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Date Created: 09/19/15
Neurons The Molecular Basis of their Electrical Excitability Viva La Complexity Consider The human brain contains gt1011 neurons 39 Each neuron makes 103 average synaptic contacts on up to 103 other neurons and receives up to 105 typically 104 synaptic inputs Conclusion the computational power of the nervous system is based on sheer numbers of and connections between individual elements as well as in the computational powers of the elements Adding modulation and plasticity to each of these connections makes it even more impressive Support Staff CNS brain and spinal cord PNS nerves what are they neuronszglia 19 volume 11 In central nervous system CNS oligodendrocytes myelinate up to 40 different axons astroglia regulates extracellular uid Kt neurotransmitters glucose NH3 guidance microglia immunelike macrophagish cells In peripheral nervous system PNS Schwann cells myelinate single axons nodes of Ranvier bare axons between myelination Axons may be 1 m long nutrients axonal transport via microtubular railways consummg ATP O O O O Nemmns Are Cells mm WV 1gam g nmlmg mwlwm mmiiimhmmlu wa a iqg m nu m mi afiw mm gtolgi almmm m 1111 pmmwag lt 24mm lt dan bjiiw 1U1 mambmm Dimigphd p d Mays3E Neuron Parts by Function Input chemoelectrical reception in dendrites soma axon Metabolic Center soma axons dendrites nucleus protein synthesis ATP production Trigger axon hillock digital output of spatiotemporal integration Transmission axon terminals transmit information to other cells Via action potentials Distinctive Feature In addition to all the genetic and biochemical complexities that all cells possess neurons have the added feature of having electrically excitable membranes a trait they share with certain other cell types eg myocytes Electrical excitability results from maintenance of membrane potential and differential distribution of many voltage and ligandsensitive ion channels in the membrane Communication between cells requires electrical responsivity Membrane Function physical barrier holds in contents chemical barrier allows for selective passage of compounds ions water amphipathic polar amp nonpolar regions composition phospholipids cholesterol integral membrane proteins capacitor 1 uFsz membrane potential typically 65 mV 7nm Integral membrane proteins receptors ion channels Ion Channels Na K Cl39 Ca2 diffuse across membranes at high rates despite their ionic nature How Integral membrane proteins form channels through which ions may pass through membranes May be single proteins or protein complexes Different channels are selective for different 10ns Pore is v small ions in single le H Ion Channels ggfce L0 rr tf ms I Q EFCEDA39fgfLm 1m eaxcgg ED E 03 H via mf nm ma Channel p39rot ein39s Intracellular fl Uid hamgg Open ion channel Closed ion chan nel fluid Ion Flux Electrical Forces electrical potential exists across cell membrane inside negative relative to outside ions are charged and thus are in uenced by electric eld consider if no A in vs out still have electrical driving force cations into the cell anions out of the cell electrochemical gradient takes into account concentration gradient and electrical potential forces may act in concert or oppose each other A little more complex Ion channels may be open or closed based on conformational change of the protein Process may be very rapid 1 lt 05 ms At a given EC gradient and time period total ion ow depends on the frequency and duration of it being open ie it s probability of being open x driving force Probability depends on allostericcovalent forces for ligandgated channels membrane potential in voltagegated channels physical deformation in mechanicallygated channels a combination of the above Blophysms Resting Em cells maintain negative intracellular potential via ion pumping membrane has conductive ion channels membrane also acts as capacitor ion separation electrochemical gradient E RT25Flnoutin Nemst 1888 R gas constant F faraday T temp zs valence E 6154logoutin body temp 1 atm Ion in mM outmM E Mm mV EG 60mV CG Net Na 15 150 62 122 in in in K 150 55 89 29 in out out Cl 9 125 71 11 out in 0 Ca2 0001 15 129 189 in in in must then multiply Net by conductance to get ow Goldman Equation 111 PKK0 PNaNao PC1C1i PKKi PNaNai PCIClo Vm F Based on electrochemical gradient generalizes to Nemst Eq for single species small effect of pump on Vm except to establishmaintain NaKATPase pump is electrogem c since it moves net charge across the membrane 3Na2K Ion exchangers N aCa2 use potential energy of one species to move the other species uphill tam Changes in Vm depolarizing VIn from rest towards 0 hwerpolarizing Vn more negative than rest repolarizing return to from Vm depolarization overshoot Vm positive graded potential changes because charge moves eg ions ow across the membrane due to changes in g Decay is according to length