AtomicPhysicsP.EwartContents1 Introduction 12 RadiationandAtoms 12.1 Width and Shape of Spectral Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.1 Lifetime Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.2 Collision or Pressure Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1.3 Doppler Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Atomic Orders of Magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.1 Other important Atomic quantities. . . . . . . . . . . . . . . . . . . . . . . . . 52.3 The Central Field Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 The form of the Central Field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Finding the Central Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 TheCentral FieldApproximation 93.1 The Physics of the Wave Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 The Periodic Table. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.3 Gross Energy Level Structure of the Alkalis: Quantum Defect. . . . . . . . . . . . . . 154 CorrectionstotheCentral Field: Spin-Orbitinteraction 174.1 The Physics of Spin-Orbit Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Finding the Spin-Orbit Correction to the Energy . . . . . . . . . . . . . . . . . . . . . 194.2.1 The B-Field due to Orbital Motion. . . . . . . . . . . . . . . . . . . . . . . . . 194.2.2 The Energy Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2.3 The Radial Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2.4 The Angular Integral: Degenerate Perturbation Theory . . . . . . . . . . . . . 214.2.5 Degenerate Perturbation theory and the Vector Model . . . . . . . . . . . . . . 224.2.6 Evaluation of_ s l_ using DPT and the Vector Model . . . . . . . . . . . . . . 234.3 Spin Orbit Interaction: Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.4 Spin-Orbit Splitting: Alkali Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5 Spectroscopic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27i5 Two-electronAtoms: Residual ElectrostaticEectsandLS-Coupling 305.1 Magnesium: Gross Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 The Electrostatic Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.3 Symmetry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.4 Orbital eects on electrostatic interaction in LS-coupling . . . . . . . . . . . . . . . . . 335.5 Spin-Orbit Eects in 2-electron Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 NuclearEectsonAtomicStructure 376.1 Hyperne Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.2 The Magnetic Field of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.3 Coupling of I and J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.4 Finding the Nuclear Spin,I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.5 Isotope Eects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 SelectionRules 427.1 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.2 Conguration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.3 Angular Momentum Rules. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 AtomsinMagneticFields 448.1 Weak eld, no spin. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448.2 Weak Field with Spin and Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468.2.1 Anomalous Zeeman Pattern. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488.2.2 Polarization of the radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498.3 Strong elds, spin and orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.4 Intermediate elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528.5 Magnetic eld eects on hyperne structure . . . . . . . . . . . . . . . . . . . . . . . . 528.5.1 Weak eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538.5.2 Strong eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 X-Rays: transitionsinvolvinginnershell electrons 569.1 X-ray Spectra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569.2 X-ray series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579.3 Fine structure of X-ray spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.4 X-ray absorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.5 Auger Eect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5910HighResolutionLaserSpectroscopy 6110.1Absorption Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.2Laser Spectroscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.3Spectral resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.4Doppler Free spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.1 Crossed beam spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.2 Saturation Spectroscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.4.3 Two-photon-spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6410.5Calibration of Doppler-free Spectra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6510.6Comparison of Doppler-free Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 65iiAtomicPhysics,P.Ewart 1 Introduction1 IntroductionThestructureofatomsandtheirbehaviourisresponsiblefortheappearanceofthevisibleworld.The small scale of atoms and the properties of nuclei and electrons required a new kind of mechanicsto describe their behaviour. Quantum Mechanics was developed in order to explain such phenomenaas the spectra of light emitted or absorbed by atoms. So far you have studied the physics of hydrogenand helium as illustrations of how to apply Quantum Theory. There was a time, a few seconds aftertheBigBang, whentheUniverseconsistedonlyofhydrogenandheliumnucleii. Ittookanother300,000yearsforatoms, assuch, toform. Things, however, havemovedonandtheuniverseisnowamuchmoreinterestingplacewithheavierandmorecomplicatedatoms. OuraimisnowtounderstandAtomicPhysics, notjusttoillustratethemathematicsof QuantumMechanics. Thisisbothinterestingandimportant, forAtomicPhysicsisthefoundationforawiderangeofbasicscience and practical technology. The structure and properties of atoms are the basis of Chemistry,and hence of Biology. Atomic Physics underlies the study of Astrophysics and Solid State Physics. Ithas led to important applications in medicine, communications, lasers etc, as well as still providing atesting ground for Quantum Theory and its derivatives, Quantum Electrodynamics. We have learnedmost about atoms from the light absorbed or emitted when they change their internal state. So thatis a good place to begin.2 RadiationandAtomsWe will make extensive use of models in this course to help us get a feel for the physics. A favouritemodel fortheoristsisthetwo-level atomi.e. onewithonlytwoeigenstates1, 2withenergyeigenvaluesE1,E2 respectively (E2> E1). The wave functions have, in general, a time dependence.1= 1(x)eiE1t/(1)2= 2(x)eiE2t/(2)Whentheatomisperturbeditmaybedescribedbyawavefunctionthatisalinearcombinationof1and2: = a1 + b2giving the probability amplitude. We observe, however, a probabilitydensity: the modulus squared; [[2. This will have a termab12ei(E1E2)t/(3)This is a time oscillating electron density with a frequency12:ei(E1E2)t/= ei12t(4)SoE2E1 = 12(5)This is illustrated schematically in gure 1.1AtomicPhysics,P.Ewart 2 RadiationandAtomsy1 y2y( ) = + t y y1 2IY( ) t I2Y( ) t Y( t) t+Oscillatingchargecloud:ElectricdipoleI I Y( + t) t2Figure 1: Evolution of the wavefunction of a system with time.So the perturbation produces a charge cloud that oscillates in space an oscillating dipole. Thisradiates dipole radiation. Whether or not we get a charge displacement or dipole will depend on thesymmetry properties of the two states1, and2. The rules that tell us if a dipole with be set upare called selection rules, a topic to which we will return later in the course.2.1 WidthandShapeofSpectral LinesThe radiation emitted (or absorbed) by our oscillating atomic dipole is not exactly monochromatic,i.e. therewill bearangeoffrequencyvaluesfor12. Thespectral lineobservedisbroadenedbyone, ormore, processes. AprocessthataectsalltheatomsinthesamewayiscalledHomoge-neous Broadening. A process that aects dierent individual atoms dierently is InhomogeneousBroadening.Examplesofhomogeneousbroadeningarelifetime(ornatural)broadeningorcollision(orpres-sure) broadening. Examples of inhomogeneous broadening are Doppler broadening and crystal eldbroadening.2.1.1 LifetimeBroadeningThis eect may be viewed as a consequence of the uncertainty principleEt (6)Since E = , E = and if the time uncertainty t is the natural lifetime of the excited atomicstate,, we get a spread in frequency of the emitted radiation 1 or 1(7)2AtomicPhysics,P.Ewart 2 RadiationandAtomsThe lifetime,, is a statistical parameter related to the time taken for the population of the excitedstatetodecayto1/eof itsinitial value. Thisexponential decayisreectedintheexperimentaldecayoftheamplitudeE(t)ofthelightwaveemitted. Thefrequency(orpower)spectrumofanexponentially decaying amplitude is a Lorentzian shape for the intensity as a function of frequencyI() 1( 0)2+ (1/)2(8)The full width at half-maximum, FWHM, is then 2_1_(9)A typical lifetime 108sec.N( ) t I( ) wTime,t frequency, wIntensityspectrum NumberofexcitedatomsExponentialdecay LorentzianlineshapeFigure 2: Decayof excitedstatepopulationN(t) leads tosimilar exponential decayof radiationamplitude, giving a Lorentzian spectrum.2.1.2 CollisionorPressureBroadeningAcollisionwithanotheratomwhiletheatomisradiating(oscillating)disruptsthephaseof thewave. Thewaveis thereforecomposedof various lengths of uninterruptedwaves. Thenumberofuninterruptedwavesdecaysexponentiallywitha1/etimec, whichisthemeantimebetweencollisions. Atatmosphericpressurethisistypically 1010sec. Theexponential decayof thecoherent oscillations again leads to a Lorentzian lineshape.N( ) t I( ) wTime,t frequency, wIntensityspectrum NumberofuncollidedatomsExponentialdecay LorentzianlineshapeFigure 3: Decayinnumber of undisturbedatoms radiatingleads todecayinamplitudeof waveundisturbedbyaphasechangingcollision. TheassociatedfrequencyspectrumisagainLorentzian.3AtomicPhysics,P.Ewart 2 RadiationandAtoms2.1.3 DopplerBroadeningAtoms in a gas have a spread of speeds given by the Maxwell-Boltzmann distribution. The Dopplershift of the light emitted is therefore dierent for the atoms moving at dierent speeds. There is thena spread of Doppler shifted frequencies leading to a broadening of the spectral line. Since dierentatoms are aected dierently this Doppler Broadening is Inhomogeneous broadening. The Dopplershift is given by: = 0_1 vc_(10)The spread, or width of the line, is thereforeD 0vc(11)Since0 1015rad s1andv 102ms1we ndD 2109rad s1(12) 109Hz (13)N(v) I( ) watomicspeed,v frequency, wDopplerbroadening DistributionofatomicspeedMaxwell-BoltzmanndistributionGaussianlineshapeFigure 4: Maxwell-Boltzmann distribution of speeds and the associated Doppler broadening giving aGaussian lineshape.2.2 AtomicOrdersofMagnitudeWe have already started getting a feel for typical values of parameters associated with spectral lines,so now is a good time to do the same for some other aspects of atoms. It is important to have someideaoftheordersofmagnitudeassociatedwithatomsandtheirstructure. Wewillbelookingatatoms more complex than Hydrogen and Helium for which we cant solve the Schrodinger Equationexactly. We have to do the best we can with approximate solutions using Perturbation Theory. Soweneedtoknowwhenperturbationtheorywillgiveareasonableanswer. Wewillbeparticularlyinterested in the energy level structure of atoms so lets start there.First a word about units. Atoms are small 1010m and their internal energies are small; 1019Joule. So a Joule is an inconvenient unit. We will use electron Volts, eV, which is 2 1019J. Wehave seen that atoms emit or absorb light in the UV or visible range say 500 nm. This must thenbe of the order of the energy changes or binding energy of the outer, most loosely bound electron.Now a wavelength 500 nm corresponds to a frequency, 61014Hz. So fromE = hwendE 40 10341014 4 1019J 2eV.4AtomicPhysics,P.Ewart 2 RadiationandAtomsWe could also check this using the size of an atom ( 1010m) using the Uncertainty Principle.p x (14)p p

x(15)E =p22m_

x_2/(2m) a few eV (16)We can compare 2 eV with thermal energy or kTi.e. the mean Kinetic energy available fromheat,140eV. This is not enough to excite atoms by collisions so atoms will mostly be in their groundstate.2.2.1 OtherimportantAtomicquantitiesAtomicsize: Bohrradiusa0 =40

