aula de hoje: potenciais e transformações de gaugemaplima/f789/2018/aula15.pdf · aula 15 aula de...

1 MAPLima

F789 Aula 15

Auladehoje:PotenciaisetransformaçõesdeGaugeEm Mecanica Classica o ponto zero de energia nao tem significado fısico. A

evolucao temporal de variaveis dinamicas, tais como, r(t),p(t) ou L(t), nao e

alterada se usarmos V (r) ou V (r) + V0 (constante no espaco e no tempo).

Isso porque F = �rV (r) e ) V0 uma constante aditiva, nao e importante.

Como sera a situacao analoga em Mecanica Quantica?

Suponha

8><

>:

|↵, t0; ti ! V (R)

^|↵, t0; ti ! V (R) = V (R) + V0

ambas com |↵i em t = t0.

Nos sabemos evoluir kets no tempo

^|↵, t0; ti = exp�� i

⇥ p2

2m+ V (R) + V0

⇤ (t� t0)

~ |↵i =

= exp⇥� iV0(t� t0)

~⇤|↵, t0; ti.

A presenca de V0 faz com que os estados |↵, t0; ti e ^|↵, t0; ti fiquem

diferentes! Eles diferem por uma fase global exp⇥� iV0(t� t0)

~⇤.

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2 MAPLima

F789 Aula 15

representação das coordenadas

Potencialconstante(independentedotempoedaposição)Para um estado estacionario, somar um potencial constante V0 na

Hamiltoniana implica em uma mudanca na dependencia temporal

do tipo

8><

>:

Se V (R) �! exp⇥� iE(t�t0)

~⇤

Se V (R) + V0 �! exp⇥� i(E+V0)(t�t0)

~⇤

ou seja, mudar de V (R) ! V (R) + V0 implica em E ! E + V0.

Efeitos observaveis, como evolucao temporal de valores esperados

(exemplos: hRi, hPi, etc.), dependem das chamadas frequencias de

Bohr (diferencas em energia) e portanto nao mudam mediante a

adicao de um potencial constante na Hamiltoniana.

Esse de fato, e um primeiro exemplo de transformacao de Gauge:

V (R) ! V (R) + V0

8><

>:

|↵, t0; ti ! exp⇥� iV0(t�t0)

~⇤|↵, t0; ti

E ! E + V0

(r0, t) ! exp⇥� iV0(t�t0)

~⇤ (r0, t)

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3 MAPLima

F789 Aula 15

Potencialdependentesódotempo

⇤O pacote de onda de cada partıcula e dividido em dois.

O que aconteceria, se V0 fosse espacialmente uniforme, mas dependente do

tempo? |↵, t0; ti = U(t, t0)|↵i, com U(t, t0) = exp⇥� i

~

Z t

t0

H(t0)dt

0⇤. Mas,

se H(t) = H + V0(t), entao

|↵, t0; ti = exp⇥� i

~H(t� t0)⇤exp

⇥� i

~

Z t

t0

V0(t0)dt

0⇤|↵i

e ) terıamos |↵, t0; ti| {z } �! exp⇥� i

~

Z t

t0

V0(t0)dt

0⇤|↵, t0; ti| {z }

V (R) V (R) + V0(t)

Pergunta importante: Sera que a adicao de um potencial espacialmente

constante, mas variavel no tempo, causa algum efeito mensuravel no sistema?

Para responder isso, imagine um experimento com um feixe de partıculas que

e separado⇤em dois: Parte passa por uma regiao de potencial V1(t) e a outra

parte por um potencial V2(t) que variam de forma diferente no tempo. Em

seguida juntamos os feixes e realizamos uma experiencia de interferencia.

Sera que perceberıamos diferencas nos resultados deste novo experimento,

conforme variassemos a dependencia temporal dos potenciais?<latexit 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4 MAPLima

F789 Aula 15

A interferencia sera do tipo cos (�1 � �2) e/ou sin (�1 � �2), onde

�1 � �2 =1

~

Z t2

t1

dt�V2(t)� V1(t)

�. Embora nenhuma forca atue na

partıcula, pois rV (t) = 0, existe um efeito mensuravel. Efeito

puramente quantico, pois quando ~ ! 0

(cos (�1 � �2)

sin (�1 � �2)oscilam

muito, destruindo qualquer efeito interessante.

