Quantum Mechanics (Uncertainity principle)

Question If a freely moving electron is localized in space to within $\Delta x_0$ of $x_0$, its wave function can be described by a wave packet $\psi(x,t)=\int_\infty^{-\infty}e^{i(kx-\omega t)}f(k)dk$, where $f(k)$ is peaked around a central value $k_0$. Which of the following is most nearly the width of the peak in $k$?
A. $\Delta k = 1/x_0$
B. $\Delta k = \frac{1}{\Delta x_0}$
C. $\Delta k = \frac{\Delta x_0}{x_0^2}$
D. $\Delta k = k_0\frac{\Delta x_0}{x_0}$
E. $\Delta k = \sqrt{k_0^2+(1/x_0)^2}$
Solution:
In quantum mechanics, the momentum $(p=\hbar{k})$ and position $(x)$ wave functions are Fourier transform pairs and the relation between $p$ and $x$ representations forms the Heisenberg uncertainty relation:
$\Delta{x}\Delta{k}\geq1 \Rightarrow \Delta k \geq \frac{1}{\Delta x}$
Answer: B

Symmetries and conservation Laws: Part 1

• Every conservation principle corresponds to symmetry in nature
• A symmetry of a particular kind exists when a certain operation leaves something unchanged.
• There is an intimate connection between symmetry and so called conserved quantities.
• Well known conserved quantity is energy and corresponding symmetry in this case is time translation.

Momentum Conservation

• Holds for all type of interactions
• Related to the invariance of physical laws under translation in space.
• Thus laws of interaction do not depend on the place of measurement so the space is homogeneous.
• This transnational uniformity of space leads to the conservation of linear momentum.
• Particle at rest have no momentum. If it  decays into two less massive particles , momentum conservation requires that the two particles travel away in exactly opposite directions.

Conservation of Energy

• Holds for all type of interactions.
• related to the invariance of physical laws under translations along the time axis i.e., homogeneity of time.
• laws of interaction do not depend on the time of measurement

Angular momentum conservation

• In addition to transnational symmetry , space also has a rotational symmetry.
• This symmetry of space gives rise to another conserved quantity , angular momentum.
• This law is also of general validity for all types of interactions.
• It is related to the invariance of the physical laws under rotation (isotropy of space).
• The orbital and spin angular momentum may be separately conserved.

Parity Conservation

• Holds for strong, nuclear and electromagnetic interactions but is violated in case of week interactions.
• related to the invariance of the physical laws under inversion of space co-ordinates. x,y,z are replaced by -x.-y,-z.
• is equivalent to combined reflection and rotation.
• physical laws do not depend on the right handedness of co-ordinate system.
• Parity operation symmetry represents discrete symmetry (reflection and rotation through 180 degree)
• Every particle with non zero mass has an intrinsic parity  π which can either be +1(even) or -1 (odd). Thus total parity of a system of n particles is the product of their intrinsic parities and the orbital parity (-1)^l.
Thus, π_tot=π₁π₂π₃.......π_n(-1)^l
• Intrinsic parity of pions is odd.

Conservation of charge

• Conservation of electric charge is related to gauge transformations which are shifts in the zeros of the scalar and vector electromagnetic potentials V and A
• Gauge transformations leave E and B unaffected since the latter are obtained by differentiating potentials , and this invariance leads to charge conservation.
• Charge and baryon number are conserved in all interactions.

Baryons

• There is another whole class of unstable particles known as 'hyperons' , whose masses are each greater then that of protons.
• The first hyperon was found in cosmic rays named as Λ⁰ hypron , a neutral decaying particle.
• Charged particles seen in the decay were identified as proton and π^- meson, indicating a process

  Λ⁰ →p+π^-

• Anti-Λ⁰ hypron decay to an anti proton and π⁺-meson.
• The family of hyprons wit greatedt number of members is Σ-family.
• First if it to be observed is Σ⁺ with mass about 2328m_e , and two prominent decat schemes are

  Σ⁺→p+π⁰ ; Σ⁺→n+π⁺

• Σ^- has just one set of decay products

  Σ^-→n+π^-

  its mass being slightly greater than Σ⁺ and is 2341 m_e.

