Kumar , profile picture
Uploaded byKumar
PPT, PDF93,174 views

Quantum mechanics

1. Quantum mechanics began with Max Planck's paper in 1900 explaining black body radiation. It extends physics to small dimensions and includes classical laws as special cases. 2. Photoelectric effect shows that light behaves as particles called photons. Einstein's equation explained it using the photon energy. 3. Compton scattering showed that photons can collide with and transfer energy to electrons. The Compton wavelength was derived from this.

Embed presentation

Downloaded 1,806 times
Quantum Mechanics
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 Society read his paper “On the theory of the Energy distribution law of the Normal Spectrum”. This was the start of the revolution of Physics i.e. the start of Quantum Mechanics.
Quantum Mechanics Quantum Physics extends that range to the region of small dimensions. It is a generalization of Classical Physics that includes classical laws as special cases. Just as ‘c’ the velocity of light signifies universal constant, the Planck's constant characterizes Quantum Physics. sec.10625.6 sec.1065.6 34 27 Jouleh ergh − − ×= ×=
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 was the conception of Quantum Mechanics in 1925. This new approach was highly successful in explaining about the behavior of atoms, molecules and nuclei.
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 electrons are called ‘photo electrons’. V Evacuated Quartz tube Metal plate Collecting plate Light ^^^^^^^^ A + _
Photo Electric Effect Experimental findings of the photoelectric effect 1. There is no time lag between the arrival of light at the metal surface 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 Photo Current 2I I Vo
Photo Electric Effect 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 is emitted. Voltage Photo Current v1 v2 v3
Einstein’s 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 atom known as photoelectric workfunction (Wo). 2. Other part is used in providing kinetic energy to the emitted electron .       2 2 1 mv 2 2 1 mvWh o +=ν This is called Einstein’s photoelectric equation.
If , no photoelectric effect maxKEWh o +=ν maxKEhh o += νν )(max ohKE νν −= oνν < o oo hc hW λ ν == o oo o A eVWW hc )( 12400 ==λ
It is in form of . The graph with on y-axis and on x-axis will be a straight line with slope oo hheV νν −= oVIf is the stopping potential, then )(max ohKE νν −= e h e h V o o νν −= cmxy += eh oV ν
Photons Einstein postulated the existence of a particle called a photon, to explain detailed results of photoelectric experiment. λ ν hc hEp == Photon has zero rest mass, travels at speed of light Explains “instantaneous” emission of electrons in photoelectric effect, frequency dependence.
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 scattered radiation. Presence of modified radiation in scattered X-rays is called Compton effect. electron scattered photon recoiled electron νhE = c h p ν = '' νhE = v θ φ φcosmv φsinmv incident photon θ ν cos ' c h θ ν sin ' c h
From Theory of Relativity, total energy of the recoiled electron with v ~ c is 22 cmKmcE o+== Similarly, momentum of recoiled electron is 22 cmmcK o−= 2 22 2 1 cm cv cm K o o − − =         − − = 1 1 1 22 2 cv cmK o 22 1 cv vm mv o − =
Now from Energy Conversation φθ νν cos 1 cos ' 22 cv vm c h c h o − +=         − − += 1 1 1 ' 22 2 cv cmhh oνν (i) From Momentum Conversation (ii) along x-axis φθ ν sin 1 sin ' 0 22 cv vm c h o − −= (iii) along y-axis and
Rearranging (ii) and squaring both sides φθ νν 2 22 222 cos 1 cos ' cv vm c h c h o − =      − (iv) φθ ν 2 22 222 sin 1 sin ' cv vm c h o − =      (v) Rearranging (iii) and squaring both sides Adding (iv) and (v) 22 22 2 222 1 cos '2' cv vm c h c h c h o − =−      +      θ νννν (vi) From equation (i) 22 1 ' cv cm cm c h c h o o − =+− νν
On squaring, we get Subtracting (vi) from (vii) (vii) 22 22 2 2 22 22 1 )'(2 '2' cv cm hm c h cm c h c h o oo − =−+−+      +      νν νννν 0)'(2)cos1( '2 2 2 =−+−− ννθ νν ohm c h )cos1( '2 )'(2 2 2 θ νν νν −=− c h hmo )cos1( ' )'( 2 θ νν νν −=− c h mo
But is the Compton Shift. λ ν c = )cos1( '' 11 θ λλλλ −=      − h cmo and ' ' λ ν c = So, )cos1( '' ' θ λλλλ λλ −=      − h cmo )cos1(' θλλλ −=∆=− cm h o λ∆ θ It neither depends on the incident wavelength nor on the scattering material. It only on the scattering angle i.e. is called the Compton wavelength of the electron and its value is 0.0243 Å.cm h o
Experimental Verification Monochromatic X-ray Source photon θ Graphite target Bragg’s X-ray Spectrometer 1. One peak is found at same position. This is unmodified radiation 2. Other peak is found at higher wavelength. This is modified signal of low energy. 3. increases with increase in .λ∆ θ λ∆
0.0243 (1- cosθ) Å=−=∆ )cos1( θλ cm h o =∆ maxλ λ∆ So Compton effect can be observed only for radiation having wavelength of few Å. Compton effect can’t observed in Visible Light is maximum when (1- cosθ) is maximum i.e. 2.λ∆ 0.05 Å For 1Å ~ 1% λ∆=λ =λ For 5000Å ~ 0.001% (undetectable)
Pair Production When a photon (electromagnetic energy) of sufficient energy passes near the field of nucleus, it materializes into an electron and positron. This phenomenon is known as pair production. In this process charge, energy and momentum remains conserved prior and after the production of pair. Photon Nucleus (+ve) − e + e
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 of the electron and positron. The corresponding maximum photon wavelength is 1.2 pm. Electromagnetic waves with such wavelengths are called gamma rays .)(γ
Pair Annihilation When an electron and positron interact with each other due to their opposite charge, both the particle can annihilate converting their mass into electromagnetic energy in the form of two - rays photon.γ γγ +→+ +− 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. γ
From conservation of energy γν 2 2 cmh o= Pair production cannot occur in empty space In the direction of motion of the photon, the momentum is conserved if θ ν cos2p c h = θ θchν θcosp θcosp p p − e + e here mo is the rest mass and 22 11 cv−=γ
γvmp o= Momentum of electron and positron is (i) 1cos ≤θ1< c vBut θν cos2cph = Equation (i) now becomes θγν cos2 cvmh o= θγν cos2 2       = c v cmh o and γν 2 2 cmh o<
But conservation of energy requires that γν 2 2 cmh o= Hence it is impossible for pair production to conserve both the energy and momentum unless some other object is involved in the process to carry away part of the initial photon momentum. Therefore pair production cannot occur in empty space.
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 it behaves as wave while in other cases like photoelectric and compton effect it behaves as particles (photon).
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 its nature, has waves associated with it. These waves are called “matter waves”. Energy of a photon is νhE = For a particle, say photon of mass, m 2 mcE =
Suppose a particle of mass, m is moving with velocity, v then the wavelength associated with it can be given by hvmc =2 λ hc mc =2 mc h =λ mv h =λ p h =λor (i) If means that waves are associated with moving material particles only. ∞=⇒= λ0v (ii) De-Broglie wave does not depend on whether the moving particle is charged or uncharged. It means matter waves are not electromagnetic in nature.
Wave Velocity or Phase Velocity When a monochromatic wave travels through a medium, its velocity of advancement in the medium is called the wave velocity or phase velocity (Vp). k Vp ω = where is the angular frequency and is the wave number. πνω 2= λ π2 =k
Group Velocity So, the group velocity is the velocity with which the energy in the group is transmitted (Vg). dk d Vg ω = The individual waves travel “inside” the group with their phase velocities. In practice, we came across pulses rather than monochromatic waves. A pulse consists of a number of waves differing slightly from one another in frequency. The observed velocity is, however, the velocity with which the maximum amplitude of the group advances in a medium.