V1(t)

V2(t)

Dimensao do

pacote menor

que a gaiola

Pode criar diferenca

de potencialzona de

interferencia

Separa o feixe

O Experimento

Gailola metalica: gerando V1(t) e V2(t) depois que a

partıcula esta dentro dela e desligando antes dela sair

Experimentodecorrelação.Potencialdependentesódotempo

5 MAPLima

F789 Aula 15

GravidadeemMecânicaQuânticaSera que existe algum efeito gravitacional quantico? Veremos que sim, mas

antes exploremos as diferencas entre mecanica classica e quantica.

Considere um corpo em queda livre

minercialr = �mgravr�grav = �mgravgz

como minercial = mgrav ! r = �gz

Uma vez que a massa nao aparece na equacao acima, gravidade em mecanica

classica e considerada uma teoria puramente geometrica.

A situacao em Mecanica Quantica nao e bem assim:

⇥�~22m

r2 +m�grav

⇤ = i~@

@t! a massa nao cancela!

De fato, a massa aparece junto com ~, na forma~m. Isso tambem pode ser

visto na formulacao de integrais de caminho de Feynman

hrn, tn|rn�1, tn�1i =r

m

2⇡i~�texp

⇥i

Z tn

tn�1

dtLclassica(r, r)

~⇤com

Lclassica(r, r)

~ =12mz2 �mgz

~ ! note ~ e m juntos e na formam

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6 MAPLima

F789 Aula 15

GravidadeemMecânicaQuânticaNo limite nao quantico, caımos na trajetoria classica definida pelo

princıpio de mınima acao

�

Z t2

t1

dtLclassica(t)

~ = �

Z t2

t1

dt⇥ 12mz2 �mgz

~⇤= 0 e m se cancela.

Lembra do teorema de Ehrenfest?

md2

dt2hRi = d

dthPi = �hrV (R)i se rV (R) = mgz entao

md2

dt2hRi = �mgz (nao depende de ~ nem de m)

Para ter efeito quantico nao trivial precisamos ter dependencia com

~ e consequentemente com m tambem.

•Ate 1975, nao se sabia de nenhum experimento direto que permitisse

usar o termo m�grav na equacao de Schrodinger.

•Queda livre de partıculas elementares ! equacao de Ehrenfest e

suficiente para mostrar que nao teremos efeitos quanticos.

• Experimento sobre “peso de protons” de V. Pound et al. tambem

nao estabeleceu efeitos gravitacionais quanticos (eles mediram

deslocamentos de frequencias que nao dependem de ~ explicitamente).<latexit 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7 MAPLima

F789 Aula 15

Efeito gravitacional na escala microscopica e difıcil de detectar. Porque?

As forcas envolvidas sao muito pequenas. Para ilustrar, considere o estado

ligado do atomo de hidrogenio. Sabemos que o tamanho tıpico do atomo

e estimado por a0 ⌘ raio de Bohr =~2

e2me. Como seria “a0” se a forca de

ligacao fosse gravitacional? Para calcular isso, troque o proton por um

neutron, o que equivale a trocar � e2

r2por � Gmemn

r2. Obterıamos com

isso, um raio de Bohr gravitacional “a0” =~2

Gm2emn

. A forca gravitacional

entre o neutron e eletron e cerca de 1039vezes menor que a forca eletrica

entre o proton e o eletron, poise2

Gmemn⇡ 2.3⇥ 10

39.

Isto fornece um raio de Bohr gravitacional de 1031cm ⇡ 10

13anos-luz.

