1 MAPLima

F789 Aula 18

Estruturahyperfinadoátomodehidrogênio:UmaaplicaçãodaTeoriadePerturbaçãoEstacionária.

• Whf = �µ0

4⇡

n q

meR3L.MI +

1

R3

⇥3(Ms.n).(MI.n)�MS.MI

⇤+

8⇡

3MS.MI�(R)

o

� Por ter spin, I, tem momento magnetico, MI , proporcional a ele e cada termo

significa:

8>>>>>>>>>>>>>>><

>>>>>>>>>>>>>>>:

µ0 ! permeabilidade magnetica

L ! Momento angular orbital;

MI =gpµn

~ I ! Momento Magnetico do proton;

MS = 2µB

~ S = qme

S ! Momento Magnetico do eletron;

n ! vetor unidade da linha reta que junta o proton ao eletron;

gp ⇠ 5, 585;

µB = q~2me

(Magneto de Bohr), me e a massa do eletron;

µn = � q~2Mp

(Magneto nuclear de Bohr), Mp e a massa do proton;

I ! Spin do proton; q<0 e a carga do eletron;�q e a do proton.

• Interpretacao da formula acima.

� Primeiro termo: interacao do momento magnetico do proton com o campo

magnetico, � µ0

4⇡

q

meR3L, criado pelo movimento do eletron, segundo a lei

de Bio-Savart B =µ0

4⇡qu⇥ r

r3. Lembre que e2 =

q2

4⇡✏0e c =

1pµ0✏0

.

2 MAPLima

F789 Aula 18

� Segundo termo: interacao dipolo-dipolo entre os momentos magneticos (spins

intrınsecos) do eletron e do proton, �µ0

4⇡

1

R3

⇥3(Ms.n).(MI.n)�MS.MI

⇤

| {z }.

ver complemento BXI

� Finalmente, o ultimo termo, � µ0

4⇡

8⇡

3Ms.MI�(R), tambem conhecido por

“termo de contacto” de Fermi, tem origem na singularidade em R = 0, do

campo criado pelo momento magnetico do proton. Na realidade, o proton

nao e uma carga pontual. Esse termo leva isso em consideracao (ver AXII).

• Ordem de Magnitude de Whf .

1o. e 2o. termos:µ0

4⇡

q2~2meMpR3

=e2~2

mec2MpR3/ me

MpWSO;

Mostre que 3o. termo (que contem �(R)) e ⇡ me

MpWD

<latexit sha1_base64="LqzWzVnYJjr/ogdnh687AXB3JLs=">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</latexit>

Estruturahyperfinadoátomodehidrogênio:UmaaplicaçãodaTeoriadePerturbaçãoEstacionária.

p

e

Ms

Mp n

Figure 1 do capítulo XII

Whf é 2000 vezes menor que Wf

3 MAPLima

F789 Aula 18

Estruturahiper-finadoníveln=1doÁtomodeHidrogênio.tO caminho logico seria estudar os efeitos de Whf nos nıveis 2s1/2, 2p1/2 e 2p3/2.

Estudaremos o efeito de Whf no nıvel 1s, cuja degenerescencia nao foi quebrada

por Wf . E mais simples e completo, pois pode ser facilmente generalizado para

os estados do nıvel 2.

tO problema:

• A degenerescencia do nıvel 1s e g1s = 4 e os estados associados dados por:

E1s = {|n = 1; ` = 0;mL = 0;mS = ±1

2;mI = ±1

2}

• Vimos que esse nıvel quadri-degenerado nao tem sua degenerescencia quebrada

pelas interacoes de estrutura fina, Wf . Tivemos apenas um deslocamento

coletivo de todos eles, conforme:

8>>>>>><

>>>>>>:

hWmvi1s = � 5

8mec2↵4

hWDi1s = +1

2mec2↵4

hWSOi1s=0!(`=0 ) h`=0|Lx,y,z|` = 0i• Na notacao espectroscopica o 1s vira 1s1/2, um unico nıvel (g1s1/2 = 4)

deslocado de � 1

8mec

2↵4do 1s.