• Neutral Σ hyperon decays as

  Σ⁰→Λ⁰+γ

  its mass is 2328m_e

• The Σ hyperon form a triplet Σ⁺,Σ^- and Σ⁰. A corresponding triplet of anti Σ hyperon also exists.
• The anti particle of Σ⁺ can not be Σ^- because the two have slightly different masses, whereas particle anti particle must have exactly same masses.
• Another group of members belonging to hyperon family is Ξ hyperon originally called cascade particles.
• Theit negative and neutral forms have been observed with decay processes

  Ξ^-→Λ⁰+π^-

  Ξ⁰→Λ⁰+π⁰

  Their masses are about 2582 m_e

• Anti Ξ hyperons have been detected.
• Togather with nucleons (p and n) , the hyperons form the family of baryons.

Baryons(B=+1, L_e=L_μ=L_τ=0)

Particle	Symbol	mean life (s)	spin	S	Y	I	I3	Mass MeV/c2
Nucleon	n	stable	1/2	0	+1	1/2	-1/2	938.3
	p	886					+1/2	936.6
Lambda	Λ0	2.6×10-10	1/2	-1	0	0	0	1116
Sigma	Σ+	8.0×10-11	1/2	-1	0	1	+1	1189
	Σ0	6×10-20					0	1193
	Σ-	1.5×10-10					-1	1197
Xi	Ξ0	2.9×10-10	1/2	-2	-1	1/2	+1/2	1315
	Ξ-	1.6×10-10					-1/2	1321
Omega	Ω-	8.2×10-11	3/2	-3	-2		0	1672

• There is a sequence of decay for Ω^- baryon

  Ω^-→Ξ⁰+π^-→Λ⁰ +π⁰ ....

  Final result of decay is proton , two electrons and two photons.

Mesons

• Mesons are particles with zero or integral spin so they are Bosons.
• The lightest meson is pion or π-meson, with other meson masses ranging beyond proton mass.
• All mesons are unstable and decays in various ways.

π-mesons

• This particle is to transmit nuclear forces, it must interact strongly with nuclei, and therefore it should be scattered and absorbed quickly by matter through which it passes.
• π mesons are thus hypothetical particles responsible for the nuclear forces and had properties pridicted by Yukawa.
• Protons and neurtons can be transferred into one anotherby emitting or absorbing one of these particles.
• There are three kinds of pions π⁺, π^- and π⁰. π^- is the anti particle of π⁺.
• These new particles can be thought of making bonds between (n,n) , (p,p) and (n,p) or (p,n)

K mesons

• These are heavier unstable particles and have a great variety of different decay modes.
• THere are six different ways that K⁺ mesons commonly decay, in each case giving two or three less massive particles

  K⁺→π⁺+π⁰

  K⁺→μ⁺+ν_μ

  K⁺→π⁺+π⁺+π⁰

  K⁺→π⁺+π⁰+π⁰

  K⁺→e⁺+ν+π⁰

• Mass of K⁺ is 966m_e.
• K^- are anti particles of K⁺ mesons and have the same decay modes with appropriate exchange of decay products for their anti particles.
• K⁰ and $\overline{K^0}$are anti particles.

Mesons (B=0) Bosons

Particle	Symbol	Mass MeV/c2	Mean life (s)	spin	S	Y	I	I3
Pion	π+	140	2.6×10-8	0	0	0	1	+1
	π0	135	8.7×10-17					0
	π-	140	2.6×10-8					-1
Kon	K+	494	1.2×10-8	0	+1	+1	1/2	+1/2
	K0	498	9×10-11
	K-	494	5×10-8					-1/2
Eta	η0	549	6×10-19	0	0	0	0	0

Leptons

• particles which are untouched by strong forces and which participates in weak interactions are called leptons.
• All these particles (leptons) also have their corresponding anti particles.
• Unlike baryons and mesons no heavier leptons have been detected.
• Electrons with its anti particle positron along with its associated neutrino and anti nutrino are leptons.
• Muon with their respective neutrino and anti particles of these are also leptons. Another pair of leptons known as tau,τ and its associated neutrino.
• All tau's are charged and decay into electrons, muons or pions along with appropriate neutrino.
• The breakdown of parity conservation in weak interaction has important consequence for leptons.
• The particles are left handed and anti particles are right handed.
• In normal weak interactions the particles (electrons, negative muons, neutrinos) behaves as if they were left handed screws; i.e., observer think they spin clockwise when they are travelling towards him.
• Anti particles are right handed screws; an observer thinks they are spinning counter clockwise as they approaches him.
• Nature of weak interactions, with its voilation of such established symmetries as parity, charge conjunction, isotopic spin and strangeness is physics greatest problems.
• Muons were first discovered in decay of charged pions

