Quantum mechanics

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Quantum mechanics

  1. 1. Quantum Mechanics
  2. 2. Introduction1. Stability of an atom2. Spectral series of Hydrogen atom3. Black body radiationThere are a few phenomenon which the classical mechanicsfailed to explain.Max Planck in 1900 at a meeting of German PhysicalSociety read his paper “On the theory of the Energydistribution law of the Normal Spectrum”. This was the startof the revolution of Physics i.e. the start of QuantumMechanics.
  3. 3. Quantum MechanicsQuantum Physics extends that range to the region of smalldimensions.It is a generalization of Classical Physics that includesclassical laws as special cases.Just as ‘c’ the velocity of light signifies universal constant, thePlancks constant characterizes Quantum Physics.sec.10625.6sec.1065.63427Joulehergh−−×=×=
  4. 4. Quantum Mechanics1. Photo electric effect2. Black body radiation3. Compton effect4. Emission of line spectraIt is able to explainThe most outstanding development in modern science wasthe conception of Quantum Mechanics in 1925. This newapproach was highly successful in explaining about thebehavior of atoms, molecules and nuclei.
  5. 5. Photo Electric EffectThe emission of electrons from a metal plate when illuminatedby light or any other radiation of any wavelength or frequency(suitable) is called photoelectric effect. The emitted electronsare called ‘photo electrons’.VEvacuatedQuartztube MetalplateCollectingplateLight^^^^^^^^ A+_
  6. 6. Photo Electric EffectExperimental findings of the photoelectric effect1. There is no time lag between the arrival of light at the metalsurface and the emission of photoelectrons.2. When the voltage is increased to a certain value say Vo, thephotocurrent reduces to zero.3. Increase in intensity increase the number of thephotoelectrons but the electron energy remains the same.3IVoltagePhotoCurrent2IIVo
  7. 7. Photo Electric Effect4. Increase in frequency of light increases the energy of theelectrons. At frequencies below a certain critical frequency(characteristics of each particular metal), no electron isemitted.VoltagePhoto Currentv1v2v3
  8. 8. Einstein’s Photo Electric ExplanationThe energy of a incident photon is utilized in two ways1. A part of energy is used to free the electron from the atomknown as photoelectric workfunction (Wo).2. Other part is used in providing kinetic energy to the emittedelectron . 221mv221mvWh o +=νThis is called Einstein’s photoelectric equation.
  9. 9. If , no photoelectric effectmaxKEWh o +=νmaxKEhh o += νν)(max ohKE νν −=oνν <ooohchWλν ==oooo AeVWWhc)(12400==λ
  10. 10. It is in form of . The graph with on y-axisand on x-axis will be a straight line with slopeoo hheV νν −=oVIf is the stopping potential, then)(max ohKE νν −=ehehV ooνν−=cmxy +=ehoVν
  11. 11. PhotonsEinstein postulated the existence of a particle called a photon,to explain detailed results of photoelectric experiment.λνhchEp ==Photon has zero rest mass, travels at speed of lightExplains “instantaneous” emission of electrons in photoelectriceffect, frequency dependence.
