ondas parciais e deslocamentos de fase slide da aula 04 ...maplima/f789/2018/aula05.pdf · mt ` (k)...
TRANSCRIPT
1 MAPLima
F789 Aula 05 A amplitude de espalhamento fica
f(k0,k) = �4⇡2X
`
mT`(k)
~2k2 Y 0` (k
0)
r2`+ 1
4⇡, onde (mostre)
Y 0` (k
0) =
r2`+ 1
4⇡P`(cos ✓). Definindo, f`(k) = �⇡
mT`(k)
~2k2 , temos
f(k0,k) = f(✓) =1X
`=0
(2`+ 1)f`(k)P`(cos ✓)
Para entender o significado fısico de f`(k), vamos estudar o comportamento
a longas distancias de hr| (+)k i, isto e:
limr!1
hr| (+)k i = 1
(2⇡)3/2
8:eik.z +
eikr
rf(✓)
9;
Do slide 7, temos eikz =X
`
(2`+ 1)i`j`(kr)P`(cos ✓) onde ja vimos que
limr!1
j`(kr) =e+i(kr�`⇡/2) � e�i(kr�`⇡/2)
2ikr= e�i`⇡/2
| {z }8:e+ikr � e�i(kr�`⇡)
2ikr
9;
i�`
assim limr!1
eikz =X
`
(2`+ 1)P`(cos ✓)8:e+ikr � e�i(kr�`⇡)
2ikr
9;
Ondas parciais e deslocamentos de fase – slide da aula 04
2 MAPLima
F789 Aula 05
Colocando os resultados das caixa verdes na caixa azul do slide anterior, temos:
limr!1
hr| (+)k i = 1
(2⇡)3/2
X
`
(2`+ 1)P`(cos ✓)8:e+ikr � e�i(kr�`⇡)
2ikr+
eikr
rf`(k)
9;
limr!1
hr| (+)k i = 1
(2⇡)3/2
X
`
(2`+ 1)P`(cos ✓)
2ik
8: [1 + 2ikf`(k)| {z }]
e+ikr
r� e�i(kr�`⇡)
r
9;
o potencial so afeta a onda esferica que sai
O coeficiente da onda esferica emergente [1 + 2ikf`(k)] = 1, se V = 0.
Se construıssemos pacotes, o pacote associado aeikr
rso existiria p/ cte t 1
e o associado ae�ikr
rso existiria p/ �1 t cte0. Assim, como o | (+)
k i
contem a informacao completa (antes e depois da colisao), e de se esperar que:I
J.dS = 0. As partıculas que entram devem sair (se nao ha ralos e fontes).
Assim, o fluxo associado aeikr
rdeve ser igual ao fluxo associado a
e�ikr
r. Isso
deve valer para cada onda parcial. Para calcular esses fluxos, defina antes
S` ⌘ 1 + 2ikf`(k) ) a aula 4 terminou aqui.<latexit 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Ondas parciais e deslocamentos de fase – slide da aula 04
3 MAPLima
F789 Aula 05
Ondas parciais e deslocamentos de fase
Assim, a partir de J =~mIm( ⇤r ) podemos escrever
�����
Zr2d⌦Im
8:S⇤
`e�ikr
r
@
@rS`
eikr
r
9;����� =
�����
Zr2d⌦Im
8:e�i(kr�`⇡)
r
@
@r
e�i(kr�`⇡)
r
9;�����
4⇡|S`|2 = 4⇡ =) |S`| = 1 (fruto da conservacao de fluxo).
Se |S`| = 1, e sugestiva a seguinte definicao: S` = e2i�`
8>>><
>>>:
�` ⌘(deslocamento
de fase
�` ! numero real
o 2 e convencao
O que deixa claro que a existencia do potencial pode no maximo infligir uma
mudanca de fase na onda esferica emergente. Como S` ⌘ 1 + 2ikf`(k),
podemos escrever f` =e2i�` � 1
2ik=
ei�` sin �`k| {z }
=sin �`ke�i�`
=sin �`
k(cos �` � i sin �`)
Ou ainda f` =1
k(cot �` � i)| {z }. Assim, f(✓) =
1
k
1X
`=0
(2`+ 1)ei�` sin �`P`(cos ✓)
esta forma sera util
Note que para deduzir isso, usamos apenas que [L2, T ]=[Lz, T ]=0
e conservacao de probabilidade.<latexit 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4 MAPLima
F789 Aula 05
O método de ondas parciais para o problema de espalhamento A amplitude de espalhamento depende dos deslocamentos de fase na forma:
f(✓) =1
k
1X
`=0
(2`+ 1)ei�` sin �`P`(cos ✓).
