERTZ IN THE YEAR 1887

• ALBERT EINSTEIN EXPLAINED PHOTOELECTRIC EFFECT ON THE BASIS OF PHOTON THEORY

PHYSICAL SIGNIFICANCE: PARTICLE NATURE OF LIGHT

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DUAL NATURE OF LIGHT

INTERFERENCE DIFFRACTION POLARIZATION

PHOTOELECTRIC EFFECT COMPTON EFFECT

HENCE LIGHT POSSES DUAL NATURELight behaves a particle under certain circumstances

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Historically, the electron was thought to behave like a Particle and then it was found that in many respects it behaved like a wave. So it really behaves like neither. Now we have given up.We say: It is like Neither.

Richard Feynman

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DE BROGLIE HYPOTHESIS

LOUIS DE BROGLIE

“ If radiation which is basically a wave can exhibit particle nature under certain circumstances, and since nature likes symmetry, then entities which exhibit particle nature ordinarily, should also exhibit wave nature under suitable circumstances”

In the Year 1924 Louis de Brogliemade the bold suggestion

The reasoning used might be paraphrased as follows

1. Nature loves symmetry2. Therefore the two great entities, matter and

energy, must be mutually symmetrical3. If energy (radiant) is undulatory and/or

corpuscular, matter must be corpuscular and/or undulatory

6nm

VoltsVforthus

nmVV

getweeandmhforngsubstitutimeV

h

mE

hThen

VdifferencePotentialabydaccelerate

EEnergyKineticwithelectronanfor

mv

h

p

hwavelengthBrogliede

1226.0100

226.1

100

226.1

10602.11011.92

10625.6

,,22

''

''

particle theof velocity theis v

particle theof mass theis m

Constant sPlanck' ish

1931

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DE BROGLIE WAVELENGTH

The Wave associated with the matter particle is called Matter Wave.The Wavelength associated is called de Broglie Wavelength.

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G P THOMSON’S EXPERIMENT

Diffraction of electrons from the Gold foil suggests dual nature of Electrons

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PHASE VELOCITYPhase velocity: The velocity with which a wave travels is called Phase velocity or wave velocity. It is denoted by vp. It is given by

v

cv p

2

Where c = velocity of light and v = is velocity of the particle.

The above equation gives the relationship between the phase velocity and

particle velocity.

It is clear from the above equation that, Phase velocity is not only greater than the velocity of the particle but also greater than the velocity of light,

which can never happen. Therefore phase velocity has no physical meaning in case of matter waves. Thus a concept of group

velocity was introduced.

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GROUP VELOCITYSince phase velocity has no meaning, the concept of group

velocity was introduced as follows.“ Matter wave is regarded as the resultant of the superposition of

large number of component waves all traveling with different velocities. The resultant is in the form of a packet called wave packet or wave group. The velocity with which this wave group travels is called group velocity.” The group velocity is represented by vg.

Vg

Particle

Vp

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PROPERTIES OF MATTER WAVES

p

h

mv

h

p

h

mv

h

P ro p e r tie s o f M a tte r W a v e s M a tte r w a v e s a re a s s o c ia te d w ith m o v in g p a r tic le . T h e y a re n o t E le c tro m a g n e tic w a v e s . W a v e le n g th o f th e m a tte r w a v e is g iv e n b y p

h

mv

h

T h e a m p litu d e o f th e m a tte r w a v e a t th e g iv e n p o in t d e te rm in e s th e p ro b a b ili ty o f f in d in g th e p a r tic le a t th a t p o in t.

T h e re is n o m e a n in g fo r P h a s e v e lo c ity in c a s e o f m a tte r w a v e s . O n ly g ro u p v e lo c ity h a s m e a n in g .

