sistema de partículas idênticas na mecânica quânticamaplima/f789/2018/aula25.pdf · quando um...
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1 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuânticasimetrizadoreseanti-simetrizadores
• O Postulado de Simetrizacao.
Quando um sistema inclui partıculas identicas, somente certos kets do seu
espaco de estados podem descrever seus estados fısicos. Os estados fısicos sao,
dependendo da natureza das partıculas identicas, completamente simetricos ou
completamente anti-simetricos com respeito a permutacao destas partıculas.
• Chamaremos
8><
>:
Bosons, as partıculas identicas que tem estados simetricos;
Fermions, as partıculas identicas que tem estados anti-simetricos.
• O estado | i nao e qualquer ket de E (espaco construıdo pelo produto tensorial
de espacos de uma partıcula). Isto e, | i 2 ES ou | i 2 EA, sub-espacos de E .
• Regra empırica:
8><
>:
spin semi-inteiro ! fermions: e�, e+, p+, n, muons;
spin inteiro ! bosons: fotons, mesons.
• Essa regra vale para composicao de partıculas. Exemplo
8><
>:
3He ! fermion;
4He ! boson.<latexit 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2 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuânticasimetrizadoreseanti-simetrizadores
• Removendo a degenerescencia de troca.
Considere |ui, um ket do espaco de N partıculas identicas (ignore, por hora,
simetrizacao). Se |ui e um estado fısico, P↵|ui tambem seria. Considere Euformado por todos os {P↵|ui}.
� A dimensao deste espaco seria algo entre 1 e N ! (numero de estados ortogonais).
� Tal dimensao e justamente a degenerescencia de troca.
� O novo postulado diz que, de fato, o estado fısico e ou totalmente simetrico
(bosons) ou totalmente anti-simetrico (fermions).
• Temos que mostrar que Eu contem um unico ket de ES ou um unico ket de EA.� Para isso, basta usar as relacoes S = SP↵ e A = ✏↵AP↵, pois se |ui e P↵|ui sao
ortogonais
8><
>:
S|ui = SP↵|ui ) o que faz deles colineares.
A|ui = ✏↵AP↵|ui ) o que faz deles colineares.
• Todos os kets de Eu = {P↵|ui} quando simetrizados ficam colineares, isso indica
que a simetrizacao de qualquer membro deste sub-espaco gera um unico ket.
• Adegenerescencia de troca foi removida. Vale tambem para os anti-simetrizados.
• Como veremos, existirao situacoes em que a simetrizacao ou a
anti-simetrizacao de |ui gerarao kets nulos.<latexit 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3 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuânticaConstruindo os kets físicos
⌅ Regra de Construcao para N Partıculas.
F Enumere as partıculas arbitrariamente e construa um ket |ui, correspondente
ao estado fısico considerado e aos numeros dados (arbitrados) as partıculas;
F Aplique S ou A em |ui, dependendo se as partıculas sao bosons ou fermions;
F Normalize o que obtido com a estrategia acima.
⌅ Aplicacao a sistemas de duas partıculas identicas.
� Suponha que uma possa ser caracterizada por um ket (de uma partıcula) |'ie a outra pelo ket (tambem de uma partıcula) |�i.
� Suponha que as partıculas ocupem kets distintos, |'i 6= |�i.F Primeira regra: |ui = |1: '; 2: �i
F Segunda regra:
8><
>:
S|ui =�12!
P↵ P↵
�|ui = 1
2
�|1: '; 2: �i+ |1: �; 2: 'i
�
A|ui =�12!
P↵ ✏↵P↵
�|ui = 1
2
�|1: '; 2: �i � |1: �; 2: 'i
�
F Para
(|';�i = cS|ui|';�i = cA|ui
) h';�|';�i = c2
4(1 + 1) = 1 ! c =
p2.
) os kets normalizados sao |';�i = 1p2
�|1: '; 2: �i+ ✏|1: �; 2: 'i
�
com ✏ = 1 ! bosons e ✏ = �1 ! fermions.<latexit 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4 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuânticaOperadoresdePermutação-SistemadeNpartículas
F Suponha agora que |'i = |�iNeste caso, a primeira regra fornece |ui = |1: '; 2: 'i ) ja simetrizado.
A simetrizacao nos leva a
8><
>:
S|ui = 12
�|1: '; 2: 'i+ |1: '; 2: 'i
�= |ui;
A|ui = 12
�|1: '; 2: 'i � |1: '; 2: 'i
�= 0.
