1 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldoátomodeHidrogênioUmaaplicaçãodaTeoriadePerturbaçãoEstacionária.

• A Hamiltoniana (perturbacao) Wz.

Coloque um campo magnetico B0 paralelo ao eixo Oz. Esse campo interage com

momentos magneticos

8><

>:

ML=q

2meL ! Momento magnetico orbital;

MI=gpµn

~ I=� gpq2Mp

I ! Momento Magnetico do proton;

MS=2µB

~ S= qme

S ! Momento Magnetico do eletron.

• A perturbacao sobre o atomo de hidrogenio (Wz) e dada por:

Wz = �B0 · (ML +MS +MI) = !0(Lz + 2Sz) + !nIz,

onde!0 (frequencia angular de Larmor) e!n sao definidos por (lembre que B0kz) :

!0 = � q

2meB0 e !n =

q

2MpgpB0, com |!0

!n| = gp

Mp

me>> 1.

• A frequencia de Larmor e muito maior que a nuclear.

• De fato, falta coisa emWz (um termo quadratico emB0�o termo diamagnetico),

mas esses termos nao atuam sobre as variaveis de spin eletronico e nuclear. Ele

simplesmente desloca o nıvel 1s como um todo. Veja complemento DVII.

• A perturbacao “vista” pelo nıvel 1s: Escolhemos esse nıvel por simplicidade

(so tem efeitos de estrutura hiper-fina). O nıvel 2 tem estrutura fina e hiper-fina

e seria mais complicado. O que aprenderemos sobre o 1s podera ser

usado para o nıvel 2.

2 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldoátomodeHidrogênioUmaaplicaçãodaTeoriadePerturbaçãoEstacionária.

figura-XII-5dotexto

• Mesmo com o mais forte dos campos magneticos que podem ser produzidos em

laboratorio, o efeito de deslocamento de linhas causado por Wz e muito menor

que a distancia entre 1s e os outros nıveis.

) O efeito de Wz pode ser tratado como uma perturbacao de um efeito maior.

• Na aula de hoje estudaremos o porque do efeito Zeeman se manifestar conforme

o diagrama abaixo (B0 6= 0 quebra a degenerescencia).

• Se B0 e muito grande, Wz pode ser da ordem

ou ate maior que o efeito de Whf . Lembre que

Wf so causa um deslocamento em 1s.

• Se B0 e muito fraco, Wz << Whf e isso pode

afetar a hierarquia da teoria de perturbacao.

• Em alguns casos, Wz e perturbacao de Whf .

Em outros,Whf e perturbacao deWz e as vezes,

ambos tem que ser tratados em pe de igualdade

(diagonalizados juntos).

• O termo de estrutura hiper-fina para 1s e AS·I.Os outros, vimos que nao contribuem.

3 MAPLima

F789 Aula 19 tNa pratica, escolheremos se Wz precisara ser diagonalizado em

E1s = {|n = 1; ` = 0;mL = 0;mS = ±1

2;mI = ±1

2i}

ou em

E1s = {|n = 1; ` = 0;mL = 0;F = 0, 1;mF i}tNote que Wz=!0(Lz+2Sz)+!nIz tem apenas operadores de momento angular.

Como Lz da zero sobre kets com ` = 0, a parte orbital e facilmente resolvida,

uma vez que hn = 1; ` = 0;mL = 0|n = 1; ` = 0;mL = 0i = 1tNosso problema, para o caso de hierarquia nao definida, passa a ser o de

diagonalizar o operador AS · I+2!0Sz+!nIz, cujas partes atuam somente

sobre graus de liberdade de spin.tPara simplificar ainda mais, desprezaremos o termo !nIz, considerando que

!n << !0 (o complemento CXII nao faz isso).tUsaremos a nomenclatura E1s.= {|mS = ±1

2;mI = ±1

2i} .

= {|F = 0, 1;mF i},

para descrever o espaco quadri-degenerado (quando sob a acao exclusiva de

H0) associado ao nıvel 1s.tAssim, dependendo da intensidade de B0, diagonalizaremos em uma

dessas bases, o operador AS · I+ 2!0Sz.