4 MAPLima

F789 Aula 18

Estruturahiper-finadoníveln=1doÁtomodeHidrogênio.tVeremos que so o termo de contato, expressao do slide 1, reproduzida abaixo,

Whf = �µ0

4⇡

n q

meR3L.MI +

1

R3

⇥3(Ms.n).(MI.n)�MS.MI

⇤+

8⇡

3MS.MI�(R)

o

ira contribuir. As razoes para os dois primeiros termos nao contribuirem sao:

� O primeiro termo envolve o produto escalar L ·MI , uma componente de L,

que respeita a propriedade h` = 0;mL = 0|L|` = 0;mL = 0i = 0

� O segundo termo, interacao dipolo-dipolo, e zero por se tratar de um nıvel

esfericamente simetrico (1s). Detalhes no complemento BXI, §3.tO termo de contato (ultimo da expressao acima), � 2µ0

3MS.MI�(R), envolvem

termos de matriz do tipo:

hn=1; `=0;mL=0;m0S ;m

0I |�

2µ0

3MS.MI�(R)|n=1; `=0;mL=0;mS ;mIi

� Na representacao das coordenadas {r}, podemos separar a parte orbital da

parte de spin para obter: Ahm0S ;m

0I |S.I|mS ;mIi, onde A e

dado por A =q2

3✏0c2gp

meMphn=1; `=0;mL=0|�(R)|n=1; `=0;mL=0i, ou

ainda A=q2

3✏0c2gp

meMp

1

4⇡|R10(0)|2=

4

3gp

me

Mpmec

2↵4�1 +

me

Mp

��3 1

~2<latexit 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5 MAPLima

F789 Aula 18

Estruturahiper-fina(termodecontato)doníveln=1deH.tObtendo A, definido por hW contatohf i = Ahm0

S ;m0I |S.I|mS ;mIi.

Juntando as definicoes do slide 1

8><

>:

W contatohf = � 2µ0

3 MS.MI�(R)

MS =2µB

~ S; µB =q~2me

MI =gpµn

~ I; µn = � q~2Mp

) temos

A = �2µ0

3

q

me

gp(�q)

2Mphn=1; `=0;mL=0|�(R)|n=1; `=0;mL=0i

Sabendo que hr|n=1; `=0;mL=0i = n=1,`=0(r) =1p4⇡

2

a3/20

e�ra0 , podemos

escrever A = �2µ0

3

q

me

gp(�q)

2Mp| n=1,`=0(r = 0)|2 =

µ0

3meMpq2gp

1

4⇡

4

a30

Usando que

8>>>><

>>>>:

q2

4⇡✏0= e2;

µ0✏0 =1c2 ;

a0 =~2

µe2 ;

µ =meMp

me+Mp

) A =4

3gp

e2

c21

meMp

⇣~2

µe2

⌘3 =4

3gp

e8

c2~6µ3

meMp

Como ↵=e2

~c , temos A=4

3gp

c2↵4

~2

⇣meMp

me+Mp

⌘3

meMp=

4

3gp

c2↵4

~2m2

eM2p

(me +Mp)3e

finalmente A=4

3gp

c2↵4

~2m2

eM2p

(me +Mp)3=

4

3gp

mec2↵4

~2me

Mp

⇣1 +

me

Mp

⌘�3.

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6 MAPLima

F789 Aula 18 tAs variaveis orbitais foram resolvidas. Sobrou um problema de dois spins 1/2’s,

I e S, acoplados por uma interacao do tipo: AS · I. Problema muito similar ao

ja resolvido de spin-orbita.tAuto-estados e auto-valores do termo de contato.

Para chegar ate aqui, usamos a base {|S =1

2; I =

1

2;mS ;mIi} de auto-vetores

comuns aos operadores S2, I2, Sz, e Iz.

Podemos agora introduzir o momento angular total F = S+ I e usar a base

{|S =1

2; I =

1

2;F ;mF i} de auto-vetores comuns aos operadores S2, I2,F2, e Fz,

com |S�I|F S+I| {z }

8>>><

>>>:

S2|S = 12 ; I=

12 ;F ;mF i= 3

4~2|S= 1

2 ; I=12 ;F ;mF i;

I2|S= 12 ; I=

12 ;F ;mF i= 3

4~2|S= 1

2 ; I=12 ;F ;mF i;

F2|S= 12 ; I=

12 ;F ;mF i=F (F+1)~2|S= 1

2 ; I=12 ;F ;mF i;

Fz|S= 12 ; I=

12 ;F ;mF i=mF~|S = 1

2 ; I=12 ;F ;mF i.

