electric fields : stark effect, dipole & quadrupole polarizability....electric fields : stark...
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Electricfields:Starkeffect,dipole&quadrupolepolarizability.
Weareofteninterestedintheeffectofanexternalelectricfieldontheenergylevels
andwavefunctionofHandotherone-electronatomssoletsconsidertheatomina
spatiallyconstantelectricfield.Theenergyshiftduetotheelectricfieldiscalledthe
Starkeffect.Iftheatomisinanexternalelectrostaticpotential Ļ(r ) theHamiltonian
becomes
ā
2
2mt
ācm2 ā
2
2Āµā2 ā Ze2
4ĻĪµ0r+ ZeĻ(rn )ā eĻ(re )
ā
āāā
ā āĪØ = EĪØ
where e > 0 andtheelectricfield F = āāĻ .Since
F isconstant Ļ = ār i
F uptoan
arbitraryconstant,whichweignore.Converting re&rn to
r &R wehave
(Hcm + H int )ĪØ = EĪØ
where
Hcm = ā
2
2mt
ācm2 +e
F iR(1ā Z )
and
H int = ā
2
2Āµā2 ā Ze2
4ĻĪµ0r+ eF ir (1+ (Z ā1)(me / mt ))
Sincethecenterofmassandtheinternalmotionareindependentthewavefunction
ĪØ(R, r ) = Ļ(
R)Ļ (r ) factorizesintoaproductwhere
ā
2
2mt
ācm2 +e
F iR(1ā Z )
ā
āāā
ā āĻ(R) = EcmĻ(
R)
and
ā
2
2Āµā2 ā Ze2
4ĻĪµ0r+ qF ir
ā
āāā
ā āĻ (r ) = EintĻ (r )
where q = e(1+ (Z ā1)(me / mt )) ,aneffectivechargeforZ ā 1 .Thecenterofmass
motionisconsideredlatterandwenowfocusontheeffectofthefieldonthe
internalmotionoftheelectron.Beforegettingintoanydetailletsconsidertheorder
ofmagnitudeoftheinternalelectricfieldbeforeapplyingtheexternalfield.This
fieldatadistance r fromthenucleusisgivenby
Ze4ĻĪµ0r
2 andaftersubstitutingthe
numericalvaluesoftheconstantsthefieldstrengthis (1.4399649 Ć1011) Zr2voltsm
with
r inĆ .ThefirstBohrorbitinHhasaradiusof0.5291772Ć sothefieldinHatthat
distancefromthenucleusis 5.1422081Ć1011 voltsm
.Laboratoryfieldsvarywidely
but107 voltsm
isreasonableandissmallerthantheinternalfieldbyafactorof
approximately104 suggestingthatperturbationtheorywillbeadequatetoestimate
thechangeinenergyoftheoneelectronatomintypicallaboratoryfields.
TheunperturbedinternalHamiltonianis
H 0 = ā
2
2Āµā2 ā Ze2
4ĻĪµ0r
where
H0Ļ nlm
0 = En0Ļ nlm
0
andEn0 = āe2Z 2
2(4ĻĪµ0 )aĀµn2
IfwemeasurelengthinmultiplesofĪ³ a0 andtheenergyinmultiplesofe2
(4ĻĪµ0 )aĀµwe
have
H 0 = ā 1
2ā2 ā Z
rwithEn
0 = āZ 2
2n2
Letschoose
F tobealongthezaxis,
F = Fz
H = ā 1
2ā2 ā Z
r+ qFr cosĪø = H 0 + qFr cosĪø = H 0 +VF
whereĪø istheanglebetweenthefieldandthepositionvector