(Tamanho algumas vezes maior que o raio estimado do universo - 1028cm)

GravidadeemMecânicaQuântica

8 MAPLima

F789 Aula 15

InterferênciaQuânticaInduzidaporGravidadeExperimentodecorrelaçãocomnêutrons(massamn)

C

Pacote de onda

de um neutron

D

A

B Pacote e reunido em

Medida de Interferencia

`1

`2

Pacote e dividido

em A: pedaco por ABD

outro por ACD

dois casos

8>>>>>><

>>>>>>:

1) ABCD em um plano horizontal: potencial gravitacional igual

em ambos os percursos.

2) ABCD rodado de � em torno de AC. Potencial gravitacional

no percurso AB e igual ao do CD, mas no percurso BD e

diferente do potencial do percurso AC.

�A

CB

D

`1

`2

� grav=mng2sin�

9 MAPLima

F789 Aula 15

PartículacarregadasujeitaàumcampoeletromagnéticoUm pouco sobre cargas interagindo com campos eletromagneticos

1) Comeca com a Forca de Lorentz md2r

dt2= F = e

⇥E+

v ⇥B

c

⇤.

2) Das Equacoes de Maxwell, percebemos a importancia de � e A e

temos

8>>>>>><

>>>>>>:

r.B = 0 ! B = r⇥A

r⇥E = � 1c@B@t ! r⇥

�E+ 1

c@A@t

�= 0

) E+ 1c@A@t = �r� ! E = �r�� 1

c@A@t

3) Equacoes de Movimento de Newton mx = e�Ex + y

Bz

c� z

By

c

�!

mx = e�� @�

@x� 1

c

@Ax

@t+

y

c

⇥@Ay

@x� @Ax

@y

⇤� z

c

⇥@Ax

@z� @Az

@x

⇤

4) Lagrangeana: construıda de tal forma qued

dt

� @L@qi

�=

@L

@qiforneca a

equacao de movimento. Verifique que a expressao serve:

L =1

2mr2 +

e

cr.A(r, t)� e�(r, t)

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10 MAPLima

F789 Aula 15

Partículacarregadasujeitaàumcampoeletromagnético5) Momento canonico (direcao x) px ⌘ @L

@x= mx+

e

cAx(r, t)

Ou seja, p = mr+e

cA(r, t) (nao e simplesmente mv)

6) Construa a Hamiltoniana Classica

H = p.r� L =1

2m

�p� e

cA�2

+ e�

7) Utilize as ferramentas as desenvolvidas para estudar o analogo quantico.

Para ilustrar, suponha inicialmente que E e B possam ser derivados de A

e � independentes do tempo. Isto e:

�(r, t) = �(r) e A(r, t) = A(r), com

8><

>:

E = �r�

B = r⇥A

H = 12m

�p� e

cA�2

+ e�

Na mecanica quantica, A e � sao funcoes do operador posicao R da partıcula.

Sendo assim, como [Ri, Pj ] = i~�ij ! [Pi, Aj ] 6= 0. Isso exige cuidado para

lidar com H. Por exemplo, a parte envolvendo o operador

�P� e

cA�2

= P2 � e

c

�A.P+P.A| {z }

�+

e2

c2A2

note simetria em A e P e que H e Hermiteano<latexit 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11 MAPLima

F789 Aula 15

A Hamiltoniana pode ser reescrita (formalismo de Heisenberg) como

H =1

2m

�P� e

cA�2

+ e� =⇧2

2m+ e�

Ja sabemos que mdR

dt= ⇧ = P� e

cA. Quanto vale m

d2R

dt2=

d⇧

dt?

Facamos a conta para a componente i = 1, isto e:

d⇧1

dt=

1

i~ [⇧1, H] =1

i~ [⇧1,⇧2

2m+ e�]

Mostre que

8>>><

>>>:

[⇧1,⇧21] = 0

[⇧1,⇧22] =

i~ec

�m

dR2dt B3 +B3m

dR2dt

�

[⇧1,⇧23] = � i~e

c

�m

dR3dt B2 +B2m

dR3dt

�

[⇧1, e�] = �i~e @�@R1

= i~eE1

e obtenha

para a componente 1 :d⇧1

dt= eE1 +

e

2c

��dRdt

⇥B�1��B⇥ dR

dt

�1

Enfim temos: md2R

dt2=

d⇧

dt= e

⇥E+

1

2c

�dRdt

⇥B�B⇥ dR

dt

�]

