Quantum Mechanics forQElectrical EngineersDennis M. Sullivan, Ph.D.Department of Electrical andDepartment of Electrical and Computer EngineeringUniversity of Idahoy1Why quantum mechanics?Why quantum mechanics?At the beginning of the 20thcentury, variousAt the beginning of the 20century, various phenomena were being observed that could not be explained by classical mechanics.1. Energy is quantizedgy q2. Particles have a wave nature2SfSource of electrons3Source of lelectrons4Conclusion:PilhCo c us o :Particles have wave propertiesproperties.5Incident lightPhotoelectronsPhotoelectric EffectIncident lightMaterialThe velocity of the escaping particles was dependent on the wavelength of the light, not the intensity as expected.6ggypKineticPhotoelectric EffectEnergy TPhotoelectric EffectPlanck postulated in 1900 that thermal radiation is emitted from a heated surface in discrete packets called quanta0fFrequency fquanta.7Einstein postulated that the energy of each photonEinstein postulated that the energy of each photonwas related to the wave frequency.Ehfω==hEhfω==h34h is Plank’s constant34156.625 104.135 10hJseV s−−=× −=× −This is the first major result: i ltdt fenergy is related to frequency8In 1924, Louis deBroglie postulated the existence ofIn 1924, Louis deBroglie postulated the existence of matter waves. This lead to the famous wave-particle duality principle. Specifically thatthe momentum of a photon is given byhp=the momentum of a photon is given bypλ=Or the more familiar formSecond major result:pk=hSecond major result:Momentum is related to wavelength9Bottom line:Everything is at the same time a particle and a wave.hpλ=Ehf=pλf10Solve using onlyenergyPhysicists formulate everything as an energy problemSolve using onlyenergy. Problem: Determine the velocity of the ball at the bottom of the1 kgof the ball at the bottom of the slope.1 meter11Whil th b ll i th t f th hill it hWhile the ball is on the top of the hill, it has potential energy.Since the acceleration of gravity is1 kg29.8 /gms=the potential energy is()()29.8 1 1 9.8PE g mass of the ball heightmkg m Js=××⎛⎞==⎜⎟⎝⎠1 meters⎝⎠12When it gets to the bottom of the hill, it no longer has potential energy, but it has kinetic energy. Since therehave been no other external forces, it must be the same,219.82vTJM==219.82TJMv==22129.8 4.31kg m mvskg s⎛⎞⋅=⋅ =⎜⎟⎝⎠1 meter13There are several methods ofThere are several methods of advanced mechanics that change everything into energy.eve yt g to e e gy.1Lagrangian mechanics1.Lagrangian mechanics2. Hamiltonian mechanics14Erwin Schrödinger was taking this approach and developed the following equation toand developed the following equation to incorporate these new ideas:2⎛⎞22222212Vtmψψ⎛⎞∂=− ∇ −⎜⎟∂⎝⎠hhThis equation is 2nd order in time and 4thorder⎝⎠This equation is 2nd order in time and 4order is space.15Schrödinger realized that this was a completelySchrödinger realized that this was a completely intractable problem. (There were no computers in 1921.) However, he saw that by considering ψto be a complex function, he could factor the above equation into two simpler equations, one of which is22iV∂∇hh22iVtmψψψ=−∇+∂hThis is the Schrödinger equation.16Th i h S h di iiThe parameter in the Schrödinger equation, ψ, is a state variable. It is not directly associated with any physical quantity itself, but all the information can be py q y ,extracted from it. Also, remember that the Schrödinger equation was only half of the “real” equation from which it was derived So to determineequation from which it was derived. So to determine if a particle is located between a and b, calculate() ()*()baPa x b x x dxψψ<< =∫a∫17This brings us to one of the basics requirementsof the state variable: it must be normalized()()*1dψψ∞=∫rrr()()ψψ−∞∫N t Th lit d f th t t t dtNote: The amplitude of the state verctor ψ does notrepresent the strength of the wave in the usual sense.It is chosen to achieve normalization18Computer simulations can show how the Schrödinger equation can model a particle like an likelectronpropagating as a wave packet.The real part is blue, the imaginary part is red.19p,gyp2021222324In quantum mechanics, physical properties are q,pypprelated to operators, call observables.Two of the fundamental operators are:1. Momentum2. Kinetic energygy25The momentum operator (in one dimension) is∂h.pix∂=∂hix∂This seems pretty strange until I rememberthat momentum is related to wavelength:ghkh.pkλ==h26We think of a wavepacket as a superposition of plane waves()pixtikx teeωω⎛⎞−⎜⎟−⎝⎠=heeWhen I apply the momentum operatorpEpEixt ixt⎛⎞ ⎛⎞−−⎜⎟ ⎜⎟⎝⎠ ⎝⎠∂hh hhepex⎜⎟ ⎜⎟⎝⎠ ⎝⎠∂−=∂hh hhh27Kinetic energy can be derived from gymomentum:22..2pKEm=2221∂−∂⎛⎞==⎜⎟hh222mix mx⎜⎟∂∂⎝⎠28If I want the expected value of the pkinetic Energy() ()222.. *2KE x x dxmxψψ∞−∞−∂=∂∫h2mx−∞∂∫293031Note that smaller values of wavelength lead llf dkiito larger values of momentum and kinetic energy:hpλ=22222phKEmmλ==322∂hLet’s look back at the Schrödinger equation()()()()2,,,2ixt xtVxxttmψψψ∂=− ∇ +∂hhV(x) is the potential. It represents the potential energy that a particle sees. Anypotential energy that a particle sees. Any physical barrier must be modeled in terms of a potential energy as seen by the particle. 33Conduction band diagram for n-typeConduction band diagram for ntype semiconductorsThis is modeled in the SE as a change in potential:34Similarly, if I want the expected value of the Potential Energy∞∫()()().. *PExVxxdxψψ∞−∞=∫35Let’s take a look at the Schrödinger equation and some of the things we might be able to determine from it.2222iVtmψψψ∂=− ∇ +∂hhTotal energy = Kinetic energy + Potential energy22hThe kinetic energy operator is22m−∇hThe potential energy operator is V36() ()2240*20.2nmKE x x dxmxψψ⎛⎞∂=⋅−⎜⎟∂⎝⎠∫h2mx∂⎝⎠373839Notice that as the wave interacts with the potential, some kinetic energy is lost but somepotential, some kinetic energy is lost but some potential energy is gained.404142() () ()40*20.nmnmPE x V x x dxψψ=⋅⋅∫The total energy stays the same!4344Now notice that part of the waveform has continued propagating in the medium andcontinued propagating in the medium and