Relation between Phase and Group Velocity dk d Vg ω = )( pkV dk d = dk dV kVV p pg += ( )λπλ π 2 2 d dV VV p pg += ( )λλ 1 1 d dV VV p pg +=
λ λ d dV VV p pg −=       − += λ λ λ d dV VV p pg 2 1 1 λ λ d dVp So, is positive generally (not always). pg VV <⇒ In a Dispersive medium Vp depends on frequency ≠ k ω generally i.e. constant
λ λ d dV VV p pg −= 0=⇒ λd dVp pg VV =⇒ In a non-dispersive medium ( such as empty space) = k ω constant pV=
Phase Velocity of De-Broglie’s waves According to De-Broglie’s hypothesis of matter waves mv h =λ wave number h mv k π λ π 22 == (i) If a particle has energy E, then corresponding wave will have frequency h E =ν then angular frequency will be h Eπ πνω 2 2 ==
Dividing (ii) by (i) (ii) h mc2 2π ω = mv h h mc k π πω 2 2 2 ×= v c Vp 2 = But v is always < c (velocity of light) (i) Velocity of De-Broglie’s waves (not acceptable)cVp > (ii) De-Broglie’s waves will move faster than the particle velocity (v) and hence the waves would left the particle behind. )( pV
Group Velocity of De-Broglie’s waves The discrepancy is resolved by postulating that a moving particle is associated with a “wave packet” or “wave group”, rather than a single wave-train. A wave group having wavelength λ is composed of a number of component waves with slightly different wavelengths in the neighborhood of λ. Suppose a particle of rest mass mo moving with velocity v then associated matter wave will have h mc2 2π ω = and h mv k π2 = where 22 1 cv m m o − =
and On differentiating w.r.t. velocity, v 22 2 1 2 cvh cmo − = π ω 22 1 2 cvh vm k o − = π ( ) 2 3 22 1 2 cvh vm dv d o − = πω (i) ( ) 2 3 22 1 2 cvh m dv dk o − = π (ii)
Wave group associated with a moving particle also moves with the velocity of the particle. o o m vm dk dv dv d π πω 2 2 . = Dividing (i) by (ii) gVv dk d == ω Moving particle wave packet or wave group≡
Davisson & Germer experiment of electron diffraction • If particles have a wave nature, then under appropriate conditions, they should exhibit diffraction • Davisson & Germer measured the wavelength of electrons • This provided experimental confirmation of the matter waves proposed by de Broglie
Davisson and Germer Experiment
0φ = 0 90φ = Current vs accelerating voltage has a maximum (a bump or kink noticed in the graph), i.e. the highest number of electrons is scattered in a specific direction. The bump becomes most prominent for 54 V at φ ~ 50° IncidentBeam
According to de Broglie, the wavelength associated with an electron accelerated through V volts is o A V 28.12 =λ Hence the wavelength for 54 V electron o A67.1 54 28.12 ==λ From X-ray analysis we know that the nickel crystal acts as a plane diffraction grating with grating space d = 0.91 Å
o oo 65 2 50180 =      − =θ Here the diffraction angle, φ ~ 50° The angle of incidence relative to the family of Bragg’s plane
From the Bragg’s equation which is equivalent to the λ calculated by de-Broglie’s hypothesis. θλ sin2d= o o o AA 65.165sin)91.0(2 =××=λ It confirms the wavelike nature of electrons
Electron Microscope: Instrumental Application of Matter Waves
Resolving power of any optical instrument is proportional to the wavelength of whatever (radiation or particle) is used to illuminate the sample. An optical microscope uses visible light and gives 500x magnification/200 nm resolution. Fast electron in electron microscope, however, have much shorter wavelength than those of visible light and hence a resolution of ~0.1 nm/magnification 1,000,000x can be achieved in an Electron Microscope.
Heisenberg Uncertainty Principle It states that only one of the “position” or “momentum” can be measured accurately at a single moment within the instrumental limit. It is impossible to measure both the position and momentum simultaneously with unlimited accuracy. or uncertainty in position uncertainty in momentum →∆x →∆ xp then 2  ≥∆∆ xpx π2 h =∴ The product of & of an object is greater than or equal to 2  xp∆x∆
If is measured accurately i.e.x∆ 0→∆x ∞→∆⇒ xp Like, energy E and time t. 2  ≥∆∆ tE 2  ≥∆∆ θL The principle applies to all canonically conjugate pairs of quantities in which measurement of one quantity affects the capacity to measure the other. and angular momentum L and angular position θ
Determination of the position of a particle by a microscope i Incident Photon Scattered Photon Recoiled electron Suppose we want to determine accurately the position and momentum of an electron along x-axis using an ideal microscope free from all mechanical and optical defects. The limit of resolution of the microscope is i x sin2 λ =∆ here i is semi-vertex angle of the cone of rays entering the objective lens of the microscope. is the order of uncertainty in the x-component of the position of the electron. x∆
The scattered photon can enter the microscope anywhere between the angular range +i to –i. We can’t measure the momentum of the electron prior to illumination. So there is uncertainty in the measurement of momentum of the electron. The momentum of the scattered photon is (according to de-Broglie) λ h p = Its x-component can be given as i h px sin 2 λ =∆ The x-component of the momentum of the recoiling electron has the same uncertainty, (conservation of momentum)xp∆
The product of the uncertainties in the x-components of position and momentum for the electron is i h i px x sin 2 sin2 . λ λ ×=∆∆ This is in agreement with the uncertainty relation. 2 .  >=∆∆ hpx x
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 15 max 105)( − ×=∆ 2  ≥∆∆ xpx 2 )()( minmax  =∆∆ xpx 120 min ..101.1 2 )( −− ×= ∆ =∆ smkg x px  MeVpcE 20== ∴ relativistic
Thus the kinetic energy of an electron must be greater than 20 MeV to be a part of nucleus Thus we can conclude that the electrons cannot be present within nuclei. Experiments show that the electrons emitted by certain unstable nuclei don’t have energy greater than 3-4 MeV.
But Concept of Bohr Orbit violates Uncertainty Principle m p E 2 2 = 2 .  ≥∆∆ px m pp E ∆ =∆ m pmv∆ = p t x ∆ ∆ ∆ = pxtE ∆∆=∆∆ .. 2 .  ≥∆∆ tE
According to the concept of Bohr orbit, energy of an electron in a orbit is constant i.e. ΔE = 0. 2 .  ≥∆∆ tE ∞→∆⇒ t All energy states of the atom must have an infinite life-time. But the excited states of the atom have life–time ~ 10-8 sec. The finite life-time Δt gives a finite width (uncertainty) to the energy levels.
Two-slit Interference Experiment Laser Source Slit Slit Detector Rate of photon arrival = 2 x 106 /sec Time lag = 0.5 x 10-6 sec Spatial separation between photons = 0.5 x 10-6 c = 150 m 1 meter
– Taylor’s experiment (1908): double slit experiment with very dim light: 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 of photons ⇒ interference pattern is due to some inherent property of each photon - it “interferes with itself” while passing from source to screen – photons don’t “split” – light detectors always show signals of same intensity – slits open alternatingly: get two overlapping single-slit diffraction patterns – no two-slit interference – add detector to determine through which slit photon goes: ⇒ no interference – interference pattern only appears when experiment provides no means of determining through which slit photon passes
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 specific trajectory to photon – cannot suppose that half went through one slit and half through other – can only predict how photons will be distributed on screen (or over detector(s)) – interference and diffraction are statistical phenomena associated with probability that, in a given experimental setup, a photon will strike a certain point – high probability ⇒ bright fringes – low probability ⇒ dark fringes
Double slit expt. -- wave vs quantum • pattern of fringes: – Intensity bands due to variations in square of amplitude, A2 , of resultant wave on each point on screen • role of the slits: – to provide two coherent sources of the secondary waves that interfere on the screen • pattern of fringes: – Intensity bands due to variations in probability, P, of a photon striking points on screen • role of the slits: – to present two potential routes by which photon can pass from source to screen wave theory quantum theory