  12. 12. Compton EffectWhen a monochromatic beam of X-rays is scattered from amaterial then both the wavelength of primary radiation(unmodified radiation) and the radiation of higher wavelength(modified radiation) are found to be present in the scatteredradiation. Presence of modified radiation in scattered X-rays iscalled Compton effect.electronscatteredphotonrecoiled electronνhE =chpν= νhE =vθφ φcosmvφsinmvincidentphotonθνcoschθνsinch
  13. 13. From Theory of Relativity, total energy of the recoiled electronwith v ~ c is22cmKmcE o+==Similarly, momentum of recoiled electron is22cmmcK o−=22221cmcvcmK oo−−=−−= 111222cvcmK o221 cvvmmv o−=
  14. 14. Now from Energy Conversationφθννcos1cos22cvvmchch o−+=−−+= 111222cvcmhh oνν (i)From Momentum Conversation(ii) along x-axisφθνsin1sin022cvvmch o−−= (iii) along y-axisand
  15. 15. Rearranging (ii) and squaring both sidesφθνν 222222cos1coscvvmchch o−=− (iv)φθν 222222sin1sincvvmch o−= (v)Rearranging (iii) and squaring both sidesAdding (iv) and (v)222222221cos2cvvmchchch o−=−+θνννν(vi)From equation (i)221cvcmcmchch oo−=+−νν
  16. 16. On squaring, we getSubtracting (vi) from (vii)(vii)22222222221)(22cvcmhmchcmchch ooo−=−+−++νννννν0)(2)cos1(222=−+−− ννθννohmch)cos1(2)(2 22θνννν −=−chhmo)cos1()( 2θνννν −=−chmo
  17. 17. Butis the Compton Shift.λνc=)cos1(11θλλλλ−=−hcmoandλνc= So,)cos1(θλλλλλλ−= − hcmo)cos1( θλλλ −=∆=−cmhoλ∆θIt neither depends on the incident wavelength nor on thescattering material. It only on the scattering angle i.e.is called the Compton wavelength of the electronand its value is 0.0243 Å.cmho
  18. 18. Experimental VerificationMonochromaticX-ray SourcephotonθGraphitetargetBragg’s X-raySpectrometer1. One peak is found at sameposition. This is unmodified radiation2. Other peak is found at higherwavelength. This is modified signal oflow energy.3. increases with increase in .λ∆ θλ∆
  19. 19. 0.0243 (1- cosθ) Å=−=∆ )cos1( θλcmho=∆ maxλλ∆So Compton effect can be observed only for radiation havingwavelength of few Å.Compton effect can’t observed in Visible Lightis maximum when (1- cosθ) is maximum i.e. 2.λ∆0.05 ÅFor 1Å ~ 1%λ∆=λ=λFor 5000Å ~ 0.001% (undetectable)
  20. 20. Pair ProductionWhen a photon (electromagnetic energy) of sufficientenergy passes near the field of nucleus, it materializes intoan electron and positron. This phenomenon is known as pairproduction.In this process charge, energy and momentum remainsconserved prior and after the production of pair.PhotonNucleus (+ve)−e+e
  21. 21. The rest mass energy of an electron or positron is 0.51MeV (according to E = mc2).The minimum energy required for pair production is 1.02MeV.Any additional photon energy becomes the kinetic energyof the electron and positron.The corresponding maximum photon wavelength is 1.2 pm.Electromagnetic waves with such wavelengths are calledgamma rays .)(γ
  22. 22. Pair AnnihilationWhen an electron and positron interact with each other dueto their opposite charge, both the particle can annihilateconverting their mass into electromagnetic energy in theform of two - rays photon.γγγ +→+ +−eeCharge, energy and momentum are again conversed. Two- photons are produced (each of energy 0.51MeV plus half the K.E. of the particles) to conserve themomentum.γ
  23. 23. From conservation of energyγν 22 cmh o=Pair production cannot occur in empty spaceIn the direction of motion of the photon, the momentum isconserved ifθνcos2pch=θθchνθcospθcosppp−e+ehere mo is the rest mass and2211 cv−=γ
  24. 24. γvmp o=Momentum of electron and positron is(i)1cos ≤θ1<cvButθν cos2cph =Equation (i) now becomesθγν cos2 cvmh o=θγν cos2 2=cvcmh oandγν 22 cmh o<
  25. 25. But conservation of energy requires thatγν 22 cmh o=Hence it is impossible for pair production to conserve boththe energy and momentum unless some other object isinvolved in the process to carry away part of the initialphoton momentum. Therefore pair production cannot occurin empty space.