EfeitoZeemandoestadofundamentaldoátomodeHidrogênio

4 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldoátomodeHidrogêniotDiferentes domınios da forca do campo.

• ~!0 << A~2 �! fraco;

• ~!0 >> A~2 �! forte;

• ~!0 ⇡ A~2 �! intermediario.tPodemos diagonalizar AS · I+ 2!0Sz exatamente. No entanto, para estudar

teoria de perturbacao faremos o seguinte:

• 2!0Sz sera tratado como perturbacao perto de AS · I;• AS · I sera tratado como perturbacao perto de 2!0Sz;

• Diagonalizacao exata e seus limites para comparar com os casos acima.tEfeito Zeeman de campo fraco.

• Nestas condicoes os melhores estados sao os auto-estados de AS · I.,

ou seja

({|F = 1;mF = �1, 0, 1i} ! A~2

4 (3-degenerado)

{|F = 0;mF = 0i} ! � 3A~2

4 (nao-degenerado)

• Partiremos desta base e diagonalizaremos 2!0Sz, conforme aprendemos

em teoria de perturbacao estacionaria.

• Isso exigira calcular elementos do tipo

8><

>:

hF = 1;mF |Sz|F = 1;m0F i;

hF = 1;mF |Sz|F = 0; 0i.hF = 0; 0|Sz|F = 0; 0i.

5 MAPLima

F789 Aula 19 tSaberemos facilmente aplicar Sz em cada ket da base {|F,mF i}, se escrevermos

cada ket dessa base como uma combinacao dos kets da base {|mS ;mIi}. Lembre

que Sz|mS ;mIi = mS~|mS ;mIi.tAs relacoes entre as bases foram feitas no capıtulo X, aula 10. Reproduzidas

ao lado:

8>>>><

>>>>:

|1,+1i = |++i|1,�1i = |��i|1, 0i = 1p

2(|+�i+ |�+i)

|0, 0i = 1p2(|+�i � |�+i)

)

8>>>><

>>>>:

|++i = |1,+1i|��i = |1,�1i|+�i = 1p

2(|1, 0i+ |0, 0i)

|�+i = 1p2(|1, 0i � |0, 0i)

tCom elas, podemos escrever

8>>>><

>>>>:

Sz|1,+1i = ~2 |++i = ~

2 |1,+1iSz|1,�1i = �~

2 |��i = �~2 |1,�1i

Sz|1, 0i = ~2

1p2(|+�i � |�+i) = ~

2 |0, 0iSz|0, 0i = ~

21p2(|+�i+ |�+i) = ~

2 |1, 0itNote que Sz nao e diagonal na representacao {|F,mF i}, pois ele nao comuta

com um dos geradores desta base, o operador F2.

tOs elementos nao diagonais vem de

(h0, 0|Sz|1, 0i = h0, 0|~2 |0, 0i = ~

2 ;

h1, 0|Sz|0, 0i = h1, 0|~2 |1, 0i = ~2 .

EfeitoZeemandoestadofundamentaldeH:campofraco

6 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campofracotA matriz que representa Sz na base {|F,mF i}, e dada por:

|1, 1i|1, 0i|1,�1i|0, 0i

Sz.=

h1, 1|h1, 0|h1,�1|h0, 0|

~2

0

BB@

1 0 0 00 0 0 10 0 �1 00 1 0 0

1

CCA

tCompare com a representacao matricial de Fz, dada por

|1, 1i|1, 0i|1,�1i|0, 0i

Fz.=

h1, 1|h1, 0|h1,�1|h0, 0|

~

0

BB@

1 0 0 00 0 0 00 0 �1 00 0 0 0

1

CCA

tNote que para todos os efeitos, Sz e diagonal em F = 1 (sozinho - caixa verde)

e em F = 0 (sozinho - caixa azul)tEssa propriedade esta ligada ao teorema de Wigner-Eckart, conforme discutido

aula passada e relembrado a seguir (detalhes no complemento DX).

