QUANTUM MECHANICS

1

QUANTUM MECHANICS CONTENTS

1 MATHEMATICAL

FUNCTIONS

AND STANDARD

INTEGRALS

1.1 One-Dimensional Delta Function 4

1.2 Three dimensional Delta Function 6

1.3 Gaussian Function 7

1.4 Fourier Integral transform 7

1.5 Polynomials 8

2 CONCEPTS OF

CLASSICAL

MECHANICS

2.1 Particle nature of radiation 14

2.1.1 Photoelectric effect 14

2.2 Compton effect 16

2.3 Wave aspect of particles 19

2.4 Wave Packets 21

2.5 Group velocity and phase velocity 21

2.6 Heisenberg Uncertainty Principle 26

2.7 Wave function 27

2.8 Fourier Transform 33

3 MATHEMATICAL

FORMULATION

& POSTULATES

OF QUANTUM

MECHANICS

Postulates of quantum mechanics 36

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4 BOUND STATES

PROBLEM

4.1 Properties of One-Dimensional Motion 43

4.2 Free Particle : Continuous state 44

4.3 The Potential step 45

4.4 Infinite square well potential 50

4.5 Barrier penetration 57

4.6 Eigen values & Eigen functions of

energy for a finite square well potential 60

4.7 Harmonic Oscillator 69

4.7.1 Anisotropic Oscillator 82

4.7.2 Isotropic Harmonic Oscillator 83

4.8 Hydrogen Atom 90

4.9 Spin 95

4.9.1 Stern Gerlach Experiment 95

5 ANGULAR

MOMENTUM

ALGEBRA

5.1 Orbital Angular Momentum 108

5.1.1 Eigen states and eigen values of the

angular momentum operator 111

5.1.2 Matrix representation of Angular momentum

117

5.1.3 Eigen functions of Orbital Angular Momentum

120

6 IDENTICAL

PARTICLES

6.1 Identical particles 131

6.2 Slater determinant 133

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7 SCATTERING

THEORY

7.1 Quantum Scattering theory 140

7.2 The Born approximation 143

8 TIME

INDEPENDENT

PERTURBATION

THEORY

8.1 Perturbation theory 148

8.1.1 Time independent perturbation

theory 148

8.1.2 Time dependent perturbation

theory 171

8.2 Variational method 177

8.3 WKB Method (Wentjel-Krammer-

Brillouin Method) 184

9 RELATIVISTIC

QUANTUM

MECHANICS

9.1 Klein Gorden Equation 192

9.2 Dirac‟s Relativistic Equation

195

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CHAPTER1

MATHEMATICAL FUNCTIONS AND

STANDARD INTEGRALS

1.1 ONE–DIMENSIONAL DELTA FUNCTION

Various Definitions of the Delta Function:

The delta function can be defined as the limit of

x when 0 (Figure

1.1.):

0

lim ,x x

…(1.1)

Where

1/ /2 /2

0, /2

xx

x

…(1.2)

The delta function can be defined also by means of the following integral

equations:

0 ,f x x dx f

…(1.3)

f x x a dx f a

…(1.4)

We should mention that the functions is not a function in the usual

mathematical sense. It can be expressed as the limit of analytical functions

such as

2

20

sin / sinlim , lim

a

x axx x

ax

…(1.5)

Or, 2 20

1limx

x

…(1.6)

The Fourier transform of x , which can be obtained from the limit of

sin /x

x

, is:

1

,2

ikxx e dk

…(1.7)

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Which in turn is equivalent to

0

1/

1/

sin /1 1lim lim

2 2

ikx ikxx

e dx e dk xx

…(1.8)

Properties of the Delta Function:

The delta function is even:

andx x x a a x …(1.9)

Here are the some of the most useful properties of the delta function:

0 0

0

, if ,

0, elsewhere

b

a

f x a x bf x x x dx

…(1.10)

0 for 0,x x …(1.11)

0,x x

1

0ax x aa

…(1.12)

f x x a f a x a …(1.14)

d

ca x x b dx a b for ,c a d c b d …(1.15)

1b

ax dx for 0a b …(1.16)

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1

'i

i i

g x x xg x

…(1.17)

Where ix is a zero of g x and ' 0ig x . Using (1.17) we can verify that

1

,x a x b x a x b a ba b

…(1.18)