  Charged pion decay:

  π⁺→μ⁺+ν_μ

  π^-→μ^-+$\overline{{\nu }_{\mu }}$

  Neutral pion decay:

  π⁰→γ+γ

• Muon decay

  μ⁺→e⁺+ν_e+$\overline{{\nu }_{\mu }}$

  μ^-→e^-+ν_μ+$\overline{{\nu }_e}$

Leptons have (B=0) (fermions) and their properties are

Particle	Symbol	Spin	Le	Lμ	Lτ	Mass (in MeV)	Mean Life
electron	e-	1/2	+1	0	0	0.511	stable
muon	μ-	1/2	0	+1	0	106	2.2×10-6
tau	τ-	1/2	0	0	+1	1784	3.4×10-25
electron neutrino	νe	1/2	+1	0	0	0	stable
μ neutrino	νμ	1/2	0	+1	0	0	stable
τ neutrino	ντ	1/2	0	0	+1	0	stable

Photons

• It only participates in EM interactions.
• Strong and weak interactions are not in photon domain of experience.
• When particle annihilate with anti particles the end product is often protons.
• Photon is its own anti particle.
• Under some circumstances it can disappear and can create particle -anti particle pair.
• Rest mass of photon is zero.
• Photon is a boson with angular momentum equal to 1.
• Spin of photon is 1

elementary particles and fundamental interactions

• Elementary particles are those microscopic elementary constituents out of which all matter in this universe is made of.
• Bound neutron is stable but unbound neutron is unstable and it decays according to equation

  $n\to p+e^{-}+\overline{{\nu }_e}$ (anti-nutrino)

  half life of free neutron is 14 min 49 sec.

Fundamental interactions

There are four fundamental interactions between particles

(1) strong

(2) electromagnetic

(3) week

(4) gravitational

Interaction	Particles affected	range	relative strength	particles exchanged	Role in universe
(1) Strons	Quarks	∼10-15m	1	gluons	Holds quarks togather to form nucleus
	Hadrons			Mesons	Holds nucleons togather to form atomic nuclei
(2) Electromagnetic	charged particles	infinite	∼10-2	photons	determine structure of atoms, molecules , solids etc., Important factor in astronomical universe.
(3) Weak	Quarks and leptons	∼10-17m	∼10-5m	Intermediate bosons	mediates transformations of quarks and leptons; helps determine composition of atomic nuclei
(4) Gravitational	all	infinite	∼10-39m	gravitons	Assemble matter into planet , galaxies and stars

• Elementary particles can be divided into four groups

  (1) photons

  (2) leptons

  (3) mesons

  (4) baryons

Anti particles

• A particle identical with proton except for negative charge, i.e., negative proton or antiproton was created by bombarding protons in a target with 6 GeV protons thereby inducing the reaction

  p+p+energy(6 GeV)→p+p⁺

• Particle and anti particle annihilate each otherto give rise to a form of energy.
• positron is the anti particle of electron.
• There must be an anti oarticle corresponding to each particle.
• From the collection of anti particles a world of anti matter could be created.

Reletionship between particle and anti particles is

property	relationship
(1) mass	same
(2) spin	same
(3) magnetic monemt	of opposite sign but same magnitude
(4) charge	of opposite sign but same magnitude
(5) mean life in free decay	same
(6) annihilation	in pair
(7) creation	in pair
(8) total isotopic spin	same
(9) intrinsic parity	same for bosons but opposite for fermions
(10) strangeness number	of opposite sign but same magnitude