Wave function ψψψ *|| 2 = The quantity with which Quantum Mechanics is concerned is the wave function of a body. |Ψ|2 is proportional to the probability of finding a particle at a particular point at a particular time. It is the probability density. Wave function, ψ is a quantity associated with a moving particle. It is a complex quantity. Thus if iBA+=ψ iBA−=*ψ 222222 *|| BABiA +=−==⇒ ψψψ then ψ is the probability amplitude.
Normalization τd |Ψ|2 is the probability density. The probability of finding the particle within an element of volume τψ d2 || Since the particle is definitely be somewhere, so 1|| 2 =∫ ∞ ∞− τψ d A wave function that obeys this equation is said to be normalized. ∴ Normalization
Properties of wave function 1. It must be finite everywhere. If ψ is infinite for a particular point, it mean an infinite large probability of finding the particles at that point. This would violates the uncertainty principle. 2. It must be single valued. If ψ has more than one value at any point, it mean more than one value of probability of finding the particle at that point which is obviously ridiculous. 3. It must be continuous and have a continuous first derivative everywhere. zyx ∂ ∂ ∂ ∂ ∂ ∂ ψψψ ,, must be continuous 4. It must be normalizable.
Schrodinger’s time independent wave equation One dimensional wave equation for the waves associated with a moving particle is From (i) ψ is the wave amplitude for a given x. where A is the maximum amplitude. λ is the wavelength ψ λ πψ 2 2 2 2 4 −= ∂ ∂ x (ii) )0,(and),( )( 2 )( x i pxEt i AetxAetx λ π ψψ === − − 
vm h o =λ 2 22 2 1 h vmo =⇒ λ 2 2 2 1 2 h vmm oo       = 22 21 h Kmo = λ 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 22 VE h mo −= λ
This is the time independent (steady state) Schrodinger’s wave equation for a particle of mass mo, total energy E, potential energy V, moving along the x-axis. If the particle is moving in 3-dimensional space then Equation (ii) now becomes ψ πψ )(2 4 2 2 2 2 VEm hx o −−= ∂ ∂ 0)( 2 22 2 =−+ ∂ ∂ ψ ψ VE m x o  0)( 2 22 2 2 2 2 2 =−+ ∂ ∂ + ∂ ∂ + ∂ ∂ ψ ψψψ VE m zyx o 
For a free particle V = 0, so the Schrodinger equation for a free particle 0 2 2 2 =+∇ ψψ E mo  0)( 2 2 2 =−+∇ ψψ VE mo  This is the time independent (steady state) Schrodinger’s wave equation for a particle in 3-dimensional space.
Schrodinger’s time dependent wave equation )( pxEt i Ae − − =  ψ Wave equation for a free particle moving in +x direction is (iii) ψ ψ 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 2 22 x p ∂ ∂ −=⇒ ψ ψ  (ii) Differentiating (i) w.r.t. t ψ ψ  iE t −= ∂ ∂ t iE ∂ ∂ =⇒ ψ ψ 
For non-relativistic case Using (ii) and (iii) in (iv) (iv) ψ ψψ V xmt i + ∂ ∂ −= ∂ ∂ 2 22 2   E = K.E. + Potential Energy txV m p E , 2 2 += ψψψ V m p E +=⇒ 2 2 This is the time dependent Schrodinger’s wave equation for a particle in one dimension.
Linearity and Superposition 2211 ψψψ aa += If ψ1 and ψ2 are two solutions of any Schrodinger equation of a system, then linear combination of ψ1 and ψ2 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) ψ1 and ψ2 follow the superposition principle
21 ψψψ +→Then Total probability will be 2 21 2 |||| ψψψ +==P due to superposition principle )()( 21 * 21 ψψψψ ++= ))(( 21 * 2 * 1 ψψψψ ++= 1 * 22 * 12 * 21 * 1 ψψψψψψψψ +++= 1 * 22 * 121 ψψψψ +++= PPP 21 PPP +≠ Probability density can’t be added linearly If P1 is the probability density corresponding to ψ1 and P2 is the probability density corresponding to ψ2
Expectation values dxxf∫ ∞ ∞− = 2 ||)( ψ Expectation value of any quantity which is a function of ‘x’ ,say f(x) is given by for normalized ψ Thus expectation value for position ‘x’ is >< )(xf dxx∫ ∞ ∞− = 2 ||ψ>< x Expectation value is the value of ‘x’ we would obtain if we measured the positions of a large number of particles described by the same function at some instant ‘t’ and then averaged the results.
dxx∫= 1 0 2 ||ψ Solution >< x 1 0 4 2 4       = x a Q. Find the expectation value of position of a particle having wave function ψ = ax between x = 0 & 1, ψ = 0 elsewhere. dxxa ∫= 1 0 32 >< x 4 2 a =
Operators ψ ψ p i x  = ∂ ∂ (Another way of finding the expectation value) For a free particle An operator is a rule by means of which, from a given function we can find another function. )( pxEt i Ae − − =  ψ Then Here xi p ∂ ∂ = ^ is called the momentum operator (i)
ψ ψ E i t  −= ∂ ∂Similarly Here t iE ∂ ∂ =  ^ is called the Total Energy operator (ii) Equation (i) and (ii) are general results and their validity is the same as that of the Schrodinger equation.
U ximt i +      ∂ ∂ = ∂ ∂ 2 2 1   If a particle is not free then This is the time dependent Schrodinger equation ^^^ .. UEKE += ^ ^ 2^ 2 U m p E o +=⇒ UU =∴ ^ U xmt i + ∂ ∂ −= ∂ ∂ 2 22 2   ψ ψψ U xmt i + ∂ ∂ −= ∂ ∂ 2 22 2  
If Operator is Hamiltonian Then time dependent Schrodinger equation can be written as U xm H + ∂ ∂ −= 2 22^ 2  ψψ EH = ^ This is time dependent Schrodinger equation in Hamiltonian form.
Eigen values and Eigen function Schrodinger equation can be solved for some specific values of energy i.e. Energy Quantization. ψψα a= ^ Suppose a wave function (ψ) is operated by an operator ‘α’ such that the result is the product of a constant say ‘a’ and the wave function itself i.e. The energy values for which Schrodinger equation can be solved are called ‘Eigen values’ and the corresponding wave function are called ‘Eigen function’. then ψ is the eigen function of a is the eigen value of ^ α ^ α
Q. Suppose is eigen function of operator then find the eigen value. The eigen value is 4. Solution. x e2 =ψ 2 2 dx d 2 2^ dx d G = 2 2^ dx d G ψ ψ = )( 2 2 2 x e dx d = x eG 2 ^ 4=ψ ψψ 4 ^ =G
Particle in a Box 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 > L Boundary conditions for ψ Ψ = 0 for x = 0 {0 for x = L 0 2 22 2 =+ ∂ ∂ ψ ψ E m x o  x = 0 x = L ∞=V ∞=V particle 0=V 0=∴V inside(i)
Equation (i) becomes p h =λ k π2 =  p k =⇒  Emo2 = 2 2 2  Em k o =⇒ (k is the propagation constant) (ii) 02 2 2 =+ ∂ ∂ ψ ψ k x (iii) General solution of equation (iii) is kxBkxAx cossin)( +=ψ (iv)
Equation (iv) reduces to Boundary condition says ψ = 0 when x = 0 (v) 0.cos0.sin)0( kBkA +=ψ 1.00 B+= 0=⇒ B kxAx sin)( =ψ Boundary condition says ψ = 0 when x = L LkAL .sin)( =ψ LkA .sin0 = 0≠A 0.sin =⇒ Lk πnLk sin.sin =⇒
Put this in Equation (v) (vi) L n k π = L xn Ax π ψ sin)( = When n # 0 i.e. n = 1, 2, 3…., this gives ψ = 0 everywhere. πnkL = Put value of k from (vi) in (ii) 2 2 2  Em k o = 2 2 2  Em L n o =      π
Where n = 1, 2, 3…. Equation (vii) concludes om k E 2 22  =⇒ 2 22 8 Lm hn o = (vii) 1. Energy of the particle inside the box can’t be equal to zero. The minimum energy of the particle is obtained for n = 1 2 2 1 8 Lm h E o = (Zero Point Energy) If momentum i.e.01 →E 0→ 0→∆p ∞→∆⇒ x But since the particle is confined in the box of dimension L. Lx =∆ max
Thus zero value of zero point energy violates the Heisenberg’s uncertainty principle and hence zero value is not acceptable. 2. All the energy values are not possible for a particle in potential 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
Using Normalization condition L xn Axn π ψ sin)( = 1sin 0 22 =∫ dx L xn A L π 1|)(| 2 =∫ ∞ ∞− dxxnψ 1 2 2 =      L A L A 2 =⇒ The normalized eigen function of the particle are L nx L xn π ψ sin 2 )( =
Probability density figure suggest that: 1. There are some positions (nodes) in the box that will never be occupied by the particle. 2. For different energy levels the points of maximum probability are found at different positions in the box. |ψ1|2 is maximum at L/2 (middle of the box) |ψ2|2 is zero L/2.
Eigen function zyx ψψψψ = L zn L yn L xn AAA zyx zyx πππ ψ sinsinsin= L zn L yn L xn L zyx πππ ψ sinsinsin 2 3         = 2 2 222 8 )( mL h nnnE zyx ++= Particle in a Three Dimensional Box zyx EEEE ++=Eigen energy