Neste caso, os valores possıveis de F sao 0 e 1.tNote que se ` 6= 0, terıamos F = J+ I = L+ S+ ItA forma do acoplamento, AS · I, sugere, como no caso spin-orbita, escrever

F2 = (S+ I)2 = S2 + I2 + 2S · I ) AS · I = A2(F2 � I2 � S2)

| {z }

Estruturahiper-fina(termodecontato)doníveln=1deH.

Diagonalnanovabase

7 MAPLima

F789 Aula 18 tAssim, temos

AS · I|F ;mF i =A2(F2 � I2 � S2)|F ;mF i =

A~22

(F (F + 1)� 3

4� 3

4)|F ;mF i

onde trocamos |S=1

2; I=

1

2;F ;mF i por |F ;mF i para simplificar.

No caso

8><

>:

AS · I|F = 1;mF i = A~2

2 (1(1 + 1)� 34 � 3

4 )|1;mF i = A~2

4 |1;mF i

AS · I|F = 0;mF = 0i = A~2

2 (0(0 + 1)� 34 � 3

4 )|0; 0i = � 3A~2

4 |0; 0i

tA quadri-degenerescencia e

parcialmente removida.

tVerificavel experimentalmente.

tA degenerescencia 2F + 1 e

essencial (relacionada com a

invariancia de rotacao do sistema.

Estruturahiper-fina(termodecontato)doníveln=1deH.

FiguraXII-3dotexto

8 MAPLima

F789 Aula 18

Estruturahiper-fina(termodecontato)doníveln=2deH.tResultados extrapolaveis para o caso ` 6= 0. Algo do tipo:

A0J · I|F ;mF i=A0

2(F2 � I2 � J2)|F ;mF i=

A~22

(F (F+1)� 3

4�J(J+1)|F ;mF i

Como |J � I| F J + I, temos

8><

>:

2s1/2 ! J = 1/2; I = 1/2 ) F = 0, 1;

2p1/2 ! J = 1/2; I = 1/2 ) F = 0, 1;

2p3/2 ! J = 3/2; I = 1/2 ) F = 2, 1.

FiguraXII-4dotexto

LambShift

9 MAPLima

F789 Aula 18

Importânciadaestruturahiper-finadoníveln=1deHtA estrutura hiper-fina do estado fundamental do atomo de hidrogenio (1s)

foi, ha 55 anos atras, a quantidade fısica conhecida experimentalmente

com o maior numero de algarismos significativos. Expressada em Hz, ela e

igual aA~2⇡

= 1.420.405.751, 768± 0, 001Hz. Esse numero e a diferenca em

energia entre os nıveis da estrutura hiper-fina F =1 e F =0, cuja frequencia

associada (de um foton absorvido/emitido na transicao) e

⌫ =EF=1 � EF=0

h=

A~2

4 � (�3A~2

4 )

h=

A~2h

=A~2⇡

.tA precisao foi obtida com um MASER (“Microwave Amplification by

Stimulated Emission of Radiation”) de hidrogenio em 1963. A ideia e colocar

atomos selecionados (estrategia de Stern-Gerlach) em F=1 em uma cavidade.

Se colocarmos essa cavidade em um campo com a mesma frequencia de

transicao (condicao de ressonancia), em condicoes geometricas especiais

(ganhos maiores que as perdas), os atomos oscilam entre os dois estados.tTem muito Hidrogenio no espaco interestelar do universo. Podemos detectar

a radiacao destes atomos quando eles caem espontaneamente de F = 1 para

F = 0. Essa transicao corresponde a um comprimento de onda de 21cm.

Muito do que sabemos do universo e baseado nessa linha de transicao.<latexit 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10 MAPLima

F789 Aula 18

Teoria de Perturbação independente do tempo: aplicações Interacao de Van der Waals

Um interacao de longo alcance entre 2 atomos de hidrogenio em seus estados

fundamentais. E possıvel mostrar que a interacao e atrativa e vai com1r6 , onde

r e distancia entre eles. Vamos demonstrar isso com teoria de perturbacao.

