P. Ewart2.1 Width and Shapeof Spectral Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.1 LifetimeBroadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.2 Collision or PressureBroadening . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1.3 Doppler Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Atomic Orders of Magnitude. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.1 Other important Atomic quantities . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 TheCentral Field Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Theformof theCentral Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Finding theCentral Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.1 ThePhysics of theWaveFunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.1.1 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.1.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.1.3 Radial wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.1.4 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Multi-electron atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2.1 Electron Congurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2.2 ThePeriodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.3 Gross Energy Level Structureof theAlkalis: QuantumDefect . . . . . . . . . . . . . . 154.1 ThePhysics of Spin-Orbit Interaction. . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Finding theSpin-Orbit Correction to theEnergy . . . . . . . . . . . . . . . . . . . . . 194.2.1 TheB-Field dueto Orbital Motion . . . . . . . . . . . . . . . . . . . . . . . . . 194.2.2 TheEnergy Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2.3 TheRadial Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2.4 TheAngular Integral: DegeneratePerturbation Theory. . . . . . . . . . . . . 214.2.5 DegeneratePerturbation theory and theVector Model . . . . . . . . . . . . . . 224.2.6 Evaluation of _ s l_ using DPT and theVector Model . . . . . . . . . . . . . . 234.3 Spin Orbit Interaction: Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.4 Spin-Orbit Splitting: Alkali Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5 Spectroscopic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27i5.1 Magnesium: Gross Structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 TheElectrostatic Perturbation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.3 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.4 Orbital eects on electrostatic interaction in LS-coupling. . . . . . . . . . . . . . . . . 335.5 Spin-Orbit Eects in 2-electron Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . 346.1 HyperneStructure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.2 TheMagnetic Field of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.3 Coupling of I and J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.4 Finding theNuclear Spin, I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.5 IsotopeEects. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407.1 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.2 Conguration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.3 Angular MomentumRules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438.1 Weak eld, no spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448.2 Weak Field with Spin and Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468.2.1 Anomalous Zeeman Pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488.2.2 Polarization of theradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498.3 Strong elds, spin and orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.4 Intermediateelds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528.5 Magnetic eld eects on hypernestructure . . . . . . . . . . . . . . . . . . . . . . . . 528.5.1 Weak eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538.5.2 Strong eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549.1 X-ray Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569.2 X-ray series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579.3 Finestructureof X-ray spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.4 X-ray absorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.5 Auger Eect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5910.1 Absorption Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.2 Laser Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.3 Spectral resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.4 Doppler Free spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.1 Crossed beamspectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.2 Saturation Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.3 Two-photon-spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6410.5 Calibration of Doppler-freeSpectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6510.6 Comparison of Doppler-free Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 65iiAtomicPhysics,P.Ewart 1 IntroductionThe structure of atoms and their behaviour is responsible for the appearance of the visible world.Thesmall scaleof atoms and theproperties of nuclei and electrons required a newkind of mechanicsto describetheir behaviour. QuantumMechanics was developed in order to explain such phenomenaasthespectraof light emitted or absorbed by atoms. Sofar you havestudied thephysicsof hydrogenand heliumas illustrations of howto apply QuantumTheory. Therewas a time, a fewseconds afterthe Big Bang, when the Universe consisted only of hydrogen and helium nucleii. It took another300,000 years for atoms, as such, to form. Things, however, have moved on and the universe isnow a much more interesting place with heavier and more complicated atoms. Our aimis now tounderstand AtomicPhysics, not just to illustrate the mathematics of Quantum Mechanics.Thisis both interesting and important, for Atomic Physics is the foundation for a wide range of basicscienceand practical technology. Thestructureand properties of atoms arethebasis of Chemistry,and henceof Biology. Atomic Physicsunderliesthestudy of Astrophysicsand Solid StatePhysics. Ithas led to important applications in medicine, communications, lasers etc, as well as still providingatestinggroundfor QuantumTheory anditsderivatives, QuantumElectrodynamics. Wehavelearnedmost about atoms fromthelight absorbed or emitted when they changetheir internal state. So thatis a good placeto begin.Wewill makeextensiveuseof models inthiscoursetohelpusget afeel for thephysics. A favouritemodel for theorists is the two-level atom i.e. one with only two eigenstates 1, 2 with energyeigenvalues E1, E2 respectively (E2> E1). Thewavefunctions have, in general, a timedependence.1= 1(x)eiE1t/(1)2= 2(x)eiE2t/(2)When the atom is perturbed it may be described by a wave function that is a linear combinationof 1 and 2: = a1 +b2 giving theprobability amplitude. Weobserve, however, a probabilitydensity: themodulus squared; [[2. This will havea termab12ei(E1E2)t/(3)This is a timeoscillating electron density with a frequency 12:ei(E1E2)t/=ei12t(4)SoE2E1 =12(5)This is illustrated schematically in gure1.1AtomicPhysics,P.Ewart 2 RadiationandAtomsy1 y2y( ) = + t y y1 2IY( ) t I2Y( ) t Y( t) t+Oscillating charge cloud: Electric dipoleI I Y( + t) t2Figure1: Evolution of thewavefunction of a systemwith time.So theperturbation produces a chargecloud that oscillates in space an oscillating dipole. Thisradiates dipoleradiation. Whether or not weget a chargedisplacement or dipolewill depend on thesymmetry properties of the two states 1, and 2. The rules that tell us if a dipole with be set uparecalled selection rules, a topic to which wewill return later in thecourse.Theradiation emitted (or absorbed) by our oscillating atomic dipoleis not exactly monochromatic,i.e. there will be a range of frequency values for12. The spectral line observed is broadened byone, or more, processes. A process that aects all the atoms in the same way is called Homoge-neous Broadening. A process that aects dierent individual atoms dierently is InhomogeneousBroadening.Examples of homogeneous broadening are lifetime (or natural) broadening or collision (or pres-sure) broadening. Examples of inhomogeneous broadening areDoppler broadening and crystal eldbroadening.This eect may beviewed as a consequenceof theuncertainty principleEt (6)SinceE =, E = and if thetimeuncertainty t isthenatural lifetimeof theexcited atomicstate, , weget a spread in frequency of theemitted radiation 1 or 1(7)2AtomicPhysics,P.Ewart 2 RadiationandAtomsThelifetime, , is a statistical parameter related to thetimetaken for thepopulation of theexcitedstate to decay to 1/e of its initial value.This exponential decay is reected in the experimentaldecay of the amplitude E(t) of the light wave emitted. The frequency (or power) spectrum of anexponentially decaying amplitudeis a Lorentzian shapefor theintensity as a function of frequencyI() 1( 0)2+(1/)2(8)Thefull width at half-maximum, FWHM, is then 2_1_(9)A typical lifetime 108sec.N( ) t I( ) wTime,t frequency, wIntensity spectrum Number of excited atomsExponential decay Lorentzian lineshapeFigure2: Decay of excited state populationN(t) leads to similar exponential decay of radiationamplitude, giving a Lorentzian spectrum.A collision with another atom while the atom is radiating (oscillating) disrupts the phase of thewave. The wave is therefore composed of various lengths of uninterrupted waves. The numberof uninterrupted waves decays exponentially with a 1/e time c, which is the mean time betweencollisions. At atmospheric pressure this is typically 1010sec. The exponential decay of thecoherent oscillations again leads to a Lorentzian lineshape.N( ) t I( ) wTime,t frequency, wIntensity spectrum Number of uncollidedatomsExponential decay Lorentzian lineshapeFigure3: Decay in number of undisturbed atoms radiating leads to decay in amplitude of waveundisturbed by a phase changing collision.The associated frequency spectrum is againLorentzian.3AtomicPhysics,P.Ewart 2 RadiationandAtomsAtoms in a gas havea spread of speeds given by theMaxwell-Boltzmann distribution. TheDopplershift of thelight emitted isthereforedierent for theatomsmovingat dierent speeds. Thereisthena spread of Doppler shifted frequencies leading to a broadening of the spectral line. Since dierentatomsareaected dierently thisDoppler BroadeningisInhomogeneousbroadening. TheDopplershift is given by: =0_1vc_(10)Thespread, or width of theline, is thereforeD 0vc(11)Since0 1015rad s1and v 102ms1wendD 2109rad s1(12) 109Hz (13)N(v) I( ) watomic speed, v frequency, wDoppler broadening Distribution of atomic speedMaxwell-BoltzmanndistributionGaussian lineshapeFigure4: Maxwell-Boltzmann distribution of speeds and theassociated Doppler broadeninggivingaGaussian lineshape.Wehavealready started gettinga feel for typical values of parameters associated with spectral lines,so nowis a good timeto do thesamefor someother aspects of atoms. It is important to havesomeidea of the orders of magnitude associated with atoms and their structure. We will be looking atatoms morecomplex than Hydrogen and Heliumfor which wecant solvetheSchrodinger Equationexactly. We have to do the best we can with approximate solutions using Perturbation Theory. Sowe need to know when perturbation theory will give a reasonable answer. We will be particularlyinterested in theenergy level structureof atoms so lets start there.First awordabout units. Atomsaresmall 1010mandtheir internal energiesaresmall; 1019J oule. So a J ouleis an inconvenient unit. Wewill useelectron Volts, eV, which is 2 1019J . Wehaveseen that atomsemit or absorb light in theUV or visiblerange say 500nm. Thismust thenbeof theorder of theenergy changes or binding energy of theouter, most loosely bound electron.Now a wavelength 500 nmcorresponds to a frequency, 61014Hz. So fromE =h wend E 4010341014 41019J 2eV.4AtomicPhysics,P.Ewart 2 RadiationandAtomsWecould also check this using thesizeof an atom( 1010m) using theUncertainty Principle.p x (14)p p

x(15)E =p22m_

x_2/(2m) a feweV (16)We can compare 2 eV with thermal energy or kT i.e. the mean Kinetic energy available fromheat,140 eV. This is not enough to exciteatoms by collisions so atoms will mostly bein their groundstate.a0 = 40

