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# HUMAN RIGHTS LAWSO 162

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ECE 162A Mat 162A Lecture 16Band Theory Chapter 13 John Bowers Bowerseceucsbedu ECEMat 162A ECEMat 162A Hartree Theory Multielectron atomic theory First consider a nuclear attractive coulombic force Ze Include the coulombic repulsion due to the average position of other electrons Z1e Assume eacn electron moves Independently or each other Schroedinger s equation becomes separable into Z separate equations for the motion of each electron 712 2 V2w V001 2 Egg m What is Vr First guess For r90 2 Vr Ze 4721901 For r infinity 2 e V 2 r 4721901 Then solve SE and calculate Vr ECEMat 162A Hartree Solution Assume Vr revious slide Calculate the solutions Wer999 9l r969 9lyr969 9quot39 with energies EaE Ey Put one electron in each state Soln is the product of one electron states V ZW0719919 19W ltr29629 29Wyr39639 39 39 Note ocBy includes space and spin a n1Z Oml OTgt i nzll0ml 0lgt ECEMat 162A Hartree Solution cont Use mm to calculate Vr Iterate Soln SE using new Vr 227 Argon 20 p39 Put one electron In each 18 39 state to assemble w 16 ll Calculate Vr Pm P gt Zr gt 5 i 8 6 I 4 2 L Zr ECEMat 162A I 1 I 1 viii lm Hartree Fock Theory Hartree theory ignores the requirement of antisymmetric wave functions Fock included this For a N electron atom N terms are added in the expression for m It is an N dimensional Slater determinant The effect is only significant for the valence electrons ECEMat 162A Results of Hartree Theory Eigenfunctions abeled by E will not be given by E0n2 Important for solid state physics Numerical calculation programs exist for Hartree and HartreeFoch S ECEMat 162A Order and Labeling of energy levels shells N1O N2IO N2 1 N3 O N3 1 N3 2 N4 1 ECEMat 162A 1s 2s 2p 3s 39 4s 3d 49 2 states up down s 2 states up down LiMg 6 states 2xmI1O1 2 states 2xml0 6 states 2xmI1O1 2 states 2xm0 10 states 2xmI21O12 6 states 2xmI I O1 Order and Labeling of energy levels shells N1O N2IO N2 1 N5 U N3 1 N3 2 N4 1 N5 O N5 1 5 m b T 800lt o E i at gt 13 23 2p 33 3p 43 3d 49 53 5p 63 4f 2 states 2 states 6 states 2 states 6 states 2 states 10 states 6 states 2 states 6 states 2 states 14 states Chemistry Inner electrons affected by nucleus other atoms Outer electrons se a potential of 2e and are affected by adjacent atoms The outer electrons are called valence electrons and are involved in bonding to other atoms Atoms with the same valence structure behave similarl chemical ECEMat 162A Periodic Table Mendeleev n 6S 7s H 3 4 Li Be 1 1 12 Na Mg 19 20 K Ca 37 38 Rb Sr 55 56 C5 Ba 87 88 Fr Ra 3d 4d 5d 2 He 5 6 7 8 9 10 2p B C N 0 F Ne 13 14 15 16 17 18 3p AI Si P 5 Cl A 31 32 33 34 35 36 4p Ga Ge As Se Br Kr 49 50 51 52 53 54 Ru 5p In Sn Sb Te l Xe 5314417 5s14d8 La 77 81 82 83 84 85 86 Lamha Hf 0s Ir 6 TI Pb BI PO A Rn Ac 7p Actinxdes d1 d2 d3 d4 d5 d6 d7 d3 d9 110 1 12 13 p4 p5 p6 58 59 60 61 62 63 64 65 66 67 68 1 4f Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Lanthanldes 5d04f2 5d04f3 5d04f4 5d04f5 046 5d04f7 5d14f7 5d04f9 5d04f10 5d04f11 69 70 71 El Tm Yb Lu 0412 5d0413 51104 5d14f14 90 91 92 93 94 95 96 97 98 99 100 101 102 103 5f Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lw Actmides 6d25f 6d 5f2 6d15f3 ad s4 6d15 5 6d 5f6 6d15f7 6d 58 6d 5f1 6d 5f11 6d 5f 2 6d 5r13 6d 5f14 ad sr f1 f2 f3 f4 f5 f6 f7 f8 f9 10 11 12 13 14 Band Theory of Solids Isolated 9 MW 2 p 23 D ECEMat 162A Band Theory of Solids Isolated solid 2p E Splitting higherfor 2s l outer electrons more ovedap 1s Il Spllttmg Increases as the separation decreases ECEMat 162A Energy Bands 7 I a W quotlllllll39 ullquotiquotiil39i7quot39quotquot39 I llIlllllli39ll39llllilquot39 39 1 Bands that originate from closed shells eg 1s 2s for boron have filled bands Shells that are partially filled result in bands that are partially filled Filled bands give insulators Partially filled bands give metals Filled bands with empty bands close by 701 to 3 eVV are called semiconductors Materials with odd number of electronsatom are always metals Interatomic distance A Energy eV EOEMat 162A Quantum Statistics How are electrons distributed versus energy Classical answer Maxwell Boltzman distribution Ae EkT However this has most electrons at low energies which violates the Pauli exclusion principle ECEMat 162A Quantum Statistics How are electrons distributed versus energy Classical answer Maxwell Boltzman