  26. 26. Wave Particle DualityLight can exhibit both kind of nature of waves and particlesso the light shows wave-particle dual nature.In some cases like interference, diffraction and polarizationit behaves as wave while in other cases like photoelectricand compton effect it behaves as particles (photon).
  27. 27. De Broglie WavesNot only the light but every materialistic particle such aselectron, proton or even the heavier object exhibits wave-particle dual nature.De-Broglie proposed that a moving particle, whatever itsnature, has waves associated with it. These waves arecalled “matter waves”.Energy of a photon isνhE =For a particle, say photon of mass, m2mcE =
  28. 28. Suppose a particle of mass, m is moving with velocity, v thenthe wavelength associated with it can be given byhvmc =2λhcmc =2mch=λmvh=λph=λor(i) If means that waves are associated withmoving material particles only.∞=⇒= λ0v(ii) De-Broglie wave does not depend on whether the movingparticle is charged or uncharged. It means matter waves arenot electromagnetic in nature.
  29. 29. Wave Velocity or Phase VelocityWhen a monochromatic wave travels through a medium,its velocity of advancement in the medium is called thewave velocity or phase velocity (Vp).kVpω=where is the angular frequencyand is the wave number.πνω 2=λπ2=k
  30. 30. Group VelocitySo, the group velocity is the velocity with which the energyin the group is transmitted (Vg).dkdVgω=The individual waves travel “inside” the group with theirphase velocities.In practice, we came across pulses rather thanmonochromatic waves. A pulse consists of a number ofwaves differing slightly from one another in frequency.The observed velocity is, however, the velocity with whichthe maximum amplitude of the group advances in amedium.
  31. 31. Relation between Phase and Group VelocitydkdVgω= )( pkVdkd=dkdVkVVppg +=( )λπλπ22ddVVVppg +=( )λλ 11ddVVVppg +=
  32. 32. λλddVVVppg −=−+=λλλddVVV ppg211λλddVpSo, is positive generally (not always).pg VV <⇒In a Dispersive medium Vp depends on frequency≠kωgenerallyi.e. constant
  33. 33. λλddVVVppg −=0=⇒λddVppg VV =⇒In a non-dispersive medium ( such as empty space)=kωconstant pV=
  34. 34. Phase Velocity of De-Broglie’s wavesAccording to De-Broglie’s hypothesis of matter wavesmvh=λwave numberhmvkπλπ 22== (i)If a particle has energy E, then corresponding wave willhave frequencyhE=νthen angular frequency will behEππνω22 ==
  35. 35. Dividing (ii) by (i)(ii)hmc22πω =mvhhmck ππω22 2×=vcVp2=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 theparticle velocity (v) and hence the waves would left theparticle behind.)( pV
  36. 36. Group Velocity of De-Broglie’s wavesThe discrepancy is resolved by postulating that a movingparticle is associated with a “wave packet” or “wavegroup”, rather than a single wave-train.A wave group having wavelength λ is composed of anumber of component waves with slightly differentwavelengths in the neighborhood of λ.Suppose a particle of rest mass mo moving with velocity vthen associated matter wave will havehmc22πω = andhmvkπ2= where 221 cvmm o−=
  37. 37. andOn differentiating w.r.t. velocity, v22212cvhcmo−=πω2212cvhvmk o−=π( ) 232212cvhvmdvd o−=πω(i)( ) 232212cvhmdvdk o−=π (ii)
  38. 38. Wave group associated with a moving particle alsomoves with the velocity of the particle.oomvmdkdvdvdππω22. =Dividing (i) by (ii)gVvdkd==ωMoving particle wave packet or wave group≡
  39. 39. Davisson & Germer experiment of electrondiffraction• If particles have a wave nature, then under appropriateconditions, they should exhibit diffraction• Davisson & Germer measured the wavelength of electrons• This provided experimental confirmation of the matter wavesproposed by de Broglie