More Related Content

PPTX
Classical mechanics vs quantum mechanics
byZahid Mehmood
 
PPT
Quantum mechanics a brief
byChaitanya Areti
 
PPTX
Introduction to quantum mechanics and schrodinger equation
byGaurav Singh Gusain
 
PPTX
Basic and fundamental of quantum mechanics (Theory)
byHalavath Ramesh
 
PPT
Quantum Mechanics Presentation
byJasmine Wang
 
PDF
Origin of quantum mechanics
byMUHAMMED ABDURAHMAN
 
PPTX
History of Quantum Mechanics
byChad Orzel
 
PPTX
Quantum mechanics I
bySaiva Bhanu Kshatriya College, Aruppukottai.
 
Classical mechanics vs quantum mechanics
byZahid Mehmood
 
Quantum mechanics a brief
byChaitanya Areti
 
Introduction to quantum mechanics and schrodinger equation
byGaurav Singh Gusain
 
Basic and fundamental of quantum mechanics (Theory)
byHalavath Ramesh
 
Quantum Mechanics Presentation
byJasmine Wang
 
Origin of quantum mechanics
byMUHAMMED ABDURAHMAN
 
History of Quantum Mechanics
byChad Orzel
 
Quantum mechanics I
bySaiva Bhanu Kshatriya College, Aruppukottai.
 

What's hot

PPTX
Fermi dirac basic part 1
bykiranmadhale1
 
PPTX
Particle in a box- Application of Schrodinger wave equation
byRawat DA Greatt
 
PPT
Superconductors
byHardik Soni
 
PPT
Statistical mechanics
byKumar
 
PPTX
SCHRODINGER EQUATION
byRai Saheb Bhanwar Singh College Nasrullaganj
 
PPTX
Sputtering process
byGanapathirao Kandregula
 
PPT
Dielectric Material and properties
byMayank Pandey
 
PPTX
De Broglie hypothesis
bySudeb Das
 
PPTX
Superconductivity a presentation
byAqeel Khudhair
 
PPTX
Postulates of quantum mechanics
byStudent
 
PDF
Brief introduction to perturbation theory
byAnamika Banerjee
 
PPTX
Classification of nuclei and properties of nucleus
byhemalathasenthil
 
PDF
Quantum
byKrupesh Anadkat
 
PPTX
Photoelectric effect ppt
bySantosh Jadhav
 
PPTX
Schrödinger wave equation
byHARSHWALIA9
 
PPTX
Quantum Physics
byStudent
 
PPT
Blackbody ppt
byThamirabharani Engineering college,Tirunelveli
 
PPT
statistic mechanics
byDharmendrapresentation
 
PPT
Band theory of solid
byKeyur Patel
 
PDF
Nuclear rections ppt
byDEPARTMENT OF PHYSICS
 
Fermi dirac basic part 1
bykiranmadhale1
 
Particle in a box- Application of Schrodinger wave equation
byRawat DA Greatt
 
Superconductors
byHardik Soni
 
Statistical mechanics
byKumar
 
SCHRODINGER EQUATION
byRai Saheb Bhanwar Singh College Nasrullaganj
 
Sputtering process
byGanapathirao Kandregula
 
Dielectric Material and properties
byMayank Pandey
 
De Broglie hypothesis
bySudeb Das
 
Superconductivity a presentation
byAqeel Khudhair
 
Postulates of quantum mechanics
byStudent
 
Brief introduction to perturbation theory
byAnamika Banerjee
 
Classification of nuclei and properties of nucleus
byhemalathasenthil
 
Quantum
byKrupesh Anadkat
 
Photoelectric effect ppt
bySantosh Jadhav
 
Schrödinger wave equation
byHARSHWALIA9
 
Quantum Physics
byStudent
 
Blackbody ppt
byThamirabharani Engineering college,Tirunelveli
 
statistic mechanics
byDharmendrapresentation
 
Band theory of solid
byKeyur Patel
 
Nuclear rections ppt
byDEPARTMENT OF PHYSICS
 

Viewers also liked

PPTX
Quantum theory
byHunter Cullen
 
PDF
Quantum Physics Summary
byTL Lee
 
PDF
Quantum Physics - Wave Function
byTL Lee
 
PDF
Wave functions
byTarun Gehlot
 
PPT
Quantum tunneling composite
byArefin Haque Tahsin
 
PPT
5 introduction to quantum mechanics
bySolo Hermelin
 
PDF
Physics barriers and tunneling
byMohamed Anwar
 
PDF
Tunneling
byRob Morien
 
PPT
The wave function
byShaun Wilson
 
PPT
Quantum theory ppt
byThamirabharani Engineering college,Tirunelveli
 
PPT
Quantum Theory
bylallen
 
PPTX
Wave particle duality
byDamien Poh
 
Quantum theory
byHunter Cullen
 
Quantum Physics Summary
byTL Lee
 
Quantum Physics - Wave Function
byTL Lee
 
Wave functions
byTarun Gehlot
 
Quantum tunneling composite
byArefin Haque Tahsin
 
5 introduction to quantum mechanics
bySolo Hermelin
 
Physics barriers and tunneling
byMohamed Anwar
 
Tunneling
byRob Morien
 
The wave function
byShaun Wilson
 
Quantum theory ppt
byThamirabharani Engineering college,Tirunelveli
 
Quantum Theory
bylallen
 
Wave particle duality
byDamien Poh
 

Similar to Quantum mechanics

PDF
Quantum Physics & Mechanicsiuytrewqasdfg
byfazayn927
 
PPTX
Ph 101-7 WAVE PARTICLES
byChandan Singh
 
PPTX
Quantum chemistry-B SC III-SEM-VI
byShri Shivaji Science College Amravati
 
PPTX
Engineering Physics - Quantum Mechanics
byutkarsha125538
 
PDF
Transition from Classical Mechanics to Quantum Mechanics.pdf
byamrita4u004
 
PPTX
Particle Properties of Wave
byBicol University
 
PPTX
Chapter_3.pptx .
byhappycocoman
 
PPT
wave particle duality.ppt explaining wave nature of light in easy way
bytejasviyadav6558
 
PDF
Particles and waves fisika dasar institut teknologi bandung
byFikriZain3
 
PPTX
Quantum mechanics S5
bysrijithsreedharan
 
PDF
Introduction quantum mechanics (chemistry)
byShri Jai Narain Misra P. G. College(K.K.C)
 
PPT
Ch.7 Modern Physics - 1.ppt
byMazyiar Sabet
 
PPTX
09 UNIT-9(Electronics and down of Modern Physics) (1).pptx
byFatimaAfzal56
 
PPTX
Quantum_Mechanics_2nd lecture: Few basic postulates
byTamonashJana1
 
PDF
Chapter 6.pdf jdwbviwdhvidhvidhvihrvirhirvhrivh
byssuser514c362
 
PPTX
Unit-III Quantum Mechanics-I B.Tech.pptx
byvijay989155
 
PPTX
Quantiuwewe e3er14e we3223 32222 m2.pptx
bycyberesearchprof
 
PPTX
60508_paticle like properties of waves.pptx
byClaireSadicon
 
PPTX
EP PPT.pptx
byDeveshSingh608809
 
PPTX
The reality of Quantum mechanics and stuff part 1
byjulietjenny1
 
Quantum Physics & Mechanicsiuytrewqasdfg
byfazayn927
 
Ph 101-7 WAVE PARTICLES
byChandan Singh
 
Quantum chemistry-B SC III-SEM-VI
byShri Shivaji Science College Amravati
 
Engineering Physics - Quantum Mechanics
byutkarsha125538
 
Transition from Classical Mechanics to Quantum Mechanics.pdf
byamrita4u004
 
Particle Properties of Wave
byBicol University
 
Chapter_3.pptx .
byhappycocoman
 
wave particle duality.ppt explaining wave nature of light in easy way
bytejasviyadav6558
 