distribution Ae EkT Quantum mechanical answer Fermi Dirac distribution ECEMat 162A PE 1 Boltzmann Fermi Dirac E gt A eE E0 kT Electron distribution Na NQ 0 a 0 quot5 Mt FM 1 1 1 i l l 0 53F 9 0 bF n as N05quot n lt9 N 9 Unfilled levels Unfilled levels Filled 39eVelS X Filled level X 0 gap gmax g 0 17 gmax Tgt0 n OfwdE NEfE ECEMat 162A Conduction A filled band has no net current flow For every electron going in one direction there is an electron going in the opposite direction ECEMat 162A Fermi Energy The fermi energy is determined by the doping level n EdE NEfE E E nN eX F C c p H Where NC is the effective number of states in the conduction band and EF is the Fermi energy ECEMat 162A Hole density p f dE NEgt1 M E E P NV CXP Where NV is the effective Jensity of states in the valence band and EV is the valence band edge ECEMat 162A Hole density p f dE NltEgtlt1 fE EV EF kT Where NV is the effective density of states in the valence band and EV is the valence band edge pZNV 39Xr E E E E nN N ex F C eX p C V p kT p kT EC EV T p NCNV eXp k ECEMat 162A How do electrons move through a lattice The potential is periodic WNQVVN loch found the solution to be 135 2 uk 306 LPx t 2 uk ew a The ukx are the Bloch functions ukx ukxa ukxna The free motion of the electron through the lattice is given by ikx aJt e ECEMat 162A KronigPenney Model WWWWAN t Approximate this potential with the rectilinear one shown below Wm m m ECEMat 162A VX Energy BandsExtended Zone The region from 7ca to le is the First Brillouin Zone The second Brillouin Zone extends from ale to 3 TIEa ECEMat 162A Energy Bands 6 V0 20 Extended zone Reduced zone Repeating zone 10 Energy bands 6 Bnlloum zone number ECEMat 162A Direct Gap vs Indirect Gap Semiconductors where the minimum in the conduction band and the maximum in the valence band are located at the same point in k space are called direm gap semiconductors Semiconductors where the minimum in the conduction band and the maximum in the valence band are located at different point in k space are called direct gap semiconductors Silicon and AlAs are indirect gap semiconductors GaAs and lnP are direct gap semiconductors F kO kO Direct Indirect A K t ECEMat 162A Direct Gap vs Indirect Gap Direct gap semiconductors Good absorption and good emission because photons can be directly absorbed or emitted Indirect gap semiconductors Low absorption and low emission rates because a phonon IS required to conserve energy and momentum ECEMat 162A Direct gt r kO Indirect Photons travel at the speed of light so the transitions are E almost vertical lines momentum change is very Small39 Electron Photon Epc ECEMat 162A Compound Semiconductors Tetrahedrally bonded in a zinc blende structure Several varieties IIIV compounds V compounds ECEMat 162A IIIV Compound Semiconductors ls H 3 4 2s Li Be 11 12 33 Na Mg 19 20 21 22 23 24 25 25 27 28 29 30 4s K Ca 3 Sc Ti V Cr Mn Fe Co Ni Cu Zn 4s13d 43131110 37 38 39 40 41 42 43 44 45 46 47 4s 53 Rb Sr 4d Y Zr Nb Mo Tc Ru Rh Pd Ag Cd 5914714 5314117 5s14d3 55041110 55141110 55 56 57 72 73 74 75 76 77 78 79 80 6s Cs Ba 5d Lang Hf Ta w Re Os Ir Pt Au Hg nides 6815619 6s15dl390 87 88 89 75 F Ra 6d Ac Actinides sl 52 d1 d2 d3 d4 d5 d5 d7 d3 d9 allquot p1 p2 p3 p4 p5 16 58 59 60 61 52 63 64 65 66 67 68 69 7o 71 1 4f Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthan39des 5d 4f2 5d04f3 5d04f4 5010475 5110MB 5d 4f7 5114 51049 5104 0 51104711571047 51104713 51104714 5d14r 9o 91 92 93 94 95 96 97 98 99 100 101 102 103 5f Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lw Actmides 6d25f 6d15f2 6d15f3 6d154 6d15 5 6d15f6 6d15f7 6d15f3 510510 adosf edos z 6d05f13 6d05f 6d15f 1 f2 f3 f4 f5 f6 f7 f8 f9 10 ll 12 13 14 ECEMat aN GaP GaAs GaSb InN InP InAS InSb AIN AIP AISb 25 38 4s 53 7s IIVi Compound Semiconductors VI 1 2 H He 3 4 5 6 7 3 9 10 Li Be 2p B c N 0 F Ne 11 12 13 14 15 16 17 18 Na Mg 3p Al 51 P 6 Ct A 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca 3 Sc Ti V Cr Mn Fe Co Ni Cu 21 4p Ga Ge As Se Br Kr 4s13d5 4313d10 37 38 39 40 41 42 43 44 45 46 47 43 49 50 51 52 53 54 Rb Sr 4d Y Zr Nb Mo Tc Ru Rh Pd Ag ca 5p In Sn Sb Te I Xe 5914774 5314117 5s14d3 55041110 55141110 55 56 57 72 73 74 75 76 77 78 79 33 31 82 83 34 85 86 Cs Ba 5d Lang Hf Ta w Re Os Ir Pt Au Hg 6p Ti Pb 8139 Po A1 Rn nides 6815619 6s15d39C s7 88 89 F Ra 6d Ac 7p Actinides sl 52 d1 d2 d3 d4 d5 d5 d7 d3 d9 allquot p1 p2 p3 p4 p5 16 58 59 60 61 62 63 64 65 66 67 68 69 70 71 