• O que permite concluir: Nao existe ket em EA para descrever um estado fısico
na qual dois fermions estao no mesmo estado individual |'i. Familiar? Esse
e o princıpio de exclusao de Pauli.
⌅ Generalizacao para um numero arbitrario de partıculas.
• Comece com N = 3. Suponha que os tres estados normalizados sejam dados
por |'i, |�i e |!i.F Regra 1: |ui = |1: '; 2: �; 3: !iF Regra 2: caso bosons.
S|ui = 1
3!
X
↵
P↵|ui =1
6
⇣|1: '; 2: �; 3: !i+ |1: !; 2: '; 3: �i+ |1: �; 2: !; 3: 'i+
+|1: '; 2: !; 3: �i+ |1: �; 2: '; 3: !i+ |1: !; 2: �; 3: 'i⌘
F Regra 3: Se os estados |'i, |�i e |!i sao ortonormais, c =p3!.
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5 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuânticaF Comentarios adicionais: caso bosons
� Suponha que duas partıculas ocupem o mesmo estado, por exemplo, |'i = |�i
Neste caso, |';';!i = 1p3
⇣|1: '; 2: '; 3: !i+ |1: '; 2: !; 3: 'i+ |1: !; 2: '; 3: 'i
⌘
� Se os tres kets sao iguais, |'i = |�i = |!i, entao |ui = |1: '; 2: '; 3: 'iF Regra 2: caso fermions.
A|ui = 1
3!
X
↵
✏↵P↵|ui =1
6
X
↵
✏↵P↵|1: '; 2: �; 3: !i(Perceba que os sinais sao definidos segundo
as mesma regras que um determinante.
Esse e o chamado determinante de Slater:
A|ui = 1
3!
������
|1: 'i |1: �i |1: !i|2: 'i |2: �i |2: !i|3: 'i |3: �i |3: !i
������
� Note que respeita o princıpio de exclusao de Pauli: A|ui e zero, sempre que
duas partıculas ocupam o mesmo estado, isso e
8><
>:
|'i = |�i, ou
|�i = |!i, ou
|'i = |!i
F Regra 3: Normalizacao c =p3!. Mais partıculas e direto.
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6 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuântica
F Comentarios adicionais: caso fermions
� Note que poderıamos ter dois estados com a mesma parte espacial, mas com
spins diferentes. Isso seria suficiente para garantir que o determinante nao
fosse nulo.
� A hipotese de uma solucao como um unico produto direto de kets de uma
partıcula e uma aproximacao forte. E a hipotese inicial (que trata partıculas
interagentes como se fossem independentes) da chamada aproximacao
Hartree-Fock). Precisamos definir bases que atuem nos espacos de estados
fısicos (A ou S).
⌅ Construcao de uma base no espaco de estados fısicos.
• Considere um sistema de N partıculas identicas.
� Comece com uma base {|uii} no espaco de estados de uma partıcula e
construa {|1: ui; 2: uj ; ...;N: upi} ⌘ E ! E base. Aprendemos isso em F689.
� O espaco fısico de partıculas identicas nao e E e sim ES ou EA.• Como determinar uma base do espaco fısico?
� Aplicando S ou A em todos os kets de E = {|1: ui; 2: uj ; ...;N: upi} podemos
obter um conjunto de vetores que definem ES ou EA.� Sera que eles formam uma base?
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7 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuântica
• Seja |'i um ket arbitrario de ES (caso EA e semelhante).
� Como E e uma base, podemos sempre escrever
|'i =X
i,j,...,p
ai,j,...p|1: ui; 2: uj ; ...;N: upi
� Mas |'i 2 ES , entao S|'i = |'i. E isso implica em
S|'i = |'i =X
i,j,...,p
ai,j,...pS|1: ui; 2: uj ; ...;N: upi
� Note que nem sempre os S|1: ui; 2: uj ; ...;N: upi sao independentes!
� De fato, os kets nao simetrizados (por exemplo, |1: ui; 2: uj ; ...;N: upi), que
possuem as mesmas funcoes (ui, uj , ..., up) de uma partıcula (mesmas em
numero e em especie), quando simetrizados dao o mesmo ket simetrizado
(ja vimos isso).
� O mesmo ocorreria se tivessemos antissimetrizando.
� Tal discussao introduz o conceito de Numero de Ocupacao: por definicao
para o ket |1: ui; 2: uj ; ...;N: upi, o numero de ocupacao nk do estado
individual |uki e igual ao numero de vezes que o estado |uki aparece na
sequencia |uii, |uji, ..., |upi. Isso e o numero de partıculas no estado |uki.