constant synaptic potentials generator potentials sensory neurons action potentials Spatial Integration of synaptic activity from different sites of the postsynaptic neurons long vs short space constants Temporal Integration of same at different times long vs short time constants Length amp Space Cms ams A Tern pora su mmanon Hecurdlhg df r Svnapzm mrem Data Lam ww constam HUD ms Show me 0 39 m 20 ms W B Spalzm suvnvnalwon iam A Qi AVA an I2 y M A E Lung engm mnmm H mm t39 J canstem I v1 1A n 0 33 39m 39T 25 m gawk g smea f L 1 now i Acner a emm TWESHGM Peak swam pctmna R5Hng F s mm pow mm Svr ww Dmrmc m swape smug Trgmsdwim i u i wi ui an i i a Seaway Emeedimg ilt i gt iQUTjTJQF 2 i 7 x i i X Peripheral branch Cell body Central branch primary afferent fiber W Central Terminal Periphery gt nervous system Wimp mm2 O quot 4 59quot 5 a H 4 A cmiiiyiem irm mm mm SHHmlH varln L analog Fl rm me Amphlnrlt l lllf ll l Jl llnwm normal quotrm Tsansausmn Salkegenmalmg Wulumlml m mm mm 39 Svnanlm annual C Achv pulenllal D O xlpnl SW A Hencplorlm s maplml B lnlagmlwe mlmn lransmlllev Ic leasel pom M 3 L n analog digital analog 2n l39 l l l mom 9 mm m F 1 4 QC 1 l ac l Jar 1 ll IA 3 n H lt 5 7n n 5 1n Twm wcu l39lsl Action Potential Na MINE12 BNa 67 mV opens fast positive feedback mechanism closes fast negative feedback mechanism inactivates spontaneously 0 K gK IKEEK EK 90 mV opens slowly closes v slowly inactivates from polarization only Am m P t m g l C mmm ts HGW as Y Vm I Imam l l 39 f l I l quot l I I I I 0 1 2 3 4 Twme1msec fan memll Iquot B mm Tum V ltage Clamp Expgrimgms Conductance mScmz 13 My i nk14 M 31 3 LJI A Na CONDUCTANCE B K CONDUCTANCE 20 44 UN 44 mV I1 0 O 23 mV 23 mV 2 mv 2 mV lL 1 0 2 4 0 2 4 6 8 Time after start of test pulse ms h 34 0 VOLTAGE m MV Some electrical EVENT 1ch LIFE De ayed rec er39s are opening and Nu Vgmfad TL channels are Delayed inuc vuh39 ng Na rec l39if iers NEH U gqfeii 7 dose Vgmed c unne 3 are deinuc va ng gt TIME IN MSEC channels open Depolarization EM Repolarization 1 g 0 E 1 m It m3 E h h M m3h ng l IJ I Il j 0 1 2 3 4 1 2 3 4 T ime units 2f 77 F J vi 1 E Squid axorl 63 C iNa m Ebark m JNaJ m4 M I ha Th HHS h ac Tn yms ngo 1 10 10 5 05 05 U 0 U 50 0 50 Membrane potential mV 50 0 50 N 3 Channel MMEHS Selectivity filter Inactivati0m Lining of pore A TWOPULSE EXPERIMENT Pulse 1 Pulse 2 Membrane B RECOVERY CURVE 1 0 INx pulse 2 Na pulse 1 0 U1 gt Recovery interval ms a Excisedpalch configurations Cell b Whole attached cell Membrane rupture l Pull 1 Pull d quot lpamh patch Q Insideout Q Outside out patch patch Test Test solution solution m D a Lilllll l UMLS ralolxgl H m MW WM I MWQEU g Open Closed Open Closed 1pA 50 ms ymaptic Tramsmissim A svrezm rP ex cvrcLM my knee erk WSW n mn Mwscm DHrmmeps smnme lauensnn Hmwmo Wexor wmmmw Fxxenw Hmov WWW mnmr mutov nwmn neum B Expe mema Setup fur recording from Ce s m the cwmm 5mm passmg Sensmv39wurun 050mm EX EMMA 3mmquot Mm W7 we 9 WDIO HEIHDH m 6 RUNFN hers MM J nzurrm mm mJSF E 5pm 9 I S n n y Heumn QUMJ39YCEHS C 15 7 5 Recarmng r P U quotQ F NI VID39V quotVUVquotEHVNY ROCOWHHQ Par 3 a arem hbers Mom mm from v e Smnd esa39 r rem mun quadv rep AP Propagation AP depolarization acts to open neighboring Na r channels but not ones that are inactivated Thus directional propagation of AP occurs in axon APs are regenerative not degenerative like graded pots In myelinated axons APs leap 39om one NoR to the next Na chs only at NoR myelin prevents leak through memb Tmylen Tconduction velocity Tdiam gt Tconduction velocity Passivg Spread Site 0f initiai depolarization Charge Extracellular M I Direction of current U Active AP Pmpagatihm Stimulation Na39 channel K39 channel 1 g a Point A sting potential J A 39 Point B Point 3 Saltatory Camdmctimm Direction of action potential propagation Na Channel Active node Node to which Inactive node of Ranvier action potentiai at resting site of action is spreading membrane potential dashed lines potential Synaptic Transmission AP invades miss snntn unslls rslmses t L NT binds its pssusymsp us gtp niumg Linens Cai sns s i lsw tun Thugs elecmhamisal samnicatim An action potential invades the presynaptic terminal Tra synthesized and Depolarization of presynaptic then stored in ermlnal causes 0 enan of voltagegated Ca channels In ux of Ca through channels a causes vescles to use with presynaptic embrane Retrieval ofveslcu membrane from plasma membrane Postsynaptlc current causes excitato or in ibito postsynatic potential that hanges the excitability of the postsynaptic c ell V39 V 39 39 Transmitter blnds to 39 Opening or closing of receptor molecules in postsynaptic channels postsynaptic membrane What Do I Do record the electrical activity of individual neurons in context while system is anctioning perturb systemic variables analyze neuralnetwork computations quantify information in spike trains using information theory record networksystemic