Electricfields:Starkeffect,dipole&quadrupolepolarizability.

Weareofteninterestedintheeffectofanexternalelectricfieldontheenergylevels

andwavefunctionofHandotherone-electronatomssoletsconsidertheatomina

spatiallyconstantelectricfield.Theenergyshiftduetotheelectricfieldiscalledthe

Starkeffect.Iftheatomisinanexternalelectrostaticpotential ϕ(r ) theHamiltonian

becomes

−

2

2mt

∇cm2 −

2

2µ∇2 − Ze2

4πε0r+ Zeϕ(rn )− eϕ(re )

⎛

⎝⎜⎞

⎠⎟Ψ = EΨ

where e > 0 andtheelectricfield F = −∇ϕ .Since

F isconstant ϕ = −r i

F uptoan

arbitraryconstant,whichweignore.Converting re&rn to

r &R wehave

(Hcm + H int )Ψ = EΨ

where

Hcm = −

2

2mt

∇cm2 +e

F iR(1− Z )

and

H int = −

2

2µ∇2 − Ze2

4πε0r+ eF ir (1+ (Z −1)(me / mt ))

Sincethecenterofmassandtheinternalmotionareindependentthewavefunction

Ψ(R, r ) = φ(

R)ψ (r ) factorizesintoaproductwhere

−

2

2mt

∇cm2 +e

F iR(1− Z )

⎛

⎝⎜⎞

⎠⎟φ(R) = Ecmφ(

R)

and

−

2

2µ∇2 − Ze2

4πε0r+ qF ir

⎛

⎝⎜⎞

⎠⎟ψ (r ) = Eintψ (r )

where q = e(1+ (Z −1)(me / mt )) ,aneffectivechargeforZ ≠ 1 .Thecenterofmass

motionisconsideredlatterandwenowfocusontheeffectofthefieldonthe

internalmotionoftheelectron.Beforegettingintoanydetailletsconsidertheorder

ofmagnitudeoftheinternalelectricfieldbeforeapplyingtheexternalfield.This

fieldatadistance r fromthenucleusisgivenby

Ze4πε0r

2 andaftersubstitutingthe

numericalvaluesoftheconstantsthefieldstrengthis (1.4399649 ×1011) Zr2voltsm

with

r inÅ.ThefirstBohrorbitinHhasaradiusof0.5291772ÅsothefieldinHatthat

distancefromthenucleusis 5.1422081×1011 voltsm

.Laboratoryfieldsvarywidely

but107 voltsm

isreasonableandissmallerthantheinternalfieldbyafactorof

approximately104 suggestingthatperturbationtheorywillbeadequatetoestimate

thechangeinenergyoftheoneelectronatomintypicallaboratoryfields.

TheunperturbedinternalHamiltonianis

H 0 = −

2

2µ∇2 − Ze2

4πε0r

where

H0ψ nlm

0 = En0ψ nlm

0

andEn0 = −e2Z 2

2(4πε0 )aµn2

Ifwemeasurelengthinmultiplesofγ a0 andtheenergyinmultiplesofe2

(4πε0 )aµwe

have

H 0 = − 1

2∇2 − Z

rwithEn

0 = −Z 2

2n2

Letschoose

F tobealongthezaxis,

F = Fz

H = − 1

2∇2 − Z

r+ qFr cosθ = H 0 + qFr cosθ = H 0 +VF

whereθ istheanglebetweenthefieldandthepositionvector

r andV = qr cosθ .In

theseunitsF ismeasuredinmultiplesof e(4πε0 )aµ

2 and q = 1+ (Z −1) / 1+ 1

mn

⎛

⎝⎜⎞

⎠⎟

withmn inmultiplesoftheelectronmass.Letsfirstconsidertheeffectoftheperturbationonthegroundstatewave-function

ψ 0 ≡ψ 100 withenergyE0 = − Z2

2.Sincethewave-functionandenergyinthefield

aresolutionsto

H 0 + VF( )ψ = Eψ

wemayexpandE&ψ inpowersofF

ψ = ψ nF

n

n=0

∞

∑ andE = EnFn

n=0

∞

∑

TheSchrodingerequationbecomes

H 0 +VF( ) ψ nF n

n=0

∞

∑ = ψ nF n

n=0

∞

∑ EmF m

m=0

∞

∑

whereψ n & En arethenthordercorrectionstothewave-functionandenergy.Wecanwritethisequationas