2 2 10

2x a x a x a a

a …(1.19)

1.2 THREE DIMENSIONAL DELTA FUNCTION

The three-dimensional form of the delta function is given in Cartesian

coordinates by

' ' ' 'r r x x y y z z

and in spherical coordinates by

2

2

1' ' cos cos ' '

1' ' '

sin

r r r rr

r rr

Since, according to (1.17) we have cos cos ' ' /sin

The Fourier transform of the three-dimensional delta function is:

'3

3

1'

2

ik r rr r d ke

and 3 3

0 00 ,d rf r r f d rf r r r f r

The following relations are often encountered:

2

2

ˆ 1. 4 , 4

rr r

r r

Where r̂ the unit vector along r.

We should mention that the physical dimension of the delta function is one

over the dimensions of its argument. Thus, if x is a distance, the physical

dimension of x is given by 1/ 1/x x L , where L is a length.

Similarly, the physical dimensions of r

is 31/L , Since,

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3

1 1 1 1r x y z

x y z L

1.3 GAUSSIAN FUNCTION

The Gaussian function is defined as

2

22

2

1

2

x x

f x e

x mean value

Standard deviation

Root mean square deviation.

1.4 FOURIER INTEGRAL TRANSFORM

If f x is a periodic function with a fundamental period L, then it can be

expanded in a Fourier series:

nik x

n

n

f x a e

…(1.1)

where 2 /nk n L . The coefficients na of the series are given by

0

1n

Lik x

na f x e dxL

…(1.2)

The Fourier transform of a function f x is defined as

1

2

ikxF k F f x f x e dx

…(1.3)

While the inverse Fourier transform is:

1

2

ikxf x F k e dkx

…(1.4)

Notice that in quantum mechanics we define the transformations slightly

differently, as follows:

/1

2

ipxk F x x e dxh

…(1.5)

and

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/1

2

ipxx k e dk

…(1.6)

Two formulas of Fourier transform theory are especially relevant.

Identity of norms: 2 2

f x dx F k dk

…(1.7)

Parseval’s theorem: * *f x g x dx F k G k dk

…(1.8)

1.5 POLYNOMIALS

(I) HERMITE POLYNOMIALS

The Hermite polynomials nH x are defined by the relation

2 2

1 0,1,2...n

n x x

n n

dH x e e n

dx

…(1.1)

The nH x are the solution to the differential equation

2

22 2 0

n n

n

d H x dH xx nH x

dx dx …(1.2)

The orthogonality relation for nH x is:

2

2 !x n

m n mne H x H x dx n

…(1.3)

Two important recurrence relations for nH x are

1 12 2 2n

n n n n

dH xnH x H x xH x nH x

dx

The first few Hermite polynomials are given below:

2

0 1 2

3 4 2

3 4

1 2 4 2

8 12 16 48 12

H x H x x H x x

H x x x H x x x

(II) LEGENDRE POLYNOMIALS

Legendre polynomials lP x are given by Rodrigue‟s formula,

211

2 !

nn

l n n

dP x x

n dx …(1.4)

The first Legendre polynomials are given below:

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2

0 1 2

3 4 2

3 4

11 3 1

2

1 15 3 35 30 3

2 8

P x P x x P x x

P x x x P x x

In terms of cos the first few Legendre polynomials are

0 1cos 1 cos cosP P

2

1cos 1 3cos2

4P 3

1cos 3cos 5cos3

8P

The orthogonality relation of the Legendre polynomials is:

1

' '1

2.

2 1P x P x dx

…(1.5)

(III) ASSOCIATED LEGENDRE FUNCTIONS

Associated legendre functions mP x are defined as

2

11 for 1 1m

mm

m

dP x x P x x

dx

…(1.6)

Where 0.m P x are the Legendre polynomials. Note that

0 0 for 1mP x P x P x m …(1.7)

The differential equation that mP x satisfies is:

2 2

2

2 21 2 1 0

1

md d mx x P x

dx dx x

…(1.8)

The first few associated Legendre functions are given below:

1 2 1 2 2 2

1 2 2

31 2 2 2 2 3 2

3 3 3

1 3 1 3 1

35 1 1 15 1 15 1

2

P x x P x x x P x x

P x x x P x x x P x x

The orthogonality relation of the associated Legendre functions is:

1

'1 0

!2cos cos sin

2 1 2 1 !

m m m mm

P x P x dx P P dm

…(1.9)

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(IV) SPHERICAL HARMONICS

The spherical harmonics are defined as:

2 1 !, 1 cos for 0

4 !

mm m imm

Y P e mm

…(1.10)

and

1 , 1 , *mm mY Y …(1.11)

The differential equation that mY satisfies is:

2

2 2

1 1 1sin 1 , 0

sin sin

mY

…(1.12)

The mY have well-defined parity given as follows:

, 1 ,m mY Y

…(1.13)

The orthonormalization relation of mY is written as

2 *

'

' ' '0 0

, , sinm m

m md Y Y d

…(1.14)

and the closure relation

*

0 1

, ', ' cos cos ' 'm m

m

Y Y

…(1.15)

1

' 'sin

Some important recurrence relations are given below:

cot , 1 1 ,i m m

me m Y m m Y

…(1.16)

1cot , 1 1 ,i m me m Y m m Y

…(1.17)

1 1

1 1, cos

2 1 2 3 2 1 2 1

m m mm m m m

Y Y Y

…(1.18)

The first few mY are given below:

1

0

1

4Y

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2

0 1

1 1

0 2 1 2 2

2 2 2

0 3 1 2

3 3

2 2 2 3 3 3

3 3

3 3cos sin

4 8

5 15 153cos 1 sin cos sin

16 8 32

7 215cos 3cos sin 5cos 1

16 64

105 35sin cos sin

32 64

i

i

i

i

i i

Y Y e

Y Y e Y e

Y Y e

Y e Y e

An important result for spherical harmonics is:

1 2 2

1

4cos 1 , ,

2 1

m m m

m

P Y Y

Where is the angle between the directions 1 1 2 2, and , . This

result is known as the spherical harmonics addition theorem.

(V) ASSOCIATED LAGUERRE POLYNOMIALS

First we shall deal with the Laguerre polynomials given by Rodrigue‟s

formula,

x xdL x e x e

dx

...(1.20)

The associated Laguerre polynomials are defined as

m

m

m

dL x L x

dx

…(1.21)

Where and m are nonnegative integers, Note that

0 0 0 for 1L x L x L x m …(1.22)

The first few associated Laguerre polynomials are given below:

1 1 2

1 2 2

1 2 2 3

3 3 3

1 2 4 2

3 18 18 6 18 6

L x L x x L x

L x x x L x x L x

The orthogonality relation of the associated Laguerre Polynomails is:

3

' '0

!

!

m x m mx e L x L x dxm

…(1.23)

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(VI) SPHERICAL BESSEL FUNCTIONS

Bessel‟s differential equation is given as

2

2 2 2

20

d dx x x J x

dx dx

…(1.24)

Where 0 . The solutions to this equation are called Bessel functions of

order l. J x are given by the series expansion.

22 4

0

1 /21

2 1 2 2 2 2.4 2 2 2 4 ! 1

k k

R

xx x xJ x

n k

…(1.25)

If 0,1,2,... 1J x J x

. If 0,1,2,..., andJ x J x are

linearly independent. In this case J x is bounded at 0x , while

J x is the unbounded Bessel function of the second kind. N x (also

called Neumann functions) are defined by

cos0,1,2,...

sin

J x J xN x

…(1.26)

These functions are unbounded at 0x . The general solution of (1.24) is:

0,1,2...

all

y x AJ x BJ x

y x AJ x BN x

…(1.27)

Where A and B are arbitrary constants. Spherical Bessel functions are

related to Bessel functions according to

/22

J x J xx

…(1.28)

Also, the Newmann spherical functions are related to the Newumann

function N x by

1 /22

n x N xx

…(1.29)

andJ x n x are given explicitly as

1 sind x

J x xx dx x

…(1.30)

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1

1 cosd xn x x

x dx x

…(1.31)

The first few andj x n x are given below:

0 0

1 12 2

2 23 2 3 2

sin cos

sin cos cos sin

3 1 3 3 1 3sin cos cos sin

3

x xj x n x

x x

x x x xj x n x

x x x x

j x x x n x x xx x x x x

The asymptotic behaviour of the andj x n x as and 0x x is

given by

0

10

2 1 !!