activity using optical imaging techniques What Can YOU Do designmanufacture microrecording electrodes arrays signal processing of optical imaging data signal processing of spike train data designmanufacture microchannel devices for chronic implantable devices for integration with the nervous system Chapter 7 University of Delaware Electrical and Computer Engineering Department ELEGPHYS667 Magnetism amp Spintronics Prof lan Appelbaum 1 Spin lifetime in metals Letls revisit electron spin in a magnetic eld The equation of motion is d ig a EihlLXB 1 When B is along 2 this equation has solutions describing Larmor precession around the magnetic eld Mac MsinOcosth 2 My MsinOsinth 3 M 0086 4 However this is a perfectcase scenario In reality there are damping effects which relax the precessing state to a static orientation along the eld direction If there were no such relaxation effects then the two spin species in a paramagnetic metal would have different Fermi Energies in a magnetic eld We know this situation does not occur We can include relaxation effects phenomenologically and separate out the z y 2 components dm 7g a a Moim d e hltMgtltBz7T1 5 dw era a a a d 74thng lt6 dMy 79 a My W WXBM E 7 These are called the Bloch equations after Felix Bloch who won the Nobel Prize in 1952 for his work on magnetic resonance T1 is called the longitudinal or spinlattice relaxation time and T2 is called the transverse or spinspin relaxation time T1 and T2 are in general not equal although T1 sets an upper bound to Tgi This is because precessional dephasing can have the same effect to measured transverse component may as coherent decay when measuring an ensemble of moments This dephasing can be the result of eg inhomogeneous magnetic elds resulting in slightly different Larmor precession frequencies across the sample Because the relaxation times determine state lifetimes they also determine the linewidths of excitation and absorption spectra of oscillating elds through the relation AE6t h s A rigorous analysis yields the following power absorption spectrum of AC elds as a function of DC eld Electron Spin Resonance ESR or Electron Paramagnetic Resonance EPR 2 7 BM 2 P B 17mwg where 7 gh ET is the eld at which the Zeeman splitting is resonant with the driving oscillating eld and B75 is the magnitude of the driving eld This is a Lorentzian function centered at the resonant eld The HWHM ABlg of this Lorentzian is ln cubic solids T1 T2 so B12 9 1 T 7 10 1 VABi2 Since E hwo 9MB 11 7 7 9h 7 7 7 B 12 11 Measuring T1 and T2 Experimental measurement of T1 can be accomplished with the inversionrecovery technique Moments aligned with a static magnetic eld 1 0 are rotated by 7r with an RF pulse of proper magnitude and duration to M 7M0 The longitudinal component of magnetic moment then recovers to its initial value over the unknown timescale T1 During this recovery after a waiting period 739 a 7r2 pulse rotates this component into the transverse plane where it precesses The amplitude of the resulting radiated eld indicates the longitudinal component at time 739 The measurement is repeated for different 739 and the exponential timescale is determined T2 can be measured by the spinecho technique lnitially aligned states are rotated into the transverse plane and allowed to precess Dephasing will occur where fast precessing states will have a phase lead over slower states After a time 739 a 7r pulse is applied ipping the order of fast and slow components of the precessing ensemble now the fast precessing moments have a phase lag Over time 739 this dispersion is compensated and all the moments line up again for a short interval The precession of this partially coherent ensemble radiates a RF signal which can be used to measure its magnitude Repeating the spin echo sequence with different 739 can be used therefore to measure T2 12 Spin diffusion For Cu T1 m 4 X 10 103 60K This is a measure of the conduction electron spin relaxation time so multiplying times its average speed the Fermi velocity should give the spin ip mean free path Asf vFT1 15 X 105ms 4 X 10103 630mm 13 However the electron does not travel in a straight line it diffuses The relevant length scale is not the mean free path but the spin ip diffusion length determined by the modi ed diffusion equation 8P 7 6213 P at 7 612 T1 va 3 14 where the classical diffusion coef cient D The steadystate solution to this equation is P ev r 15 The characteristic length scale of this solution is the soughtafter spin diffusion length Ms A iT i 16 But what is A We can recover it from the rst successful theory of metals the Drude Model 13 Drude Model The acceleration of a charge 6 in electric eld E is F GE a g g 17 If 739 is the scattering time between collisions the velocity before collision is 6E 7 18 vd a7 m 739 However the current density is given by l 1 WI l l l l I I I a Cu l 0056le 1010 0 Cu 027cm v Cu l 044lcm a A glOO 3 g H g