estrutura hyperfina do átomo de hidrogêniomaplima/f789/2018/aula18.pdf · estrutura hiper-fina do...
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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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