2 1 !!

x

x

xj x

n xx

…(1.32)

1sin

2

1cos

1

x

x

j x xx

n x xx

…(1.33)

Where 2 1 !! 13.5... 2 1 2 1

2

2

1

0

1

0

1

0 2

2

1 !

!

1 1; 1

22

4

x n

x n

n

x n

n

x x

e x dx n n

ne x dx

ne x dx n

e dx e

QUANTUM MECHANICS

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CHAPTER 2

CONCEPT OF CLASSICAL MECHANICS

1. According to classical mechanics a particle is separated by energy E and

linear momentum P and a wave is specified by amplitude (A) and

propagation vector 2

ˆk n

.

i.e. . orik r ikxk e e

2. According to classical mechanics particle and wave are mutually

exclusive.

3. According to classical mechanics all dynamical variable are continuous.

In classical mechanics energy is related with amplitude but in quantum

mechanics, energy is related with frequency.

In classical mechanics information of particle is given by position vector

and in quantum mechanics by wave function.

2.1 PARTICLE NATURE OF RADIATION

2.1.1 Photo electric effect–ejection of electrons on irradiating

any matter.

1. If 0v v no ejection of electron (where 0v threshold frequency), whatever

the intensity of radiation

2. If 0v v there is always ejection of electrons, whatever be the intensity of

radiation.

3. Kinetic energy of ejected electron depends only on the frequency of

incident radiation not on the intensity or amplitude.

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4. There is no time dilation between incidence of radiation and the ejection

of electron 810 sec .

5. The photo current varies with the intensity of incident radiation.

According to classical mechanics any radiation can ejected electron

whatever be its intensity of frequency. Now if we consider plank‟s law i.e.

E nhv and if one photon is incident the by energy conservation.

0 max 0. .Maxhv hv K E K E hv hv

If 0 max.v v K E ve

This –ve energy means this much energy is to be supplied to electron

Stopping potential is the potential at which photo current is zero.

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2.1.2 Compton effect: In the scattering of electromagnetic theory radiation

with the charged particle the wavelength of radiation after scattering is larger

than that of incident radiation wavelength.

According to classical mechanics, the radiation emitted should be of same

frequency and wave-length as the incident radiation. But it is contradictory to

Compton effect.

According to Quantum Mechanics we apply Plank‟s law, E nhv i.e. we take

radiation as comprised of particles or quanta of energy have i.e photon.

Applying law of momentum conservation along horizontal direction:

cos cos'

h hmv

… (i)

Applying law of momentum conservation along vertical direction:

0 sin sin'

hmv

… (ii)

Appling law of conservation of energy: 2 2

'e

hc hcmc m c

… (iii)

Form (i) & (ii) we get, 2 2 2 2 2

2 2

2 2 2

' '2 cos

h v h v h vvm v

c c c … (iv)

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Form (iii) we get, 2 2 2 2 2

2 2 2 2

2 2 2

' '2 2 'e e

h v h v h vvm c m c hm v v

c c c

2 2 2 2 2

2 2 2 2 2 2

2 2 2

' '2 2 'e e

h v h v h vvm c m c m v hm v v

c c c … (v)

Form (iv) & (v) get, 2

2

'2 1 cos 2 ' ' 1 cose

e

h vv hhm v v

c m c

This change in wavelength is called as Compton shift and depends only on

scattering angle and not frequency or wavelength or intensity of the incident

radiation.

here, e Compton wavelength 00.02426A

Transfer of energy increases as Q increases.

For bound state wavelength shift is negligible so we say that Compton

scattering takes place only for free particles.

Problem: In Compton scattering an incoming photon of wavelength

0

3

2

h

mc ,is scattered backward at an angle 180°. The

momentum corresponding to recoiling electron is:

(i) 20

21mc (ii)

8

21mc

(iii) 21

20mc (iv)

16

21mc

Solution: 3 2 2 7

' 1 cos ' 22 3 2

h h h h h

mc mc mc mc mc

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2 2 2 22 2

2 2

2

2 2

2 1 1 1 1 22

' ' ' ' '

1 1 1 1

'

e

h h h h hp h

h h

2 2 201 1 10

2' 3 7 21 21

c e ee c

m m mp h m

So, correct answer is (i)

Pair production:

Conversion of photon in to particle and its antiparticle.

Materialisation of radiation or conversion of energy in to mass.