4f Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthanides 5d 4f2 5d04f3 5110474 5010475 5110475 5d 4f7 511477 5d 4f9 5d 4f1 51104711511047 5604713 56104714 5d147 90 91 92 93 94 95 96 97 98 99 100 101 102 103 5f Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lw Actmides 6d25f 6d15f2 6d15f3 6d154 6d15 5 6d15f6 6d15f7 6d15f3 6d 571 6d05f11 60105712 6d05f13 611 66115714 1 f2 f3 f4 f5 f6 f7 f8 f9 10 ll 12 13 14 ECEMat 162A Semiconductor Problems pn junction in GaAs EG142 eV NC47 x1017 cm393 NV7x1O18 cm393 ni21x106 cm393 In equilibrium the Fermi level is constant across the structure Draw a pn junction in GaAs for ND1O18 cm3 and NA1O17 cm3 ECEMat 162A pn junction in GaAs EF E ECEMat 162A pn junction in GaAs x EC Rules Ev 1 Valence bands and conduction bands equadistant and parallel everywhere constant bandgap 2 Fermi energy EF is constant in equilibrium 3 Quadratic variation of band ed39e constant do inr de letion edge approximation 4 Fermi level to band edge separation appropriate for doping d2CD p nND NA EF ECEMt162A 2 a dx 8 a Binaries Bandgap always fixed Lattice constant always fixed Consequently you can grow large boules of binaries and cut identical substrates out of the boule Fermi level depends on doping ECEMat 162A Ternary Materials Binar materials alwa s have a I articular bandgap and a particular lattice constant However many times we need intermediate bandgaps not provided in nature Or we wish to make a square well or some other heterostructure Ternaries GaXAl1XAs HgXCd1XTe These are both mixtures on the group III site Mixtures on the group V site are also possmle ECEMat 162A Ternary Materials Mix two binaries together The bandgap is approximately the arithmetic avemge of u e vau Vegard s law eg GaXAl1XAs There are two types of sites group III and group V IIVI compounds also possible Ternaries cannot be grown on binary substrates in general because the lattice constants don t line up and dislocations occur Special case Ga Al1XAs because the lattice constants of Ga s and AlAs are almost equal ECEMat 162A Quaternary materials To match bandgap and lattice constan wo degrees of freedom are required Example Gaxln1XAsyP1y X is the fraction of group III sites occupied by Ga Y is the fraction of group V sites occupied by As In a bandgap chart the dots are binaries the lines are ternaries and the regions bounded by 4 lines are quaternaries ECEMat 162A Energy bandgap ev 40 3O 20 10 Lattice mismatch to silicon o 5 10 I5 25 I I o 39 ZnS ZFCIDSB AIP CdS ZnTe GaP A O O Cd8e Q Ale CdTe p o Lattice constant A GaAS ll x ulk SLE tIGe InSb 39E I IGeP l 55 60 66 04 06 SNPJNT ooom o Wavelength pm Epitaxial Layers Epitaxial layers c different compound semiconductors can be grown on top of each other Small differences in lattice constants can be accommodated lUl thin layers sualneu layers Compressive or tensile Too much accumulated strain results in dislocations ECEMat 162A lnPlnGaAslnP Square Well lnGaAs Eg V w lnP Eg135 eV How do bands line up It depends on the electron affinity Energy to bring an eleetmmfmm the conduction band edge to vacuum Bandgap Heaven Offsets in blue 3 Bandgaps in black Units are meV AEC lnAsAle 135eV m lnAs 0023 me lnAs RT p gt 30000 cmZNs ECEMat 162A Guidelines Figure out bandgaps draw horizontal where there is no depletion Figure out conduction band and valce band discontinuity Figure out Fermi level for ech terial Draw flat band first then flaten out the Fermi level Keep the band separation constant for any region where the bandgap is not changing ECEMat 162A Semiconductor Problems pn junction in GaAs EG142 eV NC47 x1017 cm393 NV7x1O18 cm393 ni21x106 cm393 In equilibrium the Fermi level is constant across the structure Draw a pn junction in GaAs for ND1O18 cm3 and NA1O17 cm3 ECEMat 162A pn junction in GaAs ECEMat 162A InPInGaAslnP Square Well N InGaAs Eg V w N InP Eg135 eV ECEMat 162A InPInGaAsInP Square Well N InGaAs E 76 eV 9 N InP Eg135 eV D AEg59 eV D AEC25 eV D AEV34 eV Draw the flatband diagram first ignoring doping ECEMat 162A InPInGaAslnP Square Well N InGaAs Eg76 eV N lnP Eg135 eV D AEg59 eV D AEC Lu cv D AEV39 eV A M J1 ECEMat 162A General comments If the composition is constant the bandgap is constant Hence the se aration of conduction band and valence band are constant If there is no field the bands are horizontal Use the depletion edge approximation either the material is depleted of free carriers and the bands are bent or there is no field and