Particles and waves fisika dasar institut teknologi bandung
byFikriZain3
 
Quantum mechanics S5
bysrijithsreedharan
 
Introduction quantum mechanics (chemistry)
byShri Jai Narain Misra P. G. College(K.K.C)
 
Ch.7 Modern Physics - 1.ppt
byMazyiar Sabet
 
09 UNIT-9(Electronics and down of Modern Physics) (1).pptx
byFatimaAfzal56
 
Quantum_Mechanics_2nd lecture: Few basic postulates
byTamonashJana1
 
Chapter 6.pdf jdwbviwdhvidhvidhvihrvirhirvhrivh
byssuser514c362
 
Unit-III Quantum Mechanics-I B.Tech.pptx
byvijay989155
 
Quantiuwewe e3er14e we3223 32222 m2.pptx
bycyberesearchprof
 
60508_paticle like properties of waves.pptx
byClaireSadicon
 
EP PPT.pptx
byDeveshSingh608809
 
The reality of Quantum mechanics and stuff part 1
byjulietjenny1
 

More from Kumar

PPT
Graphics devices
byKumar
 
PPT
Fill area algorithms
byKumar
 
PDF
region-filling
byKumar
 
PDF
Bresenham derivation
byKumar
 
PPT
Bresenham circles and polygons derication
byKumar
 
PPTX
Introductionto xslt
byKumar
 
PPTX
Extracting data from xml
byKumar
 
PPTX
Xml basics
byKumar
 
PPTX
XML Schema
byKumar
 
PPTX
Publishing xml
byKumar
 
PPTX
DTD
byKumar
 
PPTX
Applying xml
byKumar
 
PPTX
Introduction to XML
byKumar
 
PDF
How to deploy a j2ee application
byKumar
 
PDF
JNDI, JMS, JPA, XML
byKumar
 
PDF
EJB Fundmentals
byKumar
 
PDF
JSP and struts programming
byKumar
 
PDF
java servlet and servlet programming
byKumar
 
PDF
Introduction to JDBC and JDBC Drivers
byKumar
 
PDF
Introduction to J2EE
byKumar
 
Graphics devices
byKumar
 
Fill area algorithms
byKumar
 
region-filling
byKumar
 
Bresenham derivation
byKumar
 
Bresenham circles and polygons derication
byKumar
 
Introductionto xslt
byKumar
 
Extracting data from xml
byKumar
 
Xml basics
byKumar
 
XML Schema
byKumar
 
Publishing xml
byKumar
 
DTD
byKumar
 
Applying xml
byKumar
 
Introduction to XML
byKumar
 
How to deploy a j2ee application
byKumar
 
JNDI, JMS, JPA, XML
byKumar
 
EJB Fundmentals
byKumar
 
JSP and struts programming
byKumar
 
java servlet and servlet programming
byKumar
 
Introduction to JDBC and JDBC Drivers
byKumar
 
Introduction to J2EE
byKumar
 

Recently uploaded

PPTX
Ritesh_kumar_Aatmanirbhar Bharat: Make in India, Make for the World.pptx
byriteshrkgs2008
 
PDF
Security Technologys: Access Control, Firewall, VPN
byShielaLasala
 
PDF
Safeguarding AI-Based Financial Infrastructure
byYasir Naveed Riaz
 
PPTX
Basics of Identity Access Management In mordern Infrastructure
byPrinceXavier18
 
DOCX
Introduction to the World of Computers (Hardware & Software)
byALIABBAS ABASOV
 
PDF
Top Cybersecurity Threats 2025 Guide by Sureshdas
byPrintersassist
 
PDF
Innovative AI Solutions for Business Growth.pdf
byVox Works
 
PDF
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 
PDF
Unlocking the Power of Salesforce Architecture: Frameworks for Effective Solu...
byvarsha30tiwari
 
PDF
Decoding the DNA: The Digital Networks Act, the Open Internet, and IP interco...
byCSUC - Consorci de Serveis Universitaris de Catalunya
 
PDF
TPPmark2025 Kenta Inoue's answer 12/04/2025
byKenta Inoue
 
PDF
Usage Control for Process Discovery through a Trusted Execution Environment
byValerioGoretti
 
PDF
AI and Computer Architecture: 200 Years Together
byDmitry Zinoviev
 
PDF
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
PDF
API206-S: Transforming Supply Chains with Amazon Bedrock AgentCore - AWS re:I...
byChris Bingham
 
PDF
Accelerating Responsible AI Adoption in Public Sector and Private Organizations.
byYasir Naveed Riaz
 
PDF
Top 7 Manufacturing Software for Small Businesses Boosting Growth in 2025
byEnvertis Software Solutions
 
PDF
Vibe Coding vs. Spec-Driven Development [Free Meetup]
byHaim Michael
 
PDF
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
PDF
Unser Jahresrückblick – MarvelClient in 2025
bypanagenda
 
Ritesh_kumar_Aatmanirbhar Bharat: Make in India, Make for the World.pptx
byriteshrkgs2008
 
Security Technologys: Access Control, Firewall, VPN
byShielaLasala
 
Safeguarding AI-Based Financial Infrastructure
byYasir Naveed Riaz
 
Basics of Identity Access Management In mordern Infrastructure
byPrinceXavier18
 
Introduction to the World of Computers (Hardware & Software)
byALIABBAS ABASOV
 
Top Cybersecurity Threats 2025 Guide by Sureshdas
byPrintersassist
 
Innovative AI Solutions for Business Growth.pdf
byVox Works
 
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 
Unlocking the Power of Salesforce Architecture: Frameworks for Effective Solu...
byvarsha30tiwari
 
Decoding the DNA: The Digital Networks Act, the Open Internet, and IP interco...
byCSUC - Consorci de Serveis Universitaris de Catalunya
 
TPPmark2025 Kenta Inoue's answer 12/04/2025
byKenta Inoue
 
Usage Control for Process Discovery through a Trusted Execution Environment
byValerioGoretti
 
AI and Computer Architecture: 200 Years Together
byDmitry Zinoviev
 
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
API206-S: Transforming Supply Chains with Amazon Bedrock AgentCore - AWS re:I...
byChris Bingham
 
Accelerating Responsible AI Adoption in Public Sector and Private Organizations.
byYasir Naveed Riaz
 
Top 7 Manufacturing Software for Small Businesses Boosting Growth in 2025
byEnvertis Software Solutions
 
Vibe Coding vs. Spec-Driven Development [Free Meetup]
byHaim Michael
 
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
Unser Jahresrückblick – MarvelClient in 2025
bypanagenda
 