H I E E 10399 UJ Z Fl m 39 I0 39 tOO TEMPERATURE PK quot 5 LIOILH V 1 MawL A34 9 w T T Jl L q 1183 Na 16107quot 7 20M 16139 91310 4 L m w 2746 Ho 0 Mquot U 31 10 quq Hr 161 Cr 1 ID 199m 091 I 0 3 Cw 63140quot 1033 Mr Lmoquot 4 38104 P3 Mr uhoquot pm L o 10 Ll0t003 v70 an 39 Re 69 1039 100m a Ni 23 10 6 1a 14 1910 I41 601 0534 W3 j nevd 19 where n is the density of electrons Therefore j 6E i 7 20 me m T If we use Ohmls law E jp j ej i 7 21 me m T 1 ep i i 22 me m7 m 23 762 Since A UFT vpm A 24 p627 For an orderofmagnitude approximation we can substitute some values using e2 m 10 7eV cm 1030ms X 5 X105eV02 718 1 A N 10 59 cm10 7eV cm10230m 3 N10 9 398 25 But we expect this meanfree path to have units of cm To reconcile note that in units of eV cm and s resistance has units of 97 Volt 7 Volt 7 eVolts 7 eVolts 26 7 Ampere 7 Coulombsecond 7 e Coulomb 7 6216 X10 19 Since 62 m 10 7eV cm in m 1012cms1 27 Therefore m 1055m 28 This guess lOnm is a bit low since for Cu A 6607mm But it gives us a reasonable order of magnitude for the spin diffusion length in metals AAsd 3 A 1m 29gt 2 JohnsonSilsbee experiment In 1985 Silsbee and his graduate student Mark Johnson performed what could be called the rst modern spintronics experiment The motivation of this work was to inject spin polarized electrons from a ferromagnet into a metal Aluminum and detect their presence lnjecting the spins is easy drive an electron current from one ferromagnet into the Aluminumi We want to detect the presence with another ferromagneti In our explanation we approximate both ferromagnets as halfmetals conduction electrons at the Fermi Energy are of only one spin speciesi Then the injected polarization is initially 100 at the injection pointi These electrons can only ll states in the appropriate spin band so the density of electrons in that spin band risesi Consequently the Fermi level of that band also risesi Since total charge is conserved the density and hence Fermi energy of the opposite spin band decreases This is because spins of both types are being carried away by the other contact to the Aluminum If no current ows through the interface between the Aluminum and the halfmetallic ferromagnet used as spin detector equilibrium demands that its Fermi energy equal the Fermi energy of the spin band with compatible spin in the Aluminumi Measuring potential voltage between the ferromagnet and a point far away from the injection where the Fermi levels have equilibrated then Injector detecwr 7 359f 5L 50 Ioo 50 o 50 ENE 50 Field GI tells us that spin is piesent This is called a norrlocal measuiement and it is done to avoid measunrg voltages due m n legions wheze chaige cunents Rutheimoie by sweepirg an exteinal magnetic charged and the measuied voltage will charge sign By usirg a magnetic eld pexpendlcul al to the magnetization and hence spin aacis piecession Will catse the signal to decay This is called the Hanle effect ow eld the ielative magnetizations of iniectoi and detectoi can be Va Plcuvallsl 2V1 Calculation of signal magnitude It is cleai that the voltage measuied will be piopoitional to the spin piesent but how big is that voltage7 signal magnitude can be detemiined by calculating magnetization cunent I La 245 30 The rate of magnetization increase in the Aluminum is M inning dt Injection A39d eA39d At steady state this must match the spin decay rate dM M W lrelamation i yeilding an equilibrium magnetization MBITi Mequilibrium EA I d This is the consequence of lling states above the Fermi Energy Mequilibrium 7 1T1 7 B 7 ieAld 7 DEFeV So we can de ne a resistance which tells us the ratio of the measured voltage to the injection current R 7 K 7 T1 7 I 7 e2AdDEF For the free electron gas we know 3n D E 7 F 2EF Therefore 7 2T1EF 7 3n62A d What is the magnitude of this expression For a device 100mm on a side we have 7 10 93 16V 7 1023572171 10 7eV cm 107657213 Again this value is a resistance but is given in units of cms 1 not 9 Since we have derived before R 10 19cms 1 19 10 12cms 1 Our expected signal will be 10 19cms 1 10 12Qcms 1 This is consistent with the signal of approximately 60pV at 22mAi 1007M 31 32 33 34 35 36 37 38 39 40 66 13Butadiene Cisuid Symmetry 2 H39 u IMSH r I IIIJ N 1 5A1 EO707 O u wxu r q u HUhm 7 14 W1 ur Mums i 1 39 qJI Figure la RUHJH vi39lhc llgt mlcu anuu Mn x H Al u u 39 39 lu clxzuul kmrh nl lu wtmns m Ihu mm In m39cxmuuun u 1 mm m mum lur Ihu grown um x 0289 n 4 2A2 E 0526 0 4 0120 n 3 E 231 o354 n2 1A2 E 70602 ngH 532 E 0477 n E 13 EO629 0 6A HIqu I7 rIIIIpy lncl uIImmn IIIIIIIIIIIIIIII Im LOIHHQIHL U Ivydw I ImII IIIIIII IIIIIIIIIIIcIIcII CIVJHH muc 39 IIIIm III IIIIIIIII4II hy IUL39 HMU mm TIII wum w L AIIIIII IIIIMIIIII II NVquotrllyhrldlltd LII mde w q zu 7 7 LUMO u I 1 1 4 1 1 F O O 4 I H M IL 4quot 1quot 4L 1 4 4 u 4L 1gt 4 1I 1I quotV 4 1I 4 4 1l 4 1L quot1quot quot1quot 1t 1 4 4b IL 4 A 1L 4 L 1 1I w II 4 4V 1L 1 4 W2 1quot 391 4 4L II II