2me2=0.531010m (17)EH =me4(40)222 =13.6 eV (18)R =EHhc =1.097107m1(19)R is useful is relatingwavelengths to energies of transitions sincewavenumber =1 in units ofm1. =e240c 1137(20)This is a dimensionless constant that gives a measureof therelativestrength of theelectromagneticforce. It is actually also the ratio of the speed ve of the electron in the ground state of H to c, thespeed of light. =ve/c and so it is a measureof when relativistic eects becomeimportant.B =e2m 9.271024J T1(21)This is the basic unit of magnetic moment corresponding to an electron in a circular orbit withangular momentum, or onequantumof angular momentum.As well as havingorbital angular momentumtheelectron alsohas intrinsic spin and spin magneticmomentS = 2B. A proton also has spin but because its mass is 2000 larger than an electronits magnetic moment is 2000times smaller. Nuclear moments arein general 2000times smallerthan electron moments.5AtomicPhysics,P.Ewart 2 RadiationandAtomsWe can solve the problemof two bodies interacting with each other via some force e.g. a star andone planet with gravitational attraction, or a proton and one electron the hydrogen atom. If weadd an extra planet then things get dicult. If we add any more we have a many body problemwhich is impossibleto solveexactly. Similarly, for a many electron atomwearein serious diculty wewill need to makesomeapproximation to simplify theproblem. Weknowhowto do Hydrogen;wesolvetheSchrodinger equation:

22m2 Ze240r =E (22)Wecanndzeroorder solutions wavefunctions that wecanusetocalculatesmaller perturbationse.g. spin-orbit interaction.TheHamiltonian for a many electron atomhowever, is much morecomplicated.H =N

i=1_ 22m2i Ze240ri_+

i>je240rij(23)We ignore, for now, other interactions like spin-orbit. We have enough on our plate! The secondtermon ther.h.s. is themutual electrostatic repulsion of theN electrons, and this prevents us fromseparating the equation into a set of N individual equations. It is also too large to treat as a smallperturbation.Werecall that thehydrogen problemwassolved usingthesymmetry of thecentral Coulomb eld the1/r potential. This allowed us toseparatetheradial and angular solutions. In themany electroncase, for most of the time, a major part of the repulsion between one electron and the others actstowards thecentre. So wereplacethe1/r, hydrogen-like, potential with an eectivepotential duetothenucleus and thecentrally acting part of the1/rij repulsion term. Wecall this theCentral FieldU(r). Noteit will not bea 1/r potential. WenowwritetheHamiltonianH =H0 + H1(24)whereH0= i_ 22m2i +U(ri)_(25)andH1= i>je240rij

i_Ze240ri+U(ri)_(26)H1 is the residual electrostatic interaction. Our approximation is now to assume H1