2me2= 0.53 1010m (17)IonisationEnergyofHydrogenEH =me4(40)222= 13.6 eV (18)RydbergconstantR =EHhc= 1.097 107m1(19)R is useful is relating wavelengths to energies of transitions since wavenumber =1in units ofm1.Finestructureconstant =e240c 1137(20)This is a dimensionless constant that gives a measure of the relative strength of the electromagneticforce. It is actually also the ratio of the speedveof the electron in the ground state of H toc, thespeed of light. = ve/c and so it is a measure of when relativistic eects become important.BohrmagnetonB =e2m 9.27 1024JT1(21)Thisisthebasicunitofmagneticmomentcorrespondingtoanelectroninacircularorbitwithangular momentum , or one quantum of angular momentum.As well as having orbital angular momentum the electron also has intrinsic spin and spin magneticmomentS= 2B. A proton also has spin but because its mass is 2000 larger than an electronits magnetic moment is 2000 times smaller. Nuclear moments are in general 2000 times smallerthan electron moments.5AtomicPhysics,P.Ewart 2 RadiationandAtoms2.3 TheCentral FieldApproximationWe can solve the problem of 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 weaddanextraplanetthenthingsgetdicult. Ifweaddanymorewehaveamanybodyproblemwhich is impossible to solve exactly. Similarly, for a many electron atom we are in serious diculty we will need to make some approximation to simplify the problem. We know how to do Hydrogen;we solve the Schrodinger equation:

22m2 Ze240r = E (22)We can nd zero order solutions wave functions that we can use to calculate smaller perturbationse.g. spin-orbit interaction.The Hamiltonian for a many electron atom however, is much more complicated.H =N

i=1_ 22m2i Ze240ri_+

i>je240rij(23)Weignore, fornow, otherinteractionslikespin-orbit. Wehaveenoughonourplate! Thesecondterm on the r.h.s. is the mutual electrostatic repulsion of the N 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.We recall that the hydrogen problem was solved using the symmetry of the central Coulomb eld the 1/r potential. This allowed us to separate the radial and angular solutions. In the many electroncase,formostofthetime,amajorpartoftherepulsionbetweenoneelectronandtheothersactstowards the centre. So we replace the 1/r, hydrogen-like, potential with an eective potential due tothe nucleus and the centrally acting part of the 1/rijrepulsion term. We call this the Central FieldU(r). Note it will not be a 1/r potential. We now write the HamiltonianH =H0 +H1(24)whereH0=

i_ 22m2i+U(ri)_(25)andH1=

i>je240rij

i_Ze240ri+U(ri)_(26)H1is the residual electrostatic interaction. Our approximation is now to assumeH1