Applying laws of conversation of energy and momentum.

2 2 22N N N e eehv m c K m c k k m c

22N e e ehv K K K m c

Minimum energy for pair production 22 1.022em c MeV

For free space, 2 2 2

1 22 e eehv m c k k m c m c

2 22 ee em c m m c … (1)

Momentum conversation, 2 cos 2 cos 2e e

hc vp m m c

c c

22 e

vhv m c

c

Comparing (1) and (2) we get the condition that if energy is conserve then linear

momentum is not conserve. If linear momentum is conserved then energy is

not conserve i.e both are not conserve simultaneously. So, pair production

QUANTUM MECHANICS

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don‟t take place in free space due to non-conservation of energy and linear

momentum simultaneously.

PAIR ANIHILATION

Inverse of pair production or conversion of mass into energy.

In pair annihilation atleast two photons generates not less than this-more than

two photons may be generated but probability of generation of more than two

photons is very low because to generate 3 or more than 3 photons very large

K.E. is needed to be supplied to the particle and antiparticle.

2.2 WAVE ASPECT OF PARTICLES

De- Broglie‟s Hypothesis: Matter waves In 1923 de- Broglie thought that this

wave particle duality is not restricted to radiation, but, must be universal: all

material particles should also display a dual (wave – particle )behaviour.

According to de-Broglie a moving particle, whatever its nature, has wave

properties associated with it i.e. each material particle of momentum p

behaves

as a group of waves.

De-Brodlie wavelength associated with any moving particle of momentum p is

given by

h h

p mv & wave vector

pK

h

These relations are called de- Broglie relations de Broglie‟s concept was

confirmed experimentally in 1927 by Davison & Germer and later by Thomson,

who obtained interference patterns with electrons.

The general rule for the detection of wave- behaviors of any object is:

Whenever the de-Broglie wavelength associated with any object

is in the range for or exceeds, its size, the wave nature of this object is

detectable. But if its de- Broglie wavelength is too small compared to its size,

the wave nature of this object is undetectable.

For understand these things we calculate de-Broglie wavelength for

macroscopic and microscopic system.

Example: Calculate the de –Broglie wavelength for

(a) a proton of kinetic energy 70 MeV Ans. 153.4 10 meter

(b) a 10 gm bullet moving at 900 m/sec Ans. 347.4 10 m

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The wavelength associated with proton has the same order of

magnitude as the size of atomic nucleus. The wave aspect of the

proton cannot be neglected.

The ratio 192.2 10b

p

the wave aspect of the bullet lies

beyond the human observational abilities.

Problem: Find the de-Broglie wavelength in angstroms for an electron of

energy V electron volts (or for an electron accelerated through

the potential difference V)

Solution: 34

31 19

6.62 10meter

2 2 9.1 10 1.6 10

h h

p mh V

1012.28 10 12.28 150

A AVV V

The non- relativistic de-Broglie wavelength for a proton

accelerated through the potential difference V is given by

0.287A

V

The de-Broglie wavelength of a material particle at temperature

T is given by

2 2

h hE kT

mE mkT Most probable kinetic energy at

temperature T 21 1 2

2 2p

kTmv m kT

m

Problem: Show that the relativestic de-Broglie wavelength associated with

a charged particle of charge q accelerated through potential

difference V is given by

0 2

0

2 12

h

qVm qV

m c

Problem: Calculate the de- Broglie wavelength of thermal neurons at

20°C

Solution: 2Aapproximately

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2.3 WAVE PACKETS

A superposition of waves or group of waves of slightly different wave-lengths

and velocities, with amplitude and phases so chosen that they interfere

constructively in small region of space (in neighborhood of particle ) and

destructively elsewhere.

A localized wave function is called wave-packet.

If the wave function is made to vanish everywhere except in the neighborhood

of particle, it is called localized wave function. If the wave function depends on

the whole space, it is called non-localized.

Not only the wave packets useful in the description of particles that are

confined to a certain spatial region, they also play a key role in understanding

the connection between classical mechanics and quantum mechanics. The

wave-packet embody nature‟s particle like behavior and also its wave-like

behavior.

2.4 GROUP VELOCITY AND PHASE VELOCITY

Phase velocity is the average velocity of individual waves, forming the packet.

The velocity propagation of phase of individual waves that comprises wave

packet.