  40. 40. Davisson and Germer Experiment
  41. 41. 0φ =090φ =Current vs accelerating voltage has a maximum (a bump orkink noticed in the graph), i.e. the highest number of electronsis scattered in a specific direction.The bump becomes most prominent for 54 V at φ ~ 50°IncidentBeam
  42. 42. According to de Broglie, the wavelength associated with anelectron accelerated through V volts isoAV28.12=λHence the wavelength for 54 V electronoA67.15428.12==λFrom X-ray analysis we know that the nickel crystal acts as aplane diffraction grating with grating space d = 0.91 Å
  43. 43. ooo65250180= −=θHere the diffraction angle, φ ~ 50°The angle of incidence relative to the family of Bragg’s plane
  44. 44. From the Bragg’s equationwhich is equivalent to the λ calculated by de-Broglie’shypothesis.θλ sin2d=oooAA 65.165sin)91.0(2 =××=λIt confirms the wavelike nature of electrons
  45. 45. Electron Microscope: Instrumental Applicationof Matter Waves
  46. 46. Resolving power of any optical instrument is proportional to thewavelength of whatever (radiation or particle) is used toilluminate the sample.An optical microscope uses visible light and gives 500xmagnification/200 nm resolution.Fast electron in electron microscope, however, have muchshorter wavelength than those of visible light and hence aresolution of ~0.1 nm/magnification 1,000,000x can be achievedin an Electron Microscope.
  47. 47. Heisenberg Uncertainty PrincipleIt states that only one of the “position” or “momentum” can bemeasured accurately at a single moment within the instrumentallimit.It is impossible to measure both the position and momentumsimultaneously with unlimited accuracy.oruncertainty in positionuncertainty in momentum→∆x→∆ xpthen2≥∆∆ xpxπ2h=∴The product of & of an object is greater than or equal to2xp∆x∆
  48. 48. If is measured accurately i.e.x∆ 0→∆x ∞→∆⇒ xpLike, energy E and time t.2≥∆∆ tE2≥∆∆ θLThe principle applies to all canonically conjugate pairs of quantities inwhich measurement of one quantity affects the capacity to measurethe other.and angular momentum L and angular position θ
  49. 49. Determination of the position of a particle by a microscopeiIncidentPhotonScatteredPhotonRecoiled electronSuppose we want to determine accurately the position andmomentum of an electron along x-axis using an ideal microscopefree from all mechanical and optical defects.The limit of resolution of themicroscope isixsin2λ=∆here i is semi-vertex angle of thecone of rays entering the objectivelens of the microscope.is the order of uncertainty in thex-component of the position of theelectron.x∆
  50. 50. The scattered photon can enter the microscope anywhere betweenthe 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 theelectron.The momentum of the scattered photon is (according to de-Broglie)λhp =Its x-component can be given asihpx sin2λ=∆The x-component of the momentum of the recoiling electron has thesame uncertainty, (conservation of momentum)xp∆
  51. 51. The product of the uncertainties in the x-components of position andmomentum for the electron isihipx x sin2sin2.λλ×=∆∆This is in agreement with the uncertainty relation.2.>=∆∆ hpx x
  52. 52. Order of radius of an atom ~ 5 x10-15mthenIf electron exist in the nucleus thenApplications of Heisenberg Uncertainty Principle(i) Non-existence of electron in nucleusmx 15max 105)( −×=∆2≥∆∆ xpx2)()( minmax=∆∆ xpx120min ..101.12)( −−×=∆=∆ smkgxpxMeVpcE 20== ∴ relativistic
  53. 53. Thus the kinetic energy of an electron must be greater than 20MeV to be a part of nucleusThus we can conclude that the electrons cannot be presentwithin nuclei.Experiments show that the electrons emitted by certain unstablenuclei don’t have energy greater than 3-4 MeV.
