Momento Angular na Mecânica Quânticamaplima/f789/2018/aula10.pdfMomento Angular na Mecânica...

2 MAPLima

F789 Aula 10

MomentoAngularnaMecânicaQuântica• Quando |k, j,mi esta normalizado, J± gera kets normalizados, se usarmos:

J±|k, j,mi = ~pj(j + 1)�m(m± 1)|k, j,m± 1i

• Definimos um subespaco E(k, j) de vetores com o mesmo k e j e diferentes m0s.

O subespaco E(k, j) tem a dimensao 2j + 1. O espaco E pode ser considerado

como a soma direta dos espacos ortogonais E(k, j), isto e E =

X

k,j

E(k, j).

Propriedades de E(k, j)• E(k, j) e (2j+1)-dimensional.

• E(k, j) e globalmente invariante sob acao de J2, Jz e J± (de fato 8F (J), ou seja

hk0, j0,m|F (J)|k, j,mi / �k0,k�j0,j .

• Dentro do subespaco E(k, j), os elementos de matriz hk, j,m0|F (J)|k, j,mi naodependem de k. Para entender o significado de k, escolhemos um operador A,

tal que {A,J2, Jz} e im C.C.O.C. Construımos {|k, j,mi}, como solucoes de

A|k, j,mi = ak,j |k, j,mi.O ındice k distingue os ak,j com o mesmo j.

3 MAPLima

F789 Aula 10

Somandomomentosangulares.Casogeral.

Globalmente invariante sob acao de J21,J

22, J1z e J2z.

• Como somar momentos angulares na mecanica quantica? Um bom

comeco: definir o espaco de estados. Que tal: E = E1 ⌦ E2 onde, k1 e k2 sao

fixos e

8>>>>>>>><

>>>>>>>>:

E1 !P

�j1E1(k1, j1)

E2 !P

�j2E1(k2, j2)

E !

8>>><

>>>:

uniao de dois sub-sistemas,

(1) e (2). Por exemplo, duas

partıculas. Ou ainda, o

espaco de spin + espaco orbital.

) �j = soma direta em j

• E1

8><

>:

J21|k1, j1,m1i = j1(j1 + 1)~2|k1, j1,m1i ) com j1 � 0.

J1z|k1, j1,m1i = m1~|k1, j1,m1i ) com � j1 m1 j1.

J1±|k1, j1,m1i = ~pj1(j1 + 1)�m1(m1 ± 1)|k1, j1,m1 ± 1i

• E2

8><

>:

J22|k2, j2,m2i = j2(j2 + 1)~2|k2, j2,m2i ) com j2 � 0.

J2z|k2, j2,m2i = m2~|k2, j2,m2i ) com � j2 m2 j2.

J2±|k2, j2,m2i = ~pj2(j2 + 1)�m2(m2 ± 1)|k2, j2,m2 ± 1i

• E =

X

�E(k1, k2; j1, j2) com E(k1, k2; j1, j2) = E1(k1, j1)⌦ E1(k2, j2)

• E(k1, k2; j1, j2) = {|k1, k2; j1, j2;m1,m2i} ⌘ {|k1, j1,m1i ⌦ |k2, j2,m2i}• A dimensao de E(k1, k2; j1, j2) e (2j1 + 1)(2j2 + 1).

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4 MAPLima

F789 Aula 10

Somandomomentosangulares.RelaçõesdeComutação.• Considere J=J1+J2, onde J1 atua em E1 e J2 atua em E2. Misturas de J1 e J2,

inclusive a soma, atuam em E = E1 ⌦ E2.

Sabemos que se

8><

>:

[J1i, J1j ] = i~✏ijkJ1k[J2i, J2j ] = i~✏ijkJ2k[J1i, J2j ] = 0

=)

8><

>:

J = J1 + J2

J = J1 ⌦ 11 + 11⌦ J2

[Ji, Jj ] = i~✏ijkJk• A escolha do eixo z, permite escrever um base de autokets de J2 e Jz, onde

Jz = J1z + J2z.

