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