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.