H2 but this wont be true for allelements. We can also add smaller perturbations, H3 etc due to, for example, interactions withexternal elds, or eects of the nucleus (other than its Coulomb attraction). The Central FieldApproximation allows us to nd solutions of theSchrodinger equation in terms of wavefunctions oftheindividual electrons:(n, l, ml, ms) (30)The zero-order Hamiltonian H0 due to the Central Field will determine the gross structure of theenergy levels specied by n, l. Theperturbation H1, residual electrostatic, will split theenergy levelsinto dierent terms. Thespin orbit interaction, H2 further splits theterms, leadingto ne structureof theenergy levels. Nuclear eects lead to hyperne structures of thelevels.Within the approximation we have made, so far, the quantumnumbers ml, ms do not aect theenergy.The energy levels are therefore degenerate with respect to ml, ms. The values ofml, ms orany similar magnetic quantumnumber, specify thestate of theatom.There are 2l +1 values ofml i.e.2l +1 states and the energy level is said to be (2l +1)-folddegenerate. The only dierence between states of dierent ml (or ms) is that the axis of theirangular momentumpoints in a dierent direction in space. Wearbitrarily chosesomez-axis so thatthe projection on this axis of the orbital angular momentuml would have integer number of units(quanta) of . (ms does thesamefor theprojection of thespin angular momentums on thez-axis).(Atomic physicists areoften a bit casual in their useof languageand sometimes usethewords en-ergy level, and state(e.g. Excited stateor ground state) interchangeably. Thispracticeisregrettablebut usually no harmis doneand it is unlikely to changeanytimesoon.)You will know (or you should know!) how to nd the formof the wave functions (n, l, ml, ms) inthecaseof atomic hydrogen. In thiscoursewewant tounderstand thephysics themathsisdoneinthetext books. (You may liketo remind yourself of themaths after lookingat thephysics presentedhere!) Beforewelook at many electron atoms, weremind ourselves of theresults for hydrogen.Theenergy eigenvalues, giving thequantized energy levels aregiven by:En= _n,l,ml H n,l,ml_(31)= Z2me4(40)222n2(32)Notethat theenergy dependsonly onn, thePrincipal quantumnumber. Theenergy doesnot dependon l this is true ONLY FOR HYDROGEN! The energy levels are degenerate in l. We representtheenergy level structureby a (Grotrian) diagram.9AtomicPhysics,P.Ewart 3 TheCentralFieldApproximation1234nEnergy-13.6 eV0l = 0 1 2sp dFigure7: Energy level structureof hydrogen, illustrating how thebound stateenergy depends on nbut notl.For historical reasons, thestates with l =0, 1, 2arelabelled s, p, d. For l =3and abovethelabelsarealphabetical f, g, h etc.Thewavefunction has radial and angular dependence. Sincethesevary independently wecan write:n,l,ml(r, , ) =Rn,l(r)Ymll(, ) (33)Rn,l(r) and Ymllareradial and angular functions normalized as follows:_R2n,l(r)r2dr =1_ Ymll(, )2d =1 (34)Theangular functions areeigenfunctions of thetwo operatorsl2andlz:l2Ymll(, ) = l(l +1)2Ymll(, ) (35)lzYmll(, ) = mlYmll(, ) (36)Wherel(l +1) and ml aretheeigenvalues of l2andlz respectively, such thatl =0, 1, 2...(n 1) l ml l (37)The spin states of the electron can be included in the wave function by multiplying by a spinfunction s which isan eigenfunction both of s2and sz with eigenvaluess(s+1) and ms respectively.(s =1/2 and ms =1/2)Roughly speaking, the angular functions specify the shape of the electron distribution and theradial functions specify thesizeof theorbits.l = 0, s-states, are non-zero at the origin (r = 0) and are spherically symmetric. Classicallythey correspond to a highly elliptical orbit theelectron motion is almost entirely radial. Quantummechanically wecanvisualiseaspherical cloudexpandingandcontracting breathing, astheelectronmoves in space.10AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationl 1. As l increases, theorbit becomes less and less elliptical until for thehighestl =(n 1) theorbit is circular. An important case(i.e. worth remembering) is l =1 (ml =1, 0)Y11= _ 38_1/2sinei(38)Y11=+_ 38_1/2sinei(39)Y01= _ 34_1/2cos (40)|Y10( )| q,f2(a)|Y1+( )| q,f2(b)Figure8: Theangular functions, spherical harmonics, givingtheangular distribution of theelectronprobability density.If there was an electron in each of these three states the actual shapes change only slightly togivea spherically symmetric cloud. Actually wecould t two electrons in each l stateprovided theyhad opposite spins. The six electrons then ll the sub-shell (l = 1) In general, lled sub-shells arespherically symmetric and set up, to a good approximation, a central eld.As noted above the radial functions determine the size i.e. where the electron probability ismaximum.11AtomicPhysics,P.Ewart 3 TheCentralFieldApproximation2 4 62 4 6 8102 4 6 810 20Zr a /oZr a /oZr a /oGround state, n = 1, = 0 l 1st excited state, n = 2, = 0 l2nd excited state, n = 3, = 0 ln = 3, = 2 lN = 2, = 1 ln = 3, = 1 l2 1.01 0.50.4Figure9: Radial functions giving theradial distribution of theprobability amplitude.NB: Main features to remember! l =0, s-states do not vanish atr =0. l ,=0, states vanish atr =0 and havetheir maximumprobability amplitudefurther out withincreasing l. Thesize, position of peak probability, scales with n2. Thel =0 function crosses theaxis (n 1) times ie. has (n 1) nodes. l =1 has (n 2) nodes and so on. Maximuml =n 1 has no nodes (except atr =0)The parity operator is related to the symmetry of the wave function.The parity operator takesr r. It is likemirror reection through theorigin. and + (41)If this operation leaves the sign of unchanged the parity is even. If changes sign the parity isodd. Some states are not eigenstates of the parity operator i.e. they do not have a denite parity.Theparity of two states is an important factor in determining whether or not a transition betweenthemis allowed by emitting or absorbing a photon. For a dipoletransition theparity must change.12AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationIn central elds the parity is given by (1)l. So l = 0, 2 even states, s, d etc. are even parity andl = 1, 3, odd states are odd parity. Hence dipole transitions are allowed for sp or pd, but notallowed for ss or sd.Beforeconsideringdetailsabout multi-electron atomswemaketwoobservationson theCentral FieldApproximation based on theexact solutions for hydrogen.Firstly, for a central eld theangular wavefunctions will beessentially thesameas for hydrogen.Secondly, the radial functions will have similar properties to hydrogen functions; specically theywill havethesamenumber of nodes.Theenergy level-structurefor a multi-electron atomis governed by two important principles:a) Paulis Exclusion Principleandb) Principleof Least EnergyPauli forbids all the electrons going into the lowest or ground energy eld.The Least Energyprinciple requires that the lowest energy levels will be lled rst. The n = 1, l = 0 level can take amaximumof twoelectrons. Theleast energy principlethen means thenext electron must gointothenext lowest energy level i.e. n = 2. Given thatl can take values up to n = 1 and each l-value canhave two values of spin,s, there are 2n2vacancies for electrons for each value ofn. Our hydrogensolutions showthat higher n values lead toelectrons havingamost probableposition at larger valuesof r. Theseconsiderations lead to theconcept of shells, each labelled by their valueof n.Nowlook at thedistributionof theradial probabilityfor thehydrogenicwavefunctionsRn,l(r). Thelow angular momentumstates, especially s-states have the electrons spending some time inside theorbits of lower shell electrons. At theseclosedistances they areno longer screened fromthenucleusZ protonsandsothey will bemoretightly boundthantheir equivalent stateinhydrogen. Theenergylevel will depend on thedegreeto which theelectron penetrates thecoreof inner electrons, and thisdepends on l. Therefore the energy levels are no longer degenerate in l. To this approximation mlandms do not aect the energy. We can therefore specify the energy of the whole system by theenergy of the electrons determined by quantumnumbers n and l. Specifying n, l for each electrongives theelectron congurations:nlxwherex indicates thenumber of electrons with a given nl.As atomic sciencedeveloped theelements weregrouped together according to perceived similaritiesin their physical or chemical properties. Several dierent periodic tables wereproposed. Theonewe all learned in our chemistry classes has been adopted since it has a basis in the fundamentalstructureof theatoms of each element. Thephysical and chemical properties, e.g. gas, solid, liquid,reactivity etc areseen to beconsequences of atomic structure. In particular they ariselargely fromthebehaviour of theouter shell or most loosely bound electrons.Using the Pauli Principle and the Least energy Principle we can construct the congurations oftheelements in their ground states:-13AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationH: 1sHe: 1s22sLi: 1s22s2Be: 1s22s22pC: 1s22s22p2. . .Ne: 1s22s22p6Na: 1s22s22p63sEverything proceeds according to this pattern up to Argon: 1s22s22p63s23p6At this point things get a littlemorecomplicated. Weexpect thenext electron to go into the3dsub-shell. As we have seen, however, a 3d electron is very likea 3d electron in hydrogen: it spendsmost of its timein acircular orbit outsidetheinner shell electrons. An electron in a4s statehowevergoes relatively close to the nucleus, inside the core, and so ends up more tightly bound - i.e. lowerenergy, than the3d electron. So thenext element, Potassium, has thecongurationK: 1s22s22p63s23p64sCa: 4s2The3d shell nowbegins to llSc: 1s22s22p63s23p63d 4s2The3d and 4s energies arenowvery similar and at Chromiuma 3d electron takes precedenceovera 4s electronVa: 3s23p63d34s2Cr: 3d54sMn: 3d54s2Asthe3d shell llsup, thesuccessiveelements, thetransition elements, haveinterestingpropertiesas a result of thepartially lled outer shell but this isnt on thesyllabus!There are, however, two features of the periodic table that are worth noting as consequences oftheCentral Field Approximation.These elements are chemically inert and have high ionization potentials (the energyneeded to pull o a single electron). This is not because, as is sometimes (often) stated, that theyhaveclosed shells. They dont all haveclosed shells i.e. all states for each n arefull. They all dohavelled s and p sub-shellsHe: 1s2Kr: 4s24p6Ne: 1s22p6Xe: 5s25p6A: 1s22p63s23p6Rn: 6s26p6As we noted earlier this leads to a spherically symmetric charge distribution. Since electrons areindistinguishableall theelectrons takeon a common wavefunction. Thepoint is that this results ina higher binding energy for each oneof theelectrons. So it is harder for themto losean electron ina chemical bond they arechemically inert and havehigh ionization energies.14AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationThese are the next elements to the rare gases and have one electron outside the full (sp)sub-shells. This outer, or valence, electron thereforemoves in a hydrogen likecentral potential. Theelectron is generally well-screened by (Z 1) inner electrons fromthe nucleus and is easily lost toa chemical bond (ionic or co-valent). They arechemically reactiveand havelow ionization energieswhich dont changemuch fromonealkali to another.As noted already, the single outer electron in an alkali moves in a potential that is central to anexcellent approximation. We are ignoring, at this stage, any other perturbations such as spin-orbitinteraction.As an example we consider sodium.The ground stage (lowest energy level) has theconguration:(closed shells) 3s.We know the 3s penetrates the core (inner shells) and is therefore more tightly bound lowerenergy than a 3s electron would be in Hydrogen. When the electron is excited, say to 3p, itpenetrates thecoremuch less and in a 3d stateits orbit is virtually circular and very closethen =3level of Hydrogen.Thehigher excited states n > 3will followa similar pattern. Thethingto notice and this turnsout to beexperimentally useful is that thedegreeof corepenetration depends on l and very littleon n. As a result thedeviation fromthehydrogenic energy level is almost constant for a given l asn increases.Thehydrogen energy levels can beexpressed asEn =Rn2(42)For alkalis, and to somedegreefor other atoms too, theexcited stateenergies may beexpressed asEn =R(n)2(43)Wheren is an eectivequantumnumber. n diers fromtheequivalent n-valuein hydrogen by (l)i.e. n =n (l).(l) is the quantum defect and depends largely onl only. It is found empirically, and it can beshown theoretically, to be independent ofn. Thus all s-states will have the same quantum defect(s); all p-states will havethesame(p), etc and (s) > (p) > (d).For Na:(s) 1.34(p) 0.88(d) 0For heavier alkalis, the (l) generally increases as the core is less and less hydrogen-like. Theionization potential however isalmost constant asnoted previously. Although theincreasein Z leadsto stronger binding than in hydrogen theground stateelectron starts in a higher n, and this almostexactly o-sets theincreased attraction of theheavier nucleus.Everyoneknowsthat thediscretewavelengthsof light emitted, or absorbed, by anatomarediscretebecause they arise fromtransitions between the discrete energy levels. But how do we know whichenergy levels? This is wherethequantumeect is very useful.Wealso need to remember that a transition involves a changein l of 1(l =1). A transitionfroma lower statewill thereforetaketheatomto an energy level with angular momentumdiering15AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationby 1 from the initial level. For simplicity, so that the general idea becomes clear, we considertransitions fromtheground state. If theground state(level) has l =0then therewill then beonly asingleset of transitions to levels with angular momentumoneunit larger i.e. l =1. Thespectrumofabsorption lines will consist of a series of lines of increasing wavenumber (energy) corresponding totransitions to excited states with increasing principal quantumnumber n. Sincetheseexcited levelsall havethesamevalueof l they should havethesamevalueof quantumdefect. Soif wecan identifylines with the same quantumdefect they will belong to the same series i.e. have the same angularmomentumquantumnumberl. Theprocedureis then as follows.(1) Measurethewavelength of each absorption linein theseries, n.(2) Calculatethecorresponding wavenumbern =1n(3) Estimatethevalueof theseries limit . This corresponds totheionization potential or Termvaluefor theground state, Tno.(4) Find theTermvalue, Tn, or bindingenergy of an excited level fromthedierencebetweenand n:Tni =Tno n(5) TheTermvalueis given by thehydrogenic formula:Tni =R(ni)2whereni is theeectivequantumnumber given by:ni =n (l)and (l) is thequantumdefect.(6) Finally we plot(l) againstTni. This should be a straight line indicating a constant value of(l). Deviation froma straight lineindicates that wehavenot estimated (i.e. Tno) correctly.(7) Theprocedureisthentotry adierent valueof Tno until weobtainastraight line. Thisprocessis ideally suited to computation by a programthat minimizes thedeviation froma straight lineforvarious values of Tno.Once we have found the most accurate value ofTno from the spectrum we can put the energylevels on an absoluteenergy scale. Wehavethen determined theenergy level structureof theatomfor this set of angular momentumstates. Theprocedurecan becarried out usingdatafromemissionspectra. In this casea rangeof dierent series will bebeprovided fromthespectrum. It remains tond thequantumdefects for each series of lines and assign values of l to each series of levels. Fromthediscussion abovethequantumdefects will decreasefor increasing l.Figure10: Plot of quantumdefect(l) against Termvalue, showingeect of choosingTno either toolargeor too small. Thecorrect valueof Tno yields a horizontal plot.16AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionThe Central Field Approximation gives us a zero-order Hamiltonian H0 that allows us to solve theSchrodinger equation and thus nd a set of zero-order wavefunctions i. The hope is that we cantreat theresidual electrostatic interaction (i.e. thenon-central bit of theelectron-electron repulsion)as a small perturbation, H1 . Thechangeto theenergy would befound using thefunctions i.Theresidual electrostatic interaction however isnt theonly perturbation around. Magnetic inter-actionsarisewhentherearemovingcharges. Specically weneedtoconsider themagneticinteractionbetween themagnetic moment dueto theelectron spin and themagnetic eld arisingfromtheelec-tron orbit. This eld is dueto themotion of theelectron in theelectric eld of thenucleus and theother electrons. This spin-orbit interaction has an energy described by the perturbation H2. Thequestion is: which is thegreater perturbation, H1 or H2?Wemay betempted to assume H1> H2 sinceelectrostatic forces areusually much stronger thanmagnetic ones. However by setting up a Central Field we have already dealt with the major partof the electrostatic interaction. The remaining bit may not be larger than the magnetic spin-orbitinteraction. In many atoms theresidual electrostatic interaction, H1, does indeed dominatethespin-orbit. Thereis, however, a set of atoms wheretheresidual electrostatic repulsion is eectively zero;the alkali atoms. In the alkalis we have only one electron orbiting outside a spherically symmetriccore. The central eld is, in this case, an excellent approximation. The spin-orbit interaction, H2will be the largest perturbation provided there are no external elds present. So we will take thealkalis e.g. Sodium, as a suitablecasefor treatment of spin-orbit eects in atoms. You havealreadymet thespin-orbit eect in atomic hydrogen, so you will befamiliar with thequantummechanics forcalculating thesplitting of theenergy levels. Thereare, however, someimportant dierences in thecase of more complex atoms. In any case, we are interested in understanding the physics, not justdoingthemaths of simplesystems. In what follows weshall rst outlinethephysics of theelectronsspin magnetic moment interactingwith themagnetic eld, B, duetoits motion in thecentral eld(nucleus plus inner shell electrons). Theinteraction energy is found to beB so theperturbationto theenergy, E, will betheexpectation valueof thecorresponding operator B_.We then use perturbation theory to nd E. We will not, however, be able simply to use ourzero order wavefunctions 0(n, l, ml, ms) derived fromour Central Field Approximation, since theyare degenerate in ml, ms. We then have to use degenerate perturbation theory, DPT, to solve theproblem. Wewont haveto actually do any complicated maths becauseit turns out that wecan usea helpful model the Vector Model, that guides us to thesolution, and gives someinsight into thephysics of what DPT is doing.What happens to a magnetic dipole in a magnetic eld? A negatively charged object having amoment of inertia I, rotatingwith angular velocity , has angular momentum, I =. Theenergyis then E =12I2Wesupposetheangular momentumvector is at an angle to thez-axis. Therotating chargehas a magnetic moment = (44)Thesign is negativeas wehaveanegativecharge. is known as thegyromagnetic ratio. (Classically =1 for an orbiting charge, and =2 for a spinning charge).If a constant magnetic eld B is applied along the z-axis the moving charge experiences a force a torque acts on the body producing an extra rotational motion around the z-axis. The axis ofrotation of precesses around thedirection of B with angular velocity

. Theangular motion ofour rotating changeis changed by this additional precession from to +

cos(). If theangularmomentum was in theoppositedirection then thenewangular velocity would be

cos().17AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionFigure11: Illustration of the precession (Larmor precession) caused by the torque on the magneticmoment by a magnetic eld B.Thenewenergy is, thenE

=12I(

cos)2(45)=12I2+ 12I(

cos)2I

cos (46)We now assume the precessional motion

to be slow compared to the original angular velocity .