the bands are flat Depleted doped material has a quadratic bend and linearly increasing field Depleted undoped material has constant electric field and linear band bending dZCD 0p nND NA 2 ECEMat 162A 6156 8 8 Gain and Absorption ECE 162C Lecture 6 Prof John Bowers ECE 162C Energy bandgap ev 40 3O 20 10 Lattice mismatch to silicon O 5 IO 15 25 I I I I I o 39 ZnS ZrCIDSe A CdS ZnTe AIAS 0 O GaP 39 Q Ale CdTe Ch C Lattice constant A iGaAS Milk iSLE bee wmanb S I InAs InSb E I GEP I I I I l I I I HgITe 55 50 65 04 05 U39PJNT ooom o Wavelength pm Bandgap Heaven Offsets in blue 3 Bandgaps in black Units are meV AEC lnAsAle 135eV m lnAs 0023 me lnAs RT p gt 30000 cmZNs ECE 162C General comments If the composition is constant the bandgap is constant Hence the sel aration of conduction band and valence band are constant If there is no eld the bands are horizontal Use the depletion edge approximation either the material is depleted of free carriers and the bands are bent or there is no eld and the bands are at Depleted doped material has a quadratic bend and linearly increasing eld Depleted undoped material has constant electric eld and linear band bending dZCD 0p nND NA air2 8 8 ECE 162C InPInGaAsInP Square Well N InGaAs Eg76 eV l N InP Eg135 eV AEg59 eV AEC20 eV AEV39 eV Ark Adjust Fermi level to account for bias Keep bandgaps constant Calculate the absorption bandedge assuming a 100 Angstrom quantum well and me01 me and mh1 me Calculate the absorption bandedge assuming a 100 Angstrom quantum dot and me01 me and mh1 me Check the validity of your assumptions ECE 162C Three types of heterojunctions Type I Straddling Free electrons and holes reside in the same region of space Examples AlAsGaAs InPGaInAs ECE 162C Three types of heterojunctions Type I Straddling lineup Type II Staggered lineup D ECE 162C ABC and ABC have the same sign Free electrons and holes reside in different revions of SA ace Example AleInAs Three types of heterojunctions Type 1 Straddling lineup Type II Staggered lineup Type 111 Broken Gap Lineup D ABC and ABC have the same sign Free electrons and holes reside in different regions of space Example GaSbInAs ECE 162C Bands Direct bandgap Minimum of conduction band and maximum of valenc band occur at the same point in k space typically k0 de ned as F E h2k2 2 e E A ECE 162C Si Indirect gap Hence a phonon is required to conserve momentum Less likelv to occur Lower vain and absorption except at higher energies Figure 32 Illustration of bandto band of absorption and recombination processes in a direct bandgap semi conductor and b and c indirect ECE 162C C bandgap semiconductor Electronic transitions R12 absorption of a photon ECE 162C Electronic transitions O WT 1W O Rsp 0 RM absorption of a photon Rsp spontaneous emission of a photon ECE 162C Electronic transitions o c c WT lvx Wlw o o o R21 6 R12 absorption of a photon w R310 spontaneous emission of a photon R21 stimulated emission of a photon ECE 162C Electronic transitions o o o o1 WT 1m Wlw l o o o 0 RBI R12 absorption of a photon Rsp spontaneous emission of a photon R21 stimulated emission of a photon Rm nonradiative recombination Auger trap etc ECE 162C Rate Equations dt N is the electron density assumed equal to hole density G is the generation rate of electrons R is the total recombination rate ECE 162C Rate Equations dN G R dt G211 qV RRspRnrRiRsl RBN2ANCN3RSI Rz T N is the electron densitd assumed e lual to hole densitd G is the generation rate of electrons ECE 162C R is the total recombination rate Material Gain Calculated gain curves 25039 3213 for InGaAsPInP laser 7 20 39 Operating at 13 pm g 15 Gain peak moves to g 10 39 N WE shorter wavelengths 4 5 quot with higher pumping g Higher differential 50 i gain for wavelengths 10 p shorter than the gain 5 15 0 v if peak 1 200 090 092 p 094 096 ENERGY eV NKEl 1Etth6lCprl Phys 51 6095 1980 Gain n j NC a F Ed a J W Elwmnencrgy Figure 218 39 39 distributions in energy in an tx lype o u 1 io quot 8 P FINVE17f8J as and carrier High gain requires 1 upper level full f1 2 lower level empty f0 fE 1 6E EFkT ECE 162C Carrier Injection 0 In equilibrium pm 2 Fl 1 Under forward bias2 pn gtgt nl Under reverse bias 2 pm ltlt I ll ECE 162C Carrier Injection 0 In equilibrium 2 pm 2 n1 EFn EFP EF Under forward bias 2 pm gtgt 711 EF Under reverse bias 2 pm ltlt n1 EF ECE 162C Quasi Fermi Levels E E kT Fn 1 71211 EFp ElkT p nie 2 EFn EFpkT pm I VIZ e ECE 162C Quasi Fermi Levels E E