Quantum mechanics

  • 1.
  • 2.
    Introduction 1. Stability ofan 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 Society read his paper “On the theory of the Energy distribution law of the Normal Spectrum”. This was the start of the revolution of Physics i.e. the start of Quantum Mechanics.
  • 3.
    Quantum Mechanics Quantum Physicsextends that range to the region of small dimensions. It is a generalization of Classical Physics that includes classical laws as special cases. Just as ‘c’ the velocity of light signifies universal constant, the Planck's constant characterizes Quantum Physics. sec.10625.6 sec.1065.6 34 27 Jouleh ergh − − ×= ×=
  • 4.
    Quantum Mechanics 1. Photoelectric 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 was the conception of Quantum Mechanics in 1925. This new approach was highly successful in explaining about the behavior of atoms, molecules and nuclei.
  • 5.
    Photo Electric Effect Theemission 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 electrons are called ‘photo electrons’. V Evacuated Quartz tube Metal plate Collecting plate Light ^^^^^^^^ A + _
  • 6.
    Photo Electric Effect Experimentalfindings of the photoelectric effect 1. There is no time lag between the arrival of light at the metal surface 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 Photo Current 2I I Vo
  • 7.
    Photo Electric Effect 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 is emitted. Voltage Photo Current v1 v2 v3
  • 8.
    Einstein’s Photo ElectricExplanation The energy of a incident photon is utilized in two ways 1. A part of energy is used to free the electron from the atom known as photoelectric workfunction (Wo). 2. Other part is used in providing kinetic energy to the emitted electron .       2 2 1 mv 2 2 1 mvWh o +=ν This is called Einstein’s photoelectric equation.
  • 9.
    If , nophotoelectric effect maxKEWh o +=ν maxKEhh o += νν )(max ohKE νν −= oνν < o oo hc hW λ ν == o oo o A eVWW hc )( 12400 ==λ
  • 10.
    It is inform of . The graph with on y-axis and on x-axis will be a straight line with slope oo hheV νν −= oVIf is the stopping potential, then )(max ohKE νν −= e h e h V o o νν −= cmxy += eh oV ν
  • 11.
    Photons Einstein postulated theexistence of a particle called a photon, to explain detailed results of photoelectric experiment. λ ν hc hEp == Photon has zero rest mass, travels at speed of light Explains “instantaneous” emission of electrons in photoelectric effect, frequency dependence.
  • 12.
    Compton Effect When amonochromatic 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 scattered radiation. Presence of modified radiation in scattered X-rays is called Compton effect. electron scattered photon recoiled electron νhE = c h p ν = '' νhE = v θ φ φcosmv φsinmv incident photon θ ν cos ' c h θ ν sin ' c h
  • 13.
    From Theory ofRelativity, total energy of the recoiled electron with v ~ c is 22 cmKmcE o+== Similarly, momentum of recoiled electron is 22 cmmcK o−= 2 22 2 1 cm cv cm K o o − − =         − − = 1 1 1 22 2 cv cmK o 22 1 cv vm mv o − =
  • 14.
    Now from EnergyConversation φθ νν cos 1 cos ' 22 cv vm c h c h o − +=         − − += 1 1 1 ' 22 2 cv cmhh oνν (i) From Momentum Conversation (ii) along x-axis φθ ν sin 1 sin ' 0 22 cv vm c h o − −= (iii) along y-axis and
  • 15.
    Rearranging (ii) andsquaring both sides φθ νν 2 22 222 cos 1 cos ' cv vm c h c h o − =      − (iv) φθ ν 2 22 222 sin 1 sin ' cv vm c h o − =      (v) Rearranging (iii) and squaring both sides Adding (iv) and (v) 22 22 2 222 1 cos '2' cv vm c h c h c h o − =−      +      θ νννν (vi) From equation (i) 22 1 ' cv cm cm c h c h o o − =+− νν
  • 16.
    On squaring, weget Subtracting (vi) from (vii) (vii) 22 22 2 2 22 22 1 )'(2 '2' cv cm hm c h cm c h c h o oo − =−+−+      +      νν νννν 0)'(2)cos1( '2 2 2 =−+−− ννθ νν ohm c h )cos1( '2 )'(2 2 2 θ νν νν −=− c h hmo )cos1( ' )'( 2 θ νν νν −=− c h mo
  • 17.
    But is the ComptonShift. λ ν c = )cos1( '' 11 θ λλλλ −=      − h cmo and ' ' λ ν c = So, )cos1( '' ' θ λλλλ λλ −=      − h cmo )cos1(' θλλλ −=∆=− cm h o λ∆ θ It neither depends on the incident wavelength nor on the scattering material. It only on the scattering angle i.e. is called the Compton wavelength of the electron and its value is 0.0243 Å.cm h o
  • 18.
    Experimental Verification Monochromatic X-ray Source photon θ Graphite target Bragg’sX-ray Spectrometer 1. One peak is found at same position. This is unmodified radiation 2. Other peak is found at higher wavelength. This is modified signal of low energy. 3. increases with increase in .λ∆ θ λ∆
  • 19.
    0.0243 (1- cosθ)Å=−=∆ )cos1( θλ cm h o =∆ maxλ λ∆ So Compton effect can be observed only for radiation having wavelength of few Å. Compton effect can’t observed in Visible Light is maximum when (1- cosθ) is maximum i.e. 2.λ∆ 0.05 Å For 1Å ~ 1% λ∆=λ =λ For 5000Å ~ 0.001% (undetectable)
  • 20.
    Pair Production When aphoton (electromagnetic energy) of sufficient energy passes near the field of nucleus, it materializes into an electron and positron. This phenomenon is known as pair production. In this process charge, energy and momentum remains conserved prior and after the production of pair. Photon Nucleus (+ve) − e + e
  • 21.
    The rest massenergy 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 of the electron and positron. The corresponding maximum photon wavelength is 1.2 pm. Electromagnetic waves with such wavelengths are called gamma rays .)(γ
  • 22.
    Pair Annihilation When anelectron and positron interact with each other due to their opposite charge, both the particle can annihilate converting their mass into electromagnetic energy in the form of two - rays photon.γ γγ +→+ +− 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. γ
  • 23.
    From conservation ofenergy γν 2 2 cmh o= Pair production cannot occur in empty space In the direction of motion of the photon, the momentum is conserved if θ ν cos2p c h = θ θchν θcosp θcosp p p − e + e here mo is the rest mass and 22 11 cv−=γ
  • 24.
    γvmp o= Momentum ofelectron and positron is (i) 1cos ≤θ1< c vBut θν cos2cph = Equation (i) now becomes θγν cos2 cvmh o= θγν cos2 2       = c v cmh o and γν 2 2 cmh o<
  • 25.
    But conservation ofenergy requires that γν 2 2 cmh o= Hence it is impossible for pair production to conserve both the energy and momentum unless some other object is involved in the process to carry away part of the initial photon momentum. Therefore pair production cannot occur in empty space.
  • 26.
    Wave Particle Duality Lightcan 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 it behaves as wave while in other cases like photoelectric and compton effect it behaves as particles (photon).
  • 27.
    De Broglie Waves Notonly 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 its nature, has waves associated with it. These waves are called “matter waves”. Energy of a photon is νhE = For a particle, say photon of mass, m 2 mcE =
  • 28.
    Suppose a particleof mass, m is moving with velocity, v then the wavelength associated with it can be given by hvmc =2 λ hc mc =2 mc h =λ mv h =λ p h =λor (i) If means that waves are associated with moving material particles only. ∞=⇒= λ0v (ii) De-Broglie wave does not depend on whether the moving particle is charged or uncharged. It means matter waves are not electromagnetic in nature.
  • 29.
    Wave Velocity orPhase Velocity When a monochromatic wave travels through a medium, its velocity of advancement in the medium is called the wave velocity or phase velocity (Vp). k Vp ω = where is the angular frequency and is the wave number. πνω 2= λ π2 =k
  • 30.
    Group Velocity So, thegroup velocity is the velocity with which the energy in the group is transmitted (Vg). dk d Vg ω = The individual waves travel “inside” the group with their phase velocities. In practice, we came across pulses rather than monochromatic waves. A pulse consists of a number of waves differing slightly from one another in frequency. The observed velocity is, however, the velocity with which the maximum amplitude of the group advances in a medium.
  • 31.
    Relation between Phaseand Group Velocity dk d Vg ω = )( pkV dk d = dk dV kVV p pg += ( )λπλ π 2 2 d dV VV p pg += ( )λλ 1 1 d dV VV p pg +=
