1 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.E

0

(n = 4)

(n = 3)

(n = 2)

(n = 1)

�EI

4s 4p 4d 4f3s 3p 3d

2s 2p

1s

`=0 `=1 `=2 `=3

(s) (p) (d) (f)

Fig. 4, cap. 7 do texto

• Ek,` = � EI

(k + `)2= �EI

n2= En;

• A degenerescencia orbital total

e gn = n2 ! g2 = 4;

• Os spins do eletron e proton

ampliam a degenerescencia

para gS,I2 = g2 ⇤ 2 ⇤ 2 = 16;

• Cinco numeros quanticos (n, `,mL,

mS ,mI) especificam o autoket

|n, `,mL, S,mS , I,mIi simultaneo do

CCOC: H,L2, Lz,S

2, Sz, I

2, Iz.

• Nosso problema se reduz a

diagonalizar uma matriz

16⇥ 16.

2 MAPLima

F789 Aula 17

Dimensãodoproblema tO nıvel 2 do atomo de Hidrogenio

n = 2, E2 = �1

8

e2

a0

(2s ! ` = 0,mL = 0,mS = ± 1

2 ,mI = ± 12

2p ! ` = 1,mL = �1, 0, 1,mS = ± 12 ,mI = ± 1

2tNa estrutura fina introduzirıamos apenas o spin do eletron e a matriz

seria 8⇥ 8:

mS = +1

2mS = �1

2z }| {2s 2p�1 2p0 2p+1

z }| {2s 2p�1 2p0 2p+1

2s, mS = +1/22p�1, mS = +1/22p0, mS = +1/2

2p+1, mS = +1/22s, mS = �1/2

2p�1, mS = �1/22p0, mS = �1/2

2p+1, mS = �1/2

0

BBBBBBBBBB@

W11 W12 W13 W14 W15 W16 W17 W18

W21 W22 W23 W24 W25 W26 W27 W28

W31 W32 W33 W34 W35 W36 W37 W38

W41 W42 W43 W44 W45 W46 W47 W48

W51 W52 W53 W54 W55 W56 W57 W58

W61 W62 W63 W64 W65 W66 W67 W68

W71 W72 W73 W74 W75 W76 W77 W78

W81 W82 W83 W84 W85 W86 W87 W88

1

CCCCCCCCCCA

tMesmo para esse caso, a matriz e grande. Entretanto, muitos dos elementos

de matriz serao zero por simetria e a matriz sera bloco diagonal.

3 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.tA Hamiltoniana do atomo de hidrogenio foi escrita por H = H0 +W, com

W = Wf +Whf e Wf cerca de 2000 mais forte que Whf . Assim, faz sentido,

ignorar nesse momento a estrutura hiper-fina e estudar a influencia apenas

da estrutura fina. E bastante comum incluir os efeitos da estrutura hiper-

fina sobre os da estrutura fina (perturbacao da perturbacao).tNa aula que vem, estudaremos a quebra de degenerescencia no nıvel 1 do

atomo de hidrogenio devido a estrutura hiper-fina, Whf .tNa aula passada, escrevemos os termos de estrutura fina por:

H =P2

2me+ V (R)

| {z }� P4

8m3ec

2

| {z }+

1

2m2ec

2

1

R

dV (R)

dRL.S

| {z }+

~28m2

ec2�V (R)

| {z }H0 Wmv WSO WD| {z }

termos de estrutura finatVeremos que a degenerescencia (g2 = 8) do nıvel n=2 de H0 (matriz

diagonal 8⇥ 8) sera parcialmente removida por Wf (representada pela

matriz do slide 2).tLembre que a diagonalizacao de Wf fornece correcao em primeira ordem

em energia e de ordem zero nos kets.

4 MAPLima

F789 Aula 17 tWf nao acopla 2s com 2p. Para mostrar isso basta notar que os

operadores envolvidos respeitam: [L2,Wf ] = 0.