kT n l E E kT p Fp 2 E n E kT Gain occurs when ghagt0 when EF 72 EFp gt720 ECE 162C Optical Gain in Semiconductors Gain between two levels depends on Carrier density ie level of inversion amp Z AEfE21kT R12 f11 f2 Reduced density of states 1 Z l 1 pr p0 pV 0 Transition matrix element M2 ECE 162C Gain Reduced Density of States The optical gain is proportional to tne IGUUCGU density or state at the transition energy gw OC f2 f1prhw The reduced density of states J 2m 32 Bulk pr 472392 p Wire Bulk m 1 PXE Well Well 2 1 pr E F Wire E h 27 ny ECE 162C Gain Strained QW Compressive strain increases the energy gap between the heavy hole and the light hole subbands This means fewer carriers in light we uallu Lattice matched QW Compressive strain LA SW Corzine Diode Lasers and Photonic Integrated Circuit Wiley 1995 Gain NP 2 Np 06gz gzlde 1 6 1 Np dz 12ng dt vgN ECE 162C R21 R12 Gain Np Np0egz dN dN 1 p 1 p 1 g Np dz 2ng dt Vng 21 12 R21 er21f1 R12 er11f2 R21R12 Rrf2f1 ECE 162C Gain ECE 162C NPNPOegZ gZNLde 1 de p dz 12ng dt vg RZIZer21f1 R12er11 f2 R21R12Rrf2f1 Fermi 39 SGoldenRule 1 R R Np 21 12 2 Rr 77Z M21 2pr E21 127139 m7 M21 prE21f2 f1 8 Lineshape function To get total gain integrate over all possible states that contribute to the gain Within a lineshape function gmw Ignme E21dE21 1 112quot LUM E21 2 m 2 7r flTm hw E21 This Lorentzian lineshape is common but not the most accurate ECE 162C Lasing threshold Lasing occurs when the round trip gain equals the loss in the cavity Not all of the mode sees gain but only the fraction that overlaps with the gain region Hence the modal gain is related to the material gain by an effective con nement factor F ECE 162C Double Heterostructure Lasers Kroemer Carriers diffuse away so it is difficult to get hivh vain A method of con ning the carriers to a region in space is necessary Double heterostructure proposed in 1964 but not implemented until 1968 which led to the first cw lasers m ECE 162C Gain Doping pdoping in the active region increases the differential gain at the expense of increased threshold current dens y 6 E E 8 0 E 3 gonnn 15 0 V g 7 2 E a CI 2 39U Puquot funquot nn innn uv wvu vuu uuu IUUU Curfem Dansquot Alcmz Current Density Acmz LACgi dr1e12acxid SW Corzine Diode Lasers and Photonic Integrated Circuit Wiley 1995 Rate Equations Neglecting the phase of the optical eld the length dependence of the carrier and photon densities and the modal dependence the rate equations for the averaged photon and carrier densities become ECE 162C djw5i dt 18S r r dN I vgaNNzrS N Gain NP 2 Np 06gz gzlde 1 6 1 Np dz 12ng dt vgN ECE 162C R21 R12 Gain Np Np0egz dN dN 1 p 1 p 1 g Np dz 2ng dt Vng 21 12 R21 er21f1 R12 er11f2 R21R12 Rrf2f1 ECE 162C Gain ECE 162C NPNPOegZ gZNLde 1 de p dz 12ng dt vg RZIZer21f1 R12er11 f2 R21R12Rrf2f1 Fermi 39 SGoldenRule 1 R R Np 21 12 2 Rr 77Z M21 2pr E21 127139 m7 M21 prE21f2 f1 8 Lineshape function To get total gain integrate over all possible states that contribute to the gain Within a lineshape function gmw Ignme E21dE21 1 112quot LUM E21 2 m 2 7r flTm hw E21 This Lorentzian lineshape is common but not the most accurate ECE 162C Lasing threshold Lasing occurs when the round trip gain equals the loss in the cavity Not all of the mode sees gain but only the fraction that overlaps with the gain region Hence the modal gain is related to the material gain by an effective con nement factor F ECE 162C Lasing threshold r1 is the amplitude re ection coef cient R r 2 R1 is the power re ection coefficient 1 1 R is the avera e 7 ower re ection coefficient R R1 R2 r1 r2 Pg ai2L Rlee h 1 1 1 thh 2 at 1n 2L RIR2 thh az am OCT m 2 ill 1 i1ni 2L R1132 L R R 032 typical for InGaAsP or GaAs Lasers ECE 162C 06 Round trip phase Round trip phase must be a multiple of pi The round triA cavit39 len39th must be a multiA le of the wavelength m z m The spacing between modes is 12 A anL ECE 162C ECE 162A Mat 162A Lecture 9 3D Solutions Expectation values Read Chapter 38 of FrenchTaylor John Bowers Bowerseceucsbedu Engineering I Room 4163 ECEMat 162A Midterm Material Lecture EisbergResnick Chapters 16 FrenchTaylor 1589 except 55 Background of quantum theory Waveparticle basis of light Planck s postulate Planck s constant Waveparticle basis of matter Photoelectric effect Compton effect pair production Waveparticle duality Uncertaint rinci le Atom models Thompson model Rutherford model Bohr model ECEMat 162A Midterm Material Time de endent