F p H 0ψ p + 1−δ p0( )Vψ p−1 − Ep−kψ k

k=0

p

∑⎛⎝⎜

⎞⎠⎟p=0

∞

∑ = 0

andsinceitmustbetrueforanyF eachcoefficientofF p mustbezero.AccordinglywerecovertheusualRayleigh-Schrodingerperturbationequations

H 0ψ p + 1−δ p0( )Vψ p−1 = Ep−kψ k

k=0

p

∑

Thefirstfewequationsarep=0 H 0ψ 0 = E0ψ 0 ,theunperturbedequation

whereψ 0 =Z 3

πe−Zr andE0 = − Z

2

2

p=1; (H 0 − E0 )ψ 1 + (V − E1)ψ 0 = 0 ,thefirstorderequationp=2; (H 0 − E0 )ψ 2 + (V − E1)ψ 1 = E2ψ 0 ,thesecondorderequationp=3; (H 0 − E0 )ψ 3 + (V − E1)ψ 2 = E3ψ 0 + E2ψ 1 ,thethirdorderequationp=4; (H 0 − E0 )ψ 4 + Vψ 3 = E4ψ 0 + E2ψ 2 Notethatsincetheenergycannotdependonthedirectionoftheelectricfieldalloddordercorrections, E1,E3,mustbezero.WecanverifythisexplicitlyforE1asfollows.Multiplythefirstorderequationbythegroundstatewave-functionandintegrate

0 (H 0 − E0 ) 1 + 0 (V − E1) 0 = 0 Since isHermitian

0 (H 0 − E0 ) 1 = 1 (H 0 − E0 ) 0 = 0 andsothefirstordercorrectiontotheenergyistheaverageoftheperturbationoverthegroundstatewave-functionwhichiszerobysymmetry.

E1 = 0 V 0 = 2qZ 3 e−2Zrr3 dr

0

∞

∫ cos(θ )sin(θ )dθ = 00

π

∫

Consequentlythefirstorderequationbecomes

(H 0 − E0 )ψ 1 + Vψ 0 = 0 Ithappensthatonecansolvetheaboveperturbationequationsexactlytodetermineψ n & En andwewilldosobeforediscussingthemorecommonapproachofexpandingthefunctionsψ n intermsoftheeigenfunctionsof H

0 .

H 0

Wenotethattheatominthefieldhascylindricalsymmetryandlookforasolutionoftheform ψ 1 = g(r,θ )ψ 0 Since

(H 0 − E0 )g(r,θ )ψ 0 =ψ 0 − 1

2∂2g∂r2

+ ∂g∂r(Z − 1

r)+ L

2g2r2

⎛⎝⎜

⎞⎠⎟

thefirstorderequationbecomes

∂2g∂r2

− 2 ∂g∂r(Z − 1

r)− L

2gr2

= 2qr cosθ Themostgeneralformof g(r,θ ) is

g(r,θ ) = f(r)P(cosθ )

=0

∞

∑ where P(cosθ ) isaLegendrepolynomial.

Substitutingthisformintothedifferentialequationfor g andnoting

L2P(cosθ ) = (+1)P(cosθ )

resultsin

P

=0

∞

∑ f'' + 2 f

' (1r− Z )− (+1)

r2f

⎛⎝⎜

⎞⎠⎟ = 2qrP1

wherewenotethat cosθ = P1andconsequentlyonlythetermwith = 1 contributes.Accordinglywehave

f '' + 2 f ' (1

r− Z )− 2

r2f = 2qr

Thesolutiontowhichis

f = − q

Z 2r + Zr

2

2⎛⎝⎜

⎞⎠⎟

andso

ψ 1 = − q

Z 2r + Zr

2

2⎛⎝⎜

⎞⎠⎟P1(cosθ )ψ 0

TofindE2 wemultiplythesecondorderequationbyψ 0 andintegrate

0 (H 0 − E0 ) 2 + 0 V 1 = E2 Since isHermitian

0 (H 0 − E0 ) 2 = 2 (H 0 − E0 ) 0 = 0 wehave

E2 = 0 V 1 Doingtheintegrationresultsin

E2 = − 9q

2

4Z 4

andsincethedipolepolarizabilityα isdefinedbyE2 = − 12αF2 wehave

α = 9q

2

2Z 4

TofindE4 weusethep=4equation whichaftermultiplyingbyψ 0 andintegratingresultsin

E4 = 0 V 3 − E2 0 2 Notethatusingthefirstorderequationwecanwrite

0 V 3 = − 1 H 0 − E0 3 butfromthep=3equation

H 0

(H 0 − E0 )ψ 4 + Vψ 3 = E4ψ 0 + E2ψ 2

H 0 − E0( )ψ 3 = E2ψ 1 − Vψ 2

whichresultsin

0 V 3 = 1 V 2 − E2 1 1 andfinally

E4 = 1 V 2 − E2 0 2 − E2 1 1 Sothefourthordercorrectiontotheenergydependsonthefirstandsecondordercorrectionstothewavefunctionandsincewehavethefirstorderweneedthesecondordercorrectionwhichwewillgetfromthep=2equation