3d sincethe3p electron penetrates theinner coreof electrons far morethan the3d electron.The actual size of the splitting is set by the value of the radial integral(n, l). For hydrogen wenoted this wasn,l =0Z4gs2B41n3a30l(l +1/2)(l +1)(99)This applies to hydrogen-likeatoms i.e. ions with all electrons except onestripped o. For a neutralatom, however, the electrostatic force of the central eld will be less and so the dependence onZis reduced fromthe Z4in a hydrogenic ion. We can get an approximate idea of the Z-dependenceby considering theZe in our Central Field to bea step function i.e. to changeabruptly fromitseective value Zo (o for outer) outside the core to Zin (in for inner) inside the core. This is avery crudeapproximation, and wecan makeit even morecrudeby supposingZo =1(total screeningby (Z 1) electrons) and Zin =Z (no screening at all).Thetotal energy of thespin-orbit interaction will bemadeup of two parts, each with theform:ESO Z_ 1r3_(100)Theouter part, Z =1, is very small compared to thecontribution fromtheinner part, Z>> 1, soweneglect thecontribution fromtheZo bit. Theinner part is hydrogenic so:_Zr3_inner=Z4n3a30l(l +1/2)(l +1)(101)The contribution will be weighted by the fraction of the time the electron spends inside the core.This fraction is indicated by thequantumdefect or by theeectivequantumnumbern. Using theBohr theory wecan showthat thefraction insidethecoreis_ n3Z2in_/_n3Z2o_(102)So thecontribution fromtheinner part becomes:_Zinr3_inner

n3Z2on3Z2in(103)26AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionHencen,l 04gs2BZ2inZ2on3a30l(l +1/2)(l +1)(104)So approximately, thesizeof thespin-orbit splitting scales with Z2in Z2.We now need to consider atoms other than alkalis, where residual electrostatic interaction, cantbe ignored. Before we do that we dene a way of indicating the total angular momentumof atomsas a whole, and not just individual electrons with l, s, j, ml, ms, mj, etc.So far we have seen how the physical interactions aecting the energy of an atom can be orderedaccording to their strength. Weuseperturbation theory, or theVector Model, by starting with thelargest interaction the Coulomb force of the Central Field. This led to the idea of the electronconguration; n1l1n2l2n3l3 etc. with each individual electrons angular momentum labelled by alower case letter; s, p, d, f, etc. For single-electron atoms, like alkalis, we dened a total angularmomentum j = l + s. The spin-orbit interaction led to energy levels labelled by their value ofj,i.e. (l +1/2) or (l 1/2). What do we do when there is more than one electron to worry about?We anticipatetheresult of the next section whereweconsider theresidual electrostatic interactionbetween electrons.If we were to ignore any spin-orbit interaction then the energy levels are completely determinedby electrostatic forces only. In this casethespaceand spin systems within theatommust separatelyconserve angular momentum. This suggests that we can dene a total orbital angular momentumby thevector sumof theindividual orbital momenta. Wewill label such total angular momenta bycapital letters; thus thetotal orbital angular momentumis:L =

ili(105)For simplicity we will consider two-electron atoms i.e. atoms with two valence electrons outsideclosed inner shells. So L =l1 +l2. In thesameway wedenethetotal spin of theatomto beS =

isi =s1 +s2(106)Thesecan bevisualized using our vector model:l2Ll1Ss2s1Figure15: Diagramshowing thecoupling of allli to L and allsi to S.Quantummechanically theseangular momenta arerepresented by operators L2and LzL2=_l1 +l2_2(107)27AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionWith eigenvalues L(L +1)2where[l1l2[ L l1 +l2, andLz =lz1 +lz2(108)With eigenvalues ML whereL ML L.Similarly wehavetotal spin operators.S2=( s1 + s2)2(109)With eigenvalues S(S +1)2where[s1s2[ S s1 +s2, andSz = sz1 + sz2(110)With eigenvalues MS whereS MS S.ML andMS are the quantized projections of L andS respectively on the z-axis (the axis ofquantization). The total angular momentumis found by adding (or coupling) the total orbital andspin momenta: to giveJ:J2=_L + S_2(111)With eigenvalues J(J +1)2and[L S[ J L +S (112)Also:Jz = Lz + Sz(113)With eigenvalues MJ whereJ MJ J i.e. MJ is theprojection of J on thequantization axis.By convention L =0, 1, 2, 3is labelled S, P, D, F etc as with thes, p, d, f for singleelectrons. Thevalues of L and S specify a Term. The spin-orbit interaction between L and S will split each terminto levels labelled by J. For a given L therewill be(2S +1) values of J and so (2S +1) is knownas themultiplicitySpin-orbit interaction will split each terminto levels according to its valueof J. Thus themulti-plicity tells us thenumber of separateenergy levels (i.e. number of J values) except, however whenL =0, when thereis no spin-orbit splitting and thereis only a singlelevel.Eachlevel isdegenerateinJ; thereare(2J+1) valuesof MJ; theprojectionof J onanexternal axis.Each valueof MJ between J and +J constitutes a state, and thedegeneracy is raised by applyingan external magnetic eld in which thedierent orientations of J relativeto theeld direction willhavedierent energy.In general then wehavethefollowing notation for an energy level:n1l1n2l2...2S+1LJ(114)This reects thehierarchy of interactions:Central Field conguration, n1l1n2l2. . .Residual Electrostatic Terms, L =S, P, D. . .Spin-Orbit Level, J =[L S[ L +SExternal Field State, MJ =J +JTheground level of sodiumcan thereforebewritten as 1s22s22p63s2S1/2.It is common practicetodrop theconguration of theinner shells and giveonly that for theouter,valence, electrons.Thus the rst two excited levels are 3p2P1/2,3/2 and 3d2D3/2,5/2.these aresometimes referred to as excited states, but this is an example of the loose terminology used byAtomic Physicists!Wecannowdrawafull Energy Level diagramfor sodiumwithall thelevelslabelledby appropriatequantumnumbers.28AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionHydrogenn43Figure16: Na energy level diagramshowingnestructure(spin-orbit splitting) greatly exaggerated.Note that the dierence between energy levels the same n but of increasingl becomesmaller as they approach thehydrogenic energy level for thecorresponding n. Notealsothat the dierence between energy levels of the same n but dierentl becomes smallerwith increasing n.29AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingSofar wehavemadetheCentral FieldApproximation which allowsustoestimatethegrossstructureof the energy levels. This introduced quantumnumbers n and l allowing us to dene the electronconguration. Wenoted two perturbations that need to beconsidered in going beyond theCentralField Approximation. Thesewere, H1 , theresidual electrostatic interaction and H2, thespin-orbitinteraction. Perturbation theory requires that weapply thelarger perturbation rst. In thespeciccase of a one-electron atomin a Central Field we could ignore H1 . The perturbation due to spin-orbit eects then led us to wave functions labelled by the quantumnumbers j,mj. In the case ofan atomwith two electrons in a Central Field wehaveto consider rst H1, and this will lead us todierent quantumnumbers. Wehaveanticipated this result by introducingthesequantumnumbersL,S and J. In the following sections we will look more closely at the physics of how electrostaticinteractions lead to LS-coupling. It needs to bestressed at theoutset that LS-coupling is themostappropriatedescription when electrostatic eects arethedominant perturbation. Wecan then applythesmaller spin-orbit perturbation H2 to theseLSJ -labelled states.All this is conveniently visualized using the Vector Model, where the strongest interactions leadto thefastest precessional motions. Theresidual electrostatic perturbation H1 causes theindividualorbital angular momental1 and l2 tocoupletoformL =l1+l2. Similarly thespin angular momentas1 and s2 coupleto givea total spin S =s1 +s2. So l1 and l2 precess rapidly around L and s1 ands2 precess rapidly around S. L and S arethen well-dened constants of themotion.The weaker, spin-orbit interaction H2 then leads to a slower precession ofL and S around theirresultant J. Asin thesingleelectron case, theprecession rate and hencetheenergy shift dependson therelativeorientation of L and S. Theresult is a ne-structuresplittingof theterms dened byL and S according to thevalueof J.Wehavebeenassumingthat H1>> H2 but thismay not alwaysbethecase. Wehaveseenalreadythat H2 increaseswith Z2. Soour assumption that electrostatic dominatesover magnetic, spin-orbit,coupling will become less valid for heavy elements. When this happens L and S cease to be goodquantum numbers. The total angular momentum J will continue to be a good quantum number,but it will bea result of adding theangular momenta of theindividual electrons in a dierent way.In thelimit that H2>> H1, individual electron momenta li and si coupleto giveji =li +si. ThetotalJ is then thevector sumof j1 and j2. This jj-coupling is, however, not on thesyllabus.In thefollowingwewill takeMagnesiumas our exampleof a two-electron atom. Wewill look rstat thegross energy level structurearising fromtheCentral Field Approximation. Secondly, wewillconsider theresidual electrostatic eect (perturbation H1) which splits theenergy levels into termslabelled by L and S. Wewill think about why theseterms havedierent energies. Thirdly, weapplythesmaller H2, spin-orbit, perturbation and examinetheresulting ne-structuresplitting.We choose Mg as our typical two-electron atom rather than Helium because, in the Helium case,thespin-spin interaction is of comparablemagnitudeto theother perturbations. In heavier elementsthis is less so. We can, however, use the basic results found for Helium to help us understandthe behaviour of our two valence electrons in Magnesium. Magnesiumis basically sodiumwith anextra proton in the nucleus and an extra electron outside. The conguration of the ground energylevel is 1s22s22p63s2.If we restrict ourselves to exciting only one of the 3s electrons, the excitedcongurations will be( )3snl. Theelectron remaining in the3s orbit acts as a spectator electronand so the excited electron moves in a Central Field not much dierent fromthat of sodium. Thebasic energy level structure will be roughly the same as sodium, but the energy levels will be morenegativeowing to theincreased Coulomb attraction of thenucleus.What will betheeect of theperturbations H1 and H2?30AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingWewill eliminate H2 for themoment by consideringtheconguration3s4s; allowingustoconcentrateon the eects arising fromthere being two electrons with electrostatic repulsion between them. Itmay seem, at rst sight, that wecould deal with this conguration entirely within theCentral FieldApproximation. After all, the spectator, 3s, electron is a spherically symmetric charge and so alsois the 4s electron. Could their mutual interaction not be contained within the Central Field? Theanswer is no becausetheelds thetwo electrons seeis not thesame.So wewill havea left-over perturbation H1 and it doesnt haveto benon-central.In the Central Field Approximation we can nd wave functions for each electron in our 3s4sconguration; (3s) and (4s). What will bethewavefunction for thesystemas a whole? Wecouldforma product wavefunction 1(3s)2(4s). Such a wavefunction implies we can identify individualelectrons by thesubscripts 1and 2. Theparticles, arefor themoment, distinguishable. Now, can weusethis wavefunction to nd our perturbation energy dueto H1, denoted by E1?In other words, is E1 =1(3s)2(4s)[ H1[1(3s)2(4s)) ?It isfairlyobviousthat if weswappedthelabels1and2wewouldget thesameenergy. Sothestates:[1(3s)2(4s)) and[2(3s)1(4s)) aredegenerate. WethereforehavetouseDegeneratePerturbationTheory. Thismeanswehavetondcorrect linear combinationsof our zero-order wavefunctions. Thecorrect linear combination wavefunction will diagonalizetheperturbation matrix. Sowecan usetheformof theperturbation to guideus to thecorrect combinationH1 =