  54. 54. ButConcept of Bohr Orbit violates Uncertainty PrinciplempE22=2.≥∆∆ pxmppE∆=∆mpmv∆= ptx∆∆∆=pxtE ∆∆=∆∆ ..2.≥∆∆ tE
  55. 55. According to the concept of Bohr orbit, energy of an electron in aorbit is constant i.e. ΔE = 0.2.≥∆∆ tE∞→∆⇒ tAll energy states of the atom must have an infinite life-time.But the excited states of the atom have life–time ~ 10-8sec.The finite life-time Δt gives a finite width (uncertainty) to the energylevels.
  56. 56. Two-slit Interference ExperimentLaserSourceSlitSlit DetectorRate of photon arrival = 2 x 106/secTime lag = 0.5 x 10-6secSpatial separation between photons = 0.5 x 10-6c = 150 m1 meter
  57. 57. – Taylor’s experiment (1908): double slit experiment with very dimlight: interference pattern emerged after waiting for few weeks– interference cannot be due to interaction between photons, i.e.cannot be outcome of destructive or constructive combination ofphotons⇒ interference pattern is due to some inherent property of eachphoton - it “interferes with itself” while passing from source toscreen– photons don’t “split” –light detectors always show signals of same intensity– slits open alternatingly: get two overlapping single-slit diffractionpatterns – no two-slit interference– add detector to determine through which slit photon goes:⇒ no interference– interference pattern only appears when experiment providesno means of determining through which slit photon passes
  58. 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 assignspecific trajectory to photon– cannot suppose that half went through one slit and half throughother– can only predict how photons will be distributed on screen (orover detector(s))– interference and diffraction are statistical phenomenaassociated with probability that, in a given experimental setup, aphoton will strike a certain point– high probability ⇒ bright fringes– low probability ⇒ dark fringes
  59. 59. Double slit expt. -- wave vs quantum• pattern of fringes:– Intensity bands due tovariations in square ofamplitude, A2, of resultantwave on each point onscreen• role of the slits:– to provide two coherentsources of the secondarywaves that interfere on thescreen• pattern of fringes:– Intensity bands due tovariations in probability, P,of a photon striking pointson screen• role of the slits:– to present two potentialroutes by which photon canpass from source to screenwave theory quantum theory
  60. 60. Wave functionψψψ *|| 2=The quantity with which Quantum Mechanics is concerned is thewave function of a body.|Ψ|2is proportional to the probability of finding a particle at aparticular point at a particular time. It is the probability density.Wave function, ψ is a quantity associated with a moving particle. Itis a complex quantity.Thus if iBA+=ψ iBA−=*ψ222222*|| BABiA +=−==⇒ ψψψthenψ is the probability amplitude.
  61. 61. Normalizationτd|Ψ|2is the probability density.The probability of finding the particle within an element of volumeτψ d2||Since the particle is definitely be somewhere, so1|| 2=∫∞∞−τψ dA wave function that obeys this equation is said to be normalized.∴ Normalization
  62. 62. Properties of wave function1. It must be finite everywhere.If ψ is infinite for a particular point, it mean an infinite largeprobability of finding the particles at that point. This wouldviolates the uncertainty principle.2. It must be single valued.If ψ has more than one value at any point, it mean more thanone value of probability of finding the particle at that pointwhich is obviously ridiculous.3. It must be continuous and have a continuous first derivativeeverywhere.zyx ∂∂∂∂∂∂ ψψψ,, must be continuous4. It must be normalizable.