Schroedinger e uation Time independent Schroedinger equation Interpretation ofLP xt robabilit Requirements on LPxt Understand how to solve SE apply boundary conditions initial conditions How to find stationary solutions Specific cases free particle barrier infinite square well finite square well harmonic oscillator ECEMat 162A Midterm Next Thursday 1 hour 50 minutes One 85x11 crib sheet allowed one side What do you need to know Given a stationary potential find a solution to SE 40 Use separation of variables find solutions apply boundary conditions find eigenvalues apply initial conditions find time dependent solution Given a stationar otential sketch solutions to SE 30 Waveparticle duality 15 Early quantum theory 15 Square Well 2 2 2 2 h k E h K V0 E 2m 2m V0 For x lt 612 0 W00 A 111006 B 003kx a2 32 F or x lt a 2 Ax C eXpIoc D eXp Ioc Boundary condition D 0 F or x gt a 2 Ax F eXpIoc G eXp Ioc Boundary condition F 0 Solution in Appendix H 4 Equations w and dex at two interfaces 4 Unknowns ABDG Solution for stems V2 82 where 2h282 R2 mVOa2 E ma2 2122 3 Dimensional Time Independent Schroedinger Equation h2 82 82 82 x Z Vx Z X Z 2 fax2 ayz 8Z2l y y w y E WW 2 hz V2lVl 2 El 2m ECEMat 162A Free Particle in a 3D Box m wxyzVxyztxyz E 1x y Z Z hZ V21V1 2 EV 2m BoundaryCondz tions 10yz 1x0z 1xy0 0 C y 1a y Z 1x b 2 1x y c 0 ECEMat 162A Separation of Variables Voltage is separable Boundary conditions are separable So try w x y Z X JOY yZ Z ECEMat 162A 3D Particle in a Box h2YZ dZX hZXZ dZY hZXY dZZ EXYZ 2m dxz 2m dyz 2m dzz Divide by XYZ hz d2X hz d2Y hz dZZE 2mX dxz 2mY dyz 2mZ dzz F unctionan F unclianon F unclianon Cansl Ex Ey E2 E 2 2 ExX 2m dx X eikx ikx iky ikz ikxikyikz wze e 6 6 ECEMat162A Simplest to use Sines and Cosines to match boundary conditions Xx sin nxmc nx 123 a 1492 I Zz sin quot272 nz 123 c Yy sin ny 123 hz quotx772 M2 M2 E j b C 2 2 2 h nx nynz 8m a2 b2 62 ECEMat 162A Time Dependence of Solution Solve for stationary solutions wnx with energies En Px 2 Z Anwnxequot39 quot mm 2 En h Find the values for An that satisfy the intial conditions ECEMat 162A Motion of a Particle in a Box W A sum A2 M277 En 2 E0112 Special case11 A2 10thers O we 272x LP xt in 6 in 64m L L ECEMat 162A Wave functic evolut ECEMat 162 Ian solidline at Cl Proba biliry densityunc lion 139 v DfarAu wnindn t Hm e Plat Harry xprmding m b and chur the later time thZE a E C 1 14on ing E 0 at the button of the well quot2quot2 Equot EMU Therefore Time Evolution W Sim sing fl 23 LIJxt eiwt 64101 mm 2 ff f 2 f1 f2 cosE2 E1th ECEMat 162A VZ9L WWI303 xpltxgtmltxgtAltxgtm j ltA mmmgm xxw j d xpx lzxx l j ltZx xpx 1xx l j xszxM llx j ltx gt gt gt gt senleA uoneloedxa Expectation Values ltxgt Ixwx 2dx Iwxxwxdx ltle l wxx2wxdx Q l wx ihwxdx l W l wltxgtVltxgtwltxdx Calculate these for the lowest order solution for an infinite square well ECEMat 162A ECEMat 162A Expectation Values 2 x I xllJx dx 1D Infinite Square Well ENE Wquot L L ltxgt j xltx2dx f xsin2dx ltxgt fx1 cos2nL xdx 1 x2 x2 2mm xL 2mm ltxgtz7 7005 L H s L 1L2 L 7 3 L 0 Hyrdrogen like solutions 3 Dimensional Time Independent Schroedinger Equation h2 62 62 62 x Z Vx Z X Z 2max2 ayz azzw y y w y E 06 y 2 hZ V2wVw El 2m If V06 yax V0 Switch coordinate systems ECEMat 162A Spherical Coordinates rzwx2 yZZ2 x rsichosg y rsianing zrcos 1 62 1 6 1 6 V2 2 r2 6r r r2 sin2 6 62 2 sinei 6r r sm666 66 ECEMat 162A Transform 121x2 322 22 xrsin6005 yrsin63in zr0056 g g i i 6x axar 6x 66 6x 63 18 26 1 a r750 Eyrrzsin g 1 62 v2 r2 sin2 6 62 a smB a6 ECEMat 162A ECE 162A Mat 162A Lecture 11 Hydrogen like Solutions and Angular momentum ER Chapter 7 John Bowers Bowerseceucsbedu ECEMat 162A 2mR r2 Solution to SE in Spherical Coo rd mates hZ V2w Vw E w 2m 1 Ze2 4780 r Then try separation of variables WU 6 Rr 9 Substitute and divide by R ltIgt 21 isin6d r s1n67d6 If Vr6 Vr hZ 1d 2a R EU d ECEMat 162A 1 d2lt1gt d6 1er sin2 6 61 Vr E Separate I dependence Rearrange 1 6121 2mr2 sin2 6 sin2 6 d dR sin6 d 2 2ltE Vltrgtgt lt ltr2 lts1 CD d h R dr dr 9 d6 LHS is a function of only ECEMat 162A d n d6 ECEMat 162A Solution of CD 1 aich 2 2 m c1 d2 1 CD Aemw Single valued means DU5 1345 27F Which means m is an integer Separation of r and 9 2 21m2 sin26 sin26 d 2 dR sine d d m E Vr r sm6 hz D R dr dr 9 d6 016 Rearrange 2mr2 m2 1 d 1 d dR d E Vr r2 sm6 EZZ1 if U Rdr dr sin26 sin6d6 d6 LHS is a function of r only and RHS is a function of 6 only ECEMat 162A Solution of G mf 1 d sin26 sin6 d6 