• E facil mostrar que (mostre)

8><

>:

[Jz,J21] = [Jz,J2

2] = 0

[J2,J21] = [J2,J2

1] = 0

[J1z, Jz] = [J2z, Jz] = 0

• Entretanto, J2 nao comuta com J1z, nem com J2z, embora comute com a soma

[J2, Jz] = [J2, J1z + J2z] = 0.

• Faremos uso das relacoes

(J2 = J2

1 + J22 + 2J1.J2

J2 = J21 + J2

2 + 2J1zJ2z + J1+J2� + J1�J2+

onde

(J1± = J1x ± iJ1yJ2± = J2x ± iJ2y

=) J± = J1± + J2± = Jx ± iJy<latexit 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5 MAPLima

F789 Aula 10

Somandomomentosangulares.Casogeral.• Por construcao, E(k1, k2; j1, j2) = E(k1, j1)⌦ E(k2; j2), entendido por:

E(k1, k2; j1, j2) = {|k1, k2; j1, j2;m1,m2i} ⌘ {|k1, j1,m1i ⌦ |k2, j2,m2i}e globalmente invariante sob a acao de J2

1,J22, J1z e J2z. Isso ocorre,

uma vez que E(k1, j1) e globalmente invariante com respeito a qualquer

funcao de J1, E(k2, j2) e globalmente invariante com respeito a qualquer

funcao de J2 e operadores de E(k1, j1) nao atuam em E(k2, j2) e vice-versa.

• Dito isso e como J21,J

22,J

2e Jz sao operadores que podem ser escritos como

funcoes dos operadores J1 e J2, uma vez que

(J2

= J21 + J2

2 + 2J1.J2

Jz = J1z + J2z

e correto afirmar que E(k1, k2; j1, j2) tambem e globalmente invariante sob

acao de J21,J

22,J

2e Jz. Isso permite concluir que esses operadores sao, na

pior das hipoteses, bloco-diagonais em E ⌘X

�k1 ,�k2 ,�j1 ,�j2

E(k1, k2; j1, j2).

• Isso e o mesmo que dizer que, para qualquer que seja a funcao F (J1,J2),

vale: hk1, k2; j1, j2|F (J1,J2)|k01, k02; j01, j02i / �k1,k01�k2,k0

2�j1,j01�j2,j02 .

•O que precisamos fazer e diagonalizar J21,J

22,J

2e Jz no sub-espaco (bloco)

E(k1, k2; j1, j2) e perceber que esses operadores formam um CCOC

neste sub-espaco. Sera que ja sao diagonais?<latexit 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6 MAPLima

F789 Aula 10

Somandomomentosangulares.Casogeral.• A resposta e sim para J2

1,J22 e Jz, pois eles comutam com todos os operadores

que definiram E(k1, k2; j1, j2),mas, nao para J2, pois [J2, J1z] 6= 0 e [J2, J2z] 6= 0.

• Estamos atras dos auto-estados simultaneos de J21,J

22,J

2 e Jz, sabendo que serao

combinacoes dos {|k1, k2; j1, j2i}.• Na aula passada, obtivemos na forca bruta (diagonalizando J2) o caso de duas

partıculas com (somente) o grau de liberdade de spin, isto e j1 = 1/2 e j2 = 1/2.

• Como esses resultados nao dependem de k1 e k2 (ver slide 2), daqui para frente

chamaremos

8><

>:

E(k1, k2; j1, j2) de E(j1, j2)

|k1, k2; j1, j2;m1,m2i de |j1, j2;m1,m2iFizemos coisa semelhante na aula passada no caso das partıculas de spin 1/2.

• Como nao sabemos quais valores, J 0s autovalores associados a J2, (e se repetem)

na construcao de E(j1, j2), escrevemos por enquanto que procuramos por

E(k, J) tal que E(j1, j2) =X

�E(k, J) ! k pode ser util, caso J se repita.