iZe240ri+

i>je240rij

iU(ri) (115)Since our zero order wavefunctions are eigenfunctions ofU(ri), the third termwill not give us anytrouble. Therst termis also a single-electron operator so this will not causea problemeither. Weareleft with the1/rij operator and need to nd a wavefunction that will givea diagonal matrix.Wecan formtwo orthogonal functions as follows:-1 =a1(3s)2(4s) +b1(4s)2(3s) (116)2 =b1(3s)2(4s) a1(4s)2(3s) (117)It remains to nd a and b to makethediagonal matrix elements vanish i.e.1[ V [2) =0 (118)WhereV is thetwo-electron electrostatic repulsion. This perturbation does not distinguish electron1 fromelectron 2, so [a[ =[b[ and for normalization [a[, [b[ =1/2. Wenowhave1 = 12(1(3s)2(4s) +1(4s)2(3s)) (119)2 = 12(1(3s)2(4s) 1(4s)2(3s)) (120)We note now that interchanging the labels does not change 1, and simply changes the sign of2.Thesefunctions havea deniteexchangesymmetry. Weshall return to this later.Theo-diagonal matrix element is now:121(3s)2(4s) +1(4s)2(3s)[ V [1(3s)2(4s) 1(4s)2(3s)) (121)1 2 3 431AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingMultiplying out theterms in theintegral wend:13= 1(3s)2(4s)[ V [1(3s)2(4s)) =J (122)24= 1(4s)2(3s)[ V [1(4s)2(3s)) =J (123)23= 1(4s)2(3s)[ V [1(3s)2(4s)) =K (124)14= 1(3s)2(4s)[ V [1(4s)2(3s)) =K (125)So thetotal element is zero as required!Thediagonal elements nowgiveus our energy eigenvalues for thechangein energy:-1[ V [1) =E1 =J +K (126)2[ V [2) =E2 =J K (127)J is known as thedirect integral and K is theexchangeintegral. (Thewavefunctions in K dier byexchangeof theelectron label).Energy level with noelectrostatic interactionJ+K-KSingletTripletFigure17: The eect of residual electrostatic interaction is to shift the mean energy by the directintegral J and to split thesinglet and triplet terms by theexchangeintegral 2KSo the energy of any particular conguration will be split by an amount 2K. This correspondsto the two energies associated with the normal modes of coupled oscillators. For example, twopendula, or twomassesonsprings, whencoupledtogether will settleintocorrelatedor anti-correlatedmotions with dierent energies. Notethat for our 2-electrons theeect arises purely becauseof themutual electrostatic interaction. Notealso that this has nothingto do with thedistinguishability orindistinguishability of theelectrons!Our result shows that the termof the conguration described by 1, the symmetric state, has ahigher (less negative) energy than theanti-symmetric state2. To understand why weneed to thinka bit moreabout symmetry and thephysics of theinteractions between theelectrons.Degenerate Perturbation Theory forced us to choose spatial wavefunctions that have a dene sym-metry; either symmetric or antisymmetric. Why? It was because interchanging the labels on theoperator could make no physical dierence to the result,1r12=1r21. This must always be the casewhen we are dealing with identical particles. They must always be in a sate of denite symmetry;either symmetric, (+), or antisymmetric, (-).All electrons in our universe are antisymmetric, soweneed to combineour spatial wavefunctions with an appropriatespin wavefunction to ensureouroverall antisymmetric state.32AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingEach electron spin may beup or down . Wecan thereforeconstruct for two electrons combina-tions that areeither symmetric or antisymmetric:12, 12,1212 +12 : Symmetric (128)1212 12 : Antisymmetric (129)The three symmetric spin states form a triplet and combine with the antisymmetric space states,2. This triplet term energy was found to be shifted down i.e. more strongly bound. Physically,this is becausetheantisymmetric spacefunction describes astatewheretheelectrons avoid thesameregionof space. Themutual repulsionfromthee2/r12 potential isthereforereducedandtheelectronsare held more strongly. The antisymmetric state, on the other hand, is a singlet state. Since theelectrons have opposite spins, equivalent to dierent spin quantum numbers, they can occupy thesame space without violating the Pauli Principle. Increased overlap in space increases the eect ofthe1/r12 repulsion, throwing theelectrons further out in theCentral Field. This results in a lowerbinding energy. Thesinglet terms thus lieabovethetriplet termfor a given conguration.We have spent a long time considering the eect of the spin on the energy levels of the LS-coupledstates. There is, however, another aspect of the electrostatic interaction that we have overlooked;theway theenergy depends on L. TheCentral Field Approximation takes account of thesphericallyaveraged charge distribution. The value of L for a given conguration depends on the relativeorientation of the individual angular momenta l1 and l2. If either of these is zero e.g. 3s then thespatial overlap with theother electron in, say, a3p orbit will bethesamenomatter wheretheaxisofthe3p orbit points. Consider, however, what happensif both electronsarein p-orbitssay 3p4p. NowL =l1+l2 =2, 1 or 0i.e. D, P or S Terms. TheVector Model shows that theaxes of theindividualorbital angular momenta are xed in one of the three spatial orientations to give quantized valuesof L. Theelectron wavefunctions, and probability densities, arelikedoughnut shapes. Thedierentrelativeorientationsof their axeswill leadtodierent spatial overlapsof theelectrons. Consequently,theamount of mutual repulsionwill bedierent ineachcase. Thishelpsexplainwhy theelectrostaticinteraction leads to terms of dierent energy and arelabelled by L and S.l1ol2Figure18: Electron orbitalsl1, l2 have an overlap depending on their relative orientation and sodierent vector sums havedierent electrostatic energy.33AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingSofar wehaveapplied only theelectrostatic perturbation H1 tothecongurationsset by theCentralField. This electrostatic interaction led to singlet and triplet terms labelled by L and S. The 3s4sconguration in Mgwill havetoterms;1S and3S, with1S lyingabove3S in theenergy level diagram.Since there is no orbital motion there is no spin-orbit interaction.The total angular momentumJ =L +S =0 or 1. Thecompletedesignations aretherefore3s4s1S0 and 3s4s3S1Although themultiplicity, 2S +1, is 3 in thetriplet state, thereis only oneenergy level sinceJ hasonly thevalue1, in this case.Lets nowconsider theconguration 3s3p. Again S =0or 1and wehavesinglet and triplet terms.L = 1, so the terms are1P and3P.J, however, in the triplet termhas values 2, 1, 0. These threevalues arisefromeach of threerelativeorientations of L and S. This is clear in our Vector Model:L = 1S = 1S = 1S = 1L = 1 L = 1J = 2 J = 1J = 0Figure19: Vector addition of L and S to givetotal angular momentumJ.The magnetic interactions between the total spin moment and the magnetic eld established bythe total orbital moment,L, are dierent in each case leading to dierent energies for the levels ofeach J value. Thetriplet,3P, termis thereforesplit and acquires ne-structure.Thephysical argumentsweused for spin-orbit interaction in Nacarry over totheMgcaseby usingthe total angular momentum operators L, S and J instead of l, s and j. The perturbation to the[n, L, S, J) states is:ESO =_ H2___1rU(r)r_ S L_(130)Theradial integral is denoted n,l. Theangular momentumoperator is:S L = 12_J2 L2 S2_(131)and, since we have the correct representation in eigenfunctions of the operators on the R.H.S. wend theexpectation valueis given by thecorresponding eigenvalues:_J2 L2 S2_ = 12J(J +1) L(L +1) S(S +1) (132)soEJ =n,L2J(J +1) L(L +1) S(S +1) (133)Theseparation of thene-structurelevels is found by evaluating EJ forJ