  • 32.
    λ λ d dV VV p pg −=       − += λ λ λ d dV VV p pg 2 1 1 λ λ d dVp So,is positive generally (not always). pg VV <⇒ In a Dispersive medium Vp depends on frequency ≠ k ω generally i.e. constant
  • 33.
    λ λ d dV VV p pg −= 0=⇒ λd dVp pg VV=⇒ In a non-dispersive medium ( such as empty space) = k ω constant pV=
  • 34.
    Phase Velocity ofDe-Broglie’s waves According to De-Broglie’s hypothesis of matter waves mv h =λ wave number h mv k π λ π 22 == (i) If a particle has energy E, then corresponding wave will have frequency h E =ν then angular frequency will be h Eπ πνω 2 2 ==
  • 35.
    Dividing (ii) by(i) (ii) h mc2 2π ω = mv h h mc k π πω 2 2 2 ×= v c Vp 2 = But v is always < c (velocity of light) (i) Velocity of De-Broglie’s waves (not acceptable)cVp > (ii) De-Broglie’s waves will move faster than the particle velocity (v) and hence the waves would left the particle behind. )( pV
  • 36.
    Group Velocity ofDe-Broglie’s waves The discrepancy is resolved by postulating that a moving particle is associated with a “wave packet” or “wave group”, rather than a single wave-train. A wave group having wavelength λ is composed of a number of component waves with slightly different wavelengths in the neighborhood of λ. Suppose a particle of rest mass mo moving with velocity v then associated matter wave will have h mc2 2π ω = and h mv k π2 = where 22 1 cv m m o − =
  • 37.
    and On differentiating w.r.t.velocity, v 22 2 1 2 cvh cmo − = π ω 22 1 2 cvh vm k o − = π ( ) 2 3 22 1 2 cvh vm dv d o − = πω (i) ( ) 2 3 22 1 2 cvh m dv dk o − = π (ii)
  • 38.
    Wave group associatedwith a moving particle also moves with the velocity of the particle. o o m vm dk dv dv d π πω 2 2 . = Dividing (i) by (ii) gVv dk d == ω Moving particle wave packet or wave group≡
  • 39.
    Davisson & Germerexperiment of electron diffraction • If particles have a wave nature, then under appropriate conditions, they should exhibit diffraction • Davisson & Germer measured the wavelength of electrons • This provided experimental confirmation of the matter waves proposed by de Broglie
  • 40.
  • 41.
    0φ = 0 90φ = Currentvs accelerating voltage has a maximum (a bump or kink noticed in the graph), i.e. the highest number of electrons is scattered in a specific direction. The bump becomes most prominent for 54 V at φ ~ 50° IncidentBeam
  • 42.
    According to deBroglie, the wavelength associated with an electron accelerated through V volts is o A V 28.12 =λ Hence the wavelength for 54 V electron o A67.1 54 28.12 ==λ From X-ray analysis we know that the nickel crystal acts as a plane diffraction grating with grating space d = 0.91 Å
  • 43.
    o oo 65 2 50180 =      − =θ Here thediffraction angle, φ ~ 50° The angle of incidence relative to the family of Bragg’s plane
  • 44.
    From the Bragg’sequation which is equivalent to the λ calculated by de-Broglie’s hypothesis. θλ sin2d= o o o AA 65.165sin)91.0(2 =××=λ It confirms the wavelike nature of electrons
  • 45.
    Electron Microscope: InstrumentalApplication of Matter Waves
  • 46.
    Resolving power ofany optical instrument is proportional to the wavelength of whatever (radiation or particle) is used to illuminate the sample. An optical microscope uses visible light and gives 500x magnification/200 nm resolution. Fast electron in electron microscope, however, have much shorter wavelength than those of visible light and hence a resolution of ~0.1 nm/magnification 1,000,000x can be achieved in an Electron Microscope.
  • 47.
    Heisenberg Uncertainty Principle Itstates that only one of the “position” or “momentum” can be measured accurately at a single moment within the instrumental limit. It is impossible to measure both the position and momentum simultaneously with unlimited accuracy. or uncertainty in position uncertainty in momentum →∆x →∆ xp then 2  ≥∆∆ xpx π2 h =∴ The product of & of an object is greater than or equal to 2  xp∆x∆
  • 48.
    If is measuredaccurately i.e.x∆ 0→∆x ∞→∆⇒ xp Like, energy E and time t. 2  ≥∆∆ tE 2  ≥∆∆ θL The principle applies to all canonically conjugate pairs of quantities in which measurement of one quantity affects the capacity to measure the other. and angular momentum L and angular position θ
  • 49.
    Determination of theposition of a particle by a microscope i Incident Photon Scattered Photon Recoiled electron Suppose we want to determine accurately the position and momentum of an electron along x-axis using an ideal microscope free from all mechanical and optical defects. The limit of resolution of the microscope is i x sin2 λ =∆ here i is semi-vertex angle of the cone of rays entering the objective lens of the microscope. is the order of uncertainty in the x-component of the position of the electron. x∆
  • 50.
    The scattered photoncan enter the microscope anywhere between the angular range +i to –i. We can’t measure the momentum of the electron prior to illumination. So there is uncertainty in the measurement of momentum of the electron. The momentum of the scattered photon is (according to de-Broglie) λ h p = Its x-component can be given as i h px sin 2 λ =∆ The x-component of the momentum of the recoiling electron has the same uncertainty, (conservation of momentum)xp∆
  • 51.
    The product ofthe uncertainties in the x-components of position and momentum for the electron is i h i px x sin 2 sin2 . λ λ ×=∆∆ This is in agreement with the uncertainty relation. 2 .  >=∆∆ hpx x
  • 52.
    Order of radiusof 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 15 max 105)( − ×=∆ 2  ≥∆∆ xpx 2 )()( minmax  =∆∆ xpx 120 min ..101.1 2 )( −− ×= ∆ =∆ smkg x px  MeVpcE 20== ∴ relativistic
  • 53.
    Thus the kineticenergy of an electron must be greater than 20 MeV to be a part of nucleus Thus we can conclude that the electrons cannot be present within nuclei. Experiments show that the electrons emitted by certain unstable nuclei don’t have energy greater than 3-4 MeV.
  • 54.
    But Concept of BohrOrbit violates Uncertainty Principle m p E 2 2 = 2 .  ≥∆∆ px m pp E ∆ =∆ m pmv∆ = p t x ∆ ∆ ∆ = pxtE ∆∆=∆∆ .. 2 .  ≥∆∆ tE
  • 55.
    According to theconcept of Bohr orbit, energy of an electron in a orbit is constant i.e. ΔE = 0. 2 .  ≥∆∆ tE ∞→∆⇒ t All energy states of the atom must have an infinite life-time. But the excited states of the atom have life–time ~ 10-8 sec. The finite life-time Δt gives a finite width (uncertainty) to the energy levels.
  • 56.
    Two-slit Interference Experiment Laser Source Slit SlitDetector Rate of photon arrival = 2 x 106 /sec Time lag = 0.5 x 10-6 sec Spatial separation between photons = 0.5 x 10-6 c = 150 m 1 meter
  • 57.
    – Taylor’s experiment(1908): double slit experiment with very dim light: 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 of photons ⇒ interference pattern is due to some inherent property of each photon - it “interferes with itself” while passing from source to screen – photons don’t “split” – light detectors always show signals of same intensity – slits open alternatingly: get two overlapping single-slit diffraction patterns – no two-slit interference – add detector to determine through which slit photon goes: ⇒ no interference – interference pattern only appears when experiment provides no means of determining through which slit photon passes
  • 58.
    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 specific trajectory to photon – cannot suppose that half went through one slit and half through other – can only predict how photons will be distributed on screen (or over detector(s)) – interference and diffraction are statistical phenomena associated with probability that, in a given experimental setup, a photon will strike a certain point – high probability ⇒ bright fringes – low probability ⇒ dark fringes
  • 59.
    Double slit expt.-- wave vs quantum • pattern of fringes: – Intensity bands due to variations in square of amplitude, A2 , of resultant wave on each point on screen • role of the slits: – to provide two coherent sources of the secondary waves that interfere on the screen • pattern of fringes: – Intensity bands due to variations in probability, P, of a photon striking points on screen • role of the slits: – to present two potential routes by which photon can pass from source to screen wave theory quantum theory