Fermi Energy

The Fermi temperature of Cu is about 80,000 K. Which of the following is most nearly equal to the average speed of a conduction electron in Cu?
A. $2\times{10^{-2}}$ m/s
B. $2$ m/s
C. $2\times{10^{2}}$ m/s
D. $2\times{10^{4}}$ m/s
E. $2\times{10^{6}}$ m/s
Solution:
Fermi Energy: $E_F=kT_F=\frac{1}{2}mv^2$
where
$k=$ Boltzmann's constant
$T_F=$ Fermi temperature
$m=$ mass of the particle
$v=$ speed of the particle
Since
$T_F =8\times{10^4}$ K
$k = 1.38\times{10^{-23}}$ Joule/K
$m_e = 9.11\times{10^{-31}}$ kg
($k$ and $m_e$ are provided by ETS in the problem sheet)
Therefore,
$v=\sqrt{\frac{2kT_F}{m_e}}$
$v=\sqrt{\frac{2\cdot1.38\cdot8}{9.11}\cdot\frac{10^{-23}\cdot10^4}{10^{-31}}}\approx{\sqrt{4\cdot{10^{12}}}}=2\cdot{10^{6}}$

Beta decay

Beta decay can involves three processes, in all the three processes atomic number of nucleus becomes one unecay dit greater or smaller but the mass number remains the same.

Problems in explaining beta decay

(1) non conservation of energy

(2) non conservation of angular momentum

To explain this Pauli suggested that a second particle is emitted with beta particles simultaneously but the sum of kinetic energies of two particles must be always equal to the energy difference between the parent and daughter nuclei. Thus the principle of conservation of energy is not voilated even the beta particle do not carry same energy.

The maximum beta particle energy is equal to the energy difference between the parent and daughter nuclei.

Beta decay Theory (Fermi's Theory):-

Electron, positron and ν does not as such exists inside the nucleus but are formed at the time of the decay. Fermi assumed that β-decay results from some form of nutrino, electrons and nucleus. This type of interaction is known as weak interaction. Constant required to express its strength is g=10^-47erg cm³ . It is because of the weakness of this interaction that the β-decay does not take place instantly in the case when it is energitically possible.

This theory must include some relationship between the particles of initial and final nuclei. The relationship is expressed by means of matrix element. It involves the wave function of initial and final nuclear states and hence their spin, parities and the arrangement of the nucleons. When two states are very different from each other |M_if|² becomes smaller. The total available β-decay energy E₀ , which is the energy difference between initial and final states, can be divided between electrons and ν in large number of ways which affects the shape of beta ray spectrum.

The number of ways of distributing total available energy between electron and ν per unit total energy E₀ is dN/dE₀.

Now , number of ways in which electron may be given volume V and having momentum between p_e and p_e+dp_e is given by

${\mathrm{dn}}_e=4\pi p_e^{2}V\frac{d{p}_e}{h^3}$

Similarly , number of ways in which neutrino can be arranged between volume V having momentum between p_ν and p_ν+dp_ν is

${\mathrm{dn}}_{\nu }=4\pi p_{\nu }^{2}V\frac{d{p}_{\nu }}{h^3}$

Number of ways in which β-decay can lead to an electron having momentum between p_e and p_e+dp_e and ν having momentum between p_ν and p_ν+dpν is

dN=dn_e.dn_ν

$\mathrm{dN}=\frac{16{\pi }^2{V}^2}{h^6}{p}_e^2{p}_{\nu }^{2}d{p}_{e}d{p}_{\upsilon }$

Relativistic momentum of particle of rest mass m is given by

$p=\frac{{\left[E\right(E+2m{c}^2]}^{1/2}}{c}$

since mass of neutrino is almost zero

p=$\frac{E}{c}$

Therefore momentum of neutrino is

$p_{\nu }=\frac{E_{\nu }}{c}=\frac{E_0-E_e}{c}$

and hence,

${\mathrm{dp}}_{\nu }=\frac{d{E}_0}{c}$ for given E

This shows that ,$\frac{\mathrm{dN}}{d{E}_0}=\frac{16{\pi }^2{V}^2}{h^6{c}^3}{(E_0-E_e)}^2{p}_e^{2}d{p}_e$

We shall now also consider the role of coulomb's barrier in letting out electron and positron against beta decay. The coulomb barrier aids the escape of positron but hinders the escape of electrons. The effect of Coulomb barrier depends on the atomic number Z and the energy of electron or positron. The fermi factor represented by F(Z,E_e) is a complex function.