• Sera que os valores de J repetem? Quantos k0s sao necessarios?

• Quais valores de J participam dessa soma direta?

7 MAPLima

F789 Aula 10

Devoltaaocasodeduaspartículasdespin½ (aulapassada) • Achamos na aula passada, a representacao matricial de S2

na base 1({|✏1i, ✏2i}) :|++i|+�i|�+i|��i

S2 .=

h++ |h+� |h�+ |h� � |

~2

0

BB@

2 0 0 0

0 1 1 0

0 1 1 0

0 0 0 2

1

CCA , uma matriz bloco-diagonal.

• Diagonalizamos o bloco 2⇥ 2.

(S2)0 = ~2

✓1 1

1 1

◆!

(1p2(|+�i+ |�+i) para o autovalor 2~2

1p2(|+�i � |�+i) para o autovalor 0~2

• Aprendemos, portanto, que S2possuia dois autovalores distintos, 0~2 e 2~2. O

primeiro, nao-degenerado, e o segundo, 3-degenerado

8><

>:

|++i|��i1p2(|+�i+ |�+i)

E com isso, obtivemos

8>>>><

>>>>:

|1,+1i = |++i|1,�1i = |��i|1, 0i = 1p

2(|+�i+ |�+i)

|0, 0i = 1p2(|+�i � |�+i)

8 MAPLima

F789 Aula 10

Casodeduaspartículasdespin½:comentários(aulapassada) • Para o sistema de duas partıculas com spin 1/2, a base 1 (dada pelos kets

{|✏1i, ✏2i}, autokets de S21,S

22, S1z, S2z) forma um conjunto completo, isto e

descreve qualquer ket e qualquer operador relacionados a parte de spin do

sistema. A base 2 (dada pelos kets {|1, 1i, |1, 0i, |1,�1i, |0, 0i} autokets de S21,

S22,S

2, Sz) tambem forma um conjunto completo neste espaco. Note que o

espaco foi definido pela nossa escolha S1 = S2 = 1/2.

• Reconhecemos que {|S,Mi}, com S = 0 e 1 e todos os valores de M possıveis

para esses valores de S, e uma base completa, pois ela e feita de quatro kets

ortogonais que descrevem todos os vetores da base (completa e de mesma

dimensao) {|✏1i, ✏2i},

isto e

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)

• Ou seja, a presenca de k foi desnecessaria, pois J = 0 apareceu uma unica

vez, assim como J = 1. Poderıamos escrever

E(j1 = 1/2, j2 = 1/2)| {z } = E(J = 0)� E(J = 1)| {z } .

dimensao = 4

9 MAPLima

F789 Aula 10

Casodeduaspartículasdespin½:(umoutroolhar) • Sera que conseguirıamos deduzir, sem a diagonalizacao, que somente um S = 0

e um S = 1 estariam presentes em E(s1 = 1/2, s2 = 1/2)?

Lembre que podemos adicionar Sz entre os operadores S21,S

22, S1z, S2z que

definem E(s1 = 1/2, s2 = 1/2), pois ele comuta com todos. Desta forma,

podemos escrever

8>>><

>>>:

|++i corresponde ao autovalor M = +1 de Sz

|��i corresponde ao autovalor M = �1 de Sz

|+�i corresponde ao autovalor M = 0 de Sz

|�+i corresponde ao autovalor M = 0 de Sz

• Pergunta: sera que o M = 1 que apareceu acima, nao poderia ser de S = 2? A

resposta e nao, pois o sub-espaco E(s1 = 1/2, s2 = 1/2) e globalmente invariante

mediante a aplicacao de qualquer componente de S, e se aplicassemos S+ num

ket correspondente a S = 2,M = 1, obterıamos |S = 2,M = 2i. Como nao

existe nenhum ket, entre os que compoem a base 1, com esse valor de M,S = 2,

nao e aceitavel. Isso vale para qualquer outro valor de S, exceto S = 1, pois S+

aplicado em |S=1,M=1i e zero e nao gera problemas. Ao aplicarmos S�

em |S=1,M=1i, sabemos que vamos encontrar |S=1,M=0i e |S=1,M=�1i.Ou seja, seguindo essa logica, os 3 kets correspondentes a |S = 1,Mi, precisam

estar presentes. Agora falta apenas um ket com M = 0 que segundo

essa mesma logica, so pode ser o S = 0. Raciocınio sera util!