  63. 63. Schrodinger’s time independent wave equationOne dimensional wave equation for the waves associated with amoving particle isFrom (i)ψ is the wave amplitude for a given x.whereA is the maximum amplitude.λ is the wavelengthψλπψ22224−=∂∂x(ii))0,(and),()(2)( xipxEtiAetxAetx λπψψ ===−−
  64. 64. vmho=λ22221hvmo=⇒λ 22212hvmm oo =2221hKmo=λwhere K is the K.E. for the non-relativistic case(iii)Suppose E is the total energy of the particleand V is the potential energy of the particle)(2122VEhmo−=λ
  65. 65. This is the time independent (steady state) Schrodinger’s waveequation for a particle of mass mo, total energy E, potentialenergy V, moving along the x-axis.If the particle is moving in 3-dimensional space thenEquation (ii) now becomesψπψ)(242222VEmhxo −−=∂∂0)(2222=−+∂∂ψψVEmxo0)(22222222=−+∂∂+∂∂+∂∂ψψψψVEmzyxo
  66. 66. For a free particle V = 0, so the Schrodinger equation for afree particle0222=+∇ ψψ Emo0)(222=−+∇ ψψ VEmoThis is the time independent (steady state) Schrodinger’s waveequation for a particle in 3-dimensional space.
  67. 67. Schrodinger’s time dependent wave equation)( pxEtiAe−−= ψWave equation for a free particle moving in +x direction is(iii)ψψ2222px−=∂∂where E is the total energy and p is the momentum of the particleDifferentiating (i) twice w.r.t. x(i)2222xp∂∂−=⇒ψψ  (ii)Differentiating (i) w.r.t. tψψiEt−=∂∂tiE∂∂=⇒ψψ 
  68. 68. For non-relativistic caseUsing (ii) and (iii) in (iv)(iv)ψψψVxmti +∂∂−=∂∂2222E = K.E. + Potential EnergytxVmpE ,22+=ψψψ VmpE +=⇒22This is the time dependent Schrodinger’s wave equation for aparticle in one dimension.
  69. 69. Linearity and Superposition2211 ψψψ aa +=If ψ1 and ψ2 are two solutions of any Schrodinger equation of asystem, then linear combination of ψ1 and ψ2 will also be a solutionof the equation..Here are constantsAbove equation suggests:21 & aais also a solution(i) The linear property of Schrodinger equation(ii) ψ1 and ψ2 follow the superposition principle
  70. 70. 21 ψψψ +→ThenTotal probability will be2212|||| ψψψ +==Pdue to superposition principle)()( 21*21 ψψψψ ++=))(( 21*2*1 ψψψψ ++=1*22*12*21*1 ψψψψψψψψ +++=1*22*121 ψψψψ +++= PPP21 PPP +≠Probability density can’t be added linearlyIf P1 is the probability density corresponding to ψ1 and P2 is theprobability density corresponding to ψ2
  71. 71. Expectation valuesdxxf∫∞∞−= 2||)( ψExpectation value of any quantity which is a function of ‘x’ ,say f(x)is given byfor normalized ψThus expectation value for position ‘x’ is>< )(xfdxx∫∞∞−= 2||ψ>< xExpectation value is the value of ‘x’ we would obtain if wemeasured the positions of a large number of particles described bythe same function at some instant ‘t’ and then averaged theresults.
  72. 72. dxx∫=102||ψSolution>< x10424=xaQ. Find the expectation value of position of a particle having wavefunction ψ = ax between x = 0 & 1, ψ = 0 elsewhere.dxxa ∫=1032>< x42a=
  73. 73. Operatorsψψpix =∂∂(Another way of finding the expectation value)For a free particleAn operator is a rule by means of which, from a given functionwe can find another function.)( pxEtiAe−−= ψThenHerexip∂∂=^is called the momentum operator(i)
  74. 74. ψψEit −=∂∂SimilarlyHeretiE∂∂= ^is called the Total Energy operator(ii)Equation (i) and (ii) are general results and their validity is thesame as that of the Schrodinger equation.
  75. 75. Uximti +∂∂=∂∂221 If a particle is not free thenThis is the time dependent Schrodinger equation^^^.. UEKE +=^^2^2UmpEo+=⇒UU =∴^Uxmti +∂∂−=∂∂2222ψψψUxmti +∂∂−=∂∂2222
  76. 76. If Operator is HamiltonianThen time dependent Schrodinger equation can be written asUxmH +∂∂−= 222^2ψψ EH =^This is time dependent Schrodinger equation in Hamiltonianform.