Temos, obviamente,
X
k
nk = N.<latexit 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8 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuântica
� Dois kets diferentes, com os mesmos numeros de ocupacao, quando sujeitos
ao operador S, dao a mesma coisa. Daqui para frente sera denotado por
|n1, n2, ..., nk, ...i = cS| 1: u1; 2: u1; ...;n1: u1| {z }n1 em |u1i
;n1+1: u2, ..., n1+n2: u2| {z }n2 em |u2i
; ...i.
Genericamente, nesta notacao, temos nk partıculas em |uki.� Para o caso de fermions basta trocar S por A.
⌅ Algumas propriedades dos estados |n1, n2, ..., nk, ...i.• O que devemos esperar do produto escalar de dois kets, |n1, n2, ..., nk, ...i
e |n01, n
02, ..., n
0k, ...i?
� Serao diferentes de zero, somente se os numeros de ocupacao forem iguais,
isto e: nk = n0k, 8 k. Se n1 > n0
1, vai sobrar |u1i que e ortogonal a 8 |ujipara j 6= i.
• Se as partıculas estudadas sao bosons, os kets |n1, n2, ..., nk, ...i, com nk
arbitrario, mas restrito a condicao
X
k
nk = N, formam uma base.
� Ja vimos que 8 |ui simetrico, e combinacao de S|1: ui; 2: uj ; ...;N: upie cada um desses kets e por definicao um |n1, n2, ..., nk, ...i.
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9 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuântica
continuacao: ainda para bosons.
� Assim, 8 |ui simetrico, ele sera uma combinacao de |n1, n2, ..., nk, ...i,o que caracteriza {|n1, n2, ..., nk, ...i} como sendo uma base.
� Vimos tambem que |n1, n2, ..., nk, ...i e |n01, n
02, ..., n
0k, ...i sao ortogonais,
salvo se n1 = n01, n2 = n0
2, ..., nk = n0k (e nesse caso a norma e 1). O
caracteriza {|n1, n2, ..., nk, ...i} como sendo uma base ortonormal.
� Para caracterizar que temos de fato uma base a partir da simetrizacao de
um conjunto completo de kets de E , precisamos apenas mostrar que o ket
nulo nao pode ser obtido nesse processo. Se conseguirmos o ket nulo, a
base seria linearmente dependente.
|n1, n2, ..., nk, ...i = cS| 1: u1; 2: u1; ...;n1: u1| {z }n1 em |u1i
;n1+1: u2, ..., n1+n2: u2| {z }n2 em |u2i
; ...i =
= c1
n!P↵|1: u1; 2: u1; ...;n1: u1;n1+1: u2, ..., n1+n2: u2; ...i ) troca entre
iguais gera kets colineares com coeficientes positivos e iguais. Troca entre
distintos gera kets ortogonais. Apos juntar os iguais (soma direta pois
todos tem o mesmo coeficiente) temos uma soma de vetores ortogonais
e portanto, nunca igual a zero.<latexit 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10 MAPLima
F789 Aula 25
SistemadePartículasIdênticasnaMecânicaQuântica
• E se as partıculas forem fermions?
A base do estado fısico seria obtida escolhendo o conjunto de kets
|n1, n2, ..., nk, ...i nos quais, os numeros de ocupacao sao nk = 1 ou
nk = 0. Continua valendo
X
k
nk = N.
� Conforme vimos nos slides 2 e 5, para os caso de duas e tres partıculas,
respectivamente, quando mais de uma partıcula ocupam o mesmo estado
individual, a antisimetrizacao anula o ket fısico resultante. Ou seja, se um
dos numeros de ocupacao for maior ou igual a 2 em |n1, n2, ..., nk, ...i, esse
ket fica automaticamente igual a zero devido a antissimetrizacao.
� Se todos os numeros de ocupacao forem 1 ou 0, entao |n1, n2, ..., nk, ...i 6= 0.
Isso porque (escolhi os primeiros N estados individuais como ocupados)
|1, 1, ..., 1, 0, 0, 0...i| {z }N partıculas em estados distintos
= cX
↵
✏↵ P↵|1: u1; 2: u2; ...;N: uki| {z }kets ortogonais
• Quanto vale c, em
|n1, n2, ..., nk, ...i = cS| 1: u1; 2: u1; ...;n1: u1| {z }n1 em |u1i
;n1+1: u2, ..., n1+n2: u2| {z }n2 em |u2i
; ...i?
Mostre que c =pN ! para fermions e c =
pN !/n1!n2!... para bosons.
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