II 4 it I I I I I I I I I I I I I I I I I 1 2 a 4 5 a 7 a 9 m H 12 13 14 5 l6 number of cumugz ed c amms 37 Cyclopropenium Cation Symnmry D3I Chapter 9 University of Delaware Electrical and Computer Engineering Department ELEGPHYS667 Magnetism amp Spintronics Prof lan Appelbaum 1 Boltzmann Transport Equation To get a rmer grasp on GMR we need a more powerful way to model resistance in metals Therefore we need to be able to calculate the current densityj that results from an electric eld 5 j 05 Ohm s law In the freeelectron gas model of a metal we can sum the charge flux 6 vac in a direction I over all states j55Q hww k m where the 2 is for spin degeneracy and the volume integral is over all kspacei In the absence of an electric eld the distribution function is the FermiDirac 1 uhwgm mg q m r319 2m where the subscript on f0 denotes equilibriumi As usual Ek This function is symmetric under re ection in the z direction but vac k 52 clearly is not Therefore the integral of the product of these odd and even functions over the symmetric bounds is zero This is to be expected since no net current flows in the absence of an electric el i When an electric eld is imposed the distribution function must be pushed out of equilibrium resulting in a net current upon integration of the above expression We can write f0 fok 57 3 Since the rst term contributes nothing to the current our task is then to calculate 906 The Boltzmann Transport Equation BTE gives us this Our task to model GMR is to investigate the BTE for 1 3D bulk metal transport 2 Thin lm metal transport 3 Ferromagnetic metal transport Before tying everything together ferromagnetic multilayer in GMRi 11 3D bulk transport The BTE is a statement of steadystate There is a rate of excitation and a rate of relaxation In steadystate these two rates are equali d d d1 d lt4 The RHS is simple to express Using the relaxation rate approximation we introduce a phenomenological relax ation time 739 over which the electron gas regains equilibrium f f0 7 9 df E lrelamazion f 7 5 The LHS is due to an electric eld in say the z direction which provides an electrostatic force 65 on the electrons We therefore decompose the derivative n g m dt excltatwn dings dt 7 dings 7 dkgc h Now because k2 k k kg we can write k dkgc Edk giving amp dt excltatwn dk h k The BTE then says Now we can use this expression to calculate the current associated with the electric eld e hk 3 1W29kmdk 7 2 2e 8 df kldgk 39 k 6 7 8 9 10 11 t temperatures kBT ltlt EF the distribution function is unity inside the Fermi sphere and zero outside Therefore m 7606 7 hr This converts the volume integral into an integral over the Fermi surface 1 62739 k 1 Rig f Now transform to spherical coordinates kg kp cosqbsint d5 k sin 6d 9d k2 27r 7r 4 idS k dqs d9sin3t9cos2q5 1k k 0 0 3 j m 3W2 Now since the electron density is 2 3 2 ms k HMS0 M mVEIW Q we have 12 13 14 15 16 17 18 This you should recognize as Ohm s lawj 05 with a conductivity equal to precisely the expression we calculated within the classical Drude mode 2 FuchsSondheimer model Consider now a metallic thin lm with an electric eld 575 The thickness of the lm in the zdirection is t When we lose the translational invariance in bulk conductors by using thin lms we have to take this into account by using the full time derivative of the electron momentum distribution in the Boltzmann equation L 744 528 dt Teluxatwn d2 dt T 7 19gt g dug The rst term is called the convective derivative and can be rewritten vz Clearly this term and its 1 and y counterparts is zero in the bulk case because 9 does not depend on any spatial variable when full translational symmetry is present This equation has the solution 65739 df z if 1 F v 67 20 g m M lt gt lt gt where Fv is determined by boundary conditions Since the sign of 1 changes at either boundary at 2 0 and 2 t we must break up the solution into two pieces one for the distribution traveling up and the other traveling down 9 1 Feify 21 9 1 piggy 22 The FJr and F coef cients are determined by the boundary conditions We assume diffusive scattering meaning strong relaxation at the boundaries imposes gltz 0 7 o 23gt g 2 t 0 24 This determines the coef cients F FJr 71 Now we can determine the current owing in the lm with the usual expression j7473fkzvmkdk 25 where fv2 gvz g 1 We again use spherical coordinates and the g7 term reduces the volume integral to one over the sphere of radius kp We also remember that since we have broken up our solution to v gt 0 and v lt 0 cases and v wea g in spherical coordinates 9 is integrated only over the t9 07r2 hemisphere and g is integrated over the remainder After doing the trivial integration of 5032 we have 2 g 7r2 7 7r 7 7 e T k sin39lt17 64059gt dt sin39lt17 6 059 d9 26 4w2m 0 rQ where A M7517 is the mean free path We can combine these two integrals into one with the same bounds and simplify using t 7 2 6 2 67072 2eit2503h 2 Z 27 