A versao quantica da forca de Lorentz (teorema de Ehrenfest).<latexit 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Partículacarregadasujeitaàumcampoeletromagnético

12 MAPLima

F789 Aula 15

PartículacarregadasujeitaàumcampoeletromagnéticoComo fica a equacao de Schrodinger com � e A?

hr0|H|↵, t0; ti = i~ @@t

hr0|↵, t0; ti

Comecemos por

1

2mhr0|

�P� e

cA(R)

�2|↵, t0; ti =

=1

2m

�� i~r0 � e

cA(r0)

�hr0|

�P� e

cA(R)

�|↵, t0; ti =

=1

2m

�� i~r0 � e

cA(r0)

��� i~r0 � e

cA(R0)

�)hr0|↵, t0; ti

Note que o gradiente da esquerda “pega”todo mundo (inclusive o A(r0)

da direita). Com isso, a equacao de Schrodinger e dada por:

1

2m

�� i~r0 � e

cA(r0)

��� i~r0 � e

cA(r0)

�hr0|↵, t0; ti+ e�(r0)hr0|↵, t0; ti =

= i~ @@t

hr0|↵, t0; ti

Como seria a nova definicao de j, o fluxo de probabilidade? Lembram

de como definı-lo? Use@⇢

@t+r.j = 0, ⇢ =

⇤ , com = hr0|↵, t0; ti

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13 MAPLima

F789 Aula 15

PartículacarregadasujeitaàumcampoeletromagnéticoPara calcular j, o fluxo de probabilidade, comece por

@⇢

@t=@ ⇤

@t + ⇤ @

@t,

e use a equacao de Schrodinger do slide anterior para obter:

@⇢

@t=

1

i~⇥� hr0|⇧

2

2m|↵, t0; ti⇤hr0|↵, t0; ti+ hr0|↵, t0; ti⇤hr0|

⇧2

2m|↵, t0; ti.

Note que a parte em � cancelou, pois �⇤ = �. A tarefa se resume em colocar

a expressao acima na forma:@⇢

@t+r0j = 0. Depois de algum trabalho, temos:

j =~mIm( ⇤r0 )� e

mcA| |2

que poderia ter sido obtido, com

8><

>:

troca p ! p� emcA

~ir

0 � emcA(r0) ! representacao r

Quando escrevemos a funcao de onda na forma (r0, t) =p⇢ exp

� iS~�,

S real, obtivemos Im( ⇤r0 ) =⇢

~r0S. Assim, o j, pode ser escrito por:

j =~m

⇢

~r0S � e

mcA⇢ =

⇢

m

�r0S � e

cA�

Mostre, seguindo o que fizemos para A = 0, que

Zd3r0j =

h⇧im

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14 MAPLima

F789 Aula 15

TransformaçãodeGaugeEstamos prontos para estudar transformacao de Gauge. Considere primeiro:

� ! �+ � e A ! A, onde � e constante - nao depende de r e t.

Esta transformacao ja foi estudada antes (mudanca no zero da energia).

Uma transformacao mais interessante e:

8><

>:

� ! �� 1c@�@t

A ! A+r�(r, t)

Cujos campos

8>><

>>:

E = �r(�� 1c@�@t )�

1c

@@t (A+r�) = �r�� 1

c@@tA

B = r⇥ (A+r�) = r⇥A+r⇥r�| {z } = r⇥A

0

ficam inalterados e ) a trajetoria classica da partıcula carregada sob

acao destes campos nao se altera (a forca de Lorenz nao se altera).

Queremos estudar os efeitos quanticos da transformacao de Gauge.

Para isso, considere um caso particular| {z }

8><

>:

� ! �

A ! A+r�(r)

�(r, t) = �(r) nao depende de t.<latexit 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15 MAPLima

F789 Aula 15

TransformaçãodeGauge

Considere B = Bz. Tal campo pode ser derivado de

8>>>>>><

>>>>>>:

Ax = �By2

Ay = Bx2

Az = 0

pois,

B = r⇥A =�@Az

@y� @Ay

@z

�x+

�@Ax

@z� @Az

@x

�y +

�@Ay

@x� @Ax

@y

�z = Bz

ou de

8>>>>>><

>>>>>>:

Ax = �By

Ay = 0

Az = 0

pela mesma razao, ou seja B = r⇥A = Bz.