and J

1.EJ EJ

1 =n,lJ

(134)This represents an Interval Rulethat is valid so long as two conditions arefullled:(1) LS-coupling is a good description for the terms i.e. when the electrostatic perturbationH1>>H2, themagnetic, spin-orbit interaction.34AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-Coupling(2) theperturbationenergy isexpressibleas_S L_. Thisisnot thecaseinHeliumwherespin-spininteractions arecomparableto spin-orbit eects.Theenergy levels of the3s3p terms aretherefore2K3s3p P113s3p P323s3p P313s3p P30ib2bFigure20: Separation of singlet and triplet terms due to electrostatic interaction and splitting oftriplet termby magnetic spin-orbit interaction.Aswegotoatomswith higher Z, themagnetic, spin-orbit, interaction increases( Z2) and beginsto becomparableto theelectrostatic eect. LS-coupling starts to break down and will beshown bytwo features of theenergy levels: (a) Theseparation of thetriplet levels becomes comparableto theseparation of singlet and triplet terms of the same conguration. (b) The interval rule is no longerobeyed.A third indicator of thebreakdown of LS-couplingis theoccurrenceof optical transitions betweensingleand triplet states.In the LS-coupling scheme the spin states are well-dened and the dipole operator,e r, does notact on the spin part of the wave function. Thus a dipole transition will link only states with thesame value ofS. i.e. we have the selection rule S = 0. This rule is well obeyed in Mg and thelighter 2-electron atoms. So we get transitions only between singlet and singlet states, or betweentriplet and triplet. It isthen common toseparateout thesinglet ad triplet energy levelsintoseparatediagrams.1 1 1 3 3 3S P D S P Do 2 13s S2 103s3p P113s3p P32,1,03s3d D124s5sns3p23pnlintercombination line(weak)resonance line(strong)Term diagram of MagnesiumSinglet terms Triplet termsFigure21: Mg termdiagram. The separation of singlet and triplet terms shows schematically thatsinglet-triplet transitions areforbidden.35AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingThe 3s3p3P2,1,0 levels are metastable i.e. they have a long lifetime against radiative decay. The6s6p3P2,1,0 levels in Mercury are also metastable but a transition 6s6p3P1 6s2 1S0 does occur.Such a transition is known spectroscopically as an intercombination line. The probability is notlarge, compared to other allowed transitions, but this transition is a strong sourceof radiation fromexcited mercury. This is because any atoms ending up in the 6s6p3P1 level have no other way todecay radiatively. This lineis thesourceof light in Hg uorescent lamps.36AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureWe have secretly made three assumptions about the nucleus so far. We have assumed that thenucleus is stationary, has innite mass and occupies only a point in space. In fact, the nuclei of allatoms consist of spinningcharged objects and so they havea resultant magnetic moment. Secondly,their mass is not innite, so thenucleus will movewith theorbiting electrons around their commoncentreof gravity. Thirdly thenucleus has a nitesizewith someshapeover which theproton chargeis distributed. Each of these eects will make a small change in the energy levels. The magneticinteraction will lead toan eect similar tospin-orbit couplingand giveasplittingof theenergy levelsknow as hyperne structure, hfs. The kinetic energy of the moving nucleus will aect the overallenergy of the atomic states.Finally, the size and shape of the charges on the nucleus will aectthe Coulomb force on the electrons and hence aect the energy. The termhyperne structure isreserved (at least in Oxford) for magnetic eects of the nuclear spin. The mass and electron eldeects will be termed isotope eects, since they depend to some extent on the number of neutronsin a given atom.Both theprotons and theneutrons havemagnetic moments dueto their spin. Thetotal nuclear spinis labelledI. Spinning fermions tend to pair up with another similar particle with a spin in theoppositedirection. For this reason nucleii with even numbers of protons and neutrons, theso-calledeven-even nuclei have I = 0. Odd-odd or odd-even nuclei can have integer or half-integer spin:I =1/2, 1, 3/2 etc. Thesenuclei will havea magnetic dipolemomentI. Nucleii with integer spinscan also have an electric quadruple moment as well. We will be concerned only with the magneticdipolemoment. Nuclear magnetic moments aresmall compared with electronic moments. Recall theclassical relationship between magnetic moment and angular moment: = (135)In QuantumMechanics thespin and orbital moments are:-s =gsBs (136)l =glBl (137)wheregs =2 and gl =1 (Notewehaveused s, l in units of ) Now B =e2me, whereme = mass oftheelectron. So nuclear moments aregoing to bemuch smaller as thenuclear mass mN>> me. Soto keep theg-factor of theorder of unity wedenethenuclear moment byI =gIII (138)wheregI is thenuclearg-factor (1) and N is thenuclear magneton, related to B by theratio ofelectron to proton mass:N =B

memp(139)The positive sign on our denition ofI is purely a convention; we cannot tell in general whetherthemagnetic moment will beparallel or antiparallel to I, it can beeither.Themagnetic momentsof theelectronsand thenucleuswill precessaround their mutual resultant.Theimportant thingto noteis that thesmallness of I will mean theinteraction with themagneticeld of theelectrons will beweak. Theprecession ratewill thereforebeslowand angular momentaIand J will remain well-dened. Theinteraction can thereforealso betreated by perturbation theoryand theVector Model. Theperturbation can beexpressed:H3 = I Bel(140)37AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureIt remains to nd the expectation value of this operator, and we will use our vector model to ndtheanswer in terms of thetimeaveraged vector quantities.Thenuclear interaction, beingweak, will not followthemotionsof individual electron orbitsor spins.Their precessions around J aretoo fast for thenuclei to follow. Thenuclear moment will thereforeeectively seeonly thetimeaveraged component resolved along J.Bel =(scalar quantity) J (141)Theelectrons in closed shells makeno net contribution to theeld Bel. Thereforeonly theelectronsoutsidetheclosed shells contribute. Electrons having l ,=0will havea magnetic eld at thenucleusdue to both orbital and spin motions.Both of these however depend onr3and so will be smallenough to ignore for most cases. Forl = 0, s-electrons, however, the situation is dierent. As wenoted before, theseelectrons have(r) ,=0atr =0. Thespin-spin interaction with thenucleus cantherefore be strong. This is known as the Fermi contact interaction. This short range interactiondominates thecontribution to theenergy. In general it is very dicult to calculatebut for thecaseof hydrogen in its ground statean analytical result can befound.In general the orbital and spin momenta of the electrons provide a eld of order 04B1r3_.Taking a Bohr radius to givetheorder of magnitudeof r wend:Bel 04Ba30 6T (142)Putting this with a nuclear momenta of theorder of N B/2000 wend an energy perturbationE 50 MHz. So a transition involving a level with hfs will besplit by this order of magnitudeinfrequency. Since the eect scales with Z it will get up to 100 times larger in heavy atoms i.e. afewGHz. Thestructureof thesplittingwill depend on theangular momentumcoupling, and to thiswenowturn.Since the nuclear spin magnetic moment is proportional toI and the electron magnetic eld isproportional to J wecan writeH3 =AJIJand E =AJ_I J_(143)Thequantity AJ determines thesizeof theinteraction energy. Notethat it depends on other factorsaswell asJ. For examplewecould haveJ =1/2fromeither asingles-electron or asinglep-electron.The former will give a much larger value ofAJ.The factorIJ is dimensionless and will havedierent values depending on theanglebetween I and J. Wecan useour Vector Model to nd theresult of a DPT calculation as follows:I and J couplei.e. add vectorally to givea resultantF:F =I +J (144)Fromthevector trianglewendF2=I2+J2+2IJ (145)IJ = 12_F2I2J2_(146)38AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureFIJFIJIJ FFigure22: Addition of nuclear spin I and total electronic angular momentumJ to giveF.Themagnitudes of thesquared angular momenta aregiven by their eigenvalues:E =AJ2 F(F +1) I(I +1) J(J +1) (147)As with spin-orbit coupling this leads to an interval rule:EF =E(F

) E(F

1) AJF

(148)(This provides oneway of nding F and so if weknowJ wecan nd I, thenuclear spin.)Theinterval rulecanbemessedupif thereareadditional contributionstothenuclear spinmomentsuch as an electric quadruplemoment.Thenumber of levels into which thehfs interaction splits a level depends on thenumber of valuesof F. This, in turn, depends on thecouplingof I and J. Our vector model gives us a trianglerulewhereby thevector addition must yield a quantized valueof thetotal angular momentumF.So if J> I then thereare2I +1 values of F, and if I> J thereare2J +1 values.Thevalues of F will beintegers in therange: [I J[ F I +J. Theordering of thelevels i.e.whetherE(F) > E(F 1) or vice-versa depends on thesign of AJ. Thereis, in general, no way tocalculate this easily. It depends on nuclear structure, which determines the sign ofgI, and on thedirection of Bel relativeto J. (for singleunpaired s-electron Bel is always antiparallel to J).IThenuclear spin I can befound by examining, with high resolution, thestructureof a spectral linedue to a transition for a level with no, or unresolved, hfs to a hfs-split level. The selection rules ofsuch transitions turn out to be:F =0, 1 but not 00 (149)Therearethen threemethods that can beused, provided thereis sucient spectral resolution.The level separation, found fromthe frequency intervals between components of thehfs, is proportional to theF-valueof thehigher level. This allows us to nd F and then, if weknowJ for thelevel, wecan nd I.Thenumber of spectral components will givethenumber of levels of hfs.Thenumber of levels is (2I +1) for I< J and (2J +1) for I> J. (This oneworks provided weknowJ> I)39AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureThe relative intensity of the spectral components will be proportional to thestatistical weight of thehfs levels (2F +1). So if J is known wecan nd I.Usually, in practice, we may need to use evidence from more than one of these methods to besure.Conventional optical methods are often at or beyond their limit in resolving hfs.There isthe additional problemof Doppler broadening which is often larger or at least comparable to thesplitting. Laser techniques, radio frequency methods or a combination of both areused nowadays.Theenergy levels determined basically by theCentral Field aregiven by:En Z2e4mr22n2(150)Where mr is the reduced mass of the electron nucleus system. For an innitely massive nucleusmr me, theelectronmass. For real atoms, however, weneedtodeal withthemotionof thenucleusaround the common centre-of-mass. The resulting nuclear kinetic energy isp2n2mnwhere pn, mn arethe momentumand mass respectively of the nucleus. For a two-body systemwe can treat this bysubstitutingthereduced mass, mr, for me in theSchrodinger equation. Thechangein energy is thenfound simply by making thesamesubstitution in theenergy eigenvalues.Theresult (and you should beableto work this out) is to shift theenergy level up byE(mrme)/me.In the case of a two-electron systemwe will need to know explicitly the electron wavefunctions.Theenergy changeEmass is given by:Emass =p2n2mn=(p1 +p2)22mn(151)=p21 +p22 +2p1 p22mn(152)Puttingthep21, p22 terms in Schrodingers equation gives a reduced-mass-typeshift. Thep1 p2 term,however, needs tobecalculated explicitly, usually usingPerturbation Theory. Theeect of this termisknowasthespecicmassshift. Theshiftsaredetectedby changesinthefrequency of transitionssotheeectsneedtobecalculatedseparately for eachlevel involved. What will beobservedisasmalldierence in the spectral line positions for dierent isotopes. The change in mass has a decreasingrelative eect as the nuclear masses get heavier. Mass eects are therefore more important in lightnucleii.The eect of the nite size of the proton in the energy levels in hydrogen can becalculated using perturbation theory. Theenergy shift isE =_0(r)V (r)4r2dr (153)V is the change in the potential energy resulting fromthe dierence between a point charge andsome spherical distribution, either over the surface or volume of a sphere. This dierence is onlysignicant over a very small range ofr compared to a Bohr radius a0.So the wavefunctions areessentially constant over therangeof interest and so can comeoutsidetheintegral.E [(0)[2_0v(r)4rrdr (154)40AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureOnly thes-states aresignicantly aected so using[(0)[2= 1_Zna0_3(155)Theresult can bederived. For thechargeon thesurfaceof thenucleus of radius r0 wend:Ens =Z4e2r2060a30n3(156)For theground stateof hydrogen thefractional shift isE1sE1s=43_r0a0_2 51010(157)This is a small eect but important sinceuncertainty in knowing thesizeof a proton aects preciseexperiments to study QED eects.41AtomicPhysics,P.Ewart 7 SelectionRulesNow that we have a better picture of atomic energy levels and have begun to see how transitionsbetween themare so important for our study of them, it is time to revisit the subject of selectionrules. Thesearetherules that tell us whether or not an atomcan changefromoneparticular stateto another. We will consider only electric dipole transitions; the rules for magnetic dipole, electricquadrupleetc aredierent and arenot on thesyllabus(thankfully). Oneway toapproach transitionsis to usetime-dependent Perturbation Theory and its result in Fermis Golden Rule. That also isnton our syllabus but wewant to usesomeof theunderlying physics.In therst lecturewelooked at thewavefunctionsfor twostates1 and 2 involved in atransition.Physically wehadtogenerateanoscillatingdipole anacceleratedcharge inorder tocreateelectro-magneticradiation. Thesolutionsof thetime-dependent Schrodinger equation, TDSE, havetheformi =(r, , )eiEit/(158)So theperturbation matrix elements are:i[ er [j) ei(EiEj)t/(159)Thediagonal elements i =j represent stationary states and haveno oscillation term, so atoms instationary states do not absorb or emit light. As we saw in Lecture 1 the o-diagonal elements dogivean oscillation of frequency 12 =[E1E2[/.If the spatial part of the matrix element is zero, however, we will not get any radiation i.e. thetransition is forbidden. Thespatial integral will bedetermined by thequantum, numbers in thetwostates 1(n1, l1, etc) and 2(n1, l1, etc). It is thechanges in thesequantumnumbers that giveus ourselection rules. Weneedn1, l1, n2, l2...L, S, J, MJ[

