1 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica

• Explorando o fato que E(j, k) e globalmente invariante sob acao de ~J.

Lembre que 8 componente Ju pode ser escrita em funcao de Jz, J+, J� e que

esses tres operadores no maximo mudam m (resultado 2 E(j, k))

• A representacao matricial de Ju ou de qualquer F ( ~J) deve ser bloco diagonal.

hk,j,m|F ( ~J)|k0, j0,m0i E(k, j) E(k0, j) E(k0, j0)E(k, j) (2j+1)⇥(2j+1) 0 0E(k0, j) 0 (2j+1)⇥(2j+1) 0E(k0, j0) 0 0 (2j+1)⇥(2j+1)

k 6= k0 e j 6= j0

• Podemos calcula-la usando

(Jz|k, j,mi = m~|k, j,miJ±|k, j,mi = ~

pj(j+1)�m(m±1)|k, j,m±1i

ou ainda

(hk, j,m|Jz|k0, j0,m0i = m0~�kk0�jj0�mm0

hk, j,m|J±|k0, j0,m0i = ~p

j0(j0+1)�m0(m0±1)�kk0�jj0�m,m0±1

• Note que esses elementos nao dependemde k.Uma vez calculada a representacao

de Ju ou de F ( ~J), podemos usa-la para 8 sistema fısico.

2 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular

y

⇢

'

x

• Na representacao das coordenadas {|~ri}, temos Lz =~i

⇣x@

@y� y

@

@x

⌘. Para

explorar possıveis simetrias do sistema, e interessante escrever esse operador

(assim como Lx, Ly e L2) em coordenadas polares e esfericas.

• Para ilustrar, Lz em Coordenadas polares (z, ⇢,')

8><

>:

x=⇢ cos'

y=⇢ sin'

Lz=~i

⇣x(⇢,')

@⇢

@y

@

@⇢+x(⇢,')

@'

@y

@

@'�y(⇢,')

@⇢

@x

@

@⇢�y(⇢,')

@'

@x

@

@'

⌘

Usando que

8><

>:

⇢2 = x2 + y2 ! ⇢ =px2 + y2

tan' = yx ! ' = arctan y

x

e qued

d�arctan� =

1

1 + �2,

podemos escrever

8><

>:

@⇢@y = y

⇢ = sin'

@⇢@x = x

⇢ = cos'

e

8><

>:

@'@y = 1

1+y2/x2 .(1x ) =

x⇢2 = cos'

⇢

@'@x = 1

1+y2/x2 .(� yx2 ) = � y

⇢2 = � sin'⇢

• Para obter Lz=~i

⇣⇢ cos' sin'

@

@⇢+ ⇢

cos2 '

⇢

@

@'� ⇢ sin' cos'

@

@⇢+⇢

sin2 '

⇢

@

@'

⌘

e finalmente sua forma final Lz=~i

@

@'.

3 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular

'

✓

xy

z

r• Coordenadas esfericas (r, ✓,')

8><

>:

x = r sin ✓ cos'

y = r sin ✓ sin'

z = r cos ✓

A inversao das expressoes permite escrever

8>>>>>><

>>>>>>:

r =px2 + y2 + z2

✓ = arccos zpx2+y2+z2

' = arctan yx

Usando regras em cadeia do tipo@

@z=

@r

@z

@

@r+

@✓

@z

@

@✓+

@'

@z

@

@'e

lembrando que

8>>>>>><

>>>>>>:

dd� arctan� = 1

1+�2

dd� arcsin� = 1p

1��2

dd� arccos� = � 1p

1��2

e possıvel obter Lx, Ly, Lz, e L2.

A seguir fornecemos apenas os resultados.