10 MAPLima

F789 Aula 10

Casogeral:autovaloresdeJzesuasdegenerescências• Para fazer a contabilidade das degenerescencias dos autovalores de Jz em

E(j1, j2) (lembre que Jz e diagonal nesta base definida pelo C.C.O.C, J21,

J22, J1z e J2z), consideraremos j1 � j2. Sabemos que a dimensao deste sub-

espaco e (2j1 + 1)(2j2 + 1), e esse e o numero de autovalores possıveis de

Jz em E(j1, j2).• Colocada de uma outra maneira, uma vez que

Jz|j1, j2;m1,m2i = (m1 +m2)~|j1, j2;m1,m2i,quantas vezes o valor M = m1 +m2 aparece, para � j1 m1 +j1 e

� j2 m2 +j2?

• Primeiro e importante perceber que o maior valor possıvel de M e j1 + j2.

(soma dos maiores). E o menor? � j1 � j2 (soma dos menores).

• Como m1 e m2, podem variar de um em um, e de se esperar que os valores

possıveis de M, variem de um em um, entre esses extremos. Ou seja: M tem

os possıveis valores: j1 + j2, j1 + j2 � 1, j1 + j2 � 2, ...,�(j1 + j2).

• Chamaremos de gj1,j2(M) a degenerescencia de M em E(j1, j2).• Dois casos obvios: quais seriam as degenerescencias dos valores extremos

de M? Para os valores extremos

(gj1,j2(j1 + j2) = 1

gj1,j2(�j1 � j2) = 1<latexit 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sha1_base64="UzGcxUswCoeMPQsCZyqe3DbbP+Q=">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</latexit>

11 MAPLima

F789 Aula 10

(do livro texto)

• AFig.1 representa os 15 pontos possıveis associados com j1=2 e j2=1.

Note que g2,1(M) = numero de pontos nos segmentos pontilhados.

12 MAPLima

F789 Aula 10

AutovaloresdeJzesuasdegenerescências:Figura1• Note que a contagem de pontos (15) da Fig.1 e simplesmente o produto

(2j1 + 1)(2j2 + 1), a dimensao do espaco em questao, o E(j1, j2). Essadimensao e o numero de kets linearmente independentes do espaco e

a cada um deles podemos associar um unico valor de M.

• Agora, e importante notar que M = m1 +m2 e uma reta de inclinacao

� 1 na Fig. 1. Para ver isso, basta re-escrever essa expressao na forma

m2 = �m1 +M. Os segmentos pontilhados, onde contamos pontos para

calcular a degenerescencia estao sobre retas deste tipo. Razao pela

qual sua contagem faz sentido como degenerescencia para um dado M,

com todas as combinacoes de m1 e m2 que fornecem esse resultado.

• Vindo pela direita, note que a degenerescencia comeca em 1, no extremo

superior direito do retangulo, e aumenta de um em um, ate atingir um

valor maximo (no caso, 3). Ela permanece constante (no caso, no valor

3), enquanto os segmentos de reta comecarem no lado superior do

retangulo e terminarem no lado inferior. Aı, a degenerescencia decresce

de um em um, ate atingir o valor 1 novamente, agora no extremo inferior

esquerdo do retangulo.

• Note que o maior valor para degenerescencia pode ser generalizado e

e igual a 2j2 + 1. Isso vale para todos os casos onde j1 � j2.