Group velocity represents the velocity with which the wave packet propagates

as whole or the velocity of envelope . The velocity of advancement of max

amplitude for a group of wave is called group velocity.

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Note:

The envelop moves with group velocity and individual ripples moves with phase

velocity.

The phase velocity and group velocity are defined as &p b

dv v

k dk

RELATION BETWEEN GROUP VELOCITY AND PHASE VELOCITY

2

2 2,

p p p p

g p p p

d v k dv dv dvdv v k v k v p k dk d

dk dk dk dhk dp

2

2

2p p

g p p

dv dvv v v

dd

Group velocity in general may be larger or smaller than the phase velocity

depending upon the medium.

For non dispersive medium 0pdv

d (i.e. phase velocity do not depend on the

wavelength) g pv v

For dispersive medium 0pdv

d 0

pdv

d (i.e. phase velocity depends upon the

wavelength) gv may be larger and smaller than pv

For normal dispersive medium, ,p

g p

dvve v v

d

For Anomalous dispersive medium, ,p

g p

dvve v v

d

Problem: Prove that group velocity of the wave packet is classical velocity

of particle.

Solution:

,g p

dd dE Ev v

dk d k dp k p

Consider the case of a particle travelling in a constant potential

V its total energy is

QUANTUM MECHANICS

23

2

particle

2

g

pE V

m

dE pv v

dp m

This suggests we should view the center of the wave packet as

travelling like a classical particle that obeys the laws of classical

physics, the center would then follow the classical trajectory of

the particle.

Problem: Show that the phase velocity of a non-relativistic free particle is

1/2 of the group velocity.

Solution: de-Broglie wave associated with particle

g

h

mv …(i)

2

21 1

2 2

g

g

mvEE hv mv

h h …(ii)

The phase velocity is given by

pv …(iii)

21

2 2

g g

p

g

mv vhv

h mv

Problem: The phase velocity of de- Broglie wave is greater than speed of

light.

Solution:22 2

& 2 2 2g

g

h E mck k mv v

mv h h h

2

2

2

2 p gg g

mcchV V v c

mvk v

h

pv c but gv is always less than „c‟ so de Broglie waves does not

violate special theory of relativity.

QUANTUM MECHANICS

24

Exercise:

1. Show that for those waves whose angular frequency and wave number

k obey the dispersion relation 2 2 2k c constant the product of the

phase and group velocities is equal to 2,c i.e. 2

p gv v c , where „c‟ is the

speed of light.

2. Prove that

2

1p

mcv c

h

Where is the de Broglie wavelength associated with particle of mass m.

3. The angular frequency for a wave propagating inside a waveguide is given

in terms of the wave number k and width by

1/22

2 21kc

b k

.Find the

phase and group velocities of the wave.

4. The dispersion relation of a certain wave is 2 2 2c k m

Where is the angular frequency, k is the wave vector, c is the velocity-

of light and m is a constant. The group velocity vof the wave has the

following properties.

(a) v c as 0k and v c ask

(b) v c as 0k and v ask

(c) 0v as 0k and v ask

(d) 0v as 0k and v c ask

5. The angular frequency of the surface waves in a liquid is given by

3jkgk

, where

g acceleration due to gravity

density of liquid

T Surface Tension

Find the phase and group velocities for the limiting cases when the surface

waves have (a) very large wavelengths (b) very small wavelengths

g

dvdV v k

dk dk

(a) gV V

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25

(b) 3

2gV V

2.6 HEISENBERG UNCERTAINTY PRINCIPLE

According to classical mechanics a moving particle has a definite momentum

and occupies a definite position in space and it is possible to determine both its

position and momentum simultaneously. If the initial conditions 0r

& 0v

and

all forces acting on it are know the position r t

and velocity v t

(future

behaviour) are uniquely determined by means of Newton‟s second law. Thus

classical mechanics is completely deterministic.

In quantum mechanics, a particle is described by a group of waves of wave

packet, that moves with group velocity. Wave packet spread spread over space

and particle may be found any where within the wave packet. This implies that

position of particle is uncertain with the limits of wave packet. Wave packet has

velocity spread and hence there is uncertain about the velocity of momentum of

particle. This means it is impossible to know where within the wave packet the

particle is and what is its exact velocity or momentum. Thus quantum

mechanics is completelyin deterministic.