4 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular

'

✓

xy

z

r

• Servico doloroso (mas sem grandes dificuldades) leva a:

(1) Lx = i~⇣sin'

@

@✓+

cos'

tan ✓

@

@'

⌘;

(2) Ly = i~⇣� cos'

@

@✓+

sin'

tan ✓

@

@'

⌘;

(3) Lz = i~ @

@';

(4) L2= �~2

⇣ @2

@✓2+

1

tan ✓

@

@✓+

1

sin2 ✓

@2

@'2

⌘;

(5) L+ = ~ei'⇣ @

@✓+ i cot ✓

@

@'

⌘;

(6) L� = ~e�i'⇣� @

@✓+ i cot ✓

@

@'

⌘;

• Ja que temos todos os operadores na representacao das coordenadas, nosso

problema se resume a resolver:

L2Y m` (✓,') = `(`+ 1)~2Y m

` (✓,')

LzYm` (✓,') = m~Y m

` (✓,')

onde Y m` (✓,') sao as chamadas Harmonicas Esfericas.

5 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular

• Na representacao das coordenadas, temos duas equacoes diferenciais: a de

Lz e em primeira ordem em ' (tem solucao simples) e permite que na de L2,

a dependencia em ' seja facilmente retirada. Assim o desafio seria resolver

uma equacao em segunda ordem em ✓ (cuja solucao da origem aos polinomios

de Legendre). Ao inves de resolver a equacao envolvendo segundas derivadas

em ✓, usaremos os operadores L± para obter solucoes resolvendo equacoes de

primeira ordem.

• Antes, alguns comentarios:

� As integrais em volume envolvendo coordenadas cartesianas e esfericas estao

relacionadas da seguinte maneira:

dv = dxdydz = r2 sin ✓drd✓d',

onde, as cartesianas

8><

>:

�1<x<+1�1<y<+1�1<z<+1

enquanto que as esfericas

8><

>:

0 r < 10 ✓ ⇡

0 ' 2⇡

� Uma vez obtidas as Harmonicas esfericas, podemos escrever uma funcao de

base do espaco R3como k`m(r, ✓,') = fk`m(r)Y m

` (✓,').

6 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular � Nestas condicoes a normalizacao de k`m(r, ✓,'), dada por

Z Z Zr2 sin ✓drd✓d'

�� k`m(r, ✓,')��2 = 1,

pode ser separada em duas condicoes

8><

>:

R10 r2dr

��fk`m(r)��2 = 1

R Rsin ✓d✓d'

��Y m` (✓,')

��2 = 1

� Vimos que em casos especiais, a parte radial fk`m(r), pode nao depender de m

(indexaremos por fk`(r)) e ate mesmo pode nao depender de ` e m (neste caso

indexaremos apenas com k, e a parte radial sera chamada de fk(r)).

• Obtencao das Harmonicas Esfericas Y m` (✓,').

Conforme indicamos, a parte em' e relativamente simples. Comece pela equacao

de autovalor de Lz, na representacao das coordenadas, isto e:

hr|Lz|`,mi = ~i

@

@'hr|`,mi| {z } = m~ hr|`,mi| {z } com |ri = |✓,'i

Y m` (✓,') Y m

` (✓,')

~i

@

@'Y m` (✓,') = m~Y m

` (✓,')

As variaveis ✓ e ' aparecem desacopladas, sugerindo a separacao e

a forma da solucao para ' ) Y m` (✓,')=Fm

` (✓)eim'

7 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular � Note que ao exigir que Y m

` (✓, 0)=Y m` (✓, 2⇡), temos eim2⇡

= 1, que so ocorre

se m for inteiro.

� Se m e inteiro, ` tambem e inteiro. Ou seja, ao exigir que a funcao de onda seja

unicamente definida ao darmos uma volta completa no espaco R3, aprende-se

que a possibilidade ` semi-inteiro para momento angular orbital precisa ser

descartada.

• Em seguida, acharemos Y `` (✓,') e usaremos L+ (slide 4) para obter Y m

` (✓,').