Now Fermi's theory finally gives the probability of decay with the emission of an electron having a given momentum p_e by the expression which involves the nucleon-beta-neutrino force constant g , the matrix element |M_if|² and the function of Fermi factor.

Capacitor problem

Problem:
A capacitor is constructed from two square metal plates of area $L^2$ separated by a distance $d$. One half of the space between the plates is filled with a substance of dielectric constant $(\kappa_1)$. The other half is filled with another substance with constant $(\kappa_2)$. Calculate the capacitance of the device assuming that the free space capacitance is $C_0$.

Solution:
Given
$A = L^2$
$d_1 = d_2 = \frac{d}{2}$
Dielectric constant $\kappa_1$ and $\kappa_1$
$C_0$

Find $C$

Formula
In series $C = \frac{A\epsilon_1\epsilon_2}{\epsilon_{2}d_1+\epsilon_{1}d_2}$
$\epsilon = \epsilon_{0}\kappa$
$C_0 = \epsilon_{0}\frac{A}{d}$

Calculation
$C = \frac{A\epsilon_1\epsilon_2}{\epsilon_{2}d_1+\epsilon_{1}d_2} = \frac{A\epsilon_0\kappa_1\epsilon_0\kappa_2}{(\epsilon_0\kappa_2+\epsilon_0\kappa_1)d/2}$
$C = 2\frac{A\epsilon_0}{d}\frac{\kappa_1\kappa_2}{(\kappa_2+\kappa_1)}$
$C = 2C_0\bigg(\frac{\kappa_1\kappa_2}{\kappa_2+\kappa_1}\bigg)$

Lagrangian problem

Question The Lagrangian for a mechanical system is $L=a\dot q^2+bq^4$ where $q$ is a generalized coordinate and $a$ and $b$ are constants. The equation of motion for this system is
A. $\dot q=\sqrt{\frac{b}{a}}q^2$
B. $\dot q=\frac{2b}{a} q^3$
C. $\ddot q=\frac{-2b}{a} q^3$
D. $\ddot q=+\frac{2b}{a} q^3$
E. $\ddot q=\frac{b}{a} q^3$
Solution:
The Lagrangian equation of motion for the generalized coordinate $q$:
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3$
$\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q$
$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$
source :-http://physics-problems-solutions.blogspot.com

What are constraints

Sometimes motion of a particle or system of particles is restricted by one or more conditions. The limitations on the motion of the system are called constraints. The number of co-ordinates needed to specify the dynamical system becomes smaller when constraints are present in the system. Hence the degree of freedom of a dynamical system is defined as the minimum number of independent co-ordinates required to simplify the system completely along with the constraints. Thus if k is the number of constraints and N is the number of particles in the system possessing motion in three dimensions then the number of degrees of freedom are given by
n=3N-k                                  (1)
thus the above system has n degrees of freedom.
Constraints may be classified in many ways. If the condition of constraints can be expressed as equations connecting the co-ordinates of the particles and possibly the time having the form
f(r₁,r₂,......t)=0                          (2)
then constraints are said to be holonoic and the simplest example of holonomic constraints is rigid body. In case of rigid body motion the distance between any two particles of the body remains fixed and do not change with the tie. If r_i and r_j are the position vectors of the i'th and j'th particles then , the distance between the is given by
|r_i-r_j|=c_ij                                   (3)
The constraints which are not expressible in the form of equation 2 are called non-holonoic for example, the motion of a particle placed on the surface of a sphere of radius a will be described as
|r|≥a or, r-a≥0
Constraints can further be described as (i) rehonoic and(ii) scleronoous. In rehonomous constraints the equation of constraints contains time as explicit variable while in case of scleronomous constraints they are not explicitly dependent on time.

What are Legendre transformations

The transition from Lagrangian to Hamiltonian formulism corresponds to changing the variables in our mechanical functions from $(q,\stackrel{}{\stackrel{.}{q}},t)\stackrel{}{\mathrm{to}(q,p,t)}$

where , p is related to q and $\stackrel{.}{q}$ by the equation

[Math Processing Error]$p_i=\frac{\partial L(q_j,{\dot{q}}_j,t)\stackrel{}{{}_{}}}{\partial {\dot{q}}_i}$

The procedures for switching variables in this manner is provided by the legendre transformations.