d sm 9 1 1 The solution is in Appendix N Use a ower series ex ansion in cos 6 The series terminates for lm ml1 9 sin39 6 Flml cos 6 00 ECEMat 162A Solution of R 2mr2 hz The solution is in Appendix N Use a power series expansion in r The series terminates for 21 Gx is a polynomial in x ECEMat 162A E VrR r2 j f 1 1R where mZZeA39 0 472398022h2 nlll2 136eV Rm r EYanZr mo a0 472290712 0 2 me 25251 Quantum numbers Nlml are called quantum numbers The energy eigenvalue depends only on n N is called the principle quantum number The angular momentum depends on I so I is called the azimuthal quantum number The energy in a magnetic field depends on m so ml is called the magnetic quantum number ECEMat 162A 72 The first convincing verification of Schrodinger s theory was this calculations of eigenvalues in agreement with experiment just as Bohr s model n E eV E w 0 n 4 085 n 2 3 Y i 151 2 339 I l l 1 Lyman Balmer Paschen 13396 l l l l l l l l l o 1000 1300 2000 3000 5000 10000 20000 x A l l l l l l 3000 2400 1700 1000 500 200 y 1012 Hz Energy Levels of Hydrogen Fine structure splitting When the spectral lines of the hydrogen spectrum are examined at very high resolution they are found to be closely HOW to explain With Bor spaced doublets This splitting is called fine theory structure and was one of the first Sommer feld s model experimental evidences for electron spin Attempt to explain using elliptical orbits Treat How to explain with Schrodinger s theory relativistically Soon n 4 However dashed lines n3ng I 1 n3o2don t appear I i i ll quot 339 0 1expermentally Why W I 39 kn21 n 2 no 1 Selection rules l v v n 1 no 1 Figure 419 The finestructure splitting of some energy levels of the hydrogen atom The splitting is greatly exaggerated Transitions which produce observed lines of the hydrogen spectrum are indicated by solid arrows Comparison of Solutions 00 E ane Simple harmonic gto m square well oscillator Coulomb Figure 74 A comparison between the allowed energies of several binding potentials The three dimensional Coulomb potential is shown in a cross sectional View along a diameter the other potentials are one dimensional Enn12hv E n 2 gt 8 EC EMat 1 62A Examination of the solution The solution of the spherical potential has solutions for particular quantum numbers mllnE where m1 012 2 ml ml 1 9 nlLl2w ECEMat 162A Examination of the solution The solution of the spherical potential has solutions for particular quantum numbers mlnE Where lm 012 1 m m 1 n l1l2 This is equivalent to n 123 I 012n 1 ECEMat162A ml Z 10Z 11 Degeneracy of the solution n 123 2 012n 1 m1 1 110z 1z For each value of n There are n possible values ofl For each value of l There are 21 values of m For each value of n There are n2 degenerate eigenfunctions ECEMat162A Actual hydrogen atom 6 spatial coordinates XeYeze Xp yp zp What to do ECEMat 162A Actual hydrogen atom 6spatial coordinates Xe yetze Xp yp zp Switch to center of mass m 39 coordinates The electron moves about a stationary infinite mass nucleus The problem reduces to 3 spatial coordinates Xre yre zre With reduced mass u LL 2 m M m ECEMat 162A 3 spatial variables 3 quantum ECEMat 162A numbers m 911031kg M 2167210 31kg y 9051031kg A small but measurable correction Lowest energy solution n1 l0 ml0 E136 eV There is only one solution no degeneracy w 5 100 39 10 The solution is spherically symmetric 32 ZWa 60 ECEMat 162A Lowest energy solution n1 l0 ml0 E136 eV There is only one solution no degeneracy 1 Z 32 Zra0 W100 910 e The solution is spherically symmetric ECEMat 162A What is the probability of finding the electron at a distance r Second lowest energy solutions n2 E1364ol UV There are four degenerate solutions One solution is spherically symmetric einZao 1 Z 32 Zr 2 2 W200 4 2 a0 a0 One solution is cylindrically symmetric 1 Z 32 ZI Zr2a e 0 cos6 W210 45a0 a0 Two solutions are degenerate 1 Z 32 27 Zr2a i l e Ost 211 8 7r a0 a0 ECEMat 162A Radial Dependence Energy Figure 76 The qualitative behavior of the kinetic e energy E of a hydrogen atom as functions of the size l more rapidly than V decreases because K 1 1F1 becomes negligible compared to V As a result E h indicated by the mark on the Ft axis and at this si l l l l i l i l l l 0 5 10K 15 20 25 r W Figure 75 The radial probability density tor the electron in a one electron atom for n 2 i 2 3 and the values of 1 shown The triangle on each abscissa indicates the value of rquot as given by 7 29 For n 2 the