Para obter a secao de choque diferencial basta calculard�
d⌦= |f(✓)|2. Vamos
calcular a secao de choque integral, tambem chamada de secao de choque total,
se espalhamento de potenciais que nao carregam a estrutura interna do alvo.
Em geral, se integrarmos apenas nos angulos, ela chamara secao de choque
integral. Mas, se integrarmos nos angulos e somarmos sobre todos os processos
possıveis, ela sera denominada secao de choque total. A secao de choque integral
pode ser para um determinado processo, por exemplo, secao de choque integral
elastica, ou secao de choque integral da excitacao X ! A, do alvo Z. No caso
presente, de potencial Hermiteano de um corpo so, a secao de choque integral e
elastica e igual a total, pois so o processo elastico e permitido. Ela e definida
por: �total =
Zd⌦
d�
d⌦=
Zd⌦|f(✓)|2 =
Z 2⇡
0d�
Z ⇡
0sin ✓d✓|f(✓)|2 =
=2⇡
k2
Z +1
�1d(cos ✓)
X
`,`0
(2`+ 1)(2`0 + 1)ei(�`��`0 ) sin �` sin �`0P`0(cos ✓)P`(cos ✓).
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5 MAPLima
F789 Aula 05
O método de ondas parciais para o problema de espalhamento
Mas,
Z +1
�1d(cos ✓)P`0(cos ✓)P`(cos ✓) =
2
2`+ 1�``0 (mostre!) e assim :
�total =2⇡
k2
X
`,`0
(2`+ 1)(2`0 + 1)ei(�`��`0 ) sin �` sin �`02
2`+ 1�``0 e finalmente
�total =4⇡
k2
X
`
(2`+ 1) sin2 �`
Como fica o teorema optico, Imf(✓ = 0) =k�total
4⇡? Isso pode ser obtido de
Imf(✓ = 0) =1
k
1X
`=0
(2`+ 1) Imei�`| {z } sin �` P`(cos (✓ = 0))| {z } =k
4⇡
4⇡
k2
X
`
(2`+ 1) sin2 �`
| {z }sin �` 1 �total
De volta para a relacao entre f` e �`. Quando mudamos a energia, �` muda, e por
conseguinte muda f`. Note que f`, entretanto, nao muda de qualquer maneira,
f` =e2i�` � 1
2ik=) kf` = �i
e2i�`
2+
i
2
Como fica essa dependencia no plano complexo? Proximo slide.
6 MAPLima
F789 Aula 05
O método de ondas parciais para o problema de espalhamento
12
i2
2�`
12
2�`
i
i2
kf` = �ie2i�`
2kf` = �i
e2i�`
2+
i
2Im
Im
Re Re
Comentarios
• kf` precisa cair sobre a circunferencia.
• Se �` << 1 ! kf` fica na parte debaixo do cırculo da direita e e quase real.
• |kf`| e maximo quando 2�` = ⇡ ) �` = ⇡/2. Isso faz sentido, pois, lembre
que �total =4⇡
k2
X
`
(2`+ 1) sin2 �`.
Proxima tarefa: determinar os �`
7 MAPLima
F789 Aula 05
Determinação dos deslocamentos de fase
Suponha V (r) = 0 se r > R (alcance do potencial). Para r > R a funcao de
onda e combinacao de ondas esfericas. De fato, e solucao da equacao de
Schrodinger da partıcula livre, cuja forma geral e uma combinacao de funcoes
de
8><
>:
Newman, ⌘`(kr), que nao se comporta bem na origem.
e
Bessell, j`(kr), que se comporta bem na origem.
Assim, para r > R, onde o potencial e zero, a funcao de onda deve ser uma
combinacao de ⌘`(kr)P`(cos ✓) e j`(kr)P`(cos ✓). Ou se quisermos trabalhar
com funcoes de Hankel
8><
>:
h(1) = j` + i⌘` ) limr!1 h(1) = ei(kr�`⇡/2)
ikr
h(2) = j` � i⌘` ) limr!1 h(2) = � e�i(kr�`⇡/2)
ikr
A funcao de onda pode ser escrita como
hx| (+)k i = 1
(2⇡)3/2
X
`
i`(2`+ 1)A`P`(cos ✓) para r > R, onde
A`(kr) = c(1)` h(1)` (kr) + c(2)` h(2)
` (kr). Os coeficientes que multiplicam as
funcoes de Hankel sao escolhidos de tal forma que se V = 0, A`(kr) fica
j`(kr) em todos os pontos.