  77. 77. Eigen values and Eigen functionSchrodinger equation can be solved for some specific values ofenergy i.e. Energy Quantization.ψψα a=^Suppose a wave function (ψ) is operated by an operator ‘α’ suchthat the result is the product of a constant say ‘a’ and the wavefunction itself i.e.The energy values for which Schrodinger equation can be solvedare called ‘Eigen values’ and the corresponding wave function arecalled ‘Eigen function’.thenψ is the eigen function ofa is the eigen value of^α^α
  78. 78. Q. Suppose is eigen function of operator then findthe eigen value.The eigen value is 4.Solution.xe2=ψ 22dxd22^dxdG =22^dxdGψψ = )( 222xedxd=xeG 2^4=ψψψ 4^=G
  79. 79. Particle in a BoxConsider a particle of rest mass mo enclosed in a one-dimensionalbox (infinite potential well).Thus for a particle inside the box Schrodinger equation isBoundary conditions for PotentialV(x)=0 for 0 < x < L∞{ for 0 > x > LBoundary conditions for ψΨ =0 for x = 0{0 for x = L02222=+∂∂ψψEmxox = 0 x = L∞=V ∞=Vparticle0=V0=∴V inside(i)
  80. 80. Equation (i) becomesph=λkπ2=pk =⇒Emo2=22 2Emk o=⇒(k is the propagation constant)(ii)0222=+∂∂ψψkx(iii)General solution of equation (iii) iskxBkxAx cossin)( +=ψ (iv)
  81. 81. Equation (iv) reduces toBoundary condition says ψ = 0 when x = 0(v)0.cos0.sin)0( kBkA +=ψ1.00 B+= 0=⇒ BkxAx sin)( =ψBoundary condition says ψ = 0 when x = LLkAL .sin)( =ψLkA .sin0 =0≠A 0.sin =⇒ LkπnLk sin.sin =⇒
  82. 82. Put this in Equation (v)(vi)Lnkπ=LxnAxπψ sin)( =When n # 0 i.e. n = 1, 2, 3…., this gives ψ = 0 everywhere.πnkL =Put value of k from (vi) in (ii)22 2Emk o=222EmLn o= π
  83. 83. Where n = 1, 2, 3….Equation (vii) concludesomkE222=⇒ 2228 Lmhno= (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 = 12218 LmhEo= (Zero Point Energy)If momentum i.e.01 →E 0→ 0→∆p∞→∆⇒ xBut since the particle is confined in the box ofdimension L.Lx =∆ max
  84. 84. Thus zero value of zero point energy violates theHeisenberg’s uncertainty principle and hence zero value isnot acceptable.2. All the energy values are not possible for a particle inpotential well.Energy is Quantized3. En are the eigen values and ‘n’ is the quantum number.4. Energy levels (En) are not equally spaced.n = 1n = 3n = 23E1E2E
  85. 85. Using Normalization conditionLxnAxnπψ sin)( =1sin022=∫ dxLxnALπ1|)(| 2=∫∞∞−dxxnψ122= LALA2=⇒The normalized eigen function of the particle areLnxLxnπψ sin2)( =
  86. 86. Probability density figure suggest that:1. There are some positions (nodes) in the box that will never beoccupied by the particle.2. For different energy levels the points of maximum probabilityare found at different positions in the box.|ψ1|2is maximum at L/2 (middle of the box)|ψ2|2is zero L/2.
  87. 87. Eigen functionzyx ψψψψ =LznLynLxnAAA zyxzyxπππψ sinsinsin=LznLynLxnLzyx πππψ sinsinsin23=222228)(mLhnnnE zyx ++=Particle in a Three Dimensional Boxzyx EEEE ++=Eigen energy

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