r andV = qr cosĪø .In
theseunitsF ismeasuredinmultiplesof e(4ĻĪµ0 )aĀµ
2 and q = 1+ (Z ā1) / 1+ 1
mn
ā
āāā
ā ā
withmn inmultiplesoftheelectronmass.Letsfirstconsidertheeffectoftheperturbationonthegroundstatewave-function
Ļ 0 ā”Ļ 100 withenergyE0 = ā Z2
2.Sincethewave-functionandenergyinthefield
aresolutionsto
H 0 + VF( )Ļ = EĻ
wemayexpandE&Ļ inpowersofF
Ļ = Ļ nF
n
n=0
ā
ā andE = EnFn
n=0
ā
ā
TheSchrodingerequationbecomes
H 0 +VF( ) Ļ nF n
n=0
ā
ā = Ļ nF n
n=0
ā
ā EmF m
m=0
ā
ā
whereĻ n & En arethenthordercorrectionstothewave-functionandenergy.Wecanwritethisequationas
F p H 0Ļ p + 1āĪ“ p0( )VĻ pā1 ā EpākĻ k
k=0
p
āāāā
āā āp=0
ā
ā = 0
andsinceitmustbetrueforanyF eachcoefficientofF p mustbezero.AccordinglywerecovertheusualRayleigh-Schrodingerperturbationequations
H 0Ļ p + 1āĪ“ p0( )VĻ pā1 = EpākĻ k
k=0
p
ā
Thefirstfewequationsarep=0 H 0Ļ 0 = E0Ļ 0 ,theunperturbedequation
whereĻ 0 =Z 3
ĻeāZr andE0 = ā Z
2
2
p=1; (H 0 ā E0 )Ļ 1 + (V ā E1)Ļ 0 = 0 ,thefirstorderequationp=2; (H 0 ā E0 )Ļ 2 + (V ā E1)Ļ 1 = E2Ļ 0 ,thesecondorderequationp=3; (H 0 ā E0 )Ļ 3 + (V ā E1)Ļ 2 = E3Ļ 0 + E2Ļ 1 ,thethirdorderequationp=4; (H 0 ā E0 )Ļ 4 + VĻ 3 = E4Ļ 0 + E2Ļ 2 Notethatsincetheenergycannotdependonthedirectionoftheelectricfieldalloddordercorrections, E1,E3,mustbezero.WecanverifythisexplicitlyforE1asfollows.Multiplythefirstorderequationbythegroundstatewave-functionandintegrate
0 (H 0 ā E0 ) 1 + 0 (V ā E1) 0 = 0 Since isHermitian
0 (H 0 ā E0 ) 1 = 1 (H 0 ā E0 ) 0 = 0 andsothefirstordercorrectiontotheenergyistheaverageoftheperturbationoverthegroundstatewave-functionwhichiszerobysymmetry.
E1 = 0 V 0 = 2qZ 3 eā2Zrr3 dr
0
ā
ā« cos(Īø )sin(Īø )dĪø = 00
Ļ
ā«
Consequentlythefirstorderequationbecomes
(H 0 ā E0 )Ļ 1 + VĻ 0 = 0 IthappensthatonecansolvetheaboveperturbationequationsexactlytodetermineĻ n & En andwewilldosobeforediscussingthemorecommonapproachofexpandingthefunctionsĻ n intermsoftheeigenfunctionsof H
0 .
H 0
Wenotethattheatominthefieldhascylindricalsymmetryandlookforasolutionoftheform Ļ 1 = g(r,Īø )Ļ 0 Since
(H 0 ā E0 )g(r,Īø )Ļ 0 =Ļ 0 ā 1
2ā2gār2
+ āgār(Z ā 1
r)+ L
2g2r2
āāā
āā ā
thefirstorderequationbecomes
ā2gār2
ā 2 āgār(Z ā 1
r)ā L
2gr2
= 2qr cosĪø Themostgeneralformof g(r,Īø ) is
g(r,Īø ) = f(r)P(cosĪø )
=0
ā
ā where P(cosĪø ) isaLegendrepolynomial.