  • 60.
    Wave function ψψψ *||2 = The quantity with which Quantum Mechanics is concerned is the wave function of a body. |Ψ|2 is proportional to the probability of finding a particle at a particular point at a particular time. It is the probability density. Wave function, ψ is a quantity associated with a moving particle. It is a complex quantity. Thus if iBA+=ψ iBA−=*ψ 222222 *|| BABiA +=−==⇒ ψψψ then ψ is the probability amplitude.
  • 61.
    Normalization τd |Ψ|2 is the probabilitydensity. The probability of finding the particle within an element of volume τψ d2 || Since the particle is definitely be somewhere, so 1|| 2 =∫ ∞ ∞− τψ d A wave function that obeys this equation is said to be normalized. ∴ Normalization
  • 62.
    Properties of wavefunction 1. It must be finite everywhere. If ψ is infinite for a particular point, it mean an infinite large probability of finding the particles at that point. This would violates the uncertainty principle. 2. It must be single valued. If ψ has more than one value at any point, it mean more than one value of probability of finding the particle at that point which is obviously ridiculous. 3. It must be continuous and have a continuous first derivative everywhere. zyx ∂ ∂ ∂ ∂ ∂ ∂ ψψψ ,, must be continuous 4. It must be normalizable.
  • 63.
    Schrodinger’s time independentwave equation One dimensional wave equation for the waves associated with a moving particle is From (i) ψ is the wave amplitude for a given x. where A is the maximum amplitude. λ is the wavelength ψ λ πψ 2 2 2 2 4 −= ∂ ∂ x (ii) )0,(and),( )( 2 )( x i pxEt i AetxAetx λ π ψψ === − − 
  • 64.
    vm h o =λ 2 22 2 1 h vmo =⇒ λ 2 2 2 1 2 h vmm oo      = 22 21 h Kmo = λ 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 22 VE h mo −= λ
  • 65.
    This is thetime independent (steady state) Schrodinger’s wave equation for a particle of mass mo, total energy E, potential energy V, moving along the x-axis. If the particle is moving in 3-dimensional space then Equation (ii) now becomes ψ πψ )(2 4 2 2 2 2 VEm hx o −−= ∂ ∂ 0)( 2 22 2 =−+ ∂ ∂ ψ ψ VE m x o  0)( 2 22 2 2 2 2 2 =−+ ∂ ∂ + ∂ ∂ + ∂ ∂ ψ ψψψ VE m zyx o 
  • 66.
    For a freeparticle V = 0, so the Schrodinger equation for a free particle 0 2 2 2 =+∇ ψψ E mo  0)( 2 2 2 =−+∇ ψψ VE mo  This is the time independent (steady state) Schrodinger’s wave equation for a particle in 3-dimensional space.
  • 67.
    Schrodinger’s time dependentwave equation )( pxEt i Ae − − =  ψ Wave equation for a free particle moving in +x direction is (iii) ψ ψ 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 2 22 x p ∂ ∂ −=⇒ ψ ψ  (ii) Differentiating (i) w.r.t. t ψ ψ  iE t −= ∂ ∂ t iE ∂ ∂ =⇒ ψ ψ 
  • 68.
    For non-relativistic case Using(ii) and (iii) in (iv) (iv) ψ ψψ V xmt i + ∂ ∂ −= ∂ ∂ 2 22 2   E = K.E. + Potential Energy txV m p E , 2 2 += ψψψ V m p E +=⇒ 2 2 This is the time dependent Schrodinger’s wave equation for a particle in one dimension.
  • 69.
    Linearity and Superposition 2211ψψψ aa += If ψ1 and ψ2 are two solutions of any Schrodinger equation of a system, then linear combination of ψ1 and ψ2 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) ψ1 and ψ2 follow the superposition principle
  • 70.
    21 ψψψ +→Then Totalprobability will be 2 21 2 |||| ψψψ +==P due to superposition principle )()( 21 * 21 ψψψψ ++= ))(( 21 * 2 * 1 ψψψψ ++= 1 * 22 * 12 * 21 * 1 ψψψψψψψψ +++= 1 * 22 * 121 ψψψψ +++= PPP 21 PPP +≠ Probability density can’t be added linearly If P1 is the probability density corresponding to ψ1 and P2 is the probability density corresponding to ψ2
  • 71.
    Expectation values dxxf∫ ∞ ∞− = 2 ||)(ψ Expectation value of any quantity which is a function of ‘x’ ,say f(x) is given by for normalized ψ Thus expectation value for position ‘x’ is >< )(xf dxx∫ ∞ ∞− = 2 ||ψ>< x Expectation value is the value of ‘x’ we would obtain if we measured the positions of a large number of particles described by the same function at some instant ‘t’ and then averaged the results.
  • 72.
    dxx∫= 1 0 2 ||ψ Solution >< x 1 0 4 2 4       = x a Q. Findthe expectation value of position of a particle having wave function ψ = ax between x = 0 & 1, ψ = 0 elsewhere. dxxa ∫= 1 0 32 >< x 4 2 a =
  • 73.
    Operators ψ ψ p i x  = ∂ ∂ (Another wayof finding the expectation value) For a free particle An operator is a rule by means of which, from a given function we can find another function. )( pxEt i Ae − − =  ψ Then Here xi p ∂ ∂ = ^ is called the momentum operator (i)
  • 74.
    ψ ψ E i t  −= ∂ ∂Similarly Here t iE ∂ ∂ =  ^ iscalled the Total Energy operator (ii) Equation (i) and (ii) are general results and their validity is the same as that of the Schrodinger equation.
  • 75.
    U ximt i +      ∂ ∂ = ∂ ∂ 2 2 1   Ifa particle is not free then This is the time dependent Schrodinger equation ^^^ .. UEKE += ^ ^ 2^ 2 U m p E o +=⇒ UU =∴ ^ U xmt i + ∂ ∂ −= ∂ ∂ 2 22 2   ψ ψψ U xmt i + ∂ ∂ −= ∂ ∂ 2 22 2  
  • 76.
    If Operator isHamiltonian Then time dependent Schrodinger equation can be written as U xm H + ∂ ∂ −= 2 22^ 2  ψψ EH = ^ This is time dependent Schrodinger equation in Hamiltonian form.
  • 77.
    Eigen values andEigen function Schrodinger equation can be solved for some specific values of energy i.e. Energy Quantization. ψψα a= ^ Suppose a wave function (ψ) is operated by an operator ‘α’ such that the result is the product of a constant say ‘a’ and the wave function itself i.e. The energy values for which Schrodinger equation can be solved are called ‘Eigen values’ and the corresponding wave function are called ‘Eigen function’. then ψ is the eigen function of a is the eigen value of ^ α ^ α
  • 78.
    Q. Suppose iseigen function of operator then find the eigen value. The eigen value is 4. Solution. x e2 =ψ 2 2 dx d 2 2^ dx d G = 2 2^ dx d G ψ ψ = )( 2 2 2 x e dx d = x eG 2 ^ 4=ψ ψψ 4 ^ =G
  • 79.
    Particle in aBox 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 > L Boundary conditions for ψ Ψ = 0 for x = 0 {0 for x = L 0 2 22 2 =+ ∂ ∂ ψ ψ E m x o  x = 0 x = L ∞=V ∞=V particle 0=V 0=∴V inside(i)
  • 80.
    Equation (i) becomes p h =λ k π2 =  p k=⇒  Emo2 = 2 2 2  Em k o =⇒ (k is the propagation constant) (ii) 02 2 2 =+ ∂ ∂ ψ ψ k x (iii) General solution of equation (iii) is kxBkxAx cossin)( +=ψ (iv)
  • 81.
    Equation (iv) reducesto Boundary condition says ψ = 0 when x = 0 (v) 0.cos0.sin)0( kBkA +=ψ 1.00 B+= 0=⇒ B kxAx sin)( =ψ Boundary condition says ψ = 0 when x = L LkAL .sin)( =ψ LkA .sin0 = 0≠A 0.sin =⇒ Lk πnLk sin.sin =⇒
  • 82.
    Put this inEquation (v) (vi) L n k π = L xn Ax π ψ sin)( = When n # 0 i.e. n = 1, 2, 3…., this gives ψ = 0 everywhere. πnkL = Put value of k from (vi) in (ii) 2 2 2  Em k o = 2 2 2  Em L n o =      π
  • 83.
    Where n =1, 2, 3…. Equation (vii) concludes om k E 2 22  =⇒ 2 22 8 Lm hn o = (vii) 1. Energy of the particle inside the box can’t be equal to zero. The minimum energy of the particle is obtained for n = 1 2 2 1 8 Lm h E o = (Zero Point Energy) If momentum i.e.01 →E 0→ 0→∆p ∞→∆⇒ x But since the particle is confined in the box of dimension L. Lx =∆ max
  • 84.
    Thus zero valueof zero point energy violates the Heisenberg’s uncertainty principle and hence zero value is not acceptable. 2. All the energy values are not possible for a particle in potential 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
  • 85.
    Using Normalization condition L xn Axn π ψsin)( = 1sin 0 22 =∫ dx L xn A L π 1|)(| 2 =∫ ∞ ∞− dxxnψ 1 2 2 =      L A L A 2 =⇒ The normalized eigen function of the particle are L nx L xn π ψ sin 2 )( =
  • 86.
    Probability density figuresuggest that: 1. There are some positions (nodes) in the box that will never be occupied by the particle. 2. For different energy levels the points of maximum probability are found at different positions in the box. |ψ1|2 is maximum at L/2 (middle of the box) |ψ2|2 is zero L/2.
  • 87.
    Eigen function zyx ψψψψ= L zn L yn L xn AAA zyx zyx πππ ψ sinsinsin= L zn L yn L xn L zyx πππ ψ sinsinsin 2 3         = 2 2 222 8 )( mL h nnnE zyx ++= Particle in a Three Dimensional Box zyx EEEE ++=Eigen energy