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HEINSENBERG’S UNCERTAINTY PRINCIPLE

“It is impossible to determine both position and momentum of a particle simultaneously and accurately. The product of uncertainty involved in the determination of position and momentum simultaneously is greater or equal to h/2Π ”

2h

px x

2h

tE

Significance: “Probalility” replaces “Exactness”

An event which is impossible to occur according to classical physics has a finite probability of occurrence according to Quantum Mechanics

Heisenberg - 1927

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GAMMA RAY MICROSCOPE EXPERIMENTImaginary Experiment

B

Incident gamma ray

Scattered gamma ray

Recoiled electron

X-axis

A

O

To determine both position and momentum of the electron

Limit of resolution Δx is the measure of uncertainty involved in the measurement of position

Sin

x2

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During the collision, photon transfers momentum to the electron. The momentum transferred to the electron is of the order h/λ the momentum transferred to the electron is of the order If photon enters the microscope along the path OA and OB are of the order ,

Since photon can enter the microscope anywhere between the paths OA or OB, there exist uncertainty in the determination of momentum of the electron given by 

Taking the product of equations 1 and 2 we get

From more sophisticated theory it can be shown thatWhich is nothing but HUP.

Sinh

Sinh

Sinh

Sinh

Sinh

px

2

hSinh

Sinpx x

2

2

2h

px x

GAMMA RAY MICROSCOPE EXPERIMENT

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SCHRODINGER’S CAT

“You w

ill see

me t

here,”

said th

e cat,

But v

anish

ed

Lewis

Carroll

In Alic

e in W

onder Lan

d

A Paradox

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TWO SLIT EXPERIMENT The only mystery

Double slit

Interference patternExperiment conducted with Bullets, Light and electrons

No matter how many times we repeat the experiment for electrons we get consistent results-”Interference it there are no observers, and No interference pattern if there are observers”- This is confirmed.

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Never in human history had such a thing been encountered before. I mean who ever heard of Nature behaving one way when you are looking and completely different way when you are not looking? And yet that is precisely what happens, at least in the world of electrons. It seems that when we observe we disturb whenever it is that we are trying to observe

Heisenberg’s Comment on two slit experiment

This is a very strange result, since it seems to indicate that the observation plays a decisive role in the event and that reality varies, depending upon whether we observe it or not

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TIME INDEPENDENT SCHROEDINGER EQUATION

Erwin Schroedinger

Consider a particle of mass ‘m’, moving with a velocity ‘v’ along + ve X-axis. Then the according to de Broglie Hypothesis, the wave length of the wave associated with the particle is given by

mv

h

A wave traveling along x-axis can be represented by the equation

xktieAtx ,

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Where Ψ(x,t) is called wave function. The differential equation of matter wave in one dimension is derived as

08

2

2

2

2

VE

h

m

dx

d

The above equation is called one-dimensional Schroedinger’s wave equation in one dimension.In three dimensions the Schroedinger wave equation becomes

08

08

2

22

2

2

2

2

2

2

2

2

VEh

m

VEh

m

zyx

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According to Max Born’s interpretation of the wavefunction, the only quantity that has some meaning is

2 2

PHYSICAL INTERPRETATION OF WAVE FUNCTION

2

2

The state of a quantum mechanical system can be completely understood with the help of the wave function ψ. But wave function ψ can be real or imaginary. Therefore no meaning can be assigned to wavefunction ψ as it is.

which is called as probability density.

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dv

We know that electron is definitely found somewhere in the space. The wavefunction ψ, which satisfies the above condition, is called normalized wavefunction.

Thus if ψ is the wavefunction of a particle within a small region of volume dv, then

dV2 gives the probability of finding the particle within

the region dv at the given instant of time.

V

1dv 2

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Nature of Eigenvalues and EigenfunctionsA physical system can be completely described with the help of the wave function ψ. In order to get wavefunction, first we have to set up a Schrodinger wave equation representing the system. Then, Schrodinger wave equation has to be solved to get wavefunction ψ as a solution. But Schrodinger wave equation, which is a second order differential equation, has multiple solutions. All solutions may not represent the physical system under consideration. Those wavefunction, which represent the physical system under consideration, are acceptable, and are called Eigenfunctions.A wavefunction ψ can be acceptable as wavefunction if it satisfies the following conditions.

1.  ψ should be single valued and finite everywhere.2. ψ and its first derivatives with respect to its variables are continuous everywhere.

The solution of the Schrodinger wave equation gives the wavefunction ψ. With the knowledge of ψ we can determine the Energy of the given system. Since all wavefunctions are not acceptable, all the values of energies are not acceptable. Only those values of energy corresponding to the Eigenfunctions are acceptable, and are called Eigenvalues.