Substitutingthisformintothedifferentialequationfor g andnoting
L2P(cosĪø ) = (+1)P(cosĪø )
resultsin
P
=0
ā
ā f'' + 2 f
' (1rā Z )ā (+1)
r2f
āāā
āā ā = 2qrP1
wherewenotethat cosĪø = P1andconsequentlyonlythetermwith = 1 contributes.Accordinglywehave
f '' + 2 f ' (1
rā Z )ā 2
r2f = 2qr
Thesolutiontowhichis
f = ā q
Z 2r + Zr
2
2āāā
āā ā
andso
Ļ 1 = ā q
Z 2r + Zr
2
2āāā
āā āP1(cosĪø )Ļ 0
TofindE2 wemultiplythesecondorderequationbyĻ 0 andintegrate
0 (H 0 ā E0 ) 2 + 0 V 1 = E2 Since isHermitian
0 (H 0 ā E0 ) 2 = 2 (H 0 ā E0 ) 0 = 0 wehave
E2 = 0 V 1 Doingtheintegrationresultsin
E2 = ā 9q
2
4Z 4
andsincethedipolepolarizabilityĪ± isdefinedbyE2 = ā 12Ī±F2 wehave
Ī± = 9q
2
2Z 4
TofindE4 weusethep=4equation whichaftermultiplyingbyĻ 0 andintegratingresultsin
E4 = 0 V 3 ā E2 0 2 Notethatusingthefirstorderequationwecanwrite
0 V 3 = ā 1 H 0 ā E0 3 butfromthep=3equation
H 0
(H 0 ā E0 )Ļ 4 + VĻ 3 = E4Ļ 0 + E2Ļ 2
H 0 ā E0( )Ļ 3 = E2Ļ 1 ā VĻ 2
whichresultsin
0 V 3 = 1 V 2 ā E2 1 1 andfinally
E4 = 1 V 2 ā E2 0 2 ā E2 1 1 Sothefourthordercorrectiontotheenergydependsonthefirstandsecondordercorrectionstothewavefunctionandsincewehavethefirstorderweneedthesecondordercorrectionwhichwewillgetfromthep=2equation
(H 0 ā E0 )Ļ 2 + VĻ 1 = E2Ļ 0 using
Ļ 1 = ā q
Z 2r + Zr
2
2āāā
āā āP1(cosĪø )Ļ 0 ,V = qr cosĪø andE2 = ā 9q
2
4Z 4
andwritingĻ 2 = g(r,Īø )Ļ 0 resultsin
ā2gār2
ā 2 āgār(Z ā 1
r)ā L
2gr2
+ 2q2
Z 2r2 + Zr
3
2āāā
āā ācos2Īø = 9q
2
2Z 4
using
g(r,Īø ) = f(r)P(cosĪø )
=0
ā
ā
resultsin
P
=0
ā
ā f'' + 2 f
' (1rā Z )ā (+1)
r2f
āāā
āā ā +
2q2
Z 2r2 + Zr
3
2āāā
āā ācos2Īø = 9q
2
2Z 4
andsince
cos2Īø = P12 = 2
3P2 +
13
only = 0&2 contribute.Letting A = q2
Z 4 resultsinthetwodifferentialequations
f2" + 2 f2
' (1rā Z )ā 6
r2f2 +
4AZ 2
3r2 + Zr
3
2āāā
āā ā= 0
and
f0" + 2 f0
' (1rā Z )+ 2AZ
2
3r2 + Zr
3
2āāā
āā ā= 9A2
Thesolutionsbeing
f2 =A24
2Z 2r4 +10Zr3 +15r2( ) and f0 = A24
Z 2r4 + 6Zr3 +18r2( ) so
Ļ 2 = f0Ļ 0 + f2P2 (cosĪø )Ļ 0 Asseenabove
E4 = 1 V 2 ā E2 0 2 ā E2 1 1 anddoingthevariousintegrationsresultsin
1 V 2 = ā 2529
32q4
Z10 , 0 2 = 8116
q2
Z 6 , 1 1 = 438q2
Z 6
andfinally
E4 = ā q4
Z10355564
.Sincethesecondhyper-polarizabilityĪ³ isdefinedby
E = E0 ā12Ī±F2 ā 1
24Ī³ F 4 onehasĪ³ = 10665
8q4
Z10
WiththeseresultswecancalculatethedipolemomentĀµ oftheoneelectronatominducedbytheelectricfieldsincebydefinition
Āµ = ā dEdF
=Ī±F + 16Ī³ F 3
Inadditiontothedipolemomenthighermomentsareinducedbythefield.Forexamplethezzcomponentoftheinducedquadrupolemomentisgivenby
Īzz ā” Ī =Ļ r2P2 (cosĪø )Ļ
Ļ Ļ
ThenumeratorisĻ r2P2 (cosĪø )Ļ = 2 0 r2P2 2 + 1 r2P2 1( )F2 + and
0 r2P2 2 = 1178
q2
Z 8 & 1 r2P2 1 = 24 q2
Z 8
andsince
Ļ Ļ = 1+ 2 0 2 + 1 1( )F2 + theinducedquadrupolethroughorderF2 issimplythenumerator
Ī = 2134
q2F2
Z 8
WhatarethesemomentsintheSIsystem?