Note que os dois A0s se relacionam por Anovo = Avelho +r��Bxy

2| {z }

�

�(r)<latexit sha1_base64="ROHTfGjc5B+yop5WIOtHUHYywEk=">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</latexit><latexit sha1_base64="ROHTfGjc5B+yop5WIOtHUHYywEk=">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</latexit><latexit sha1_base64="ROHTfGjc5B+yop5WIOtHUHYywEk=">AAAKBnic3VZLb9NAEJ6WAkmAksKRy4qKQkUbORUSXCr1ceGEitSXVEfR2tkkpn7h3URxrJy48Fe4cAAhrvwGbpz5I8xOFpqkMUrggISltWd3Xt9+O7OyE/ueVJb1bWHxytLVa9cLxdKNm7eWb5dX7hzLqJO44siN/Cg5dbgUvheKI+UpX5zGieCB44sT53xf60+6IpFeFB6qNBa1gLdCr+m5XOFStLK0DGtggwIBPXxnsA8RhCDBgwauJTgYDHBkaOVAE6U9mm+TNLpuQxs4xmDQR4sKftlYZAaHqPfx6+I3gBgzMXo3KIukfMzk9aCLVg2y0SsDyiOghZoQo+kY2kOipgS7UMcsGtMm2jXRn6NFhghT1GewRf42IbKNfUr249a9XOs+WVs4szFriJgmMYzvVe/LQ82G0f2MtoazPYw1yluIcRxkhhOzOo6H/Oi4o/zuonabYjmob8GjCfQ25uQ4095Dnoe4B1N1KeXaxFGaIUqaE6VvEA8RrefUQ484eDw39t5vs86KPY+B3kzY0z/EnsdYby7seQykM2HvE/ZtrLc87XhdjldwBB3Te8zU8Gz9p3tueq9Zc3fV5A0So6z7hJn+CEjWHPbhOyLe+IVbR3hF2sFf9Vs+dxWzk+n8vUAcirh7jXgE4ZLE5/BemLxVdbaHtD6+Y0m+idm3i74R4Q/MHZNMjVSn+ZAfnypmNGqIfl0cA9rhPJ5dwtE2vror8ni96BYb9x+aW90x9S7Qb/Km7o3c1Rf+6yPssn8y/tfMOrpPleTg6XA8qYuzTPAE1qFUL69aFYsedlmoGmEVzHNQL3+1G5HbCUSoXJ9LeVa1YlXLeKI81xeDkt2RIubuOW+JMxRDHghZy+g3ZsAe4EqDNaMER6gYrY56ZDyQMg0ctAy4astJnV6cpjvrqOazWuaFcUeJ0B0manZ8piKm/4lYw0uEq/wUBe4mHmJlbpsn3FX456RJqE5u+bJwvFWpWpXqyyerO1uGjgLcg/tIaxWewg48hwM4ArfwpvCu8KHwsfi2+L74qfh5aLq4YHzuwthT/PIDgPvp5A==</latexit><latexit sha1_base64="ROHTfGjc5B+yop5WIOtHUHYywEk=">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</latexit>

16 MAPLima

F789 Aula 15

TransformaçãodeGauge

Para B = Bz, considere a forca de Lorentz

F = e�v ⇥B

c

�=

e

c

������

x y zvx vy vz0 0 B

������=

eB

c(vyx� vxy) =

=eB

cv?(sin'x� cos'y| {z }) = �eB

cv?(e')

� e'

@ forca na direcao z e a forca sempre aponta ? a v e no plano xy.

) a partıcula realiza uma espiral (movimento circular no plano xy).