13 MAPLima

F789 Aula 10

AutovaloresdeJzesuasdegenerescências:Figura1• Inspirado no caso, j1 = 2 e j2 = 1, e possıvel generalizar (j1 � j2) :

8j1, j2

8>>>>>>>><

>>>>>>>>:

Para M = j1 + j2 ! gj1,j2(j1 + j2) = 1;

Para M = �j1 � j2 ! gj1,j2(�j1 � j2) = 1;

2j2 + 1 ! maximo valor de gj1,j2(M)

(comeca em M = j1 � j2,

termina em M = j2 � j1;

Permite contar grupos com degenerescencia maxima: 2(j1 � j2) + 1;

gj1,j2(�M) = gj1,j2(M).

• Com isso em mente, podemos escrever:

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

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

gj1,j2(j1+j2)=1 ) g2,1(3)=1

gj1,j2(j1+j2�1)=2 ) g2,1(2)=2...

gj1,j2(j1�j2)=2j2 + 1 ) g2,1(1)=3... =2j2 + 1 ) g2,1(0)=3

gj1,j2(j2�j1)=2j2 + 1 ) g2,1(�1)=3...

gj1,j2(�j1�j2+1)=2 ) g2,1(2)=2

gj1,j2(�j1�j2)=1 ) g2,1(3)=1

• E colocar isso numa figura (Fig. 2 do livro texto), no proximo slide.

14 MAPLima

F789 Aula 10

(do livro texto)

• A Fig.2 representa a degenerescencia associada a cada M para o caso j1 = 2

e j2=1. Contagem dos pontos nos segmentos pontilhados da Fig 1.

15 MAPLima

F789 Aula 10

Somandomomentosangulares.AutovaloresdeJ2.• Vimos que: Jz|j1, j2;m1,m2i = (m1 +m2)~|j1, j2;m1,m2i, com Jz = J1z + J2z.

Os autovalores de Jz,M~, sao tais que M = m1 +m2 com

8><

>:

�j1 m1 +j1e

�j2 m2 +j2

Concluımos que M tem os valores: j1 + j2, j1 + j2 � 1, j1 + j2 � 2, ...,�(j1 + j2).

• Note que os M 0s sao

8>>>>>>>><

>>>>>>>>:

inteiros, se j1 e j2

8><

>:

ambos inteiros ou

ambos semi-inteiros

semi-inteiros, se j1 e j2

8><

>:

um e inteiro e o

outro e semi-inteiro

• Com isso, o que podemos esperar de J? !

8>>>>>><

>>>>>>:

J = J1 + J2

J2|J,Mi = J(J + 1)~2|J,Mi

�J M +J pulando de 1 em 1

• Os J 0s

(serao inteiros, se os M 0s forem inteiros ou

serao semi-inteiros, se os M 0s forem semi-inteiros<latexit 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16 MAPLima

F789 Aula 10

Somandomomentosangulares.AutovaloresdeJ2.• Entre os valores de M: j1+j2, j1+j2�1, j1+j2�2, ...,�(j1+j2), o maior valor

de M e j1 + j2 ) nenhum J, com valor maior que j1 + j2 esta presente.

• Para J=j1+j2

(associamos um sub-espaco E(k, J)={|J,Mi},�JM+J ;

e somente um sub-espaco, pois M e nao-degenerado

• O vetor associado ao valor M = j1 + j2 � 1 aparece uma vez no E(k, J), mas

vimos que esse autovalor e bi-degenerado. Assim, o sub-espaco associado a

J = j1 + j2 � 1 tambem precisa estar presente. E aqui de novo, so uma vez.

• Ate agora E(j1, j2) = E(k = 1, J = j1 + j2)� E(k = 1, J = j1 + j2 � 1)�+...

• Para continuar de forma geral, denote Pj1,j2(J) o numero de sub-espacos

associados a um dado J. Isto e, o numero de valores diferentes de k para esse

valor de J (arbitrariamente, chamei de k = 1 os casos acima, onde apareceu

somente um sub-espaco de cada).