Consider a function of only two variables f(x,y), so that differential of f has the form

df=udx+vdy

where , $u=\frac{\mathrm{df}}{\mathrm{dx}}$ and $v=\frac{\mathrm{df}}{\mathrm{dy}}$                      (1)

To change the basis of description from x,y to a new set of variables u,y , so that differential quantities are expressed in terms of differential du and dy. Let g be the function of u and y defined by the equation

g=f-ux

differential of g is given as

dg=df-udx-xdu

or,

dg=vdy-xdu

which is exactly in the desired form. The quantities x and v are now functions of variables u and y given by the relations

$x=-\frac{\partial g}{\partial u},v=\frac{\partial g}{\partial y}$

which are exactly converse of equation 1

magnetic properties of matter an introduction

All substances possess magnetic properties and most general definition of magnetism defines it as a particular form of interactions originating between moving electrically charged particles.

Magnetic interaction relates spatially separate material objects and it is transmitted by means of magnetic field about which we have already studied .This magnetic field is important characteristics of EM form of matter.

We already know that source of magnetic field is a moving electric charge i.e. an electric current. On atomic scale, there are two types of macroscopic current associated with electrons.
a) Orbital current is which electron in an atom moves about the nucleus in closed paths constituting electric currents loops
b) Spin currents related to the internal degrees of freedom of the motion of electrons and this can only be understood through quantum mechanics.

Like electrons in an atom, atomic nucleus may also have magnetic properties like magnetic moment but it is fairly smaller then that of electrons.

Magnetic moment m is nothing but the quantitative measure of the magnetism of a particle.

For an elementary closed loop with a current i in it, the magnitude |m| of a magnetic moment vector equals the current times the loop area S i.e.
|m|=iS and direction of m can be determined using right hand rule.

All micro structural elements of matter electrons, protons and neutrons are elementary carriers of magnetic moment and combination of these can be principle sources of magnetism

Thus magnetic properties are inherent to all the substances i.e. they are all magnets

An external magnetic field has an influence on these atomic orbital and spin currents and two basic effects of an external field are observed
i) First is diamagnetic effect which is consequences of faraday's law of induction. According to the Lenz law’s, a magnetic field always sets up an induced current with its magnetic field direction opposite to an initial field .Therefore diamagnetic moment created by the external field is always negative related to this field
ii) Second effect occurs if there is a resultant non zero magnetic moment in the atom i.e. there is a spin magnetic moment and orbital magnetic moment .In this case external field will attempt to orient the intrinsic atomic magnetic moment in its own direction .As a result a positive moment parallel to the field is created and this is called paramagnetic moment.

Because of the universality of the diamagnetic effect, all substances possess diamagnetic.

However, diamagnetism is by no means actually observed in all matter. This is because in many instances the diamagnetic effect is masked by the more powerful paramagnetic effect.

Thus in paramagnetic substances we actually always observe a difference effect produced by the prominent Para magnetism and weaker diamagnetism.

Polarization

Longitudinal wave- has the same property with respect to any plane through its line of polarization.
Transverse waves- behaves differently in different planes. Light waves are transverse in nature and the viberation in them are at right angles to the direction in which wave is traveling.
Plane Polarized - Since the viberations constituting the beam of light are confined only to one definite plane through the axis of the beam, the light is generally said to be plane polarized. Properties of plane polarized light beam wrt two planes, one containing the viberation and other at right angles to it

Note that

• Vibrations of polarized light are linear --- light is plane polarized 
• Vibrations of polarized light are circular --- light is circularly polarized 
• Vibrations of polarized light are elliptic --- light is elliptically polarized 
• circular and elliptical vibrations are the resultant of two linear vibrations perpandicular to each other differing in phase by π/2. 
• The elliptically polarized light results from compounding together two mutually perpendicular plane polarized coherent light waves of unequal amplitude but differing in phase by π/2.
• Circularly polarized light arises from compounding together two mutually perpendicular coherent vibrations of equal amplitudes and periods but differing in phase by  π/2.