H2butthiswontbetrueforallelements. Wecanalsoaddsmallerperturbations,H3etcdueto, forexample, interactionswithexternal elds, oreectsof thenucleus(otherthanitsCoulombattraction). TheCentral FieldApproximation allows us to nd solutions of the Schrodinger equation in terms of wave functions ofthe individual electrons:(n, l, ml, ms) (30)Thezero-orderHamiltonianH0duetotheCentralFieldwilldeterminethegrossstructureoftheenergy levels specied by n, l. The perturbationH1, residual electrostatic, will split the energy levelsinto dierent terms. The spin orbit interaction,H2 further splits the terms, leading to ne structureof the energy levels. Nuclear eects lead to hyperne structuresof the levels.Within the approximation we have made,so far,the quantum numbersml, msdo not aect theenergy.Theenergylevelsarethereforedegeneratewithrespecttoml, ms. Thevaluesof ml, msorany similar magnetic quantum number, specify the stateof the atom.Thereare2l + 1valuesof mli.e. 2l + 1statesandtheenergylevel issaidtobe(2l + 1)-folddegenerate. Theonlydierencebetweenstatesof dierent ml(or ms) isthat theaxisof theirangular momentum points in a dierent direction in space. We arbitrarily chose somez-axis so thatthe projection on this axis of the orbital angular momentumlwould have integer number of units(quanta) of . (ms does the same for the projection of the spin angular momentums on thez-axis).(Atomic physicists are often a bit casual in their use of language and sometimes use the words en-ergy level, and state (e.g. Excited state or ground state) interchangeably. This practice is regrettablebut usually no harm is done and it is unlikely to change anytime soon.)3.1 ThePhysicsoftheWaveFunctionsYou will know (or you should know!) how to nd the form of the wave functions(n, l, ml, ms) inthe case of atomic hydrogen. In this course we want to understand the physics the maths is done inthe text books. (You may like to remind yourself of the maths after looking at the physics presentedhere!)Before we look at many electron atoms, we remind ourselves of the results for hydrogen.3.1.1 EnergyThe energy eigenvalues, giving the quantized energy levels are given by:En=_n,l,ml H n,l,ml_(31)= Z2me4(40)222n2(32)Note that the energy depends only on n, the Principal quantum number. The energy does not dependonlthisistrueONLYFORHYDROGEN!Theenergylevelsaredegenerateinl. Werepresentthe energy level structure by a (Grotrian) diagram.9AtomicPhysics,P.Ewart 3 TheCentralFieldApproximation1234nEnergy-13.6eV0l = 012spdFigure 7: Energy level structure of hydrogen, illustrating how the bound state energy depends onnbut notl.For historical reasons, the states with l = 0, 1, 2 are labelled s, p, d. For l = 3 and above the labelsare alphabetical f, g, h etc.3.1.2 AngularMomentumThe wavefunction has radial and angular dependence. Since these vary independently we can write:n,l,ml(r, , ) = Rn,l(r)Ymll(, ) (33)Rn,l(r) andYmllare radial and angular functions normalized as follows:_R2n,l(r)r2dr = 1_ Ymll(, )2d = 1 (34)The angular functions are eigenfunctions of the two operators l2and lz:l2Ymll(, ) = l(l + 1)2Ymll(, ) (35)lzYmll(, ) = mlYmll(, ) (36)Wherel(l + 1) andmlare the eigenvalues of l2and lzrespectively, such thatl = 0, 1, 2...(n 1) l ml l (37)Thespinstatesof theelectroncanbeincludedinthewavefunctionbymultiplyingbyaspinfunction s which is an eigenfunction both of s2and sz with eigenvalues s(s+1) and ms respectively.(s = 1/2 andms = 1/2)Roughlyspeaking, theangularfunctionsspecifytheshapeof theelectrondistributionandtheradial functions specify the size of the orbits.l =0, s-states, arenon-zeroattheorigin(r =0)andaresphericallysymmetric. Classicallythey correspond to a highly elliptical orbit the electron motion is almost entirely radial. Quantummechanically we can visualise a spherical cloud expanding and contracting breathing, as the electronmoves in space.10AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationl 1. Asl increases, the orbit becomes less and less elliptical until for the highestl = (n 1) theorbit is circular. An important case (i.e. worth remembering) isl = 1 (ml = 1, 0)Y11= _38_1/2sin ei(38)Y11= +_38_1/2sin ei(39)Y01=_34_1/2cos (40)|Y10( )| q,f2(a)|Y1+( )| q,f2(b)Figure 8: The angular functions, spherical harmonics, giving the angular distribution of the electronprobability density.If therewasanelectronineachof thesethreestatestheactual shapeschangeonlyslightlytogive a spherically symmetric cloud. Actually we could t two electrons in eachl state provided theyhad oppositespins. The sixelectronsthen llthesub-shell(l = 1)Ingeneral,lledsub-shellsarespherically symmetric and set up, to a good approximation, a central eld.As notedabovetheradial functions determinethesizei.e. wheretheelectronprobabilityismaximum.11AtomicPhysics,P.Ewart 3 TheCentralFieldApproximation3.1.3 Radial wavefunctions24624681024681020Zra /oZra /oZra /oGroundstate, n=1, =0 l 1stexcitedstate, n=2, =0 l2ndexcitedstate, n=3, =0 ln=3, =2 lN=2, =1 ln=3, =1 l2 1.01 0.50.4Figure 9: Radial functions giving the radial distribution of the probability amplitude.NB: Main features to remember! l = 0, s-states do not vanish atr = 0. l ,= 0, states vanish atr = 0 and have their maximum probability amplitude further out withincreasingl.The size, position of peak probability, scales with n2.Thel = 0 function crosses the axis (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)3.1.4 ParityTheparityoperatorisrelatedtothesymmetryof thewavefunction. Theparityoperatortakesr r. It is like mirror reection through the origin. and + (41)Ifthisoperationleaves the signof unchangedthe parityis even. If changessigntheparityisodd. Some states are not eigenstates of the parity operator i.e. they do not have a denite parity.The parity of two states is an important factor in determining whether or not a transition betweenthem is allowed by emitting or absorbing a photon. For a dipole transition the parity must change.12AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationIn central elds the parity is given by (1)l. Sol = 0, 2 even states, s, d etc. are even parity andl = 1, 3,oddstatesareoddparity. Hencedipoletransitionsareallowedforsporpd,butnotallowed for ss or sd.3.2 Multi-electronatoms3.2.1 ElectronCongurationsBefore considering details about multi-electron atoms we make two observations on the Central FieldApproximation based on the exact solutions for hydrogen.Firstly, for a central eld the angular wave functions will be essentially the same as for hydrogen.Secondly, theradial functionswill havesimilarpropertiestohydrogenfunctions; specicallytheywill have the same number of nodes.The energy level-structure for a multi-electron atom is governed by two important principles:a) Paulis Exclusion Principle andb) Principle of Least EnergyPauli forbidsall theelectronsgoingintothelowestorgroundenergyeld. TheLeastEnergyprinciple requires that the lowest energy levels will be lled rst. Then = 1, l = 0 level can take amaximum of two electrons. The least energy principle then means the next electron must go into thenext lowest energy level i.e. n = 2. Given thatlcan take values up ton = 1 and eachl-value canhave two values of spin, s,there are 2n2vacancies for electrons for each value ofn. Our hydrogensolutions show that higher n values lead to electrons having a most probable position at larger valuesofr. These considerations lead to the concept of shells, each labelled by their value ofn.Now look at the distribution of the radial probability for the hydrogenic wave functions Rn,l(r). Thelow angular momentum states, especially s-states have the electrons spending some time inside theorbits of lower shell electrons. At these close distances they are no longer screened from the nucleusZ protons and so they will be more tightly bound than their equivalent state in hydrogen. The energylevel will depend on the degree to which the electron penetrates the core of inner electrons, and thisdepends onl. Therefore the energy levels are no longer degenerate inl. To this approximationmlandmsdonotaecttheenergy. Wecanthereforespecifytheenergyofthewholesystembytheenergyoftheelectronsdeterminedbyquantumnumbersnandl. Specifyingn, lforeachelectrongives the electron congurations: nlxwherex indicates the number of electrons with a givennl.3.2.2 ThePeriodicTableAs atomic science developed the elements were grouped together according to perceived similaritiesin their physical or chemical properties. Several dierent periodic tables were proposed. The oneweall learnedinourchemistryclasseshasbeenadoptedsinceithasabasisinthefundamentalstructure of the atoms of each element. The physical and chemical properties, e.g. gas, solid, liquid,reactivity etc are seen to be consequences of atomic structure. In particular they arise largely fromthe behaviour of the outer shell or most loosely bound electrons.UsingthePauliPrincipleandtheLeastenergyPrinciplewecanconstructthecongurationsofthe elements 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 little more complicated. We expect the next electron to go into the 3dsub-shell. As we have seen, however, a 3d electron is very like a 3d electron in hydrogen: it spendsmost of its time in a circular orbit outside the inner shell electrons. An electron in a 4s state howevergoes relatively close to the nucleus, inside the core, and so ends up more tightly bound - i.e. lowerenergy, than the 3d electron. So the next element, Potassium, has the congurationK: 1s22s22p63s23p64sCa: 4s2The 3d shell now begins to llSc: 1s22s22p63s23p63d 4s2The 3d and 4s energies are now very similar and at Chromium a 3d electron takes precedence overa 4s electronVa: 3s23p63d34s2Cr: 3d54sMn: 3d54s2As the 3d shell lls up, the successive elements, the transition elements, have interesting propertiesas a result of the partially lled outer shell but this isnt on the syllabus!Thereare, however, twofeaturesoftheperiodictablethatareworthnotingasconsequencesofthe Central Field Approximation.Raregases Theseelementsarechemicallyinertandhavehighionizationpotentials(theenergyneeded to pull o a single electron). This is not because, as is sometimes (often) stated, that theyhave closed shells. They dont all have closed shells i.e. all states for eachn are full. They all dohave lled s and p sub-shellsHe: 1s2Kr: 4s24p6Ne: 1s22p6Xe: 5s25p6A: 1s22p63s23p6Rn: 6s26p6As we noted earlier this leads to a spherically symmetric charge distribution. Since electrons areindistinguishable all the electrons take on a common wave function. The point is that this results ina higher binding energy for each one of the electrons. So it is harder for them to lose an electron ina chemical bond they are chemically inert and have high ionization energies.14AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationAlkalis Thesearethenextelementstotheraregasesandhaveoneelectronoutsidethefull(sp)sub-shells. This outer, or valence, electron therefore moves in a hydrogen like central potential. Theelectron is generallywell-screened by(Z 1)innerelectrons fromthenucleus and iseasilylosttoa chemical bond (ionic or co-valent). They are chemically reactive and have low ionization energieswhich dont change much from one alkali to another.3.3 GrossEnergyLevel StructureoftheAlkalis: QuantumDefectAs 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. Asanexampleweconsidersodium. Thegroundstage(lowestenergylevel)hastheconguration:(closed shells) 3s.Weknowthe3spenetratesthecore(innershells)andisthereforemoretightlyboundlowerenergythana3selectronwouldbeinHydrogen. Whentheelectronisexcited, sayto3p, itpenetrates the core much less and in a 3d state its orbit is virtually circular and very close the n = 3level of Hydrogen.The higher excited statesn > 3 will follow a similar pattern. The thing to notice and this turnsout to be experimentally useful is that the degree of core penetration depends onl and very littleonn. As a result the deviation from the hydrogenic energy level is almost constant for a givenl asn increases.The hydrogen energy levels can be expressed asEn = Rn2(42)For alkalis, and to some degree for other atoms too, the excited state energies may be expressed asEn = R(n)2(43)Where n is an eective quantum number. n diers from the equivalent n-value in hydrogen by (l)i.e. n = n (l).(l)isthequantumdefectanddependslargelyonlonly. Itisfoundempirically, anditcanbeshowntheoretically, tobeindependentof n. Thusall s-stateswill havethesamequantumdefect(s); all p-states will have the same(p), etc and(s) > (p) > (d).For Na:(s) 1.34(p) 0.88(d) 0Forheavieralkalis, the(l)generallyincreasesasthecoreislessandlesshydrogen-like. Theionization potential however is almost constant as noted previously. Although the increase in Z leadsto stronger binding than in hydrogen the ground state electron starts in a highern, and this almostexactly o-sets the increased attraction of the heavier nucleus.Everyone knows that the discrete wavelengths of light emitted, or absorbed, by an atom are discretebecause they arise from transitions between the discrete energy levels. But how do we know whichenergy levels?This is where the quantum eect is very useful.We also need to remember that a transition involves a change inl of 1 (l = 1). A transitionfrom a lower state will therefore take the atom to an energy level with angular momentum diering15AtomicPhysics,P.Ewart 3 TheCentralFieldApproximationby 1fromtheinitial level. Forsimplicity, sothatthegeneral ideabecomesclear, weconsidertransitions from the ground state. If the ground state (level) has l = 0 then there will then be only asingle set of transitions to levels with angular momentum one unit larger i.e. l = 1. The spectrum ofabsorption lines will consist of a series of lines of increasing wavenumber (energy) corresponding totransitions to excited states with increasing principal quantum numbern. Since these excited levelsall have the same value of l they should have the same value of quantum defect. So if we can identifylines with the same quantum defect they will belong to the same series i.e. have the same angularmomentum quantum numberl. The procedure is then as follows.(1) Measure the wavelength of each absorption line in the series,n.(2) Calculate the corresponding wavenumbern = 1n(3) Estimate the value of the series limit . This corresponds to the ionization potential or Termvalue for the ground state,Tno.(4) Find the Term value, Tn, or binding energy of an excited level from the dierence between andn:Tni = Tno n(5) The Term value is given by the hydrogenic formula:Tni =R(ni)2whereniis the eective quantum number given by:ni= n (l)and(l) is the quantum defect.(6) Finally we plot(l) againstTni. This should be a straight line indicating a constant value of(l). Deviation from a straight line indicates that we have not estimated (i.e. Tno) correctly.(7) The procedure is then to try a dierent value of Tno until we obtain a straight line. This processis ideally suited to computation by a program that minimizes the deviation from a straight line forvarious values ofTno.Oncewehavefoundthemostaccuratevalueof Tnofromthespectrumwecanputtheenergylevels on an absolute energy scale. We have then determined the energy level structure of the atomfor this set of angular momentum states. The procedure can be carried out using data from emissionspectra. In this case a range of dierent series will be be provided from the spectrum. It remains tond the quantum defects for each series of lines and assign values ofl to each series of levels. Fromthe discussion above the quantum defects will decrease for increasingl.Figure 10: Plot of quantum defect(l) against Term value, showing eect of choosingTnoeither toolarge or too small. The correct value ofTnoyields a horizontal plot.16AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-Orbitinteraction4 CorrectionstotheCentral Field: Spin-OrbitinteractionThe Central Field Approximation gives us a zero-order HamiltonianH0that allows us to solve theSchrodingerequationandthusndasetofzero-orderwavefunctionsi. Thehopeisthatwecantreat the residual electrostatic interaction (i.e. the non-central bit of the electron-electron repulsion)as a small perturbation,H1 . The change to the energy would be found using the functionsi.The residual electrostatic interaction however isnt the only perturbation around. Magnetic inter-actions arise when there are moving charges. Specically we need to consider the magnetic interactionbetween the magnetic moment due to the electron spin and the magnetic eld arising from the elec-tron orbit. This eld is due to the motion of the electron in the electric eld of the nucleus and theotherelectrons. Thisspin-orbitinteractionhasanenergydescribedbytheperturbationH2. Thequestion is: which is the greater perturbation,H1 orH2?We may be tempted to assumeH1>H2since electrostatic forces are usually much stronger thanmagneticones. HoweverbysettingupaCentralFieldwehavealreadydealtwiththemajorpartoftheelectrostaticinteraction. Theremainingbitmaynotbelargerthanthemagneticspin-orbitinteraction. In many atoms the residual electrostatic interaction,H1, does indeed dominate the spin-orbit. There is, however, a set of atoms where the residual electrostatic repulsion is eectively zero;the alkali atoms. In the alkalis we have only one electron orbiting outside a spherically symmetriccore. Thecentraleldis, inthiscase, anexcellentapproximation. Thespin-orbitinteraction,H2will be the largest perturbation provided there are no external elds present. So we will take thealkalis e.g. Sodium, as a suitable case for treatment of spin-orbit eects in atoms. You have alreadymet the spin-orbit eect in atomic hydrogen, so you will be familiar with the quantum mechanics forcalculating the splitting of the energy levels. There are, however, some important dierences in thecase of more complex atoms. In any case,we are interested in understanding the physics,not justdoing the maths of simple systems. In what follows we shall rst outline the physics of the electronsspin magnetic moment interacting with the magnetic eld, B, due to its motion in the central eld(nucleus plus inner shell electrons). The interaction energy is found to beBso the perturbationto the energy, E, will be the expectation value of the corresponding operator B_.WethenuseperturbationtheorytondE. Wewill not, however, beablesimplytouseourzero order wavefunctions0(n, l, ml, ms) derived from our Central Field Approximation, since theyaredegenerateinml, ms. Wethenhavetousedegenerateperturbationtheory,DPT,tosolvetheproblem. We wont have to actually do any complicated maths because it turns out that we can usea helpful model the Vector Model, that guides us to the solution, and gives some insight into thephysics of what DPT is doing.4.1 ThePhysicsofSpin-OrbitInteractionWhat happenstoamagneticdipoleinamagneticeld? Anegativelychargedobject havingamoment of inertia I, rotating with angular velocity, has angular momentum, I = . The energyis thenE =12I2We suppose the angular momentum vector is at an angle to thez-axis. The rotating chargehas a magnetic moment = (44)The sign is negative as we have a negative charge. is known as the gyromagnetic ratio. (Classically = 1 for an orbiting charge, and = 2 for a spinning charge).IfaconstantmagneticeldBisappliedalongthez-axisthemovingchargeexperiencesaforceatorqueactsonthebodyproducinganextrarotationalmotionaroundthez-axis. Theaxisofrotation of precesses around the direction ofBwith angular velocity