plots are redrawn with abscissa and ordinate scales expanded by a factor of 10 to show the behavior of Pr near the origin Note that in the three cases for which i lam n 1 the maximum of Pquotr occurs at rBohr nzaOIZ which is indicated by the location of the dashed line Table 72 Some Quantum Numbers n 1 N m Radial P 0395 7 39 for the OneElectron Atom Eigenl unctions 312 l v 7 100 J an 1 232 Zr V 79 2494121 200 4 2n do do 1 2 3 Zr erz 11300 i gt 27 18 i 2 2 0quot 0 Sly3T5 a0 a0 0 W110 2L2 lt gtB lt6 ae zra m cos 6 8I7r do 0 0 32 1 E 675 317W sin 0 4 1quotquot 81 o 10 0 1 2 32 zzr2 11 320 m z AWN cos2 0 71 81x6TE 10 32 w 2 E a e ZYZ n COS 6 210 4XZ do do 1 Z 3221 739 I W21i1j gt i39e Artzausinoaiw 8V7 do an 1 3111 o 1 z W 2212 s i 11321 fltigt 78 Z393quotquotsirt 0605 0 81quot 8km do 5 1 z 3 22er l132r2 2 E 3 sm2 0511quot l z 70 an L 110Z Figure 75 The radial probability density for the electron in a oneelectron atom for n 11 2 3 and the values of 1 shown The triangle on each abscissa indicates the value of m as given by 7291 For n 2 the plots are redrawn with abscissa and ordinate scales expanded by a factor of 10 to show the behavior of Pultr near the origin Note that in the three cases for which I II n 1 the maximum oi Pr occurs at rmH nlauZ which is indicated by the location of the dashed line 2 Polar Dependence z z E lmi3 1 Kim i2 13l7l10 dodo 7 7 2 O ml0 I1mI 1 l Z m 2 z 1 23 m3 I4mt4 Figure 79 Polar diagrams of the EOEMat162A densities for I 01234m directional dependence of the one electron probability fl N 39l N2 N3 n2lml0 n3l1ml0 L2 n3l2ml0 L98 SEIJJSNEIG Ail1IEVEOHd 1 39398 Classical Angular Momentum ECEMat 162A gt gt gt Lzrxp Lx ypz Zpy Ly 2 2px po LZ 2 2px po Angular momentum Cartesian coordinates Classical Quantum Mechanical L rgtlt p L N p szypZ Zp 1 2 ihy2zg y x 82 8y Ly pr xpz a a Ly ihz x L2 3 ypx ax az L ihx3 y3 8y 8x ECEMat 162A Angular Momentum in Spherical ECEMat 162A Coordinates gt 2 rxp 2 irx LAX ihsin6 cot cos i 66 M LA ih cos icot65in 3 y 86 5 L 472 What is the 2 component of angular momentum 7239 Calculate the I oo 2d dgmd j expectation value Fir r i W 21 j Rnlr lmleimz Zmgz 69 zz1 ihieim hmleimlquot 69 h oo 7 27 Z2 j Rn rRnrr2drI ml mld6 j d hm 0 0 0 12 hm So the 2 component of angular momentum has the average value given above ECEMat 162A What is the total squared angular momentum Calculate the expectation value 00 7t 27 P Irzdr Ida Id gift 0 0 r lmleiml 2 3sinl9iaT sm6 86 86 sm 6 6 By 11 1hzz P 111h2 2 h2 ECEMat 162A Vector picture of angular momentum The arrow has length 22 l While the vertical component has length 21012 The average value of LxLy is zero The energy ofthe atom does not depend on m ie orientation EOEMato 2ag Momentum Quantization We showed that the average value of LZ is mh That doesn t mean that LZ is quantized However since a zz1 ih eim hmle 69 iml 12 hm A 52 Lil1 h2 ew hzmlze W P 2 Wm2 The average ofza set can only equal the average of the square of the set if all values are equal Hence LZ IS quantized ECEMat 162A In general if the quantity f has the value F in the quantum state described by w then fwFw Where is the operator corresponding to f ECEMat 162A Note LAxw 7 ngy LAygy 7i lyw So LX and Ly are not quantized ECEMat 162A L L z hL L L ith y Z L L z hL Z x y Under what conditions ca two V Ve observable properties of a quantum systs an amuse s5swws fv a given quantum state ECEMat 162A If two operators commute then the eigenvalues associated with those operators are simultaneous eigenvalues If two operators do not commute then the eigenvalues associated with those two operators typically exhibit an uncertainty relation ECEMat 162A ECE 162A Mat 162A Lecture 14Identical particles multielectron atoms ER Chapter 9 John Bowers Bowerseceucsbedu ECEMat 162A Magnetic moments CLASSICALLY an electron moving in a loop produces a current 0 e l period dis tan ce 27zr perlod veloczly v 8V 1 27zr A current in a loop produces a magnetic dipole moment ev 2 evr u 2 current gtlt area 21A 2 7zr ECEMat162A 7Z7 2 Bohr Magneton Classically L mrv evr eL 7 E If eh 23 2 Bohr Magneton U 2m 927gtlt10 Am L Ll i The correct quantum mechanical result is g is the orbital 7 factor and lb b 1 ggzz ECEMat 162A Dipole in a Magnetic N The effect of a magnetic field on a magnetic dipole is to exert a torque fz x The potential energy is lowest when the dipole is aligned with the magnetic moment AE2 f3 ECEMat 162A

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