Escreva a Hamiltoniana H (considere os protons fixos). Que tal: H = H0 + V,

onde

8>>>>>>>>>>>>>><

>>>>>>>>>>>>>>:

H0 = � ~22m

(r12+r2

2)

| {z }�e

2

r1� e

2

r2| {z }energia cinetica

dos eletrons

atracao proton-

eletron proximo

V =e2

r|{z}� e

2

|r� r1|� e

2

|� r� r2|| {z }+

e2

|r2 � (r1 � r)|| {z }repulsao

dos nucleos

atracao eletron

longe - nucleo

repulsao

entre eletrons

r1 r

r2

11 MAPLima

F789 Aula 18

Teoria de Perturbação: Interação Van der Waals O estado fundamental de H0 e dado pelo produto de dois kets iguais (ambos

relativos a menor energia do atomo isolado). Isto por que H0 e a soma de dois

atomos que nao interagem. Na representacao das coordenadas seria:

U(0)0 = U

(0)100(r1)U

(0)100(r2)

Para descobrirmos o comportamento do potencial entre os atomos para r

grande, comecamos por expandir V em potencias deri

r. Tome, por exemplo:

1

|r� ri|=

1pr2 � 2r.ri + r

2i

=1

rp1� 2r. rir + (

rir )

2e use a expansao de

1p1 + x

p/ x << 1, isto e1

p1 + x

⇡ 1�x

2+

3x2

8+ . . .Feito isso, teremos:

1

|r� ri|⇡

1

r

✓1�

1

2

�� 2r.

rir+ (

ri

r)2�+

3

8

�� 2r.

rir+ (

ri

r)2�2

+ . . .

◆

=1

r

✓1 +

r.rir

�1

2

�rir

�2+

3

2

� r.rir

�2+O

�(ri

r)3�◆

Mostre que se tomassemos apenas os dois primeiros termos da expansao acima,

obterıamos V = 0, com V dado pela expressao do slide 10.<latexit 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sha1_base64="CEgikznGXdD135htcSfiDel/9gs=">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</latexit><latexit 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12 MAPLima

F789 Aula 18

Teoria de Perturbação: Interação Van der Waals

Aplicando1

|r� ri|=

1

r

✓1 +

r.rir

�1

2

�rir

�2+

3

2

� r.rir

�2+O

�(rir)3�◆

para os 3 casos em V =e2

r�

e2

|r� r1|�

e2

|� r� r2|+

e2

|r2 � (r1 � r)|,

temos: V = �e2

r

✓�

1

2

�r1r

�2+

3

2

� r.r1r

�2�

1

2

�r2r

�2+

3

2

� r.r2r

�2+

+1

2

� |r1 � r2|

r

�2�

3

2

� r.(r1 � r2)

r

�2◆

e, apos notar que so os termos

cruzados dos dois ultimos termos contribuem, obtemos:

V = �e2

r

✓�

1

2

�2r1.r2r2

�+

3

2

�2(r.r1)(r.r2)r2

�◆. Escolha r = z e

finalmente obtenha V =e2

r3�x1x2 + y1y2 + z1z2 � 3z1z2

�, ou melhor

V =e2

r3�x1x2 + y1y2 � 2z1z2

�+O(

1

r4) interacao entre dois dipolos -

ver Jackson p.141.

Estamos prontos para aplicar teoria de perturbacao

13 MAPLima

F789 Aula 18

Teoria de Perturbação: Interação Van der Waals

É um parâmetro Acontece que quando vamos fazer as contas em 1a ordem, temos que

V (r) = E(1)(r) = hU (0)100(r1)U

(0)100(r2)|V |U (0)

100(r1)U(0)100(r2)i = 0,

pois

8>>>>>>>>>>><

>>>>>>>>>>>:

V (r) e obtida com integracoes independentes em r1 e r2;

V = e2

r3

�x1x2 + y1y2 � 2z1z2

�e uma funcao ımpar em r1 e em r2;

U0)100(r1) e uma funcao par em r1;

U0)100(r2) e uma funcao par em r2.

Contribuicao de 2a ordem fornece:

V (r) = E(2) =e4

r6

X

k 6=0

|hk(0)|(x1x2 + y1y2 � 2z1z2|0(0)i|2

E(0)0 � E(0)

k

Uma interacao que varia com1

r6e e atrativa, pois E(0)

0 � E(0)k e sempre

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