. The angular motion ofour rotating change is changed by this additional precession from to +

cos(). If the angularmomentum was in the opposite direction then the new angular velocity would be

cos().17AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionFigure 11: Illustration of the precession (Larmor precession) caused by the torque on the magneticmoment by a magnetic eldB.The new energy 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 since the 3p electron penetrates the inner core of electrons far more than 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-like atoms i.e. ions with all electrons except one stripped o. For a neutralatom, however, theelectrostaticforceofthecentraleldwillbelessandsothedependenceonZis reduced from theZ4in a hydrogenic ion. We can get an approximate idea of theZ-dependenceby considering theZein our Central Field to be a step function i.e. to change abruptly from itseective valueZo(o for outer) outside the core toZin(in for inner) inside the core. This is avery crude approximation, and we can make it even more crude by supposing Zo = 1 (total screeningby (Z 1) electrons) andZin = Z(no screening at all).The total energy of the spin-orbit interaction will be made up of two parts, each with the form:ESO Z_ 1r3_(100)The outer part,Z = 1, is very small compared to the contribution from the inner part,Z>> 1, sowe neglect the contribution from theZo bit. The inner part is hydrogenic so:_Zr3_inner=Z4n3a30l(l + 1/2)(l + 1)(101)Thecontributionwill beweightedbythefractionofthetimetheelectronspendsinsidethecore.This fraction is indicated by the quantum defect or by the eective quantum numbern. Using theBohr theory we can show that the fraction inside the core is_ n3Z2in_/_n3Z2o_(102)So the contribution from the inner part becomes:_Zinr3_inner

n3Z2on3Z2in(103)26AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionHencen,l 04gs2BZ2inZ2on3a30l(l + 1/2)(l + 1)(104)So approximately, the size of the spin-orbit splitting scales withZ2in 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 momentum of atomsas a whole, and not just individual electrons withl, s, j, ml, ms, mj, etc.4.5 SpectroscopicNotationSofarwehaveseenhowthephysical interactionsaectingtheenergyofanatomcanbeorderedaccording to their strength. We use perturbation theory, or the Vector Model, by starting with thelargestinteractiontheCoulombforceoftheCentral Field. Thisledtotheideaoftheelectronconguration; n1l1n2l2n3l3etc. witheachindividual electronsangularmomentumlabelledbyalower case letter; s, p, d, f, etc. For single-electron atoms, like alkalis, we dened a total angularmomentumj=l + s. Thespin-orbitinteractionledtoenergylevelslabelledbytheirvalueof j,i.e. (l + 1/2)or(l 1/2). Whatdowedowhenthereismorethanoneelectrontoworryabout?We anticipate the result of the next section where we consider the residual electrostatic interactionbetween electrons.Ifweweretoignoreanyspin-orbitinteractionthentheenergylevelsarecompletelydeterminedby electrostatic forces only. In this case the space and spin systems within the atom must separatelyconserveangularmomentum. Thissuggeststhatwecandeneatotalorbitalangularmomentumby the vector sum of the individual orbital momenta. We will label such total angular momenta bycapital letters; thus the total orbital angular momentum is:L =

ili(105)Forsimplicitywewill considertwo-electronatomsi.e. atomswithtwovalenceelectronsoutsideclosed inner shells. SoL = l1 +l2. In the same way we dene the total spin of the atom to beS =

isi = s1 +s2(106)These can be visualized using our vector model:l2Ll1Ss2s1Figure 15: Diagram showing the coupling of allli toL and allsi toS.Quantum mechanically these angular momenta are represented by operatorsL2andLzL2=_l1 + l2_2(107)27AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionWith eigenvaluesL(L + 1)2where [l1l2[ L l1 +l2, andLz = lz1 + lz2(108)With eigenvaluesML where L ML L.Similarly we have total spin operators.S2= ( s1 + s2)2(109)With eigenvaluesS(S + 1)2where [s1s2[ S s1 +s2, andSz = sz1 + sz2(110)With eigenvaluesMS where S MS S.MLandMSarethequantizedprojections of LandSrespectivelyonthez-axis (theaxis ofquantization). The total angular momentum is found by adding (or coupling) the total orbital andspin momenta: to giveJ:J2=_L +S_2(111)With eigenvaluesJ(J + 1)2and[L S[ J L +S (112)Also:Jz =Lz +Sz(113)With eigenvaluesMJ where J MJ Ji.e. MJis the projection ofJon the quantization axis.By convention L = 0, 1, 2, 3 is labelled S, P, D, F etc as with the s, p, d, f for single electrons. Thevalues ofL andSspecify a Term. The spin-orbit interaction betweenL andSwill split each terminto levels labelled byJ. For a givenL there will be (2S + 1) values ofJand so (2S + 1) is knownas the multiplicitySpin-orbit interaction will split each term into levels according to its value ofJ. Thus the multi-plicity tells us the number of separate energy levels (i.e. number ofJvalues) except, however whenL = 0, when there is no spin-orbit splitting and there is only a single level.Each level is degenerate in J; there are (2J+1) values of MJ; the projection ofJ on an external axis.Each value ofMJbetween Jand +Jconstitutes a state, and the degeneracy is raised by applyingan external magnetic eld in which the dierent orientations ofJrelative to the eld direction willhave dierent energy.In general then we have the following notation for an energy level:n1l1n2l2...2S+1LJ(114)This reects the hierarchy 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 +JThe ground level of sodium can therefore be written as 1s22s22p63s2S1/2.It is common practice to drop the conguration of the inner shells and give only that for the outer,valence, electrons. Thusthersttwoexcitedlevelsare3p2P1/2,3/2and3d2D3/2,5/2. thesearesometimesreferredtoasexcitedstates,butthisisanexampleofthelooseterminologyusedbyAtomic Physicists!We can now draw a full Energy Level diagram for sodium with all the levels labelled by appropriatequantum numbers.28AtomicPhysics,P.Ewart 4 CorrectionstotheCentralField: Spin-OrbitinteractionHydrogenn43Figure 16: Na energy level diagram showing ne structure (spin-orbit splitting) greatly exaggerated.Notethatthedierencebetweenenergylevelsthesamenbutof increasingl becomesmaller as they approach the hydrogenic energy level for the correspondingn. Note alsothatthedierencebetweenenergylevelsofthesamenbutdierentl becomessmallerwith increasingn.29AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-Coupling5 Two-electronAtoms: Residual ElectrostaticEectsandLS-CouplingSo far we have made the Central Field Approximation which allows us to estimate the gross structureoftheenergylevels. Thisintroducedquantumnumbersnandlallowingustodenetheelectronconguration. We noted two perturbations that need to be considered in going beyond the CentralField Approximation. These were,H1, the residual electrostatic interaction andH2, the spin-orbitinteraction. Perturbation theory requires that we apply the larger perturbation rst. In the speciccase of a one-electron atom in a Central Field we could ignoreH1. The perturbation due to spin-orbiteectsthenledustowavefunctionslabelledbythequantumnumbersj, mj. Inthecaseofan atom with two electrons in a Central Field we have to consider rstH1, and this will lead us todierent quantum numbers. We have anticipated this result by introducing these quantum numbersL, SandJ. Inthefollowingsectionswewilllookmorecloselyatthephysicsofhowelectrostaticinteractions lead to LS-coupling. It needs to be stressed at the outset that LS-coupling is the mostappropriate description when electrostatic eects are the dominant perturbation. We can then applythe smaller spin-orbit perturbationH2 to these LSJ-labelled states.AllthisisconvenientlyvisualizedusingtheVectorModel, wherethestrongestinteractionsleadto the fastest precessional motions. The residual electrostatic perturbationH1 causes the individualorbital angular momenta l1 and l2 to couple to form L = l1+l2. Similarly the spin angular momentas1ands2couple to give a total spinS = s1 + s2. Sol1andl2precess rapidly aroundL ands1ands2 precess rapidly aroundS. L andSare then well-dened constants of the motion.The weaker,spin-orbit interactionH2then leads to a slower precession ofL andSaround theirresultant J. As in the single electron case, the precession rate and hence the energy shift dependson the relative orientation of L and S. The result is a ne-structure splitting of the terms dened byL andSaccording to the value ofJ.We have been assuming thatH1>>H2 but this may not always be the case. We have seen alreadythatH2 increases with Z2. So our assumption that electrostatic dominates over magnetic, spin-orbit,couplingwillbecomelessvalidforheavyelements. WhenthishappensLandSceasetobegoodquantumnumbers. Thetotal angularmomentumJwill continuetobeagoodquantumnumber,but it will be a result of adding the angular momenta of the individual electrons in a dierent way.In the limit thatH2>>H1, individual electron momentaliandsicouple to giveji = li + si. ThetotalJis then the vector sum ofj1 andj2. This jj-coupling is, however, not on the syllabus.In the following we will take Magnesium as our example of a two-electron atom. We will look rstat the gross energy level structure arising from the Central Field Approximation. Secondly, we willconsider the residual electrostatic eect (perturbationH1) which splits the energy levels into termslabelled by L and S. We will think about why these terms have dierent energies. Thirdly, we applythe smallerH2, spin-orbit, perturbation and examine the resulting ne-structure splitting.5.1 Magnesium: GrossStructureWechooseMgasourtypical two-electronatomratherthanHeliumbecause, intheHeliumcase,the spin-spin interaction is of comparable magnitude to the other perturbations. In heavier elementsthis is less so. Wecan, however, usethebasicresults foundfor Heliumtohelpus understandthebehaviourofourtwo valenceelectronsinMagnesium. Magnesium isbasicallysodiumwithanextra proton in the nucleus and an extra electron outside. The conguration of the ground energylevel is1s22s22p63s2. If werestrictourselvestoexcitingonlyoneof the3selectrons, theexcitedcongurations will be ( )3snl. The electron remaining in the 3s orbit acts as a spectator electronand so the excited electron moves in a Central Field not much dierent from that of sodium. Thebasic energy level structure will be roughly the same as sodium, but the energy levels will be morenegative owing to the increased Coulomb attraction of the nucleus.What will be the eect of the perturbationsH1 andH2?30AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-Coupling5.2 TheElectrostaticPerturbationWe will eliminateH2 for the moment by considering the conguration 3s4s; allowing us to concentrateontheeectsarisingfromtherebeingtwoelectronswithelectrostaticrepulsionbetweenthem. Itmay seem, at rst sight, that we could deal with this conguration entirely within the Central 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 because the elds the two electrons see is not the same.So we will have a left-over perturbationH1 and it doesnt have to be non-central.IntheCentral FieldApproximationwecanndwavefunctions for eachelectroninour 3s4sconguration;(3s) and (4s). What will be the wavefunction for the system as a whole?We couldform a product wavefunction1(3s)2(4s). Such a wavefunction implies we can identify individualelectrons by the subscripts 1 and 2. The particles, are for the moment, distinguishable. Now, can weuse this wave function to nd our perturbation energy due toH1, denoted by E1?In other words, is E1 = 1(3s)2(4s)[H1[1(3s)2(4s)) ?It is fairly obvious that if we swapped the labels 1 and 2 we would get the same energy. So the states:[1(3s)2(4s)) and [2(3s)1(4s)) are degenerate. We therefore have to use Degenerate PerturbationTheory. This means we have to nd correct linear combinations of our zero-order wavefunctions. Thecorrect linear combination wave function will diagonalize the perturbation matrix. So we can use theform of the perturbation to guide us to the correct combinationH1 =