giving 625739 k 2 12 3 it t72z z 24w2m QSW 0 3m 9 1 7 624059608hm d9 28 Now since this is a function of 2 ie the current density is different near the boundaries we need to average over the crosssection 7t 1 3A W 3 j gO zdz 005 17 50 3m 6006lt17 4 59d6gt 29 where 00 is the Drude conductivity eigni By making the variable substitution X 15039 this integral can be For a thick lm A ltlt t and 6471 A 0 Then we can do either form of the integral analytically and rewritten as 3A J N 800 17 31 For a thin lm A gtgt t and the result is 3 t A jwfaolxln 32 These results assume perfectly diffusive scattering at the boundaries 1f the scattering is instead partially specular with fraction p then the thick lm limit is a 3A B 1 7 i 1 7 33 a0 8lt p lt gt which reduces to the bulk conductivity when the scattering is perfectly speculari This makes sense because perfectly specular re ecting boundaries are tantamount to periodic boundary conditions repeating the lm on both sides and extending the effective thickness to in nity 3 Boltzmann Equation for Ferromagnetic Metals For ferromagnets we have to solve two equations one for each spin species In each equation we must include the scattering time as we have done before but in general these will be different for each spini In addition we must also account for relaxation of spin up down into spin down up over the spin relaxation time 739 dfo 97 m i 91 idEvgch 7 T T 34 dfo 971 91 i 9T idEvgch 7 T 7 35 These are two linear equations with two unknownsi The solutions for g and 91 are readily found with some algebra 1 1 1 1 dfo 7 7 m EMU 9T 36 7t TTL and 1 1 1 1 dfo F t WW t W vgch 7 91 37 This is just like the FuchsSondheimer Boltzmann equation except now is replaced with the term in brackets Since this term does not depend on k we can skip to the last step of the determination of the conductivity 2 1 1 1 1 i ne 7 n 7 aalail71 17imj 38 TTTl 7m U 7m This is equivalent to a resistor network shown in the gure where as usual p olt 1 For small pm most of the current passes through path a corresponding to weak spin mixingi For a large pm current must pass through path 12 corresponding to strong spin mixing 4 Boltzmann Results for GMR MagnetoElectronics Electron Spin SternGerlach Experiment Magnetic materials Spindependent DOS NMFM Interface JohnsonSilsbee Expt NMFM Multilayers Giant Magnetoresistance GMR FMlnsulatorFM Magnetic Tunnel Junc ons Beam of magnets Classical Bar Magnet in a Magnetic Field W a w sunny w Ivlvv gang 0 Waging n rng muwmm V I1 Luff quot 39 qiitri i i my ll 3 B Eiield W s if it v B Field SPIN l w 3d band 2 f 4E baud 0m wrimvuip nmsrrqusrnEs l Immurm msan mum y Munmam mmm 5m amnmeth unwolnunu uwouvrvxm 6 4 I I m m u a 4 1 a 2 m x a 4 1 n z m n a o 2 a a b 23 Scheme afdensiljv qf39smtes DOS in Iranxi C firm meraIJ 11 Ni b Co and 3 Fe 1013 mamas suns snusriw w amm m snrm GuamEm n Johnson and Silsbee PRL 55 1790 1985 and PRB 37 5326 1988 II39IIECT39DI deter Tor c H UNBIASED Va 9V Johnson and Silsbee PRL 55 1790 1985 and PRB 37 5326 1988 G E 39l quotI39 m I 1 I a L L ara Q E G g F39 u 50 5 30 an quotan Field 6 Anisotropic MagnetoResistance T T T35 resislance high resxstance J I i39 p Figure 1I4 I39hc m igin ul AMR pparallel U 2 0 ppcrpcndicular 7H 45 0 5 l0 Baibich et al PRL 61 Glant MagnetoRe51stance 2472 1988 RREH l 50 Magne c fiend k6 RKKY coupling RudermanKittel KasuyaYoshida Parkin et al PRL 64 2304 1990 A pllill lll39 I r 1h clmesa IREII RKKY oscillations of AIUR in multilayers spin a spacing TwoCurrent Model pl lt pl 1 l plZ plZ plZ pTZ parallel antiparallel Magnetic Tunnel Junctions barrier barrier Current nA 35 30 25 20 15 FM1C0 Scan FM2 NiFe Direction O gt lt n I 1 I n I 100 0 100 Magnetic Field gauss Magnetoresistance ARR RAPRP ARR RP V v RPquot nu RAVW 1 1 AWR JHN 391Hu HHLNH 1 n1 nu 2 2 2 21391 211 211 11 2 112112 112112 4 Z1 12112 11 Polarization P W 2 ARR T Julliere s Formula Chapter 2 University of Delaware Electrical and Computer Engineering Department ELEGPHYS667 Magnetism amp Spintronics Prof lan Appelbaum Now that we have established the empirical fact that the electron has an intrinsic magnetic moment which is called spin let s take a look at the theory which predicts its existence a priori Since electron spin is fundamentally a consequence of relativistic quantum mechanics we need to study the relativistic version of the Schrodinger equation called the Dirac Equation 1 Schrodinger Equation 11 Classical background Without assuming too much let s rst construct the nonrelativistic Schrodinger equation This is a wave equation so let s look at how the classical wave equation can be constructed from the general form of a ldimensional plane wave um t eiltkHEgti 1 The wavenumber k 27rA is related to the angular frequency w 27w by the dispersion relation For classical waves such as electromagnetic waves we have AV c so the dispersion relation is w kc 2 and therefore zdt eik17kct 3 where c is the