EnergyandelectricfieldintheSIandausystemsarerelatedby ESI =
e2
4ĻĪµ0aĀµ
ā
āā
ā
ā ā Eau
and FSI =
e4ĻĪµ0aĀµ
2
ā
āā
ā
ā ā Fau soif
212SI SI SIE FĪ±= ā then
ESI =
e2
4ĻĪµ0aĀµ
ā
āā
ā
ā ā Eau = ā 1
2Ī± au Fau
2 e2
4ĻĪµ0aĀµ
ā
āā
ā
ā ā = ā 1
2Ī± SI Fau
2 e4ĻĪµ0aĀµ
2
ā
āā
ā
ā ā
2
andso
3
0 04SI au aĪ± Ī± ĻĪµ= .Thequantity04
SIĪ±ĻĪµ
isusuallyreportedandiscalledthe
polarizabilityvolume.Noteitās 3 30 30 0 1481847 10au aua . x mĪ± Ī±ā= .Thepolarizability
volumeoftheHatomintheSIsystemisthen
Ī± SI
4ĻĪµ0
= 0.1481847x10ā30 92( ) qau
Z 4
āāā
āā ā
m3 .
Thethirdorderfunctionisgivenby ( ) ( )3 3 0100 1 3 100F dP ePĪ¦ = + Ī¦ 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 cosĪø thesecondorderchangeinthegroundstateenergyofHisgivenbythegeneralformula
E100
( 2 ) = ' V100 ,nlmVnlm,100
E100 ā Enlmnlmā whereV100,nlm = Ļ 100 rF cosĪø Ļ nlm andEnlm = ā 1
2n2
whichassumesthattheunperturbedfunctionsarecompletewithrespecttotheperturbedsolutions.TherearetwoclassesofeigenfunctionsfortheHatom,thosethatdescribetheboundstates, 0E < ,andthosethatdescribethecontinuumstates,
0E > .Sincewehavesolvedthedifferentialequationexactlyweknowtheexact
secondorderenergy, ( )2 2100
94
E F= ā ,soletsseewhattheerrorisifweuseonlythe
boundwavefunctionstocalculatethesecondordercorrection.Accordingly
E100( 2 ) =
100 Fr cosĪø nlm2
E10 ā En
0nlmā = F 2
100 r cosĪø nlm2
E10 ā En
0nlmā = 2F 2
100 r cosĪø nlm2
ā1+ 1n2
āāā
āā ā
nlmā
Therequiredmatrixelementis
100 r cosĪø nlm = dĻ
0
2Ļ
ā« sinĪø dĪø0
Ļ
ā« Y00 cosĪø Yl
m dr R10 r( )0
ā
ā« r3Rnl r( )
andtheangularfactorcanbeeasilyevaluatedbyobservingthat 01
43
cos YĻĪø = 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
Ļ Ļ Ļ Ļ
Ļ Īø Īø Īø Ļ Īø Īø Ī“ Ī“= =ā« ā« ā« ā« so
100 r cosĪø nlm = 1
3Ī“1lĪ“0m dr R10 r( )
0
ā
ā« r3Rn1 r( ) = 13Ī“1lĪ“0m R10 r Rn1
2
10 12 2100
22
213 1
n( )
n
R r RE F
n
ā
=
= āā ā+ āā āā ā
ā .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
āā
+=
ā= ā = ā
+ā
SincetheexactE1s(2) = ā 9
4F2 theboundstatewave-functionsthenaccountfor~81%
oftheexactresult,asignificanterror!Theneedtoincludethecontinuumsolutionsintheexpansionmakessensephysicallybecauseintheelectricfieldtheelectron
feelsthepotential 1 Fr cosr
Īøā + andif 0cosĪø < 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