(H 0 − E0 )ψ 2 + Vψ 1 = E2ψ 0 using

ψ 1 = − q

Z 2r + Zr

2

2⎛⎝⎜

⎞⎠⎟P1(cosθ )ψ 0 ,V = qr cosθ andE2 = − 9q

2

4Z 4

andwritingψ 2 = g(r,θ )ψ 0 resultsin

∂2g∂r2

− 2 ∂g∂r(Z − 1

r)− L

2gr2

+ 2q2

Z 2r2 + Zr

3

2⎛⎝⎜

⎞⎠⎟cos2θ = 9q

2

2Z 4

using

g(r,θ ) = f(r)P(cosθ )

=0

∞

∑

resultsin

P

=0

∞

∑ f'' + 2 f

' (1r− Z )− (+1)

r2f

⎛⎝⎜

⎞⎠⎟ +

2q2

Z 2r2 + Zr

3

2⎛⎝⎜

⎞⎠⎟cos2θ = 9q

2

2Z 4

andsince

cos2θ = P12 = 2

3P2 +

13

only = 0&2 contribute.Letting A = q2

Z 4 resultsinthetwodifferentialequations

f2" + 2 f2

' (1r− Z )− 6

r2f2 +

4AZ 2

3r2 + Zr

3

2⎛⎝⎜

⎞⎠⎟= 0

and

f0" + 2 f0

' (1r− Z )+ 2AZ

2

3r2 + Zr

3

2⎛⎝⎜

⎞⎠⎟= 9A2

Thesolutionsbeing

f2 =A24

2Z 2r4 +10Zr3 +15r2( ) and f0 = A24

Z 2r4 + 6Zr3 +18r2( ) so

ψ 2 = f0ψ 0 + f2P2 (cosθ )ψ 0 Asseenabove

E4 = 1 V 2 − E2 0 2 − E2 1 1 anddoingthevariousintegrationsresultsin

1 V 2 = − 2529

32q4

Z10 , 0 2 = 8116

q2

Z 6 , 1 1 = 438q2

Z 6

andfinally

E4 = − q4

Z10355564

.Sincethesecondhyper-polarizabilityγ isdefinedby

E = E0 −12αF2 − 1

24γ F 4 onehasγ = 10665

8q4

Z10

Withtheseresultswecancalculatethedipolemomentµ oftheoneelectronatominducedbytheelectricfieldsincebydefinition

µ = − dEdF

=αF + 16γ F 3

Inadditiontothedipolemomenthighermomentsareinducedbythefield.Forexamplethezzcomponentoftheinducedquadrupolemomentisgivenby

Θzz ≡ Θ =ψ r2P2 (cosθ )ψ

ψ ψ

Thenumeratorisψ r2P2 (cosθ )ψ = 2 0 r2P2 2 + 1 r2P2 1( )F2 + and

0 r2P2 2 = 1178

q2

Z 8 & 1 r2P2 1 = 24 q2

Z 8

andsince

ψ ψ = 1+ 2 0 2 + 1 1( )F2 + theinducedquadrupolethroughorderF2 issimplythenumerator

Θ = 2134

q2F2

Z 8

WhatarethesemomentsintheSIsystem?

EnergyandelectricfieldintheSIandausystemsarerelatedby ESI =

e2

4πε0aµ

⎛

⎝⎜

⎞

⎠⎟ Eau

and FSI =

e4πε0aµ

2

⎛

⎝⎜

⎞

⎠⎟ Fau soif

212SI SI SIE Fα= − then

ESI =

e2

4πε0aµ

⎛

⎝⎜

⎞

⎠⎟ Eau = − 1

2α au Fau

2 e2

4πε0aµ

⎛

⎝⎜

⎞

⎠⎟ = − 1

2α SI Fau

2 e4πε0aµ

2

⎛

⎝⎜

⎞

⎠⎟

2

andso

3

0 04SI au aα α πε= .Thequantity04

SIαπε

isusuallyreportedandiscalledthe

polarizabilityvolume.Noteit’s 3 30 30 0 1481847 10au aua . x mα α−= .Thepolarizability

volumeoftheHatomintheSIsystemisthen

α SI

4πε0

= 0.1481847x10−30 92( ) qau

Z 4

⎛⎝⎜

⎞⎠⎟

m3 .