Polarization by reflection

• It is the simplest method of obtaining plane polarized light. 
• This method was first discovered by Etiennie Louis Malus.
• When a beam of light is reflected from the surface of a transparent medium the reflected light is partially polarized and the degree of polarization varies with the angle of incidence.
• Percentage of polarized light in reflected beam is greatest when it is incident at an angle known as angle of polarization for the medium which is equal to 57.5 degree for glass and varies slightly with the wavelength of incident light.
• Complete polarization is possible only with monochromatic light.
• Reflected light is said to be plane polarized in the plane of incident.

Brewster's Law

• This Law states that there is a simple relation between the angle of maximum polarization and the refractive index of the medium. This relation known as Brewster's Law is given by
  μ=tan i
  where i is the angle of incidence
  and μ is the index of refraction
• Using this law it can be showen that 'when light is incident at angle of maximum polarization the reflected ray is at right angles to the refracted ray. From Brewster's Law
  μ=tan i = (sin i)/(cos i)
  From Snell's Law
  μ = (sin i)/(sin r)
  where r is the angle of refraction
  Therefore , (sin i)/(cos i) = (sin i)/(sin r)
  (sin i)/(sin (90- i)) = (sin i)/(sin r)
  or we have
  90-i=r
  this gives i+r=90 degree
  showing that at maximum angle of polarization reflected and refracted rays are at right angles.
• Brewster's Law is obeyed even when light is reflected at the surface of rarer medium.
• Light reflected from both the upper and lower surfaces of a glass plate will be polarized in the plane of incidence.

Double Refraction

• Certain crystals split a ray of incident light into two refracted rays , one which gives the fixed image and follows all the laws of refraction and this ray is known as ordinary ray (o ray). Other ray gives an image that rotates with rotation of crystal and this ray is know as extra ordinary ray (e ray).
• Both e and o rays are plane polarized.
• When the crystal is rotated about the incident ray as an axis , the o-ray remains fixed but the e-ray revolves around it.
• The index of refraction for e-ray is therefore a function of direction.
• There is always one direction in the crystal for which there is no distinction between the o and e rays and this direction is called optic axis.
• e ray and o ray are parallel to each other.
• Plane of polarization of both e and o rays are at right angles to each other.
• A class of crystals in which there is a single direction known as optic axis along which all waves are transmitted with one uniform velocities while in any other direction there are two velocities are called uni-axial crystals for example calcite and tourmaline crystals are uni-axial.
• Crystals having two optical axis that is they have two directions of uniform velocity are bi-axial crystals for example topaz, mica etc. 

Law of Malus

• Intensity of incident polarized ray is equal to the sum of the intensities of two refracted rays
  I_o = a²sin²θ
  and , I_e = a²cos²θ
  I_o+I_e=a²=I

Nicole Prism

• Nicole prism is a polarizer.
• In the Nicole prism the o-ray is eliminated by total internal reflection so that we are left only with e-ray         whose vibrations are in the principle plane.

How to use Gauss's Law to find electric field

W all know that Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space. Also Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux. Now we'll go through the main steps which we can employ for applying Gauss's Law

1. First identify the symmetry properties of the charge distribution. By this we mean that the point at which the field is to be determined must lie on a surface and this surface must have enough symmetry which allows integrals involved to be evaluated properly.
2. Determine the direction of the electric field and a surface on which the magnitude of electric field is constant. 
3. Now choose the Gaussian surface accordingly for example if the problem has spherical symmetry then Gaussian surface would usually be spherical and for cylindrical symmetry problem Gaussian surface would be cylindrical.
4. Calculate the flux through the Gaussian surface.
5. Now calculate the charge enclosed inside the chosen Gaussian surface.
6. Equate the two sides of Gauss's law in order  to find the expression for the magnitude of the electric field in that region of space.

How to use Gauss's Law to find electric field

W all know that Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space. Also Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux. Now we'll go through the main steps which we can employ for applying Gauss's Law

1. First identify the symmetry properties of the charge distribution. By this we mean that the point at which the field is to be determined must lie on a surface and this surface must have enough symmetry which allows integrals involved to be evaluated properly.
2. Determine the direction of the electric field and a surface on which the magnitude of electric field is constant. 
3. Now choose the Gaussian surface accordingly for example if the problem has spherical symmetry then Gaussian surface would usually be spherical and for cylindrical symmetry problem Gaussian surface would be cylindrical.
4. Calculate the flux through the Gaussian surface.
5. Now calculate the charge enclosed inside the chosen Gaussian surface.
6. Equate the two sides of Gauss's law in order  to find the expression for the magnitude of the electric field in that region of space.