Isso e decorrente de

8><

>:

[L2,P2] = 0

[L2, f(R)] = 0

[L2,L.S] = 0

e

8><

>:

h2s|[L2, O|2pi = 0

com

O = P2, f(R), ou L.S

Como os autovalores de L2 aplicado em |2si e |2pi sao distintos

concluı-se que h2s|O|2pi = 0 e ) h2s|Wf |2pi = 0.tDesta forma, a matriz, 8⇥ 8 do slide 2 fica:

mS = +1

2mS = �1

2z }| {2s 2p�1 2p0 2p+1

z }| {2s 2p�1 2p0 2p+1

2s, mS = +1/22p�1, mS = +1/22p0, mS = +1/2

2p+1, mS = +1/22s, mS = �1/2

2p�1, mS = �1/22p0, mS = �1/2

2p+1, mS = �1/2

0

BBBBBBBBBB@

W11 0 0 0 W15 0 0 00 W22 W23 W24 0 W26 W27 W28

0 W32 W33 W34 0 W36 W37 W38

0 W42 W43 W44 0 W46 W47 W48

W51 0 0 0 W55 0 0 00 W62 W63 W64 0 W66 W67 W68

0 W72 W73 W74 0 W76 W77 W78

0 W82 W83 W84 0 W86 W87 W88

1

CCCCCCCCCCA

Estruturafinadoníveln=2doÁtomodeHidrogênio.

5 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.tNote que o nao acoplamento entre 2s e 2p pode ser deduzido partir

de

(2s e par

2p e ımpare Wf e par, pois na inversao

8>>><

>>>:

R ! �R

P ! �P

L ! L

S ! StNo nosso livro texto a matriz Wf e organizada na forma (bloco-diagonal)

2s" 2s# 2p�1" 2p0" 2p+1" 2p�1# 2p0# 2p+1#2s "2s #

2p�1 "2p0 "