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MOTION OF AN ELECTRON IN ONE DIMENSIONAL POTENTIAL WELL (PARTICLE IN A BOX)

Consider an electron of mass m, moving along positive x-axis between two walls of infinite height, one located at x=0 and another at x=a. Let potential energy of the electron is assumed to be zero in the region in-between the two walls and infinity in the region beyond the walls.

axxforV

axforV

&0

00

X-axis

X=0 X=a

V= V=0 V=

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Region beyond the walls:The Schrodinger’s wave equation representing the motion of the particle in the region beyond the two walls is given by

08

2

2

2

2

E

h

m

dx

d

The only possible solution for the above equation is ψ=0.Since ψ=0 , the probability of finding the particle in the region x<0 and x>a is Zero . i.e., particle cannot be found in region beyond the walls.

Region between the two walls: The Schrodinger’s wave equation representing the motion of the particle in the region between the two walls is given by

0008

2

2

2

2

VEh

m

dx

d

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ah

mEwhere

dx

d

h

mE

dx

d

18

10

08

2

22

2

2

2

2

2

2

2

Solution of the equation 1 is of the form

2cossin xBxA

Where A and B are unknown constants to be determined. Since particle cannot be found inside the walls

IIaxatand

Ixat

0,

0,0

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The equations are called boundary conditions. Using the I boundary condition in equation 2, we get

0

0cos0sin0

B

BA

Therefore equation 2 becomes 3sin xA

a

n

nwhere

na

aA

aA

,......2,1,0

0sin0

sin0

Using condition II in equation 3 we get

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Therefore correct solution of the equation 1 can be written as

4sin xa

nAn

The above equation represents Eigenfunctions. Where n=1,2,3,..(n=0 is not acceptable because, for n=o the wavefunction ψ becomes zero for all values of x. Then particle cannot be found anywhere)

Substituting for in equation 1a we get

2

228

h

Em

a

n

Therefore energy Eigenvalues are represented by the equation

58 2

22

ma

hnEn

58 2

22

ma

hnEn

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Where n=1,2,3,.. It is clear from the above equation that particle can have only desecrated values of energies. The lowest energy that particle can have corresponds to n=o , and is called zero-point energy. It is given by

68 2

2

int ma

hE pozero

Normalization of wave function:

We know that particle is definitely found somewhere in space

a

n dx0

21

xa

n

an

sin2

Therefore Normalised wavefunction is given by

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aAor

aA

aA

xa

n

n

ax

A

dxa

nA

dxxa

nA

a

a

a

22

10002

12

sin22

12

cos12

1

1sin

2

2

0

2

0

2

0

22

Therefore Normalized wave function is given by

xa

n

an

sin2

xa

n

an

sin2

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Wavefunction, Probability density and energy of the particle in different energy levels and at

different positionsFor n=1 energy is given by

Kma

hEn

2

2

8

Wavefunction and probability density for different values of x is given by

0 0 0

0 0 0

xx

a

n

an

sin2

xa

n

ap n

22sin

2

2

aa

2

a

2

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Similarly for n=2

Kma

hEn 4

8

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2

x

0 0 0

0 0

a 0 0

xa

n

an

sin2

xa

n

ap n

22sin

2

4

aa

2a

2

4

3aa

2

2

a

a

2

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Wavefunctions, probability density and energies are as shown in the figure.

n=1

n=2

1

P1

2

P2

x=0 x=a/2 x=a

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Free Particle  Consider a particle of mass m moving along positive x-axis.

Particle is said to be free if it is not under the influence of any field or force. Therefore for a free particle potential energy can be considered to be constant or zero. The Schrodinger wave equation for a free particle is given by.

008

2

2

2

2

E

h

m

dx

d

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10

08

2

22

2

2

2

2

2

2

2

h

mEwhere

dx

d

h

mE

dx

d

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The solution of the equation 1 is of the form

3cossin xBxA

Where A and B are unknown constants to be determined. Since there are no boundary conditions A, B and can have any values.

Energy of the particle is given by

48 2

22

m

hE

Since there is no restriction on there is no restriction on E.

Therefore energy of the free particle is not quantised. i.e., free particle can have any value of energy.

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