8 MAPLima
F789 Aula 05
Para r > R e R muito grande, vimos que
limr!1
hx| (+)k i = 1
(2⇡)3/2
X
`
(2`+ 1)P`(cos ✓)
2ik
8:S`
eikr
r� e�(ikr�`⇡)
r
9;
que precisa ser comparado com
limr!1
hx| (+)k i = 1
(2⇡)3/2
X
`
i`(2`+ 1)P`(cos ✓) limr!1
A`(kr) =
=1
(2⇡)3/2
X
`
i`(2`+ 1)P`(cos ✓) limr!1
8:c(1)` h(1)
` (kr) + c(2)` h(2)` (kr)
9; =
=1
(2⇡)3/2
X
`
i`(2`+ 1)P`(cos ✓)8:c(1)`
ei(kr�`⇡/2)
ikr� c(2)`
e�i(kr�`⇡/2)
ikr
9; =
=1
(2⇡)3/2
X
`
(2`+ 1)P`(cos ✓)
2ik
8:2c(1)`
eikr
r� 2c(2)`
e�i(kr�`⇡)
r
9;
(usei que i` = ei⇡`/2). Comparacao direta fornece:
c(1)` = 12S` =
12e
i2�` e c(2)` = 12
Devolvendo isso em A`, temos A` =12e
i2�`h(1) + 12h
(2). Mostre que
(para r > R) A`(kr) = ei�`�cos �` j`(kr)� sin �` ⌘`(kr)
�
Determinação dos deslocamentos de fase
9 MAPLima
F789 Aula 05
Determinação dos deslocamentos de fase
A partir de A`(kr) = ei�`�cos �` j`(kr)� sin �` ⌘`(kr)
�podemos
calcular �` ⌘� r
A`
dA`
dr
���r=R
= kR8:j0`(kR) cos �` � ⌘0`(kR) sin �`
j`(kR) cos �` � ⌘`(kR) sin �`
9;
onde j0` e ⌘0` sao derivadas com respeito a kr. Se aprendermos como
obter �`, podemos inverter esta equacao e calcular �`. A inversao e
relativamente simples. Basta agrupar os coeficientes dos sin �` e cos �`.
�`
kR
�j`(kR) cos �` � ⌘`(kR) sin �`
�= j0`(kR) cos �` � ⌘0`(kR) sin �` ou ainda
� �`
kRj` � j0`
�cos �` =
� �`
kR⌘` � ⌘0`
�sin �` que fornece:
tan �` =kRj0`(kR)� �`j`(kR)
kR⌘0`(kR)� �`⌘`(kR))
(Se acharmos �`
teremos achado �`.
Para determinar �`, precisamos resolver a equacao de Schrodinger na
regiao do potencial,
d2u`
dr2+�k2 � 2m
~2 V � `(`+ 1)
r2�u` = 0 onde u` = rA`(r) e u`(r)|r=0 = 0
Integre esta equacao ate r = R, calcule �`
��solucaointerna
e iguale a �`
��solucaoexterna
Note que �` e um numero real. Tendo ele, calcule tan �`.
10 MAPLima
F789 Aula 05
Aplicação: Espalhamento da esfera dura
R
|rA0|
r
O problema tridimensional de barreira infinita ou esfera dura e definido pelo
potencial V
8><
>:
1 para r R
0 para r > R
Como a esfera e impenetravel, a funcao de onda precisa se anular em r = R.
Isto ja e suficiente para achar �` (nao e preciso calcular �` neste caso).
A`(kr)��r=R
= ei�`�cos �` j`(kR)� sin �` ⌘`(kR)
�= 0 ) tan �` =
j`(kR)
⌘`(kR)
Caso ` = 0
8><
>:
j0(kR) =sin kRkR
⌘0(kR) = � cos kRkR
=) tan �0 = � tan kR ) �0 = �kR
A parte radial da funcao de onda varia da seguinte forma:
A0(kr) = ei�08:cos �0 sin kr
kr+
sin �0 cos kr
kr
9; = ei�0
sin (kr + �0)
krou
A0(kr) = ei�0sin k(r �R)
kr(note que se nao houvesse V, A0 = j0 =
sin kr
kr)
11 MAPLima
F789 Aula 05
Aplicação: Espalhamento da esfera dura Ainda para a esfera dura, estudaremos tan �` para altas (kR >> 1) e
baixas (kR << 1) energias.