'e'v?

vk

B = Bz

v

x

y

z

v?

e'

e''

'

'

x

y

17 MAPLima

F789 Aula 15

TransformaçãodeGauge

mostre isso usando a transformação de Gauge para redefinir a Lagrangeana e consequentemente p.

Considere agora a equacao de Hamiltondpx

dt= �@H

@x

onde H =1

2m

�p� e

cA�2

+ e �|{z} =1

2m

�p2 � e

cp.A� e

cA.p+

e2

c2A2

�

0

dpx

dt= �@H

@x

8><

>:

= 0 se usarmos A = (�By, 0, 0)

6= 0 se usarmos A = (�By2 ,

Bx2 , 0)

Isto mostra que em geral o momento canonico nao e invariante de Gauge.

E importante, entretanto que ⇧ seja invariante de Gauge, pois ele

e responsavel pela trajetoria que depende apenas das condicoes iniciais

e da forca de Lorentz (que nao depende de �).

Como ⇧ e p estao relacionados por ⇧ = p� e

cA ! p e

e

cA devem mudar

de forma a nao alterar ⇧| {z } .

18 MAPLima

F789 Aula 15

TransformaçãodeGaugeE razoavel esperar que os valores esperados de quantidades classicas

invariantes mediante transformacoes de Gauge tambem sejam invariantes.

Assim, esperamos que hRi e h⇧i nao mudem mediante transformacao de

Gauge, mas hPi mude. Em outras palavras:

Se

8><

>:

|↵i um estado na presenca de A

f|↵i um estado na presenca de eA = A+r�

entao, esperamos que:

8><

>:

h↵|R|↵i = he↵|R|e↵i

h↵|P� ecA|↵i = he↵|P� e

ceA|e↵i

com h↵|↵i = he↵|e↵i = 1.

Vamos construir um operador G tal que |e↵i = G|↵i e que quando inserido

em he↵|R|e↵i = h↵|G†RG|↵i = h↵|R|↵i e ) G†RG = R.

Ou, de forma semelhante, para o momento cinematico:

he↵|P� e

ceA|e↵i = h↵|G†�P� e

cA� e

cr�

�G|↵i = h↵|

�P� e

cA�|↵i

e ) G†�P� e

cA� e

cr�

�G =

�P� e

cA�

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19 MAPLima

F789 Aula 15

TransformaçãodeGauge

(c.q.f.)

(c.q.f.)

Tiramos do chapeu que G = exp� ie~c�(R)

�e a solucao. Inspirada na

condicao h↵|↵i = he↵|e↵i = 1 ! G†G = 1

Podemos verificar se funciona para o caso

G†RG = exp�� ie

~c�(R)�R exp

� ie~c�(R)

�= R pois, [R,�(R)] = 0

De forma semelhante, podemos verificar:

G†�P� e

cA� e

cr�

�G = exp

�� ie

~c�(R)��P� e

cA� e

cr�

�exp

� ie~c�(R)

�

= exp�� ie

~c�(R)�P exp

� ie~c�(R)

��e

cA� e

cr�

| {z }[A,�] = 0 e [g(�),�] = 0

falta exp�� ie

~c�(R)�P exp

� ie~c�(R)

�que pode ser calculado com auxılio

da expressao

[P, G(R)] = �i~rG(R) ! PG(R)�G(R)P = �i~ exp� ie~c�(R)

� ie~cr�

ou seja PG(R)�G(R)P = Ge

cr� ! G†�PG(R)�G(R)P

�= G†G

e

cr�

G†PG = P+e

cr� ! G†�P� e

cA� e

cr�

�G = P� e

cA

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20 MAPLima

F789 Aula 15

TransformaçãodeGaugeOlhemos a equacao de Schrodinger para |↵, t0; ti com o potencial vetor A. Se⇥ 1

2m

�P� e

cA�2

+ e�⇤|↵, t0; ti = i~ @

@t|↵, t0; ti e satisfeita, a expectativa e que

^|↵, t0; ti satisfaca a equacao:⇥ 1

2m

�P� e

cA� e

cr�

�2+ e�

⇤ ^|↵, t0; ti= i~ @

@t^|↵, t0; ti.