2p+1 "2p�1 #2p0 #

2p+1 #

0

BBBBBBBBBB@

W11 W12 0 0 0 0 0 0W21 W22 0 0 0 0 0 00 0 W33 W34 W35 W36 W37 W38

0 0 W43 W44 W45 W46 W47 W48

0 0 W53 W54 W55 W56 W57 W58

0 0 W63 W64 W65 W66 W67 W68

0 0 W73 W74 W75 W76 W77 W78

0 0 W83 W84 W85 W86 W87 W88

1

CCCCCCCCCCA

tAssim, podemos separar o problema, calculando os elementos de matrizes

envolvendo so o nıvel 2s, matriz 2⇥ 2 e so o nıvel 2p, matriz 6⇥ 6, ja que

Wf e bloco-diagonal nessa base.<latexit 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sha1_base64="QmtjTKXSP/1ZbQg2JY57GWjtbIU=">AAAOEnicrVfNchtFEO4ECEFI4AC3XAZcpBJiu1ay/o6huHCgKENwnCrL5RqNRvKS/WNnZHBUeoZc8ipcOEBRXDlx40hx4pATL0BPz66l2V1vVIBUu5rp/rq/7p6dndY4CXylPe/3a9dfefW1G6/ffKPxZrP11ttbt955pOJ5KuShiIM4fTzmSgZ+JA+1rwP5OEklD8eBPBo/+cToj85lqvw4+kpfJPIk5LPIn/qCaxTFt268B3dgBAnMQcNd2IF9uAcLlAjwIcV7ABI+QkkPlvgdAcsujfLv8L6AzyGmGYNv0Iv5jfGK4E/65egjRv8BjkLURog1cjtKCd8BRRIzSvCeIGJCc4X3FH/NbILen2JUE7I3KI46TXEyim2EuTQqYzT2S9KNcTRDmwilAj0YBpXpVngb0V+XLNZ7EZNkmBeY2QpnMpsUvLvRSIr3CE6p0ilaM5iWUOv8O1lVfIorQhmjHM6pPopqvazNbkHaKdp9ScgUUTM4QybjP0X7b1G3W8KNHNuDDW0PKmw/q7EtolzLhxtZPsxWsar+q2fjvz7ppvox+lQ0DtD2nGJhlzj71DNcVU5PuI9P7ebrHaN+hpYR2XHMhF+u+JS0IY3vYtYBzs3u2kWUj9IZjg0ywMyq9sO/v8x+GMH72TXPnsp8DapRE9JGa8gVLqFa7EKb4qzzm6O9l+Byn/c38HkVfznidby3MbIcRVUt8tVZ37NWz+ECrRf0PmlkVXXzsDvEaqp9N0pZVnvIM7tau8qmDpOz1EfjbRDv/Zd4Wu3w9WqNaP9OaW/XV1WUvqbKdne2iZvhyuTzDs7vYNxVl43HIjsFy07JklXYrqTWah+/y7VZ15n1HIZ96DvagTMbZu+XKpauw9J1WLoFlq7D0nVYugWWdY6ew9FzOHoFjp7D0XM4erWZ9B2WvsPSJ5bVrO/MBs6sjmPgcAwcjoHDMXA4Bg7HoJZj6HAMHY6hwzF0OIYOx5A4rtohzDlD71Ek7H86GT+mE9HHsynvVEzHFdJJabs4++bg1KvFhDF7eky+zYlm7ASdXQLjCOgEtJ1enHWHFpl3kCrrCt1ztnjalyOV1DXF6O08q1JMEf592be+IE2QnWc7FSd5hzz6FI0inHuOyxqPSaXHfsFjn1Cu16/hD8h77HKOm3UXdT2DiVTSOnJC2u5pL+c63dr29jz6sPKgnQ22IfscnG79NprEYh7KSIuAK3Xc9hJ9suCp9kUgl43RXMmEiyd8Jo9xGPFQqpMF/aVZsg9RMmHTOMUr0oyk6xYLHip1EY4RGXJ9poo6I6zSHc/1dHiy8KNkrmUkLNF0HjAdM/P/iE38VAodXOCAi9THWJk44ykXGv9FNbAI7WLK5cGjzl7b22t/0d1+0MnKcRNuwwe4udq4YR/Ap9ggH4JoPmt+3/yx+VPreeuH1s+tXyz0+rXM5l1wPq1f/wGDYabh</latexit><latexit 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6 MAPLima

F789 Aula 17

Estruturafinadoníveln=2sdoÁtomodeHidrogênio.tA representacao matricial de Wf na subcamada 2s, exige calculo de integrais

e o reconhecimento dos termos nao-acoplaveis por spin. Em especial, nota-se

que Wmv e WD nao dependem do operador de spin. Portanto, e claro que

h2s " |Wmv|2s #i = h2s # |Wmv|2s "i = h2s " |WD|2s #i = h2s # |WD|2s "i = 0

tO problema se reduz ao calculo de

8><

>:

h2s " |Wmv|2s "i = h2s # |Wmv|2s #i

h2s " |WD|2s "i = h2s # |WD|2s #i

tReescritas por8><

>:

hWmvi2s = hn=2; `=0;mL=0|� P4

8m3ec

2 |n=2; `=0;mL=0i

hWDi2s = hn=2; `=0;mL=0| ~2q8m2

ec2�V (R)|n=2; `=0;mL=0i

e, por sua vez, escrita na forma de integrais exclusivamente radiais, pois

Y 0

0(✓,') =

1p4⇡

. O complemento BXII indica como fazer essas integrais.

Aqui apresentamos apenas os resultados:

(hWmvi2s = � 13

128mec2↵4

hWDi2s = 1

16mec2↵4

Nao deixe de deduzir em casa. Note que h2s8 "# |WSO|2s8 "#i=0, pois `=0.tO efeito (sobre os nıveis 2s) da estrutura fina e deslocar de5

128mec

2↵4.