No caso, temos P1SzP1 =1

2P1FzP1, com P1 =

+1X

m1=�1

|1,m1ih1,m1|.

7 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campofraco

tO teorema diz:

8><

>:

Em um dado auto-subespaco de momento angular total

todas as matrizes que representam vetores (operadores

vetoriais) sao proporcionais.

• Esse sub-espaco, E(k, J) = {|k, J,mji}, e de autokets de J2e Jz com os

mesmos J e k.

• A constante de proporcionalidade (teorema de projecao), e dada por (dentro

do sub-espaco E(k, J)) :

V =hJ ·Vik,JhJ2ik,J

J,

onde a notacao hAik,J indica que esse “valor medio” nao depende de mJ .

Note que V e diagonal nesse subespaco.

• No presente caso (campo fraco), temos J=F e V=S, gerando:

hF · SiF=1hF2iF=1

J=12

�F (F+1)+S(S+1)�I(I+1)

�

F (F+1)=

12

�1(1+1)+

12 (

12+1)�

12 (

12+1)

�

1(1+1)=

1

2tSob ponto de vista pratico, a matriz que representa a perturbacao no espaco

degenerado, F = 1, ja esta diagonalizada e, portanto, basta somar aos

elementos da diagonal na matriz que representa H0 +Whf , para obter os

termos em energia ate primeira ordem.<latexit 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8 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campofraco

figura-XII-5dotexto

tOs auto-estados em ordem zero e auto-valores em primeira ordem do atomo

de hidrogenio, quando submetido a um campo magnetico fraco.

Auto-estados Auto-valores

|F = 1;mf = 1i A~24

+ ~!0

|F = 1;mf = 0i A~24

+ 0

|F = 1;mf = �1i A~24

� ~!0

|F = 0;mf = 0i � 3A~24

+ 0

) valido quando ~!0 << A~2.

tO tratamento feito e valido enquanto a diferenca ~!0 entre dois nıveis

Zeeman adjacentes for muito menor que diferenca (com campo zerado) entre

os nıveis F = 1 e F = 0 (estrutura hiper-fina).tA comparacao entre o tratamento perturbativo (de fato, as frequencias de Bohr

que aparecem na evolucao temporal de hFi e hSi) e o modelo vetorial do atomo

(secao 2.c.) fica para a casa.

9 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campofortetO efeito Zeeman de campo forte, W = Whf +Wz, com Wz >> Whf .

Agora a estrategia e comecar com W = Wz e tratar Whf como perturbacao.

• Auto-estados e auto-valores do termo de Zeeman. Lembre que esse termo

e diagonal na base {|mS ;mIi} ) 2!0Sz|mS ;mIi = 2mS~!0|mS ;mIi.

Sz so atua no eletron e mS = ±1

2, ou seja, os auto-valores sao ± ~!0.

assim, temos

(2!0Sz|+,±i = +~!0|+,±i2!0Sz|�,±i = �~!0|�,±i

! dois sub-espacos, cada qual

duplamente degenerado

(|+,±i ! +~!0

|�,±i ! �~!0

• As correcoes em primeira ordem podem ser obtidas diagonalizando AS · Inestes dois sub-espacos. Novamente o assunto e mais simples do que parece,

pois AS · I e diagonal em cada sub-espaco. Para ver isso, lembre que

[F2, Fz] = 0 e da seguinte propriedade de F689: Se

(A|'ni = an|'ni[A,B] = 0

!

h'm|[A,B]|'ni=0!(am�an)h'm|B|'ni=0 e h'm|B|'ni=0, caso am 6=an.

Como, AS · I = A2(F2 � S2 � I2) tambem comuta com Fz, temos

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10 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campoforte

• Os termos cruzados

(h+;�|AS · I|+;+i = 0

h+;+|AS · I|+;�i = 0pois

(Fz|++i = 1~|++iFz|+�i = 0~|+�i

| {z }autovalores distintos.

• Da mesma forma

(h�;�|AS · I|�; +i = 0

h�; +|AS · I|�;�i = 0pois

(Fz|�+i = 0~|�+iFz|��i = �1~|��i

| {z }autovalores distintos.