iZe240ri+

i>je240rij

iU(ri) (115)Since our zero order wavefunctions are eigenfunctions ofU(ri),the third term will not give us anytrouble. The rst term is also a single-electron operator so this will not cause a problem either. Weare left with the 1/rijoperator and need to nd a wavefunction that will give a diagonal matrix.We can form two 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 nda andb to make the diagonal matrix elements vanish i.e.1[ V [2) = 0 (118)WhereVis the two-electron electrostatic repulsion. This perturbation does not distinguish electron1 from electron 2, so [a[ = [b[ and for normalization [a[, [b[ = 1/2. We now have1 =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 change1,and simply changes the sign of2.These functions have a denite exchange symmetry. We shall return to this later.The o-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 the terms in the integral we nd:1 3 = 1(3s)2(4s)[ V [1(3s)2(4s)) = J (122)2 4 = 1(4s)2(3s)[ V [1(4s)2(3s)) = J (123)2 3 = 1(4s)2(3s)[ V [1(3s)2(4s)) = K (124)1 4 = 1(3s)2(4s)[ V [1(4s)2(3s)) = K (125)So the total element is zero as required!The diagonal elements now give us our energy eigenvalues for the change in energy:-1[ V [1) = E1 = J + K (126)2[ V [2) = E2 = J K (127)J is known as the direct integral and K is the exchange integral. (The wavefunctions in K dier byexchange of the electron label).EnergylevelwithnoelectrostaticinteractionJ+K-KSingletTripletFigure 17: Theeectofresidual electrostaticinteractionistoshiftthemeanenergybythedirectintegral J and to split the singlet and triplet terms by the exchange integral 2KSotheenergyof anyparticularcongurationwill besplitbyanamount2K. Thiscorrespondstothetwoenergies associatedwiththenormal modes of coupledoscillators. For example, twopendula, or two masses on springs, when coupled together will settle into correlated or anti-correlatedmotions with dierent energies. Note that for our 2-electrons the eect arises purely because of themutual electrostatic interaction. Note also that this has nothing to do with the distinguishability orindistinguishability of the electrons!Our result shows that the term of the conguration described by1,the symmetric state,has ahigher (less negative) energy than the anti-symmetric state 2. To understand why we need to thinka bit more about symmetry and the physics of the interactions between the electrons.5.3 SymmetryDegenerate Perturbation Theory forced us to choose spatial wavefunctions that have a dene sym-metry; eithersymmetricorantisymmetric. Why? Itwasbecauseinterchangingthelabelsontheoperator 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;eithersymmetric, (+), orantisymmetric, (-). All electronsinouruniverseareantisymmetric, sowe need to combine our spatial wavefunctions with an appropriate spin wavefunction to ensure ouroverall antisymmetric state.32AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingEach electron spin may be up or down . We can therefore construct for two electrons combina-tions that are either symmetric or antisymmetric:12, 12,1212 + 12 : Symmetric (128)1212 12 : Antisymmetric (129)Thethreesymmetricspinstatesformatriplet andcombinewiththeantisymmetricspacestates,2. Thistriplettermenergywasfoundtobeshifteddowni.e. morestronglybound. Physically,this is because the antisymmetric space function describes a state where the electrons avoid the sameregion of space. The mutual repulsion from the e2/r12 potential is therefore reduced and the electronsareheldmorestrongly. Theantisymmetricstate, ontheotherhand, isasinglet state. Sincetheelectronshaveoppositespins, equivalenttodierentspinquantumnumbers, theycanoccupythesame space without violating the Pauli Principle. Increased overlap in space increases the eect ofthe 1/r12repulsion, throwing the electrons further out in the Central Field. This results in a lowerbinding energy. The singlet terms thus lie above the triplet term for a given conguration.5.4 Orbital eectsonelectrostaticinteractioninLS-couplingWe have spent a long time considering the eect of the spin on the energy levels of the LS-coupledstates. Thereis, however, anotheraspectoftheelectrostaticinteractionthatwehaveoverlooked;the way the energy depends on L. The Central Field Approximation takes account of the sphericallyaveragedchargedistribution. Thevalueof Lfor agivencongurationdepends ontherelativeorientationoftheindividualangularmomental1andl2. Ifeitheroftheseiszeroe.g. 3sthenthespatial overlap with the other electron in, say, a 3p orbit will be the same no matter where the axis ofthe 3p orbit points. Consider, however, what happens if both electrons are in p-orbits say 3p4p. NowL = l1 +l2 = 2, 1 or 0 i.e. D, P or S Terms. The Vector Model shows that the axes of the individualorbitalangularmomentaarexedinoneofthethreespatialorientationstogivequantizedvaluesofL. The electron wavefunctions, and probability densities, are like doughnut shapes. The dierentrelative orientations of their axes will lead to dierent spatial overlaps of the electrons. Consequently,the amount of mutual repulsion will be dierent in each case. This helps explain why the electrostaticinteraction leads to terms of dierent energy and are labelled byL andS.l1ol2Figure 18: Electronorbitals l1, l2haveanoverlapdependingontheir relativeorientationandsodierent vector sums have dierent electrostatic energy.33AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-Coupling5.5 Spin-OrbitEectsin2-electronAtomsSo far we have applied only the electrostatic perturbationH1 to the congurations set by the CentralField. This electrostatic interaction led to singlet and triplet terms labelled byL andS. The 3s4sconguration in Mg will have to terms;1S and3S, with1S lying above3S in the energy level diagram.Sincethereisnoorbital motionthereisnospin-orbitinteraction. Thetotal angularmomentumJ = L +S = 0 or 1. The complete designations are therefore3s4s1S0 and 3s4s3S1Although the multiplicity, 2S + 1, is 3 in the triplet state, there is only one energy level sinceJhasonly the value 1, in this case.Lets now consider the conguration 3s3p. Again S = 0 or 1 and we have singlet and triplet terms.L = 1, sothetermsare1Pand3P. J, however, inthetriplettermhasvalues2, 1, 0. Thesethreevalues arise from each of three relative orientations ofL andS. This is clear in our Vector Model:L =1S=1S=1S=1L =1 L =1J=2J=1J=0Figure 19: Vector addition ofL andSto give total angular momentumJ.Themagneticinteractionsbetweenthetotalspinmomentandthemagneticeldestablishedbythe total orbital moment, L, are dierent in each case leading to dierent energies for the levels ofeachJvalue. The triplet,3P, term is therefore split and acquires ne-structure.The physical arguments we used for spin-orbit interaction in Na carry over to the Mg case by usingthetotal angularmomentumoperatorsL, SandJinsteadofl, sandj. Theperturbationtothe[n, L, S, J) states is:ESO =_ H2___1rU(r)r_ S L_(130)The radial integral is denotedn,l. The angular momentum operator is:S L =12_ J2 L2S2_(131)and, sincewehavethecorrectrepresentationineigenfunctionsoftheoperatorsontheR.H.S. wend the expectation value is given by the corresponding eigenvalues:_ J2 L2S2_ =12J(J + 1) L(L + 1) S(S + 1) (132)soEJ=n,L2J(J + 1) L(L + 1) S(S + 1) (133)The separation of the ne-structure levels is found by evaluating EJforJ