ierin

1, l

1, n

2, l

2...L

, S

, J

, M

J_,=0 (160)The o-diagonal matrix elements represent the atomin a super-position of two eigenstates 1 and2. Any linear combination of solutions of theTDSE will also bean eigenstate[NB: this is not truefor the time-independent Schrodinger equation].Now the spatial wavefuctions all have a deniteparity, either even, (+) or odd, (-). The dipole operator er has odd parity i.e. changing r to rchanges thesign.If weconsider, for themoment, a singleelectron making thejump fromonestateto another thenthecontribution to theintegral fromoneparticular location (x, y, z) is:enl(x, y, z)_ix +jy +kz_(161)If theparity of thetwo states is thesame, then theproduct will bethesameat theopposite point(x, y, z), The operator, however, will have the opposite sign and so when we integrate over allspacetheresult is zero. Therefore, to get a non-zero dipolematrix element, theparity must change.Theparity, as wenoted before, is given by (1)l, wherel is theelectrons orbital angular momentumquantumnumber. So wehaveour rst selection rule:l =1 (162)Thismakesphysical sensebecauseaphoton hasangular momentumof oneunit of . Soconservationof angular momentumdemands l =1, for oneelectron to jump.42AtomicPhysics,P.Ewart 7 SelectionRulesCould morethan oneelectron jump? Consider an atomwith two electrons atr1 and r2 with initialand nal congurations 1s2p (n1l1n2l2) and 3p3d (n3l3n4l4). thematrix element is then1(1s)2(2p)[ r1 +r2[1(3p)2(3d)) (163)= 1(1s)[ r1[1(3p)) 2(2p)[2(3d)) +2(2p)[ r2[2(3d)) 1(1s)[1(3p)) (164)=0 (165)owing to theorthogonality of theeigenfunctions i.e. i(nl)[i(n

l

)) =0So this means theconguration can changeby only a singleelectron jump.Theangular momentumdoes not depend at all on n, so n can changeby anything.Our basic rules aretherefore:n =anything (166)l = 1 (167)And thephoton carries oneunit of angular momentum.Thebasic rulesapply tothetotal angular momentumJ. Soprovided thechangesin J in atransitionallowfor oneunit ofto betaken by thephoton wecan makea vector triangleruleto giveJ =0, 1 but not 00 (168)J1 J1J J2 1=J J2 1=hhFigure23: Selection rules reect conservation of angular momentum including for the emit-ted/ absorbed photon.Similar arguments apply to changes in L i.e.L =0, 1 but not 00 (169)TheruleforS, wehavealready notedS =0 (170)because the dipole operator er does not operate on the spin part of the wavefunction. Physicallythis is thesensiblenotion that theelectron spin plays no part in thespatial oscillation.Finally wenoted that changes in MJ aregoverned by theruleMJ =0, 1 (171)Thereis a peculiar rule, with no obvious physical interpretation, that MJ cannot changefrom00if J = 0.These rules will be apparent only in the presence of an external eld that raises thedegeneracy in MJ. Weconsider theeects of external magnetic elds in thenext section.43AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsAtoms havepermanent magnetic dipolemoments and so will experiencea forcein an external mag-netic eld. Theeect of magnetic elds on atoms can beobserved in astrophysics, eg. in theregionsof sunspots, and have some very important applications in basic science and technology. Magneticelds are used to trap atoms for cooling to within a few nanoKelvin of absolute zero; they play animportant role in the operation of atomic clocks and their action on the nuclear spin is the basisof magnetic resonance imaging for medical diagnostics. The basic physics is by now familiar; themagnetic moment of theatom, (atom), will precess around theaxis of an applied eld, Bext.Theprecessional energy is given by:Hmag =atom Bext(172)The rst question we have to ask is how big is this perturbation energy compared to that of theinternal interactions in the atom. We have a hierarchy of interactions in decreasing strength: Cen-tral Field, residual electrostatic, spin-orbit, hyperne. These are represented by the HamiltoniansH0, H1, H2, and H3 respectively. Thehyperneinteraction is weak compared to that with even themost modest laboratory magnets. So we neglect hfs for the present, but we shall return to it later.At the other end we are unlikely to compete with the Central Field energies of the order of 10eV.(Check this for yourself by estimating _ Hmag_ using B for the atoms magnetic moment and atypical valueof B availablein a laboratory).For thesamereason_ Hmag_ is usually less than residual electrostatic energies ( 1eV) sowethenhave to decide whether our external magnetic perturbation is bigger or smaller than the internalmagnetic, spin-orbit, energy.We can recognise two limiting cases determined by the strength ofthe external eld and the size of the spin-orbit splitting. We dene a weak eld as one for whichatom Bext is less than thene-structuresplittingenergy. Conversely, a strongeld is onewheretheexternal perturbation exceeds thespin-orbit energy.J ust to get the basic physics straight, and to get a feel for what happens, we consider rst thesimple case where there is no spin-orbit interaction. We select an atomwhose magnetic moment isdue only to orbital motion and nd the eect ofBext on the energy levels. Next we introduce thespin to theproblemand thecomplication of spin-orbit interaction. Then weconsider what happensin a strong eld and nd, pleasingly, that things get simpler again. Finally we look at weak andstrong eld eects on hypernestructure.An atom with no spin could be a two-electron atom in a singlet state e.g.Magnesium 3s3p1P1.When theeld and its perturbation areweak theatomic states will betheusual [n, l, s, L, S, J, MJ)states. In this case, however J =L and weneed only thequantumnumbers L and ML . Thephysicsis pictured well by thevector model for an angular momentumL and associated magnetic momentL =glBLRecall thatgLB represents thegyromagnetic ratio for an atomsized circulating charge.In a magnetic eld Bext along thez-axis L (and L) executes Larmor precession around theelddirection with energy:EZ= L Bext(173)= gLBLBext(174)Now LBext is the projection ofL onto Bext and in our vector model this is quantized by integervalues ML . HenceEZ =gLBBextML(175)44AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsBextLimLMLFigure24: Eect of external magnetic eld Bext on an atomwith no spin i.e. magnetic dipole dueonly to orbital motion.Thereare(2L+1) values of ML: 1 ML L correspondingto thequantized directions of L intheeld Bext. Thus each energy level for a given L is split into (2L +1) sub-levels separated byE

Z =BBext(176)i.e. independent of L; so all levels areequally split.Thesplitting will beobserved in transitions between levels. Theselection rules on J and MJ willapply to L and ML in this case(J =L) soML =0, 1 (177)If we look at our example level 3s3p1P1 by a transition to say 3s3d1D2 we see 3 states in the1P1level and 5 states in the1D2 level.21100-1-1-2MLD =-1 MLD =0 MLD =+1 ML(w- w (w Dw)O O O+ Dw)Figure25: Normal Zeemaneect givesthenormal Lorentz triplet duetoselectionrulesML =0, 1and splitting of all levels into equally spaced sub-levels ML.45AtomicPhysics,P.Ewart 8 AtomsinMagneticFields9 transitions areallowed but they form3 singlelines of frequency 0, 0Z whereZ =E