7 MAPLima

F789 Aula 17

Estruturafinadoníveln=2pdoÁtomodeHidrogênio.tA representacao matricial de Wf na subcamada 2p, exige calculo de integrais

e o reconhecimento dos termos nao-acoplaveis por spin. Em especial, nota-se

que Wmv e WD nao dependem do operador de spin. Portanto, e claro que

h2p " |Wmv|2p #i = h2p # |Wmv|2p "i = h2p " |WD|2p #i = h2p # |WD|2p "i = 0

tO problema se reduz ao calculo de

8><

>:

h2p " |Wmv|2p "i = h2p # |Wmv|2p #i

h2p " |WD|2p "i = h2p # |WD|2p #i

tReescritas por8><

>:

hWmvi2p = hn=2; `=1;mL|� P4

8m3ec

2 |n=2; `=1;m0Li

nos dois casos, mL = 0,±1 e m0L = 0,±1

hWDi2p = hn=2; `=1;mL| ~2q8m2

ec2�V (R)|n=2; `=1;m0

Li

tComo [Lz,Wmv] = [Lz,WD] = 0

8><

>:

hWmvi2p / �mL,m0L

hWDi2p / �mL,m0L

O complemento BXII indica, novamente, como fazer essas integrais (basta mL=0).

Aqui apresentamos os resultados:

(hWmvi2p = � 7

384mec2↵4

hWDi2p = 0 (devido ao �(R)).

lembre que limr!0

Rn,`=1(r) / r` = 0.<latexit 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8 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.tE o termo WSO? Precisamos calcular os varios elementos de matriz do tipo:

hWSOi2p(mL,m0L,mS ,m

0S) = hn=2; `=1;mL;mS |⇠(R)L.S|n=2; `=1;m0

L;m0Si

onde

8><

>:

mL e m0L = ±1, 0 ! associados ao momento angular orbital Lz;

mS e m0S = ±1/2 ! associados ao momento angular spin Sz;

⇠(R) =e2

2m2ec

21

R3 ! Operador radial (nao depende de ✓ e ').tNa representacao das coordenadas podemos separar a parte radial do elemento

de matriz acima das partes de spin e de momento angular orbital. Ficaria assim:

⇠2phn=2; `=1;mL;mS |L · S|n=2; `=1;m0L;m

0Si onde ⇠2p e um numero obtido

pela integral (ver BXII

) : ⇠2p =e2

2m2ec

2

Z 1

0

1

r3|R21(r)|2r2dr =

1

48~2mec2↵4.

tTendo resolvido a parte radial, nosso problema passa ser diagonalizar L · S na

base {|`;S;mL;mSi}, autokets de L2,S2, Lz e Sz, para o caso `=1 e S=1/2.tNo capıtulo 9, aprendemos uma outra base que descreve esse sub-espaco de

dimensao (2 ⇤ 1 + 1) ⇤ (2 ⇤ 1/2 + 1) = 6. E a de autokets de L2,S2,J2, e Jz,

dada por {|`;S; J ;mJi, com |`� S| J `+ S}. No caso, J = 1/2 ou 3/2.

Note que a dimensao deste sub-espaco e igual a: 2 ⇤ 1/2+1+2 ⇤ 3/2+1=6.tAprendemos a ir de uma base para outra (coeficientes de Clebsch-Gordan,

ver complemento AX). Por que isso e interessante?

9 MAPLima

F789 Aula 17

Estruturafinadoníveln=2pdoÁtomodeHidrogênio.tA segunda base e melhor. Porque? Por ja ser diagonal uma vez que

J2= (L+ S)2 = L2

+ S2+ 2L · S e ) L · S =

J2 � L2 � S2

2operadores que

definem a base junto com Jz.tOu seja, o operador spin-orbita WSO = ⇠2pL · S =1

2⇠2p(J

2 � L2 � S2) e diagonal

na base {|`;S; J ;mJi, J = 1/2 ou 3/2}, isto e

⇠2pL · S|`;S; J ;mJi =1

2⇠2p

⇥J(J + 1)� 1(1 + 1)� 1/2(1/2 + 1)

⇤~2|`;S; J ;mJi

)

8><

>:

|`=1;S=1/2; J=1/2;mJi auto-estados de ⇠2pL·S com auto-valor�⇠2p~2

|`=1;S=1/2; J=3/2;mJi auto-estados de ⇠2pL·S com auto-valor+1

2⇠2p~2tA degenerescencia entre J 0s e quebrada, embora a degenerescencia entre m0Js de

mesmo J continue mantida (os auto-valores nao dependem de mJ).tAssim, a degenerescencia g2p = 6 do nıvel 2p e quebrada de forma que dois nıveis

com J=1/2 descem em�⇠2p~2 e quatro nıveis com J=3/2, sobem em +1

2⇠2p~2.