andJ

1.EJ EJ

1 = n,lJ

(134)This represents an Interval Rule that is valid so long as two conditions are fullled:(1) LS-couplingis agooddescriptionfor the terms i.e. whenthe electrostatic perturbationH1>>H2, the magnetic, spin-orbit interaction.34AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-Coupling(2) the perturbation energy is expressible as_S L_. This is not the case in Helium where spin-spininteractions are comparable to spin-orbit eects.The energy levels of the 3s3p terms are therefore2K3s3p P113s3p P323s3p P313s3p P30ib2bFigure 20: Separationof singletandtriplettermsduetoelectrostaticinteractionandsplittingoftriplet term by magnetic spin-orbit interaction.As we go to atoms with higher Z, the magnetic, spin-orbit, interaction increases ( Z2) and beginsto be comparable to the electrostatic eect. LS-coupling starts to break down and will be shown bytwo features of the energy levels: (a) The separation of the triplet levels becomes comparable to theseparation of singlet and triplet terms of the same conguration. (b) The interval rule is no longerobeyed.A third indicator of the breakdown of LS-coupling is the occurrence of optical transitions betweensingle and triplet states.In the LS-coupling scheme the spin states are well-dened and the dipole operator, e r,does notactonthespinpartofthewavefunction. Thusadipoletransitionwill linkonlystateswiththesamevalueof S. i.e. wehavetheselectionruleS=0. ThisruleiswellobeyedinMgandthelighter2-electronatoms. Sowegettransitionsonlybetweensingletandsingletstates,orbetweentriplet and triplet. It is then common to separate out the singlet ad triplet energy levels into separatediagrams.1 1 1 3 3 3S P D S P Do 2 13s S2103s3p P113s3p P32,1,03s3d D124s5sns3p23pnlintercombinationline(weak)resonanceline(strong)TermdiagramofMagnesiumSingletterms TriplettermsFigure 21: Mg term diagram. The separation of singlet and triplet terms shows schematically thatsinglet-triplet transitions are forbidden.35AtomicPhysics,P.Ewart 5 Two-electronAtoms: ResidualElectrostaticEectsandLS-CouplingThe 3s3p3P2,1,0levels are metastable i.e. they have a long lifetime against radiative decay. The6s6p3P2,1,0levels in Mercury are also metastable but a transition 6s6p3P1 6s2 1S0does occur.Suchatransitionisknownspectroscopicallyasanintercombinationline. Theprobabilityisnotlarge, compared to other allowed transitions, but this transition is a strong source of radiation fromexcitedmercury. Thisisbecauseanyatomsendingupinthe6s6p3P1levelhavenootherwaytodecay radiatively. This line is the source of light in Hg uorescent lamps.36AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructure6 NuclearEectsonAtomicStructureWehavesecretlymadethreeassumptionsaboutthenucleussofar. Wehaveassumedthatthenucleus is stationary, has innite mass and occupies only a point in space. In fact, the nuclei of allatoms consist of spinning charged objects and so they have a resultant magnetic moment. Secondly,their mass is not innite, so the nucleus will move with the orbiting electrons around their commoncentre of gravity. Thirdly the nucleus has a nite size with some shape over which the proton chargeisdistributed. Eachoftheseeectswill makeasmall changeintheenergylevels. Themagneticinteraction will lead to an eect similar to spin-orbit coupling and give a splitting of the energy levelsknowashypernestructure, hfs. Thekineticenergyofthemovingnucleuswill aecttheoverallenergyof theatomicstates. Finally, thesizeandshapeof thechargesonthenucleuswill aectthe Coulomb force on the electrons and hence aect the energy. The term hyperne structure isreserved(atleastinOxford)formagneticeectsofthenuclearspin. Themassandelectroneldeects will be termed isotope eects,since they depend to some extent on the number of neutronsin a given atom.6.1 HyperneStructureBoth the protons and the neutrons have magnetic moments due to their spin. The total nuclear spinislabelledI. Spinningfermionstendtopairupwithanothersimilarparticlewithaspinintheopposite direction. For this reason nucleii with even numbers of protons and neutrons, the so-calledeven-evennuclei haveI =0. Odd-oddorodd-evennuclei canhaveintegerorhalf-integerspin:I = 1/2, 1, 3/2 etc. These nuclei will have a magnetic dipole momentI. Nucleii with integer spinscan also have an electric quadruple moment as well. We will be concerned only with the magneticdipole moment. Nuclear magnetic moments are small compared with electronic moments. Recall theclassical relationship between magnetic moment and angular moment: = (135)In Quantum Mechanics the spin and orbital moments are:-s = gsBs (136)l = glBl (137)wheregs = 2 andgl = 1 (Note we have useds, l in units of ) NowB =e2me, whereme= mass ofthe electron. So nuclear moments are going to be much smaller as the nuclear massmN>> me. Soto keep theg-factor of the order of unity we dene the nuclear moment byI= gIII (138)wheregIis the nuclearg-factor (1) andNis the nuclear magneton, related toBby the ratio ofelectron to proton mass:N= B

memp(139)Thepositivesignonourdenitionof Iispurelyaconvention; wecannottellingeneralwhetherthe magnetic moment will be parallel or antiparallel toI, it can be either.The magnetic moments of the electrons and the nucleus will precess around their mutual resultant.The important thing to note is that the smallness ofIwill mean the interaction with the magneticeld of the electrons will be weak. The precession rate will therefore be slow and angular momenta Iand Jwill remain well-dened. The interaction can therefore also be treated by perturbation theoryand the Vector Model. The perturbation can be expressed:H3 = I Bel(140)37AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureItremainsto nd the expectation value of this operator,and we will use our vector modelto ndthe answer in terms of the time averaged vector quantities.6.2 TheMagneticFieldofElectronsThe nuclear interaction, being weak, will not follow the motions of individual electron orbits or spins.Their precessions aroundJare too fast for the nuclei to follow. The nuclear moment will thereforeeectively see only the time averaged component resolved alongJ.Bel = (scalar quantity) J (141)The electrons in closed shells make no net contribution to the eldBel. Therefore only the electronsoutside the closed shells contribute. Electrons havingl ,= 0 will have a magnetic eld at the nucleusduetobothorbital andspinmotions. Bothof thesehoweverdependonr3andsowill besmallenoughtoignoreformostcases. Forl= 0, s-electrons, however, thesituationisdierent. Aswenoted before, these electrons have(r) ,= 0 atr = 0. The spin-spin interaction with the nucleus canthereforebestrong. ThisisknownastheFermicontactinteraction. Thisshortrangeinteractiondominates the contribution to the energy. In general it is very dicult to calculate but for the caseof hydrogen in its ground state an analytical result can be found.Ingeneral theorbital andspinmomentaof theelectronsprovideaeldof order 04B1r3_.Taking a Bohr radius to give the order of magnitude ofr we nd:Bel 04Ba30 6T (142)Putting this with a nuclear momenta of the order ofN B/2000 we nd an energy perturbationE 50 MHz. So a transition involving a level with hfs will be split by this order of magnitude infrequency. Since the eect scales withZit will get up to 100 times larger in heavy atoms i.e. afew GHz. The structure of the splitting will depend on the angular momentum coupling, and to thiswe now turn.6.3 CouplingofIandJSincethenuclear spinmagneticmoment is proportional toI andtheelectronmagneticeldisproportional toJwe can writeH3 = AJIJ and E = AJ_I J_(143)The quantity AJdetermines the size of the interaction energy. Note that it depends on other factorsas well as J. For example we could have J = 1/2 from either a single s-electron or a single p-electron.Theformerwill giveamuchlargervalueof AJ. Thefactor IJisdimensionlessandwill havedierent values depending on the angle betweenIandJ. We can use our Vector Model to nd theresult of a DPT calculation as follows:IandJcouple i.e. add vectorally to give a resultantF:F= I +J (144)From the vector triangle we ndF2= I2+J2+ 2IJ (145)IJ =12_F2I2J2_(146)38AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureFIJFIJIJ FFigure 22: Addition of nuclear spinIand total electronic angular momentumJto giveF.The magnitudes of the squared angular momenta are given by their eigenvalues:E =AJ2F(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 one way of ndingFand so if we knowJwe can ndI, the nuclear spin.)The interval rule can be messed up if there are additional contributions to the nuclear spin momentsuch as an electric quadruple moment.The number of levels into which the hfs interaction splits a level depends on the number of valuesof F. This, in turn, depends on the coupling of Iand J. Our vector model gives us a triangle rulewhereby the vector addition must yield a quantized value of the total angular momentumF.So ifJ> Ithen there are 2I + 1 values ofF, and ifI> Jthere are 2J + 1 values.The values of F will be integers in the range: [I J[ F I + J. The ordering of the levels i.e.whetherE(F) > E(F 1) or vice-versa depends on the sign ofAJ. There is, in general, no way tocalculatethiseasily. Itdependsonnuclearstructure,whichdeterminesthesignof gI,andonthedirection ofBel relative toJ. (for single unpaired s-electronBel is always antiparallel toJ).6.4 FindingtheNuclearSpin,IThe nuclear spinIcan be found by examining, with high resolution, the structure of 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 0 0 (149)There are then three methods that can be used, provided there is sucient spectral resolution.Interval rule The level separation, found from the frequency intervals between components of thehfs, is proportional to theF-value of the higher level. This allows us to ndFand then, if we knowJfor the level, we can ndI.Numberofspectrallines The number of spectral components will give the number of levels of hfs.The number of levels is (2I +1) for I< Jand (2J +1) for I> J. (This one works provided we knowJ> I)39AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureRelativeIntensities The relative intensity of the spectral components will be proportional to thestatistical weight of the hfs levels (2F + 1). So ifJis known we can ndI.Usually, inpractice, wemayneedtouseevidencefrommorethanoneof thesemethodstobesure. Conventional optical methodsareoftenatorbeyondtheirlimitinresolvinghfs. Thereisthe additional problem of Doppler broadening which is often larger or at least comparable to thesplitting. Laser techniques, radio frequency methods or a combination of both are used nowadays.6.5 IsotopeEectsMassEects The energy levels determined basically by the Central Field are given by:En Z2e4mr22n2(150)Wheremristhereducedmassoftheelectronnucleussystem. Foraninnitelymassivenucleusmr me, the electron mass. For real atoms, however, we need to deal with the motion of the nucleusaroundthecommoncentre-of-mass. Theresultingnuclearkineticenergyisp2n2mnwherepn, mnarethemomentumandmassrespectivelyofthenucleus. Foratwo-bodysystemwecantreatthisbysubstituting the reduced mass, mr, for me in the Schrodinger equation. The change in energy is thenfound simply by making the same substitution in the energy eigenvalues.The result (and you should be able to work this out) is to shift the energy level up byE(mrme)/me.Inthecaseofatwo-electronsystemwewillneedtoknowexplicitlytheelectronwavefunctions.The energy change Emass is given by:Emass =p2n2mn=(p1 +p2)22mn(151)=p21 +p22 + 2p1 p22mn(152)Putting thep21, p22 terms in Schrodingers equation gives a reduced-mass-type shift. Thep1 p2 term,however, needs to be calculated explicitly, usually using Perturbation Theory. The eect of this termis know as the specic mass shift. The shifts are detected by changes in the frequency of transitionsso the eects need to be calculated separately for each level involved. What will be observed is a smalldierence 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.FieldEects Theeectof thenitesizeof theprotonintheenergylevelsinhydrogencanbecalculated using perturbation theory. The energy shift isE =_0(r)V (r)4r2dr (153)V is the change in the potential energy resulting from the dierence between a point charge andsomespherical distribution, eitheroverthesurfaceorvolumeofasphere. Thisdierenceisonlysignicantoveraverysmall rangeof rcomparedtoaBohrradiusa0. Sothewavefunctionsareessentially constant over the range of interest and so can come outside the integral.E [(0)[2_0v(r)4rrdr (154)40AtomicPhysics,P.Ewart 6 NuclearEectsonAtomicStructureOnly the s-states are signicantly aected so using[(0)[2=1_Zna0_3(155)The result can be derived. For the charge on the surface of the nucleus of radiusr0 we nd:Ens =Z4e2r2060a30n3(156)For the ground state of hydrogen the fractional shift isE1sE1s= 43_r0a0_2 5 1010(157)This is a small eect but important since uncertainty in knowing the size of a proton aects preciseexperiments to study QED eects.41AtomicPhysics,P.Ewart 7 SelectionRules7 SelectionRulesNowthatwehaveabetterpictureofatomicenergylevelsandhavebeguntoseehowtransitionsbetweenthemaresoimportantforourstudyofthem,itistimetorevisitthesubjectofselectionrules. These are the rules that tell us whether or not an atom can change from one particular stateto another. We will consider only electric dipole transitions;the rules for magnetic dipole,electricquadruple etc are dierent and are not on the syllabus (thankfully). One way to approach transitionsis to use time-dependent Perturbation Theory and its result in Fermis Golden Rule. That also isnton our syllabus but we want to use some of the underlying physics.In the rst lecture we looked at the wavefunctions for two states 1 and 2 involved in a transition.Physically we had to generate an oscillating dipole an accelerated charge in order to create electro-magnetic radiation. The solutions of the time-dependent Schrodinger equation, TDSE, have the formi = (r, , )eiEit/(158)So the perturbation matrix elements are:i[ er [j) ei(EiEj)t/(159)The diagonal elementsi = jrepresent stationary states and have no oscillation term, so atoms instationary states do not absorb or emit light. As we saw in Lecture 1 the o-diagonal elements dogive an oscillation of frequency12 = [E1E2[/.Ifthespatialpartofthematrixelementiszero,however,wewillnotgetanyradiationi.e. thetransition is forbidden. The spatial integral will be determined by the quantum, numbers in the twostates1(n1, l1, etc) and2(n1, l1, etc). It is the changes in these quantum numbers that give us ourselection rules. We needn1, l1, n2, l2...L, S, J, MJ[