speed of light The wave equation is a partial differential equation in I an t so let s look at a few derivatives First with respect to 1 ii ik IJ 4 d2 11 ik2 11 5 and with respect to ii iii ich 6 2 IJ ik2c2 11i 7 Now notice that d2 1 d2 1 E f 8 This is the classical wave equation 1 2 Matter Waves For matter waves we have a different dispersion relation since we have the deBroglie hypothesis 10 F116 9 and the Einstein law E ha 10 We know from classical mechanics that 1 1 E EmUQ 102 11 since pmv1 Applying DeBroglie s and Einstein7s insight gives us 1 ho 152k 12 1 2 w hk 1 13 which is the dispersion relation for matter waves we are looking for Now this gives a plane wave 2 1m t We gt 14 which has a second derivative with respect to 1 d2 Emam 713111 15 as before However unlike the classical case in which we need two derivatives in t to recover the k2 term this time the rst derivative with respect to It gives it to us d fik2 11zt 7 72 11 16 Now we can construct our matter wave equation by equating 1 d2 2m d i l nt zma l nt 17 which is re arranged into its well known form F12 d2 d i l nt zha l nt 18 the timedependent Schrodinger Equation for a free particle no potential energy We can interpret this equation as the quantum analogue of 2 p E 7 19 gm If we make the identi cations d E 717 20 A 2 dt and h d 7 71 21 p A 239 dz This is justi ed because if these differential operators act on our wavefunction we get eigenvalues d E A 1311 howl 22 Einstein s Law and h d iii hk l 23 p A 239 dz DeBroglie s Hypothesisi When a potential is present the Schrodinger Equation becomes E2 d2 7 W Va Mm ih 11xt 24 which is equivalent to quantummechanically saying 02 Emmi g VI Ekinetic Epotential 25 13 Separation of Variables The variables I and t in the timedependent equation can be decoupled by using the method of separation of variables This assumes the wavefunction for arbitrary timeindependent potential is the product of two timedependent and spatially dependent functions 141 WIWO 26 Then our Schodinger equation is 1 W gt wlt W 7 711w W 27gt 72mdz2 I I 72 dt I Since the derivatives act on only one of the variables we can divide by to nd if wasquot at 77 V I 2717i 28 2m W l lttgt l where primes denote spatial derivatives and the dot is a time derivative Now since the LHS depends only on I and the RHS depends only on t they each must be equal to the same onstant which we can immediately identify as E the eigenvalue of energy This gives us two decoupled equations n wt Eat 29gt and E2 i wm WIWW ENE 30 The rst one can be immediately solved aw 671 31 which is not surprising since Eh wi quot The second equation is the timeindependent Schrdinger equation and can be written explicitly as an eigenvalue equation Hw Ew 32 where H the differential operator in the Schrodinger equation is known for historical reasons as the Hamiltonian 2 Dirac Equation 2 As we saw the nonrelativistic Schodinger Equation starts with E 2177 and ends with h2 d2 d egg 7 dial 33 This is not consistent with relativity which says ma 20 34gt Although this looks completely different this expression is consistent with the nonrelativistic limit 22 22 2 Em52 124m6212164um62 35 In the nonrelativistic limit the kinetic energy is much smaller than the rest mass energy 72102 and the dynamics are determined by the former The problem with starting with this relativistic form of the energy is that it results in a relativistic Schrodinger Equation which does not treat space the spatial derivative contained in p and time on the right hand side equivalently breaking the major axiom of relativity saying these dimensions should be treated on equal footing m522 1025241 271 11 36 If somehow the term in the squareroot was a perfect square space and time would be treated equally Dirac rst showed how this was done 3 721204 p252 107213 Z ajpjc2 37 13971 where j indexes each spatial xyz dimension The only way for this to occur is if the ajls obey the following rules a 1i0123 38 and 1in ajai 0i j 39 This is known as a Clifford Algebra and obviously cannot be satis ed by scalar quantities However it can be satis ed by four matrices and the simplest representation is 4 X 4 a0 l 3 l 40gt I is the 2 X 2 identity and m where the ajls are the 2 X 2 Pauli matrices This results in a 4 X 4 Dirac Hamiltonian operator matrix 3 1072102 Z ajpjc 42 j1 and the wavefunction 1 is likewise a 4component spinor a spinor is like a vector only its rotational transormation properties are different Dirac interpreted this result as being twofold in addition to the relativistic wave equation for the electron 2 components we get another equation for a particle which was not discovered at the time which was rst called a hole and later a positron The 2 components for the electron were interpreted as the amplitude of the wavefunction in the spin up and down directions Carl Anderson discovered this particle in 1932 and received the Nobel Prize four years