Baixa energia
Para kR << 1
8><
>:
j`(kr) =(kr)`
(2`+1)!!
⌘`(kr) = � (2`�1)!!(kr)`+1
=) tan �` = � (kR)2`+1
(2`+ 1)!!(2`� 1)!!
Para ` = 0 ! tan �0 = �kR ⇡ �0 mesmo resultado que antes que era exato.
Note que e justo desprezar os �` para ` > 0, pois tan �` / (kr)2`+1.
Se �0 = �kR ) d�
d⌦= |f(✓)|2 com f(✓) =
1
k
X
`
(2`+ 1)ei�` sin �`P`(cos ✓)
Truncando a soma, em ` = 0, temos f(✓) ⇡ 1
kei�0 sin �0 ! |f(✓)|2 =
sin2 �0k2
E ) d�
d⌦=
k2R2
k2= R2 =) �total =
Zd�
d⌦d⌦ = R2
Zd⌦ = 4⇡R2
A secao de choque e quatro vezes maior que a secao de choque geometrica.
Baixa energia, comprimento de onda grande, mecanica quantica difere da
mecanica classica. Sera que para altas energias teremos �total = ⇡R2?
12 MAPLima
F789 Aula 05
p
r r sin ✓⇡ � ✓
R
Alta energia
Para kR >> 1 precisamos somar as contribuicoes de �` que diferem de
zero. Lembre que soma em ` vai ate infinito, mas a partir de um dado `,
�` ⇡ 0 e a soma pode ser truncada. ` esta associado momento angular.
Classicamente L = |r⇥ p| = rp sin ✓ onde r sin ✓ e o “braco de alavanca”,
indicado na figura.
Note que classicamente quando r sin ✓ > R, a partıcula, com momento
angular superior a Rp, deixa de colidir com a esfera rıgida. Com isso em
mente, somaremos em ` ate `max = kR. Desta forma, a secao de choque
total pode ser escrita por
�total =4⇡
k2
`=kRX
`=0
(2`+ 1) sin2 �`
Aplicação: Espalhamento da esfera dura
13 MAPLima
F789 Aula 05
Aplicação: Espalhamento da esfera dura Considerando sin2 �` =
tan2 �`1 + tan2 �`
e que tan �` =j`(kR)
⌘`(kR), temos:
sin2 �` =j2` (kR)
j2` (kR) + ⌘2` (kR). Para kR >> 1
8><
>:
j`(kR) = 1kR sin (kR� `⇡/2)
⌘`(kR) = � 1kR cos (kR� `⇡/2)
e assim sin2 �` =sin2 (kR� `⇡/2)
sin2 (kR� `⇡/2) + cos2 (kR� `⇡/2)= sin2 (kR� `⇡/2)
Agora basta realizar a soma: �total =4⇡
k2
`=kRX
`=0
(2`+ 1) sin2 (kR� `⇡/2)
Para realizar esta soma, considere dois termos consecutivos da serie
�total =4⇡
k2
`=kRX
`=0
(2`+ 1) sin2 �`, isto e
8><
>:
` ! (2`+ 1) sin2 �`e
`0 = `+ 1 ! (2(`+ 1) + 1) sin2 �`+1
some-os para obter (2`+ 1) (sin2 �` + sin2 �`+1)| {z }+2 sin2 �`+1| {z }1 devido ao `⇡/2 < 1
Para finalmente obter �total =4⇡
k2
8:
`=kRX
`=02 em 2
(2`+ 1) +`=kRX
`=02 em 2
2 sin2 �`+1
9;
14 MAPLima
F789 Aula 05
Aplicação: Espalhamento da esfera dura Note agora que a primeira soma e de uma progressao aritmetica e a segunda
e uma soma de numeros menores do que um
�total =4⇡
k2
8:
`=kRX
`=02 em 2
(2`+ 1)
| {z }
+2`=kRX
`=02 em 2
sin2 �`+1
| {z }
9;
numero determos ⇥
valormedio este vai com kR
Soma de termos da progressao aritmetica e igual akR
2⇥ 1 + 2kR+ 1
2/ (kR)2
2.
A segunda soma e linear em kR e pode ser desprezada para kR, suficientemente
grande. Note que se trocassemos todos os sin2 �`+1 por 1 a soma seria kR/2.
A secao de choque total, e, portanto �total =4⇡
k2(kR)2
2⇡ 2⇡R2!! Isso e o dobro
da geometrica. Surpreso?
• Um potencial com uma descontinuidade e sempre quantico!
• Mesmo comprimentos de onda minusculos percebem a descontinuidade.