ierin

1, l

1, n

2, l

2...L

, S

, J

, M

J_,= 0 (160)7.1 ParityThe o-diagonal matrix elements represent the atom in a super-position of two eigenstates1and2. Any linear combination of solutions of the TDSE will also be an eigenstate [NB: this is not trueforthetime-independentSchrodingerequation]. Nowthespatial wavefuctionsall haveadeniteparity,either even,(+) or odd,(-). The dipole operator erhas odd parity i.e. changingrto rchanges the sign.If we consider, for the moment, a single electron making the jump from one state to another thenthe contribution to the integral from one particular location (x, y, z) is:enl(x, y, z)_ix + jy + kz_(161)If the parity of the two states is the same, then the product will be the same at the opposite point(x, y, z), The operator, however, will have the opposite sign and so when we integrate over allspace the result is zero. Therefore, to get a non-zero dipole matrix element, the parity must change.The parity, as we noted before, is given by (1)l, where l is the electrons orbital angular momentumquantum number. So we have our rst selection rule:l = 1 (162)This makes physical sense because a photon has angular momentum of one unit of . So conservationof angular momentum demands l = 1, for one electron to jump.42AtomicPhysics,P.Ewart 7 SelectionRules7.2 CongurationCould more than one electron jump?Consider an atom with two electrons atr1andr2with initialand nal congurations 1s2p (n1l1n2l2) and 3p3d (n3l3n4l4). the matrix 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 the orthogonality of the eigenfunctions i.e. i(nl)[i(n

l

)) = 0So this means the conguration can change by only a single electron jump.The angular momentum does not depend at all onn, son can change by anything.Our basic rules are therefore:n = anything (166)l = 1 (167)And the photon carries one unit of angular momentum.7.3 AngularMomentumRulesThe basic rules apply to the total angular momentum J. So provided the changes in J in a transitionallow for one unit ofto be taken by the photon we can make a vector triangle rule to giveJ = 0, 1 but not 0 0 (168)J1 J1J J2 1=J J2 1=hhFigure 23: Selection rules reect conservation of angular momentumincludingfor the emit-ted/absorbed photon.Similar arguments apply to changes in L i.e.L = 0, 1 but not 0 0 (169)The rule forS, we have already notedS = 0 (170)because the dipole operator erdoes not operate on the spin part of the wavefunction. Physicallythis is the sensible notion that the electron spin plays no part in the spatial oscillation.Finally we noted that changes inMJare governed by the ruleMJ= 0, 1 (171)There is a peculiar rule, with no obvious physical interpretation, that MJcannot change from 0 0if J=0. Theseruleswill beapparentonlyinthepresenceof anexternal eldthatraisesthedegeneracy inMJ. We consider the eects of external magnetic elds in the next section.43AtomicPhysics,P.Ewart 8 AtomsinMagneticFields8 AtomsinMagneticFieldsAtoms have permanent magnetic dipole moments and so will experience a force in an external mag-netic eld. The eect of magnetic elds on atoms can be observed in astrophysics, eg. in the regionsof 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 animportantroleintheoperationofatomicclocksandtheiractiononthenuclearspinisthebasisofmagneticresonanceimagingformedical diagnostics. Thebasicphysicsisbynowfamiliar; themagnetic moment of the atom, (atom), will precess around the axis of an applied eld,Bext.The precessional energy is given by:Hmag = atom Bext(172)Therstquestionwehavetoaskishowbigisthisperturbationenergycomparedtothatoftheinternal interactions in the atom. We have a hierarchy of interactions in decreasing strength: Cen-tralField, residualelectrostatic, spin-orbit, hyperne. ThesearerepresentedbytheHamiltoniansH0,H1,H2, andH3respectively. The hyperne interaction 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.AttheotherendweareunlikelytocompetewiththeCentralFieldenergiesoftheorderof10eV.(Checkthisforyourself byestimating_ Hmag_usingBfortheatomsmagneticmomentandatypical value ofBavailable in a laboratory).For the same reason_ Hmag_ is usually less than residual electrostatic energies ( 1eV) so we thenhavetodecidewhetherourexternal magneticperturbationisbiggerorsmallerthantheinternalmagnetic, spin-orbit, energy. Wecanrecognisetwolimitingcasesdeterminedbythestrengthoftheexternaleldandthesizeofthespin-orbitsplitting. Wedeneaweakeldasoneforwhichatom Bext is less than the ne-structure splitting energy. Conversely, a strong eld is one where theexternal perturbation exceeds the spin-orbit energy.Justtogetthebasicphysicsstraight, andtogetafeel forwhathappens, weconsiderrstthesimple case where there is no spin-orbit interaction. We select an atom whose magnetic moment isdueonlytoorbitalmotionandndtheeectof Bextontheenergylevels. Nextweintroducethespin to the problem and the complication of spin-orbit interaction. Then we consider what happensin a strong eld and nd, pleasingly, that things get simpler again. Finally we look at weak andstrong eld eects on hyperne structure.8.1 Weakeld,nospinAnatomwithnospincouldbeatwo-electronatominasingletstatee.g. Magnesium3s3p1P1.When the eld and its perturbation are weak the atomic states will be the usual [n, l, s, L, S, J, MJ)states. In this case, however J = L and we need only the quantum numbers L and ML . The physicsis pictured well by the vector model for an angular momentumL and associated magnetic momentL = glBLRecall thatgLBrepresents the gyromagnetic ratio for an atom sized circulating charge.In a magnetic eldBextalong thez-axisL(andL) executes Larmor precession around the elddirection with energy:EZ= L Bext(173)= gLBLBext(174)NowLBextis the projection of L ontoBextand in our vector model this is quantized by integervaluesML . HenceEZ = gLBBextML(175)44AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsBextLimLMLFigure 24: Eect of external magnetic eldBexton an atom with no spin i.e. magnetic dipole dueonly to orbital motion.There are (2L+1) values ofML: 1 ML L corresponding to the quantized directions ofL inthe eldBext. Thus each energy level for a givenL is split into (2L + 1) sub-levels separated byE

Z = BBext(176)i.e. independent ofL; so all levels are equally split.The splitting will be observed in transitions between levels. The selection rules onJandMJwillapply toL andML in this case (J = L) soML = 0, 1 (177)Ifwelookatourexamplelevel3s3p1P1byatransitiontosay3s3d1D2wesee3statesinthe1P1level and 5 states in the1D2 level.21100-1-1-2MLD =-1 MLD =0 MLD =+1 ML(w- w (w Dw)O O O+ Dw)Figure 25: Normal Zeeman eect gives the normal Lorentz triplet due to selection rules ML = 0, 1and splitting of all levels into equally spaced sub-levelsML.45AtomicPhysics,P.Ewart 8 AtomsinMagneticFields9 transitions are allowed but they form 3 single lines of frequency0, 0Z whereZ = E