later Paul Dirac received the Nobel prize earlier in 1933 along with Erwin Schrodinger 3 Pauli Matrices If we are interested only in the electron7 we can use only the 2 components corresponding to those states But we need to derive the elements of the Pauli matrices Consider a general Hamiltonian H11 H12 43 l H21 H22 l lt gt and corresponding Schrodinger equation H112 E112 H11 H12 1 1 E 44 H21 H22 w w gt This equation has a nontrivial solution only if the following is true H11 E H12 7 det H21 H22 7 E J 7 0 45 we can expand this equation as H11 EH22 E H21H12 0 46 E2 H11 H22E H11H22 H21H12 0 47 This quadratic equation can be solved easily with the quadratic equation E 17 i V112 4amp6 H11 H22 i H11 H222 4H11H22 H21H12 48 2a 2 H H H H 2 E 112 22i 11 4 22 7H11H227H21H12 49 7H11H22iH 12H11H22H 2 2 4 E 7 H11H22 H21H12 50 27 2 kwi lt5 H H H H 2 Ei H21H12 52 31 Pause Now lets imagine we have a spin 12 electron in a DC B eld We know from our earlier studies E igsgg iMBB 53 Lets de ne two orthogonal wavefunctions with projections l2h and ilZfi with B along 2 1 0 wup0l7wdm1l 54 with associated Energy eigenvalues 7MBB and MBB This means that H11 H12 1 7 1 H21 ngllol BBlol 55gt and H11 H12 0 7 0 H21 Hallll rBBlll 56gt This can only be true if H11 7MBB and H22 MBB 32 back to the grind The results of the above section mean that the rst term in Eq 52 is zero Therefore H 7H 2 E2 11 4 22 H21H12l 57 Now if we have an arbitrary eld such that B Bgci By B22 we know that E2 7 M B B B3 as giving H 7 H 2 H21H12 7 23B B B 59 Now we know that H11 and H22 are iMBB H 7 H 2 7 B 7 B 2 H21H12 W H21H12 60 233 H21H12 533 B Bil 61 H21H12 B Bi 62 Because H must be Hermitian to restrict its eigenvalues E to real values this has a solution H12 1330 I iBy 63 HQ1 H12 33 i 23 64 Therefore 7 B B1 7 2311 H 23 l B iBy 73 l 65 l 0 0 l 0 7i Blelo71lBwl1 olByli all M 7MBU B ayByasz 7MBEH 67 The as are the 2 X 2 Pauli matrices we are after and they tell us how a quantized spin couples to a eld in the z y or 2 directions 4 TimeDependent Spin Mechanics Now lets use the Pauli matrices to determine the mechanics of a spin in a magnetic eld consisting of a DC component in the 2 direction and an AC component in the 2 direction As seen previously our Hamiltonian is 7M5 H 68 With a eld in the X and z directions B B2 Acos wti the dot product with the vector of matrices amp gives B 0 0 Acoswt B A coswt H7H 0 7BllAcoswt 0 TMTAcoswt 7B 69 Now we need to use this Hamiltonian in the Schrodinger equation to solve for the spin wavefunction which will tell us about the mechanics of state evolution in time given this eld con guration H 11 39h d 11 70 Z 7 dt To solve this equation we make a guess or ansatz that the solution will look generally like Q 01 in 02te7iw2t 71 Therefore if 7 7MA and E13 iMBB E1 7 coswt I C1te i 1 7 hi C1te i 1 72 7coswt E2 C2 te i 2 7 2 dt C2 te i 2 This matrix equation represents two coupled differential equations 7 coswtCQE iW Elcle iwlt ih Cie iwlt 7iC1wle iw1 73 and I I I I 7 coswtCle W EQCQE M ih C eiwzt 7 ngwge M 74 where the prime denotes time derivative and the explicit time dependence in the Cs is assumed First we note that since haul E1 and hid E2 the second terms on each side are equal and cancel out Thus substituting coswt elm 7 e we are left with 7702 em 67110 671102 ihcie7iw1t 75 and 7701 em 67110 671101 ihC e7iwgt 76 2 Writing we mg 7 ml we have C elm e7zwtgt e7zwot moi 77 and 701 m 7210 m z T 6 6 6 0 zhC2 78 Distributing C emiwonjreizwwo mo 79 and 7701 lt6iwwote7iw7wotgt mo 80 Since we are interested in behavior near resonance ie when w m we the terms like eiiw wo one but the other term like eil 0 these terms and are left with will be close to will be oscillating very fast and average out to zero Therefore we discard VTCZEZ39W W mo 81 and V01 fiwfwo I Te zhC2 82 Solving the Equation 81 for C2 02 e i 01 83 and substituting into Equation 82 gives 731e7iw7wot m2 cie7iw7wot 7 w 7 w00ie7iw7wot 84 7 which simpli es to C1 772 C1 7 2w 7w0C1 85 Rearranging gives 2 Ci7iw7w0CiC1 0 86 This is a linear secondorder differential equation in CL The form of this equation comes up many times in physics and elecronicsi Take for example the LRC circuit with characteristic equation R 1 i ILILCI 0 87 or mass on a damped spring77 Fma X EX 5X 0 88 m m 41 Resonance When we are at resonance w moi The damping term disappears and we have 3972 C 7C 0 89 1 4amp2 1 This has solution A C it 90 1 cos 2h which we substitute into Equation 81 to get 7 I l zhC2 7 2 cos 2ht 91 This can be easily integrated to nd C2 7i cos tdt fisint 92 The probability of nding the spin in the up state 1 J is 7 Cf C1 cos2 t 93 whereas the probability of nding the spin in the down state 1 J is 7 C C2 s1n2 t 94 Under resonant excitation the spin apparently ip ops back and forth between up and down states at a rate determined by the intensity of the excitation A in 7 iuA
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