• Ou seja, AS · I e diagonal em ambos os sub-espacos.tComo todos os termos fora da diagonal sao nulos, precisamos apenas calcular

elementos do tipo hmS ;mI |AS · I|mS ;mIi.

Para isso lembre que S · I = SzIz +1

2

�S+I� + S�I+

�e que os dois ultimos

termos�S+I� + S�I+

�nao contribuem para termos da diagonal, uma vez que

eles mudam os valores de mI e mS .Podemos, desta forma, escrever:

hmS ;mI |AS · I|mS ;mIi = hmS ;mI |ASzIz|mS ;mIi = A~2mSmI .

• Em seguida, construiremos uma tabela similar ao caso do campo fraco.<latexit 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11 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campoforte

figura-XII-7

dotexto

~!0

E

tOs auto-estados em ordem zero e auto-valores em primeira ordem do atomo

de hidrogenio, quando submetido a um campo magnetico forte.

Auto-estados Auto-valores

|+;+i + ~!0 +A~24

|+;�i + ~!0 �A~24

|�; +i � ~!0 �A~24

|�;�i � ~!0 +A~24

) valido quando ~!0 >> A~2.

tInterpretacao sobre a separacaoA~22

devido ao campo forte entre os dois

estados {|+;±i}⇣ou entre {|�;±i}: Vimos que a Hamiltoniana total poderia

ser escrita por 2!0Sz +AIzSz = 2(!0 +A2Iz)Sz. O campo magnetico B0

parece estar fortalecido (caso mI =+1/2) ou enfraquecido (mI =�1/2)por um campo magnetico interno devido a Whf .

12 MAPLima

F789 Aula 19

EfeitoZeemandoestadofundamentaldeH:campointermediário

figura-XII-9dotexto

Note as retas cheias correspondentes ao primeiro bloco.

tJa temos todos os elementos para escrever a matriz que representa a

perturbacao total, Whf = 2!0Sz +AS · I, na base {|F,mF i}.� No que diz respeito ao termo AS · I, o calculo

e direto, uma vez que esse termo pode ser escrito

porA2

�F2 � S2 � I2

�. Nesta base a matriz que

que representa esse termo e dada por:

|1, 1i |1,�1i |1, 0i |0, 0i

AS · I .=

h1, 1|h1,�1|h1, 0|h0, 0|

0

BBB@

A~2

4 0 0 0

0A~2

4 0 0

0 0A~2

4 0

0 0 0 � 3A~2

4

1

CCCA

� Por outro lado, obtemos em aula:

|1, 1i |1,�1i |1, 0i |0, 0i

2!0Sz.=

h1, 1|h1,�1|h1, 0|h0, 0|

0

BB@

+~!0 0 0 0

0 �~!0 0 0

0 0 0 +~!0

0 0 +~!0 0

1

CCA

tAo somar, a matriz fica bloco diagonal. O segundo bloco (mF = 0)

nao e diagonal.

13 MAPLima

F789 Aula 19 tAssim a matriz fica:

|1, 1i |1,�1i |1, 0i |0, 0i

Whf.=

h1, 1|h1,�1|h1, 0|h0, 0|

0

BBB@

A~2

4 +~!0 0 0 0

0A~2

4 �~!0 0 0

0 0A~2

4 +~!0

0 0 +~!0 � 3A~2

4

1

CCCA

� Os termos diagonais tem auto-energias e auto-estados

dados por

(E1 =

A~2

4 +~!0 ! |1,+1i = |++iE2 =

A~2

4 �~!0 ! |1,�1i = |��i

� Diagonalizando o bloco menor (faca em casa), obtemos:

� = �A~24

±r⇣A~2

2

⌘2+ ~2!2

0

Esses resultados correspondem as curvas da figura ao

lado.

� Note que !0 >> 1 ) � = �A~24

± ~!0 (conhecido).

EfeitoZeemandoestadofundamentaldeH:campointermediário

figura-XII-9dotexto

Note as retas cheias correspondentes ao primeiro bloco.