L+Y`` (✓,') = 0 =) ~ei'

⇣ @

@✓+ i cot ✓

@

@'

⌘F `` (✓)e

i`'= 0

⇣ @

@✓� ` cot ✓

⌘F `` (✓) = 0 ) dF `

`

d✓(✓)� ` cot ✓F `

` (✓) = 0 ) dF ``

F ``

= ` cot ✓d✓

) dF ``

F ``

= `cos ✓

sin ✓d✓ = `

d sin ✓

sin ✓integrando dos dois lados, temos:

lnF `` = ` ln sin ✓ + cte|{z} ) lnF `

` = ln (c`(sin ✓)`) ) F `

` = c`(sin ✓)`

chame ln c`, para facilitar e por poder depender de `

Assim, temos finalmente Y `` (✓,')=c`(sin ✓)

`ei`', onde c` e a constante

de normalizacao.<latexit 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8 MAPLima

F689 Aula 24

Representação das coordenadas do momento angular • Podemos agora obter Y m

` (✓,') com auxılio do operador L�.

Para isso, basta lembrar que deduzimos que

L±Ym` (✓,') = ~

p`(`+ 1)�m(m± 1)Y m±1

` (✓,'),

que com auxılio do slide 4 permite escrever:

e±i'~⇣± @

@✓+ i cot ✓

@

@'

⌘Y m` (✓,') =

p`(`+ 1)�m(m± 1)Y m±1

` (✓,')

Assim, para obter Y `�1` (✓,') a partir de Y `

` (✓,') basta usar

Y m±1` (✓,') =

e±i'~⇣± @

@✓ + i cot ✓ @@'

⌘Y m` (✓,')

p`(`+ 1)�m(m± 1)

com sinal inferior e m = `.

• Essa formula de recorrencia pode ser usada para obter, sucessivamente, todas as

Harmonicas Esfericas para um dado ` e m. Basta aplica-la variando m de ` ate

� `+ 1, e fazendo uso da deduzida expressao para Y `` (✓,').

• Ortonormalizacao:

Z 2⇡

0d'

Z ⇡

0sin ✓d✓ Y m0

`0?(✓,')Y m

` (✓,') = �``0�mm0

• Relacao de completeza.

As harmonicas esfericas formam uma base capaz de escrever qualquer funcao

de ✓ e ' ) f(✓,') =1X

0

X

m=�`

c`mY m` (✓,'). Quanto vale c`m?

9 MAPLima

F689 Aula 24

Relação de Completeza • Usando a relacao de ortonormalidade das Harmonicas esfericas, obtemos

c`m =

Z 2⇡

0d'

Z ⇡

0sin ✓d✓ Y m

`?(✓,')f(✓,')

A relacao de completeza pode ser expressa por

1X

`=0

X

m=�`

|`,mih`,m| = 11,

que na representacao das coordenadas fica

1X

`=0

X

m=�`

hr|`,mih`,m|r0i = hr|r0i,

ou ainda

1X

`=0

X

m=�`

Y m` (✓,')Y m

`?(✓0,'0

) = hr|r0i = �(✓ � ✓0)�('� '0)

sin ✓

onde usamos a delta de Dirac em coordenadas esfericas, isto e:

h~r|~r 0i = �(~r � ~r 0) = �(x� x0

)�(y � y0)�(z � z0) =�(r � r0)�(✓ � ✓0)�('� '0

)

r2 sin ✓

Isso permite escrever f(✓,') =

Z Zd'0

sin ✓0d✓0�(✓ � ✓0)�('� '0

)

sin ✓f(✓0,'0

)

f(✓,') =

Z Zd'0

sin ✓0d✓0h 1X

`=0

X

m=�`

Y m` (✓,')Y m

`?(✓0,'0

)

if(✓0,'0

) que pode

ser escrito por f(✓,')=1X

`=0

X

m=�`

c`mY m` (✓,') com c`m definido acima.

10 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica: Simetrias

) a inversao leva M para M0Fig. 2, cap. 6 do texto

• Paridade. O que acontece quando trocamos ~r ! �~r ?