Z/ = BBext/ (178)This is the so-called normal Lorentz triplet. The splitting of the line at0 into 3 components is theNormal Zeeman Eect. (This is why we used Z as the subscript on EZ)It was explained classically, and fully, by Lorentz, long before Quantum Mechanics.8.2 WeakFieldwithSpinandOrbitSince, again, the eld is weak there is only a small perturbation so we can use the zero-order wave-functions dened by the spin-orbit interaction, [n, L, S, J, MJ)Theapplicationofanexternaleldwill,again,causetheorbitalmagneticmomentandangularmomentum L and L, to precess. The spin moment and angular momentum will also precess aroundBext. The perturbation energy operator is then:H4 = gLBLBext +gSBSBext(179)Theproblemwenowhaveisthat LandSareactuallyprecessingatafasterratearoundtheirresultantJ. TheeectofthisisthattheoperatorsLBextandSBextnolongercorrespondtoobservables that are constants of the motion.BextSLJ{{LBextsJJBextFigure 26: Precessionof sandl aroundmutual resultant Jresultsinprojectionsof LandSontheeldaxis(z-axis)notbeingconstantsofthemotion. SoMLandMSarenotgoodquantum numbers.Wewill useourVectorModel tofollowthephysicsandtohelpusndthesolution. Fromthediagramweseethattheprecessionof SandLaroundtheresultantJcausestheprojectionsof SandL on the eld axis (z-axis) to vary up and down. The diagram suggests that sinceJremains ata constant angle to thez-axis perhaps we could use the total magnetic momentJ(=L + S) tocalculate the interaction energy i.e._JBext_(180)There is a snag, however, to this cunning plan. The problem is that Jis not parallel or antiparalleltoJ. The reason is that theg-factors for orbit and spin are dierent;gL = 1 andgS = 2. ThusL = BL and S = 2BS (181)The vector triangle ofL,S, andJis not the same as the triangle ofL,SandJ.46AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsAll is not lost, however, because the very fact that the spin-orbit precession is so fast means thatwe can nd an eective magnetic moment for the total angular moment that doeshave a constantprojection on the eld axis,Bext. We resolve the vectorJinto a component along theJdirectionand a component perpendicular to J. As J precesses around Bext the projection of the perpendicularcomponent ofJonz will average to zero. The component along theJaxis is a constant, which wecall the eective magnetic moment,ei.e.e= gJBJ (182)SJLmsmLmTotalmeffZBextFigure 27: Themagneticmomentduetothetotal angularmomentumJdoesntlieontheJ-axissince the vector sum of spin and orbital magnetic moments Jis not parallel to Jowningto theg-factor for spin being 2 g-factor for orbital momentum. The precession ofs, laround J results in a variation of the projection of Jon the z-axis. An eective magneticmomentedoes, however, have a constant projection on thez-axis.gJis called the Landeg-factor.We can now proceed to nd the perturbation energy:EAZ = gJB_J Bext_(183)JBext is just the projection ofJon thez-axis, given byJz.EAZ = gJBBext_ Jz_(184)andusingour [n, L, S, J, MJ)wavefunctionswewritetheexpectationvalueofJzasitseigenvalueMJ. HenceEAZ = gJBBextMJ(185)Lande calculated the form ofgJquantum mechanically, but we can use our Vector Model to get thesame result.Our vector model replaces expectation values by time averages of quantities that are constants ofthe motion. So we look at our vectorsL, SandJto nd the eective constant projections on thez-axis. We see from the vector diagram thatL has a constant projection on theJaxisLJ|J| . This isa vector in the Jdirection so the scalar|LJ||J|is multiplied by the unit vectorJ|J|, to give|LJ|J|J|2= LJ.47AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsSimilarly, the projection ofSonJis given by[SJ[J[J[2= SJ(186)ThevectorsLJandSJhaveconstanttime-averagedprojectionsontheaxisof Bext. Sothetotal,time-averaged energy of interaction with the eld is:EAZ= gLBLJBext +gSBSJBext(187)= gLB[LJ[[J[2JBext +gSB[SJ[[J[2JBext(188)The cosine rule gives the values ofLJandSJin terms ofJ2, L2andS2. So withgL= 1 andgS = 2EAZ = B_3J2L2+S22[J[2JzBext(189)Now, as per our Vector Model method, we replace the vector (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 we nd:gJ=[3J(J + 1) L(L + 1) +S(S + 1)]2J(J + 1)(191)Thisisanimportantresult(i.e. rememberitandhowtoderiveit; itisaderivationbelovedofFinals examiners.) Its real importance comes from seeing thatgJis the factor that determines thesplitting of the energy levels.EAZ = gJBMJBext(192)gJdepends onL, SandJand so by measuring the splitting of energy levels we can determine thequantum numbersL,SandJfor the level.Wewillmeasurethesplittingfromtheseparationofspectralcomponentsofatransitioninthemagnetic eld. The energy level splitting will be dierent in dierent levels so the pattern observedwill not be the simple Lorentz triplet. Consequently, this is known as the Anomalous Zeeman eect.Hence the subscript is AZ on the energy shift EAZ for spin-orbit coupled atoms in aB-eld.8.2.1 AnomalousZeemanPatternAs an example of the Anomalous Zeeman eect we consider an atom with spin-orbit coupling andsodiumisourfavouriteexample. Atransitionbetweenthe3p2P1/2,3/2levelstothegroundlevel3s2S1/2 illustrates the main features. We look only at the pattern of lines in2P1/2 2S1/2 and leavethe2P3/2 2S1/2 as an exercise.FirstweneedtoknowgJforeachlevel. Puttinginthevaluesof L, S, Jfor2S1/2level wendgJ(3s2S1/2)=2. Wecouldhaveguessedthat! Sincethereisnoorbitalmotionthemagneticmoment is due entirely to spin and theg-factor for spin, gS= 2. Putting in the numbers for2P1/2wendgJ(2P1/2)=2/3. Wecannowdrawtheenergylevelssplitintotwosub-levelslabelledbyMJ= +1/2, 1/2.48AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsmj1/2-1/21/2-1/2D1D1s pp s3p P21/23s S21/2Figure 28: Anomalous Zeemaneect inthe3s3p2P1/2level of Na. Transitions aregovernedbyselection rules MJ= 0, 1.Thetransitionsaregovernedbytheselectionrules: MJ=0, 1, leadingtoapatternoffourlines disposed about the zero-eld transition frequency,0.8.2.2 PolarizationoftheradiationThe magnetic eld introduces an axis of symmetry into an otherwise perfectly spherically symmetricsystem. This has an eect on the polarization of the radiation emitted or absorbed; the oscillating orrotating dipole will have a particular handedness in relation to the eld, orz-axis. The transitionmatrix element 1(r, , )[ er [2(r, , )) represents the change in position of the electron and thiscan be resolved into a motion along the z-axis:1[ ez [2) and a circular motion in thex, yplane1[ e(x iy) [2). The-dependence of these integrals can be written 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 zerounlessm1m21 = 0 i.e. m = 1The MJ= 0 transition is thus identied with a linear motion parallel to theBeld. Viewed atright angles to the eld axis this will appear as linearly polarized light,-polarization.Viewed along the z-axis, however, no oscillation will be apparent so the -component is missing inthis direction.Nowconsiderthe(x iy) component arisingfromm= 1transitions. Thisappearsasacircular motioninthe x, yplanearoundthe z-axis. Thusm= 1transitionsgivecircularlypolarized+, (right/left) light viewed alongz, and plane polarized in thex, y plane.It is possible using appropriate optical devices to determine the handedness of the-polarizedemissions alongthe z-axis. The +andtransitions areeither at higher or lower frequencyrespectivelythan0dependingonthesignof theoscillatingcharge. If youcanremember theconventional rules of electromagnetism you can work out the sign of the electron. Lorentz (who couldremember these rules) was able to use this eect to show that electrons were negatively charged.49AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsZYXqfBext-rq1q2fqff21LightpolarizationsppD = 0f = fm1 2D = + 1f = fm1 2--r-rFigure 29: Polarization of Zeeman components viewed along eld axis (z-axis) and along a perpen-dicular axis (x- ory-axis).8.3 Strongelds,spinandorbitThe word strong in this context means that the interaction of the orbit and spin with the externaleld is stronger than the interaction of the orbit with the spin. The magnetic momentsLandSwill then precess much more rapidly around Bext than they do around each other. Since L and S arenow essentially moving independently aroundB, the vectorJbecomes undened it is no longer aconstant of the motion. Our spin-orbit coupled wavefunctions [n, L, S, J, MJ) are no longer a goodbasisfordescribingtheatomicstate; J, MJarenotgoodquantumnumbers. InsteadwecanuseLMLandSMStosetupabasissetof functions [n, L, S, ML, MS). Thespin-orbitinteractionisnow relatively small and we ignore it to this approximation. The energy shifts are now given by theexpectation of the operatorH4:H4 = gLBL Bext +gSB S Bext(195)Now we dont have the complication of coupled spin and orbital motion. Using our [n, L, S, ML, MS)basis functions we nd_ H4_ = EPB = (ML + 2MS)BBext(196)The energy levels are thus split into (2L + 1) levels labelled byML , each of which will be split intolevels labelled byMS.50AtomicPhysics,P.Ewart 8 AtomsinMagneticFieldsConsider our 3p2P1/2 level of Na.2P1/2mLmL10-11/21/21/2,-1/2-1/2-1/2Figure 30: The splitting of a level in a strong eld ignoring the eect of spin-orbit interaction.We can now include the spin-orbit eect as a small perturbation on these states [n, L, S, ML, MS).Again, we can use our Vector Model to nd the expectation value of the spin-orbit operatorS L.BextLSLBextLSmLimSZY XLFigure 31: Strongeldeectonorbital andspinangularmomenta: l andsprecessmorerapidlyaround the external eld relative to their mutual precession. The projections oflandson thex-y plane average to zero. As a result only the projections on thez-axis remain todene the energy in the eld.SinceLandSareprecessingrapidlyaroundthez-axistheircomponentsinthex, y-planewillaverage to zero. We are left only with thez-component so:_S L_ = MSML(197)The six states (ML = 1, 0, 1,MS = +1/2, 1/2) have spin orbit shifts of (/2, 0, /2, /2, 0, /2)as shown on the extreme right of the strong eld energy level diagram.The energy level shifts EPB, ignoring the relatively 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. The selection rules operating are