• Considere um valor de M qualquer (lembre que a esse valor corresponde um

unico vetor em cada E(k, J), devido a necessidade de J � |M |). Desta forma:

(1) gj1,j2(M) = Pj1,j2(J = |M |) + Pj1,j2(J = |M |+ 1) + Pj1,j2(J = |M |+ 2)...

(2) gj1,j2(M = J) = Pj1,j2(J) + Pj1,j2(J + 1) + Pj1,j2(J + 2)...

(3) gj1,j2(M = J + 1) = Pj1,j2(J + 1) + Pj1,j2(J + 2)...

(2)� (3) ! Pj1,j2(J) = gj1,j2(M = J)� gj1,j2(M = J + 1)<latexit 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17 MAPLima

F789 Aula 10

� Como gj1,j2(�M)=gj1,j2(M) ! Pj1,j2(J)=gj1,j2(M=�J)�gj1,j2(M=�J�1)

F Explorando o resultado: Pj1,j2(J) = gj1,j2(M = J)� gj1,j2(M = J + 1)

• Se J > j1 + j2, quanto vale Pj1,j2(J)? 0 pois, gj1,j2(M)=0 para M=J > j1+j2.

Lembre que o maior valor de M e j1 + j2.

• E se J = j1 + j2? Pj1,j2(J) = 1 pois, gj1,j2(j1 + j2) = 1 e gj1,j2(j1 + j2 + 1) = 0.

• E se J = j1 + j2 � 1? Pj1,j2(J)=1 pois, gj1,j2(j1+j2 � 1)=2 e gj1,j2(j1 + j2)=1.

Lembrando da escada de um degrau, os valores continuam sendo 1 ate j1 � j2.

• E se J = j1 � j2? Pj1,j2(J)=1 pois, gj1,j2(j1�j2)=n e gj1,j2(j1�j2+1)=n�1.

No slide 13 vimos que n = 2j2 + 1, se j1 � j2.

• Vimos que a partir de j1�j2, gj1,j2(J) fica constante ate M=�(j1�j2). Isso

faz Pj1,j2(J)=0, para 0 J < j1 � j2.

• Isso tudo permite concluir que os valores de J que contribuem estao no intervalo

|j1 � j2| J j1 + j2.

• O sub-espaco E(j1, j2) tambem pode ser descrito pelos kets {|j1, j2; J,Mi}, mas

os valores de J estao restritos ao intervalo acima. Isso pode ser expressado da

seguinte forma: E(j1, j2)=j1+j2X

�J=|j1�j2|

E(k, J) (k aparece uma unica vez para cada J).

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Somandomomentosangulares.AutovaloresdeJ2.

Ignoraremos k daqui para frente

18 MAPLima

F789 Aula 10

Somandomomentosangulares.Comentários:

• E(j1, j2)=j1+j2X

�J=|j1�j2|

E(J) com E(J) = {|j1, j2; J,Mi, |j1 � j2| J j1 + j2}

• Note que para cada par de valores J,M existe um unico vetor em E(j1, j2).Portanto, os operadores J2, e Jz} formam um C.C.O.C. em E(j1, j2).

• Qual sera o numero de pares (J,M) em E(j1, j2)? Que tal

j1+j2X

|j1�j2|

(2J + 1)?

Se adotarmos, como temos feito, j1 � j2 e J = j1 � j2 + i, a soma fica:

2j2X

i=0

[2j1 � 2j2 + 2i+ 1] =

2j2X

i=0

[2j1 � 2j2 + 1] +

2j2X

i=0

[2i] =

= [2j1 � 2j2 + 1] (2j2 + 1)| {z }+2

z}|{2j22

(2j2 + 1)| {z }, ou seja

j1+j2X

J=|j1�j2|

(2J + 1) = [2j1 � 2j2 + 1 + 2j2](2j2 + 1) = (2j1 + 1)(2j2 + 1),

conforme esperado (dimensao de E(j1, j2)).

Valor médio de i

número de possibilidades

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