Coordenadas cartesianas

8><

>:

x ! �x

y ! �y

z ! �z

Coordenadas esfericas

8><

>:

r ! r

✓ ! ⇡ � ✓

' ! '+ ⇡

Complemento AVI demonstra que Y m` (⇡ � ✓,'+ ⇡) = (�1)

`Y m` (✓,')

As harmonicas esfericas de ` par (ımpar) sao pares (ımpares).

• Conjugacao complexa Y m`

?(✓,') = (�1)

mY �m` (✓,') (ver complemento AVI).

11 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica: base padrão

• Base padrao.

Suponha k`m(~r), tal que

8><

>:

L2 k`m(~r) = `(`+ 1)~2 k`m(~r)

Lz k`m(~r) = m~ k`m(~r)

L± k`m(~r)=~p`(`+ 1)�m(m± 1) k`m±1(~r)

A dependencia em ✓ e ' e conhecida k`m(~r) = Rk`m(r)Y m` (✓,') e isso fornece

L± k`m(~r)=~p`(`+1)�m(m±1)Rk`m±1(r)Y

m±1` (✓,')=Rk`m(r)L±Y

m` (✓,')=

=Rk`m(r)~p`(`+ 1)�m(m± 1)Y m±1

` (✓,') ) Rk`m±1(r) = Rk`m(r).

Onde se concluı que Rk`m(r) nao depende de m ) use Rk`(r).

• Assim, a base padrao permite escrever k`m(~r) = Rk`(r)Ym` (✓,'), com

Zd3r ?

k`m(~r) k0`0m0(~r)=

Z 1

0r2R?

k`(r)Rk0`0(r)

Z 2⇡

0d'

Z ⇡

0sin ✓d✓Y m

`?(✓,')Y m0

`0 (✓,')

| {z }�``0�mm0

Como

Zd3r ?

k`m(~r) k0`0m0(~r)=�kk0�``0�mm0 )Z 1

0r2R?

k`(r)Rk0`(r) = �kk0 .

• Sera que Rk`(r) depende de ` ? Suponha Rk(r)Ym` (✓,') e olhe r!0.

12 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica: Medidas

• Base padrao (continuacao)

Se limr!0

Rk(r) = cte, a presenca de Y m` (✓,') faz com que essa constante dependa

do caminho, exceto para ` = 0 (neste caso Y 00 (✓,') e constante e, portanto, nao

depende de ✓ e '). Ou seja, Rk(r) nao e diferenciavel em~r=0. Para resolver isso,

impomos que

(Rk`(r)=0 para ` 6=0

Rk`(r)=cte para `=0) isso, por si so, indica a dependencia em `.

� Quem cuida de possıveis valores diferentes de zero de (~r) na origem e oRk`=0(r).

• Calculo de previsoes fısicas com respeito a medidas de L2e Lz.

Considere uma partıcula, cujo estado e descrito pela funcao de onda:

h~r| i = (~r) = (r, ✓,')

Aprendemos que uma medida de

8>>>>>><

>>>>>>:

L2fornece `(`+ 1)~2, com ` inteiro positivo

ou zero, isto e, `(`+ 1)~2 = 0, 2~2, 6~2, ....

Lz fornece m~, comm inteiro e� ` m `,

isto e, m~ = 0,±~, ...,±`~.• Como calcular a probabilidade de se obter um desses resultados

(`,m ou ambos) a partir de (r, ✓,') ?

13 MAPLima

F689 Aula 24 • Lembrando que podemos expandir qualquer (~r) em termos de k`m(~r) com

k`m(~r) = Rk`(r)Ym` (✓,')

8><

>:

L2 k`m(~r) = `(`+ 1)~2 k`m(~r)

Lz k`m(~r) = m~ k`m(~r)

L± k`m(~r)=~p`(`+ 1)�m(m± 1) k`m±1(~r)

vale: (~r) =X

k

1X

`=0

X

m=�`

ck`mRk`(r)Ym` (✓,'), com ck`m =

Zd3r ?

k`m(~r) (~r)

ou melhor ck`m =

Z 1

0r2dr R?

k`(r)

Z 2⇡

0d'

Z ⇡

0d✓ sin ✓ Y m

`?(✓,') (r, ✓,')

• Quanto vale a probabilidade de medirmos um certo par (`,m) ?