Z/ =BBext/ (178)This is theso-called normal Lorentz triplet. Thesplitting of thelineat0 into 3 components is theNormal Zeeman Eect. (This is why weused Z as thesubscript on EZ)It was explained classically, and fully, by Lorentz, long beforeQuantumMechanics.Since, again, theeld is weak thereis only a small perturbation so wecan usethezero-order wave-functions dened by thespin-orbit interaction, [n, L, S, J, MJ)The application of an external eld will, again, cause the orbital magnetic moment and angularmomentumL and L, toprecess. Thespin moment and angular momentumwill alsoprecess aroundBext. Theperturbation energy operator is then:H4 =gLBLBext +gSBSBext(179)The problem we now have is that L andS are actually precessing at a faster rate around theirresultantJ. The eect of this is that the operators LBext andSBext no longer correspond toobservables that areconstants of themotion.BextSLJ{{LBextsJJBextFigure26: Precession ofs andl around mutual resultant J results in projections ofL andS onthe eld axis (z-axis) not being constants of the motion. So ML andMS are not goodquantumnumbers.We will use our Vector Model to follow the physics and to help us nd the solution. From thediagramwe see that the precession ofS and L around the resultantJ causes the projections ofSand L on theeld axis (z-axis) to vary up and down. Thediagramsuggests that sinceJ remains ata constant angleto thez-axis perhaps wecould usethetotal magnetic momentJ (= L +S) tocalculatetheinteraction energy i.e._JBext_(180)Thereis a snag, however, to this cunningplan. Theproblemis that J is not parallel or antiparallelto J. Thereason is that theg-factors for orbit and spin aredierent; gL =1 and gS =2. ThusL =BL and S =2BS (181)Thevector triangleof L, S, and J is not thesameas thetriangleof L,S and J.46AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsAll is not lost, however, becausethevery fact that thespin-orbit precession is so fast means thatwe can nd an eective magnetic moment for the total angular moment that does have a constantprojection on theeld axis, Bext. WeresolvethevectorJ into a component along theJ directionand acomponent perpendicular toJ. AsJ precessesaround Bext theprojection of theperpendicularcomponent of J on z will averageto zero. Thecomponent alongtheJ axis is a constant, which wecall theeectivemagnetic moment, e i.e.e =gJBJ (182)SJLmsmLmTotalmeffZBextFigure27: The magnetic moment due to the total angular momentum J doesnt lie on the J-axissincethevector sumof spin and orbital magnetic moments J is not parallel to J owningto the g-factor for spin being 2 g-factor for orbital momentum. The precession ofs,laround J resultsin avariation of theprojection of J on thez-axis. An eectivemagneticmomente does, however, havea constant projection on thez-axis.gJ is called theLandeg-factor.Wecan nowproceed to nd theperturbation energy:EAZ =gJB_J Bext_(183)JBext is just theprojection of J on thez-axis, given by Jz.EAZ =gJBBext_Jz_(184)and using our [n, L, S, J, MJ) wavefunctions we write the expectation value of Jz as its eigenvalueMJ. HenceEAZ =gJBBextMJ(185)Landecalculated theformof gJ quantummechanically, but wecan useour Vector Model to get thesameresult.Our vector model replaces expectation values by timeaverages of quantities that areconstants ofthe motion. So we look at our vectors L,S and J to nd the eective constant projections on thez-axis. Weseefromthevector diagramthatL has a constant projection on theJ axisLJ|J| . This isa vector in theJ direction so thescalar|LJ||J|is multiplied by theunit vectorJ|J|, to give|LJ|J|J|2=LJ.47AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsSimilarly, theprojection of S on J is given by[SJ[J[J[2=SJ(186)The vectors LJ and SJ have constant time-averaged projections on the axis ofBext. So the total,time-averaged energy of interaction with theeld is:EAZ= gLBLJBext +gSBSJBext(187)= gLB[LJ[[J[2JBext +gSB[SJ[[J[2JBext(188)The cosine rule gives the values ofLJ and SJ in terms ofJ2,L2and S2. So with gL = 1 andgS =2EAZ =B_3J2L2+S22[J[2JzBext(189)Now, as per our Vector Model method, wereplacethevector (operators) by their magnitudes (eigen-values);EAZ = [3J(J +1) L(L +1) +S(S +1)]2J(J +1)BBextMJ(190)Comparing this with our previous expression wend:gJ = [3J(J +1) L(L +1) +S(S +1)]2J(J +1)(191)This is an important result (i.e. remember it and how to derive it; it is a derivation beloved ofFinals examiners.) Its real importance comes fromseeing thatgJ is the factor that determines thesplitting of theenergy levels.EAZ =gJBMJBext(192)gJ depends on L,S and J and so by measuring the splitting of energy levels we can determine thequantumnumbers L, S and J for thelevel.We will measure the splitting from the separation of spectral components of a transition in themagnetic eld. Theenergy level splitting will bedierent in dierent levels so thepattern observedwill not bethesimpleLorentz triplet. Consequently, this is known as theAnomalous Zeeman eect.Hencethesubscript is AZ on theenergy shift EAZ for spin-orbit coupled atoms in a B-eld.As an example of the Anomalous Zeeman eect we consider an atomwith spin-orbit coupling andsodium is our favourite example.A transition between the 3p2P1/2,3/2 levels to the ground level3s2S1/2 illustrates themain features. Welook only at thepattern of lines in2P1/2 2S1/2 and leavethe2P3/2 2S1/2 as an exercise.First we need to knowgJ for each level. Putting in the values of L, S, J for2S1/2 level wendgJ(3s2S1/2) = 2. We could have guessed that! Since there is no orbital motion the magneticmoment is due entirely to spin and the g-factor for spin,gS = 2. Putting in the numbers for2P1/2we ndgJ(2P1/2) = 2/3. We can now draw the energy levels split into two sub-levels labelled byMJ =+1/2, 1/2.48AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsmj1/2-1/21/2-1/2D1D1s pp s3p P21/23s S21/2Figure28: Anomalous Zeeman eect in the 3s3p2P1/2 level of Na. Transitions are governed byselection rules MJ =0, 1.The transitions are governed by the selection rules: MJ = 0, 1, leading to a pattern of fourlines disposed about thezero-eld transition frequency, 0.Themagnetic eld introduces an axis of symmetry into an otherwiseperfectly spherically symmetricsystem. This has an eect on thepolarization of theradiation emitted or absorbed; theoscillatingorrotating dipolewill havea particular handedness in relation to theeld, or z-axis. Thetransitionmatrix element 1(r, , )[ er [2(r, , )) representsthechangein position of theelectron and thiscan be resolved into a motion along the z-axis: 1[ ez [2) and a circular motion in the x, y plane1[ e(x iy) [2). The-dependenceof theseintegrals can bewritten as:ez) _20ei(m1m2)d (193)e(x iy)) _20ei(m1m2)d (194)The z-component is zero unless [m1 m2[ = 0 i.e. m = 0. Similarly the (x, y) component is zerounless m1m21=0 i.e. m =1TheMJ =0 transition is thus identied with a linear motion parallel to theB eld. Viewed atright angles to theeld axis this will appear as linearly polarized light, -polarization.Viewed alongthez-axis, however, no oscillation will beapparent so the-component is missinginthis direction.Now consider the (x iy) component arising from m = 1 transitions. This appears as acircular motion in the x, y plane around the z-axis. Thus m = 1 transitions give circularlypolarized +, (right/ left) light viewed along z, and planepolarized in thex, y plane.It is possibleusingappropriateoptical devices to determinethehandedness of the-polarizedemissions along thez-axis. The+and transitions are either at higher or lower frequencyrespectively than0 depending on the sign of the oscillating charge. If you can remember theconventional rulesof electromagnetismyou can work out thesign of theelectron. Lorentz (whocouldremember theserules) was ableto usethis eect to showthat electrons werenegatively charged.49AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsZYXqfBext-rq1q2fqff21LightpolarizationsppD = 0f = fm1 2D = + 1f = fm1 2--r-rFigure29: Polarization of Zeeman components viewed along eld axis (z-axis) and along a perpen-dicular axis (x- ory-axis).Theword strong in this context means that theinteraction of theorbit and spin with theexternaleld is stronger than the interaction of the orbit with the spin. The magnetic moments L and Swill then precess much morerapidly around Bext than they doaround each other. SinceL and S arenowessentially moving independently around B, thevectorJ becomes undened it is no longer aconstant of the motion. Our spin-orbit coupled wavefunctions [n, L, S, J, MJ) are no longer a goodbasis for describing the atomic state;J,MJ are not good quantum numbers. Instead we can useLML andSMS to set up a basis set of functions [n, L, S, ML, MS).The spin-orbit interaction isnowrelatively small and weignoreit to this approximation. Theenergy shifts arenowgiven by theexpectation of theoperator H4:H4 =gLBL Bext +gSBS Bext(195)Nowwedont havethecomplication of coupled spin and orbital motion. Usingour [n, L, S, ML, MS)basis functions wend_ H4_ =EPB =(ML +2MS)BBext(196)Theenergy levels arethus split into (2L +1) levels labelled by ML , each of which will besplit intolevels labelled by MS.50AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsConsider our 3p2P1/2 level of Na.2P1/2mLmL10-11/21/21/2,-1/2-1/2-1/2Figure30: Thesplitting of a level in a strong eld ignoring theeect of spin-orbit interaction.Wecan nowincludethespin-orbit eect as a small perturbation on thesestates [n, L, S, ML, MS).Again, wecan useour Vector Model to nd theexpectation valueof thespin-orbit operator S L.BextLSLBextLSmLimSZY XLFigure31: Strong eld eect on orbital and spin angular momenta: l ands precess more rapidlyaround the external eld relative to their mutual precession. The projections ofl and son thex-y planeaverageto zero. As a result only theprojections on thez-axis remain todenetheenergy in theeld.Since L andS are precessing rapidly around the z-axis their components in the x, y-plane willaverageto zero. Weareleft only with thez-component so:_S L_ =MSML(197)Thesix states (ML =1, 0, 1,MS =+1/2, 1/2) havespin orbit shifts of (/2, 0, /2, /2, 0, /2)as shown on theextremeright of thestrong eld energy level diagram.Theenergy level shifts EPB, ignoringtherelatively small spin-orbit eect, lead to a set of evenlyspaced energy levels. Transitions between these strong eld levels are illustrated by the 3p2P1/2 3s2S1/2 transition in Na. Theselection rules operating areML =0, 1 and S =0.The ML rule derives fromthe conservation of angular momentum(including the photon). TheS = 0 rule is due, again, to the fact that the dipole operator cannot change the spin. The spinplays nopart, soweexpect toseeapattern of lines similar totransitions in an atomwith nospin i.e.theNormal Zeeman triplet. Theallowed transitions do indeed t this simplepattern. Thesplittingin thestrong eld is known as thePaschen-Back eect, henceour EPB.51AtomicPhysics,P.Ewart 8 AtomsinMagneticFields10-110-11/.21/21/2-1/2-1/2-1/2MSML0 1/20-1/2sps3/21/2-1/2-3/21/2-1/21/2-1/2MJDmL-1 0 +1D1s pp sD2s s pp s sD1D23/21/21/22P2SJH + H + B + Bo so weak strongQuantumNumbersFine structure Anomalous Zeeman Paschen-BackEffect EffectFigure32: Diagram showing splitting of Na 3p-levels from zero eld, through weak to strong eldshowing changefromAnomalous Zeeman to Paschen-Back eect.When the external eld is neither strong nor weak the situation is very complicated. The VectorModel wont work and the result can be calculated quantum mechanically in only the simplest ofcases eg.ground state of atomic hydrogen.The energy levels can be found experimentally.Thetransition from weak to strong elds for the2P1/2,3/2 term is shown