Row reduction method and rank of a matrix

1. Matrices are just a display of set of numbers and it does not have any value For example
is a 2 by 3 matrix having 2 rows and 3 columns.
Aij represents a matrix element of i’th row and jth column for example here A₁₂=6 and A₂₁=-2

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Reference book:

Superconductivity fact file (part 2)

• Transition between normal and superconducting state is thermodynamically reversible.
• London's equation is j=-CA/4π(λ_L)² , where λ_L is constant with dimensions of length and A is the vector potential.
• London equation accounts for Meissner effect . In a pure SC state the only field allowed is exponentially damped as we go from an external surface  B(x)=B(0)exp(-x/λ_L)  where λ_Lis the London penetration depth and is the measure of penetration of magnetic field.
• An applied magnetic field will penetrate a thin film fairly uniformly if the thickness is much less than λ_L. Thus in a thin film Meissner effect is not complete.
• Coherence length ξ is the measure of the distance within which SC electronic concentration can not change drastically in spatially varying magnetic field.
• Coherence length is a measure of the range over which we should average A to obtain j.
• 
  

Superconductivity fact file (part 1)

• Bulk superconductor in a week magnetic field will act as a perfect diamagnet , with zero magnetic induction in the interior.
• Nonmagnetic impurities have no marked effect on the SC transition temperature.
• A sufficiently strong magnetic field will destroy SC. At critical temperature critical field is zero H_C(T_C)=0
• Values of H_C are always low for type I superconductors.
• For a given H_C the area under magnetization curve is same for type II SC as for type I SC.
• In all SC entropy decreases markedly on cooling below transition temperature.
• Superconducting state is the more ordered state.
• Contribution to the heat capacity in the SC state is an exponential form with an argument proportional to -1/T
• In SC the important interaction is electron-electron interaction which orders the electrons in K space with respect to the fermi gas of the electrons.
• The argument of the exponential factor in the electronic heat capacity of a SC is found to be -E_g/2kT
• The transition in zero magnetic field from the superconducting state to the normal state is the second order phase transition, not involving any latent heat but discontinuity in heat capacity.
• Energy gap decreases continuity to zero as the temperature is increased to transition temperature.
• For photons of energy less than energy gap , the resistivity of the superconductor vanishes at absolute zero.
• As the temperature is increased not only does the gap decreases , but the resistivity for photon with energy below the energy gap no longer vanishes except at zero frequency.

The Schrödinger wave equation

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Free electron model of atom and energy bands in solids

Electric and Magnetic fields

CSIR NET and GATE physics

We will now discuss electric and magnetic field vectors (E and B)at a point in the absence of charge.

Now let us place a charge q at point (x,y,z) in space. If this charge experiences a force as given by Lorentz force equation then we can associate vectors E and B with this point (x,y,z) in space.

Thus at any time t vectors E(x,y,z,t) and B(x,y,z,t) gives forces experienced by any charge q at point (x,y,z) with a condition that placing this charge at point (x,y,z) in space does not disturb the position or motion of all other charges responsible for the generation of the field.

So, every point in space is associated with vector E and B which are functions of x,y,z and t.

Since E(or B) can be specified at every point in space , we call it a field.

A field is that physicsl quantity which takes on different values at different points in space for example velocity field of a flowing liquid.

Electromagnetic fields as we know are produced by complex formulas but the relationships between values of the fields at one point and the values of the feld at neighbour points are vary simple and can form differential equations which can completely describe the field.

To understand and visualize the behaviour of field we can consider the field as a function of position and
time. We can also create a mental picture of field by drawing the vectors at many points in space each of which gives strength and direction of field at that point.

Flux is one property of field and flux of a vector field through a surface is defined as the average value of normal component of the vector times the area of the surface.

Another property is the circulation of the vector field and for any vector field circulation around any imagined closed curve is defined as the average tangential component of the vector multiplied by the circumfrance of the loop.

With just the idea of flux and circulation we can define all the laws of electricity and magnetism.
Refrence:- The Feynman lectures on physics Vol 2