Basta calcular a amplitude de probabilidade de se obter a trinca (k, `,m) e

tratar k como sendo a degenerescencia do par, isto e

PL2,Lz(`,m) =

X

k

��� h k`m| i| {z }

���2=

X

k

���ck`m���2

soma sobre a degenerescencia amplitude de probabilidade

• E se medıssemos somente L2? Qual a chance de obter `(`+ 1)~2. Que tal

PL2(`) =X

k

X

m=�`

��� h k`m| i| {z }

���2=

X

m=�`

PL2,Lz(`,m) =

X

k

X

m=�`

���ck`m���2

soma sobre a degenerescencia amplitude de probabilidade

Momento Angular na Mecânica Quântica: Medidas

14 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica: Medidas

• E se medıssemos somente Lz? Qual a chance de obter m~? Que tal

PLz (m) =X

`�|m|

PL2,Lz(`,m) =

X

k

X

`�|m|

���ck`m���2

• Sera que precisamos expandir em k?Lembre da expressao inicial

(~r) =X

k

1X

`=0

X

m=�`

ck`mRk`(r)Ym` (✓,'), com ck`m =

Zd3r ?

k`m(~r) (~r)

e rescreva

(~r) =1X

`=0

X

m=�`

hX

k

ck`mRk`(r)i

| {z }

Y m` (✓,'), com a`m(r)| {z } =

Zd⌦Y m

`?(✓,') (~r),

a`m(r) obtido sem auxılio de Rk`(r)

ou a`m(r)=

Z 2⇡

0d'

Z ⇡

0d✓ sin ✓ Y m

`?(✓,') (r, ✓,'). Como a`m(r)=

X

k

ck`mRk`(r),

temos ck`m =

Z 1

0r2R?

k`(r)a`m(r) e��a`m(r)

��2 =X

kk0

ck`mc?k0`mRk`R?k0`.

• Note que

Z 1

0r2��a`m(r)

��2dr=X

kk0

ck`mc?k0`m

Z 1

0r2Rk`R

?k0`

| {z }=X

k

���ck`m���2=PL2,Lz

(`,m)

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15 MAPLima

F689 Aula 24

Momento Angular na Mecânica Quântica: Medidas • De forma semelhante, podemos escrever

PL2(`) =X

m=�`

PL2,Lz(`,m) =

X

m=�`

Z 1

0r2��a`m(r)

��2dr

PLz (m) =X

`�|m|

PL2,Lz(`,m) =

X

`�|m|

Z 1

0r2��a`m(r)

��2dr

• Ou seja, obtivemos PL2,Lz(`,m),PL2(`), e PLz (m) de um sistema no estado (~r)

em funcao de a`m(r) =

Zd⌦ Y m

`?(✓,') (~r), (uma integral do produto de uma

Harmonica Esferica pelo estado do sistema). Nao foi necessario usar as Rk`(r)

para obter essas quantidades.

• E se tivessemos interessados em apenas medir Lz. Sera que poderıamos, alemde

evitar o uso das Rk`(r), evitar tambemouso explıcito das Harmonicas Esfericas?

Para tanto, bastaria expandir (r, ✓,') =X

m

bm(r, ✓)eim'

p2⇡

e usar que

bm(r, ✓)=1p2⇡

Z 2⇡

0d' e�im' (r, ✓,') (verifique por substituicao direta e uso de

Z 2⇡

0d'

eim'

p2⇡

e�im0'

p2⇡

=�mm0).ObtenhaPLz(m)=

Z 1

0

Z ⇡

0|bm(r, ✓)|2r2sin ✓d✓dr.

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