In 1927 Heisenberg proposed a very interesting principle of far reaching

importance know as the principle of indeterminacy of uncertainty principle.

This principle is a direct consequence of dal nature of matter. In its original

form, Heisenberg uncertainty principle states that

If the „x‟ component of momentum of a particle is measured with a uncertainty

xp then its x positions cannot be measured simultaneously more accurately

then ;2 2

x

x

x x pp

Similarly for other components 2

yy p

&2

zz p

.

This principle indicates that although it is possible to measure the momentum

or position of a particle accurately but it is not possible to measure momentum

and position simultaneously to an arbitrary accuracy.

Heisenberg uncertainty principle can be generalize to any pair of

complementary or canonically conjugate dynamical variables.

“It is impossible to measure two complementary or canonically conjugate

dynamical variables simultaneously to arbitrary accuracy”.

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26

If E is the uncertainty in determining the energy of the system and t is the

uncertainty in determining the energy time to which this determination refers.

Then 2

E t

Similarly, ,2

J J

uncertainty in angular momentum and, uncertainty

in angular position perpendicular to the component of any angular momentum

measured.

Note: We cannot localize a microscopic particle without giving to it a rather

large momentum. We can measure the position without disturbing it.

In quantum mechanics one can not get a definite answer about the position

and momentum of a particle, only a probability answer is possible.

APPLICATION OF UNCERTAINTY PRINCIPLE

The principle of uncertainty explains a large number of facts which could not

be explained by classical ideas some of its application are

1. THE NON EXISTENCE OF THE ELECTRON IN THE NUCLEUS

Radius of the nucleus of any atom is of the order to 1410 meters, than 142 10x meters According to Heisenberg uncertainty principle

2

x p

or 34

20

15

1.054 105.275 10

2 10p x

q

S. l. unit.

This is the minimum uncertainty in momentum than the momentum of

electron must be at least comparable with its magnitude p p

The kinetic energy,

2 4029

31

5.20515 101.5289 10

2 2 9.1 10

pT

m

joule

79.7 10 97eV MeV

This means if electron exist inside the nucleus their kinetic energy must

be of the order of 97 MeV but experimental observation show that no

electron in atom possess energy greater than 4 Me V, clearly the electron

do not exist in the nucleus.

2. RADIUS OF BOHR’S FIRST ORBIT

If the uncertainties in position and momentum in first orbit are x and

p respectively. Then according to Heisenberg uncertainty principle

x p px

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27

The uncertainty in kinetic energy,

2

2

1

2 2

pt

m m x

The uncertainty in potential energy, 2ze

Vx

So that uncertainty in total energy,

2 2

2

1

2

zeE T V

m xx

The energy will be minimum if

0

d E

d x

&

2

2

d Eve

d x

2 2

3 2

10

d E ze

d x m x x

22

2

d Ex ve

mze d x

So the radius of the first orbit is given by q

2.7 WAVE FUNCTION

According to classical mechanics, the state of a system is specified at any time

„t‟ by two dynamical variables the position r t

and the momentum p t

.Any

other physical quantity relevant to the system, can be calculate in terms of

these two dynamical variables.

Similarly all the information about a quantum mechanical system is confined

in the quantity ,r t .Known as wave function or state function all the

dynamical variables can be calculated from this.

PHYSICAL INTERPREATATION

itself has no physical interpretation only square of its norm 2

,r t has

meaning, according to Born‟s probabilistic interpretation, the square the norm

of ,r t probability density amplitude

2

. .p r y r y

position probability density in state .r y

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28

The quantity 2

3.r t d r

represents the probability of finding the particle at

time t, in a volume element 3d r located between rand

23 3; , ,r dr p r t d r r t d r

NORMALIZATION

The total probability of finding the system some where in space is equal to

unity

2

3, 1r t d r

…(1)

A wave function is said to normalized if this satisfy above relation (1)

Every acceptable wave function can be normalized by multiplying it by an

appropriate constant.

A wave function is normalized it means must go to zero as ,x z

Note:

Those wave function for which position probability density not constant can be

normalized by condition * 1d

If position probability density constant, then wave function can not be

normalized by this condition they normalized by the condition.