A degenerescencia de cada J e 2J + 1,)(J=1/2 ! 2⇥ 1/2 + 1=2

J=3/2 ! 2⇥ 3/2 + 1=4

10 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.tAs degenerescencias dos nıveis J = 1/2 e J = 3/2 sao essenciais, pois Wf e

invariante mediante rotacao (ver aula 7, onde discutimos que o operador que

roda kets ao redor de n em ' e do tipo: eiJ·n'

~ ).tComentarios finais sobre a estrutura fina do nıvel n=2:

• No nıvel 2s (` = 0, S = 1/2), J assume um unico valor J = 1/2.

• Se Wmv e WD eram matrizes unidades no espaco {|`;S;mL;mSi}, por nao

causarem acoplamento no espaco de momento angular. Consequentemente,

qualquer transformacao unitaria da matriz unidade e uma matriz unidade,

ou seja na representacao {|`;S; J ;mJi} tambem nao ha acoplamento.

• Notacao espectroscopica: n`J

8>>>>>><

>>>>>>:

n ! numero quantico principal;

` ! momento angular orbital;

J ! momento angular total;

s=1/2!momento angular spin (nao consta,

pois e igual para todos).

• E usual

8>>><

>>>:

` = 0 ! s

` = 1 ! p

` = 2 ! d

` = 3 ! f, etc.

No caso

8><

>:

2s1/2 ! �5/128mec2↵4

2p1/2 ! �5/128mec2↵4

2p3/2 ! �1/128mec2↵4

(�7/384� 1/48)

(�7/384 + 1/96)

11 MAPLima

F789 Aula 17 • No final das contas 2s1/2 e 2p1/2 continuam degenerados.

• So WSO e responsavel pela separacao 2p1/2 e 2p3/2.

• Resultados verificaveis experimentalmente. Ex.: fotons de Lyman (� = 1216A).

Estes fotons sao da transicao 2p ! 1s. Olhar atento mostra 2p1/2 ! 1s1/2.

• Vimos que os dois nıveis com mesmo J tem a mesma energia (2s1/2 e 2p1/2).

Isso vem da solucao da equacao de Dirac:

En,J = mec2

h1 + ↵2

⇣n� J � 1

2+

p(J + 1/2)2 � ↵2

⌘�2i� 12

Note que a energia so depende de n e J.

Estruturafinadoníveln=2doÁtomodeHidrogênio.

FiguraXII-2dotexto

12 MAPLima

F789 Aula 17

Estruturafinadoníveln=2doÁtomodeHidrogênio.• A expansao da solucao da equacao de Dirac em ↵

En,J = mec2h1 + ↵

2⇣n� J � 1

2+

p(J + 1/2)2 � ↵2

⌘�2i� 12

⇡ mec2 � 1

2mec

2↵2 1

n2| {z }

�mec2

2n4

⇣n

J + 1/2� 3

4

⌘↵4

| {z }

vem de H0 vem de Wf

(n=2, J=1/2 ! � 5

128mec2↵4

n=2, J=3/2 ! � 1128mec

2↵4

• Note que o calculo do nıvel 2s1/2 e identico ao 2p1/2, pois a energia nao

depende de `.

• Lamb Shift: a degenerescencia entre 2s1/2 e 2p1/2 e quebrada devido as

flutuacoes do campo eletromagnetico no vacuo (natureza quantica do campo

eletromagnetico. Isso da inıcio a chamada

eletrodinamica quantica (ver detalhes no

complemento KV ).

| Aula que vem, Whf :

FiguraXII-3dotexto