Thethirdorderfunctionisgivenby ( ) ( )3 3 0100 1 3 100F dP ePΦ = + Φ where

( )6 5 4 3 21 6 64 344 852 1590 3180480

d r r r r r r= + + + + + and

( )6 5 4 31 2 18 63 8240

e r r r r= + + + andallowstheenergytobecalculatedthrough6th

orderorF6.

SeriessolutionforE2 UsuallyonecannotsolvethedifferentialequationsforthevariousperturbationcorrectionstothewavefunctionandenergyaswehavedoneaboveandonecalculatesthesecorrectionsusingtheeigenfunctionsandeigenvaluesoftheunperturbedHamiltonian.ForexamplewhentheperturbationisV = rF cosθ thesecondorderchangeinthegroundstateenergyofHisgivenbythegeneralformula

E100

( 2 ) = ' V100 ,nlmVnlm,100

E100 − Enlmnlm∑ whereV100,nlm = ψ 100 rF cosθ ψ nlm andEnlm = − 1

2n2

whichassumesthattheunperturbedfunctionsarecompletewithrespecttotheperturbedsolutions.TherearetwoclassesofeigenfunctionsfortheHatom,thosethatdescribetheboundstates, 0E < ,andthosethatdescribethecontinuumstates,

0E > .Sincewehavesolvedthedifferentialequationexactlyweknowtheexact

secondorderenergy, ( )2 2100

94

E F= − ,soletsseewhattheerrorisifweuseonlythe

boundwavefunctionstocalculatethesecondordercorrection.Accordingly

E100( 2 ) =

100 Fr cosθ nlm2

E10 − En

0nlm∑ = F 2

100 r cosθ nlm2

E10 − En

0nlm∑ = 2F 2

100 r cosθ nlm2

−1+ 1n2

⎛⎝⎜

⎞⎠⎟

nlm∑

Therequiredmatrixelementis

100 r cosθ nlm = dφ

0

2π

∫ sinθ dθ0

π

∫ Y00 cosθ Yl

m dr R10 r( )0

∞

∫ r3Rnl r( )

andtheangularfactorcanbeeasilyevaluatedbyobservingthat 01

43

cos Yπθ = so

2 2

1 00 0 0 0

0 00 1

1 13 3 l m

m ml ld sin d cos d sin d Y Y Y Y

π π π π

φ θ θ θ φ θ θ δ δ= =∫ ∫ ∫ ∫ so

100 r cosθ nlm = 1

3δ1lδ0m dr R10 r( )

0

∞

∫ r3Rn1 r( ) = 13δ1lδ0m R10 r Rn1

2

10 12 2100

22

213 1

n( )

n

R r RE F

n

∞

=

= −⎛ ⎞+ −⎜ ⎟⎝ ⎠

∑ .Notethatonlythe znp statescontributetothe

summation.Theradialintegralcanbeevaluatedandtheseriessummed(W.Gordon,AnnalenderPhysik1929,394,1031)resultingin

( ) ( )( )

2 6992 2 21 2 6

2

12 1 82253 1

n

s nn

n nE F . F

n

−∞

+=

−= − = −

+∑

SincetheexactE1s(2) = − 9

4F2 theboundstatewave-functionsthenaccountfor~81%

oftheexactresult,asignificanterror!Theneedtoincludethecontinuumsolutionsintheexpansionmakessensephysicallybecauseintheelectricfieldtheelectron

feelsthepotential 1 Fr cosr

θ− + andif 0cosθ < theperturbationwilleventually

overwhelmtheCoulombtermandinsteadofthepotentialgoingtozeroatlargeritwilleventuallybecomelowerthanthe1s energyandtheelectronwillbeabletotunnelintothecontinuum.ThisisillustratedinthefollowingfigurewithF = 1/ 20au

0 2 4 6 8 10-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

potential

r(a0)

-1/r

-0.05r

-1/r -0.05r

H atom in electric field (1/20 au)

2.76a0 7.24a0

1s energy

electron tunnels