* * 3 'p pr r d r p p

WELL BEHAVED WAVE FUNCTION OR PHYSICAL ACCEPTABLE

(ADMISSIBLE) WAVE FUNCTION

A wave function is said to be physically acceptable if this satisfied following

conditions

1. ,r t

must be finite.

2. ,r t

must be continuous and single valued.

Since position probability density can have only one value at particular

time and place, and is continuous.

3. , ,x y z

must be finite, single valued and continuous

Since momentum, is continuous single valued and finite.

4. must be square intertribal, (i.e. * 3d r must be finite)

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29

Note: If V any where than , ,x y z

is not continuous

ORTHONORMALITY CONDITION

If a system has two wave function ,i & j then the condition of orthonormality

is defined as *

i j ijd

If i j (the wave function is said to be normalisable)

If i j (the wave function are said to be mutually orthogonal.)

Problem: The wave function of a particle is given as 1

x

ax ea

(i) Find the probability of finding the particle a x a .

(ii) Find the value of „b‟ so that the probability of finding the

particle in the range b x b is 0.5.

Solution: (i) Probability0

2 22 2

*

0

0

1 1 1 1

2 2

ax x

a a ax x a aa a

a a

a

e edx e ds e dx

a a a a

a a

2

2 2 2 2

2 2

1 1 1 1 11 1 2 2 1 1

2 2 2

ee e e e

e e

(ii) 2

* 1 10.5

2 2

xa a

a

b b

dx e dx

2

2

0

0

1 1 2 12

22 2

x

a

b

b x

ae

e dxa a

a

2 / 2 /1 22 2 2

2 2

b a b a b ae e n b n

a

Problem: The wave function of a particle confined in a box of length L is

2

sin if 0x

x x LL L

0 if elsewhere

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30

Calculate the probability of finding the particle in the region

02

Lx

Solution: 1/2

Problem: If the probability that x lies between x&x + dx is

axp x dx ae dx , where 0 , 0x a . Then the probability

that x lies between 1x and 2 2 1x x x is

(a) 21 axaxe e (b) 21 axaxa e e

(c) 2 21ax axaxe e e (d) 21 1 axax axe e e

Solution: (a)

Problem: The wavefunction of the certain particle is 2cosA x if

2 2x

(i) Find the value of normalization constant A

(ii) Find the probability that the particle be found between

0, /4x x

Solution: (i) 8

3 (ii) 0.462

Problem: The normalized state of a free particle is represented by w wave

function.

2

022,0

xik x

ax Ne

Find the normalization constant

Solution: 1/2 1/4

1N

a

Problem: Consider a one dimensional particle which is confined within

the region 0 x a & whole wave function is

, sin i txx t e

a

Calculate the probability of finding the particle in the interval

3

4 4

a ax

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31

Note: First calculate normalization constant and then calculate probability.

2

0.822

Solution: Which among the fallowing function represent physically

acceptable wave functions

(i) 3sin x , (ii) 4 x (iii) 5x

(iv) 2x

(i) 3sin x , since its and its derivatives are finite, single,

continuous everywhere and square integrable.

Note:

4 x is continuous but not finite and not square integrable

5x is not finite, single valued and not square integrable

2x is not finite and not square integrable. So, these are not physically

acceptable wave functions.

Problem: Which of the following wave function cannot be solution of

Schrodinger‟ equation (or well behaved) why not?

(a) secA x , (b) tanA x (c) 2xAe

(d) 2xAe

Solution: (a), (b) & (c)

Note:

(a) secA x

(b) tanA x are discontinuous and be come infinite at x odd multiple of

2

(c) 2xAe is not finite

Problem: Which of the following functions represent acceptable wave

function of a particle in the range x

(a) tan 0x A x A

(b) cos , constantx B x B

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32

(c) 2

cos , 0, 0

D

xx C e C D

(d) 2

, 0Fxx E x e E F

Solution: (d)

Problem: Which of the wave function in figure shown below cannot have

physical significances in the interval shown and why not?

Solution: (b), (c) (d) (f)

Note: Since, (b) is double valued

(c) has a discontinuous derivative

(d) goes to infinity, discontinuous

(f) is discontinuous

Problem: Which of the wave function in figure shown below cannot have

physical significance in the interval shown and why not?

QUANTUM MECHANICS

33

Solution: (c) & (e)

Note: Since (c) is discontinuous & goes to infinity

(e) is not single valued