Verifiquemos se a seguinte relacao e verdadeira ^|↵, t0; ti = exp� ie~c�(R)

�|↵, t0; ti?

Para tanto, insira esta expressao na equacao acima de Schrodinger para ^|↵, t0; ti,

e multiplique a equacao toda pela esquerda por exp�� ie

~c�(R)�.

Note que exp�� ie

~c�(R)�⇥�

P� e

cA� e

cr�

�2⇤exp

� ie~c�(R)

�=

exp�� ie

~c�(R)��P� e

cA� e

cr�

�exp

� ie~c�(R)

�⇥

⇥ exp�� ie

~c�(R)��P� e

cA� e

cr�

�exp

� ie~c�(R)

�e aplique

exp�� ie

~c�(R)��P� e

cA� e

cr�

�exp

� ie~c�(R)

�= P� e

cA duas vezes. A

equacao que sobra e a equacao de Schrodinger para |↵, t0; ti com A e isso

demonstra a relacao entre ^|↵, t0; ti e |↵, t0; ti.<latexit 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21 MAPLima

F789 Aula 15

TransformaçãodeGauge

Na representacao das coordenadas a relacao entre ^|↵, t0; ti e |↵, t0; ti fica

e ↵(r0, t) = exp

� ie~c�(r

0)� ↵(r

0, t), onde �(r0) e uma funcao real do vetor

posicao r0. Em termos de ⇢ e S, a funcao de onda pode ser escrita por

e ↵, (r0, t) = exp

� ie~c�(r

0)�p⇢ exp

� iS~�=

p⇢ exp

� i~ (S +

e

c�(r0)

�

Ou seja, basta

8><

>:

⇢! ⇢

S ! S +ec�

Como fica o fluxo de probabilidade j =⇢

m

�rS � e

cA�?

Mediante a transformacao de Gauge A ! A+r�, temos

j =⇢

m

�r(S +

e

c�)� e

c(A+r�)

�=

⇢

m

�rS � e

cA�e

) o fluxo de probabilidade e invariante mediante transformacao de Gauge.<latexit 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22 MAPLima

F789 Aula 15

Variáveldinâmicafundamental:AouB?Partículaprisioneiraemcabocoaxial

Dois Cilindros metalicos, um dentro do outro.z

⇢a⇢b

L

Partıcula confinada entre ⇢a e ⇢b

Condicoes de contorno para (z, ⇢,')

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

>>>>>>>>>>:

(0, ⇢,') = 0

(L, ⇢,') = 0

(z, ⇢a,') = 0

(z, ⇢b,') = 0

O espectro de energia e encontrado

resolvendo a equacao � ~22m

r2 = E

em coordenadas cilındricas e condicoes

de contorno abaixo.

23 MAPLima

F789 Aula 15

OEspectrodeEnergia

m ⌘ m-esima raiz

A equacao de Schrodinger � ~22m

r2 = E em coordenadas cilındricas fica:

@2

@⇢2+

1

⇢

@

@⇢+

1

⇢2@2

@'2+@2

@z2= �2mE

~2 ! com (z, ⇢,') = R(⇢)Q(')Z(z)

temos

8>>>>>><

>>>>>>:

d2Z(z)dz2 = �k20Z ! Z(z) = exp (±ik0z)

8><

>:

Z(0) = Z(L) = 0

Z(z) = sin k0z;

k0 = `⇡L , ` inteiro

d2Q(')d'2 = �⌫2Q ! Q(') = exp (±i⌫')

(Q(0) = Q(2⇡)

⌫ inteiro

Para ⇢ temos:@2R

@⇢2+

1

⇢

@R

@⇢� ⌫2

⇢2R+

�2mE

~2 � k20�R = 0

se tomarmos

(x = k⇢

k =q

2mE~2 � k20

! d2R

dx2+

1

x

dR

dx+ (1� ⌫2

x2)R = 0 com

solucoes conhecidas R = AJ⌫(k⇢) +BN⌫(k⇢). Aplique R(k⇢a) = R(k⇢b) = 0, e

obtenha

8><

>:

AJ⌫(k⇢a) +BN⌫(k⇢a) = 0

AJ⌫(k⇢b) +BN⌫(k⇢b) = 0

J⌫(k⇢a)N⌫(k⇢b)� J⌫(k⇢b)N⌫(k⇢a) = 0| {z }

Isso quantiza k e a energia E`m⌫ =~22m

�k2m⌫ + (

`⇡

L)2�

<latexit sha1_base64="VTkPyO8G6sVPEpNsT35l6LSEbjw=">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</latexit><latexit sha1_base64="VTkPyO8G6sVPEpNsT35l6LSEbjw=">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</latexit><latexit 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sha1_base64="VTkPyO8G6sVPEpNsT35l6LSEbjw=">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</latexit>

24 MAPLima

F789 Aula 15

CampomagnéticoconstantenointeriordocaboApartículanãovêoB.SóvêoA!

Suponha agora um campo magnetico B = B0z no interior do cilindro ⇢ < ⇢a.

Quanto vale o potencial vetor para ⇢ < ⇢a e ⇢ > ⇢a?

Caso ⇢ > ⇢aZsuperfıciecircular deraio ⇢>⇢a

B.da = B0⇡⇢2a =

Zsuperfıciecircular deraio ⇢>⇢a

(r⇥A).da =

Ilinha

fechadasobre o anelde raio ⇢>⇢a

A.d`

mas, d` = ⇢d' e ) A.d` = ⇢A�d' ! B0⇡⇢2a = ⇢A�

I

linhafechada

sobre o anel

d' = 2⇡⇢A�

) A� =B0⇢2a2⇢

! Mostre que B = r⇥A = 0.

Caso ⇢ < ⇢aZsuperfıciecircular deraio ⇢>⇢a

B.da = B0⇡⇢2 =

Zsuperfıciecircular deraio ⇢<⇢a

(r⇥A).da =

Ilinha

fechadasobre o anelde raio ⇢<⇢a

A.d`

de novo, d` = ⇢d' e ) A.d` = ⇢A�d' ! B0⇡⇢2 = ⇢A�

I

linhafechada

sobre o anel

d' = 2⇡⇢A�

) A� =B0⇢

2! Mostre que B = r⇥A = B0z.

25 MAPLima

F789 Aula 15

NovoEspectrodeEnergiaApartículanãovêoB.SóvêoA!

A solucao do problema, com campo magnetico constante no interior do cilindro

⇢ < ⇢a, sera obtida com

(A⇢ = Az = 0 e A� = B0⇢

2 p/ ⇢ < ⇢a ! B = B0z

A⇢ = Az = 0 e A� = B0⇢2a

2⇢ p/ ⇢ > ⇢a ! B = 0

Note que, embora a partıcula nao sinta o campo B na regiao ⇢ > ⇢a, ela sente

A, pois A 6= 0 nesta regiao.

Para encontrar os autovalores para este novo problema, precisamos trocar

r por r� ie

~cA. Lembrando que r em coordenadas cilındricas e:

r = ⇢@

@⇢+ z

@

@z� '

1

⇢

@

@'temos que trocar

@

@'! @

@'� ie

~cB0⇢2a2⇢

Tal troca resulta em uma mudanca no espectro de energia! (surpreendente, pois

o campo e zero onde a partıcula pode estar.) A equacao de Schrodinger em

coordenadas cilındricas e:

@2

@⇢2+

1

⇢

@

@⇢+

1

⇢2@2

@'2+@2

@z2= �2mE

~2 e usando a troca sugerida com

a =e

~cB0⇢2a2

, temos@2

@⇢2+

1

⇢

@

@⇢+

1

⇢2@2

@'2+@2

@z2� a2

⇢2 � 2ai

⇢2@

@'= �2mE

~2

26 MAPLima

F789 Aula 15

OpotencialvetorAéavariáveldinâmicafundamental

E`m⌫0 =~22m

�k2m⌫0 + (

`⇡

L)2�.

m ⌘ m-esima raiz

@2

@⇢2+

1

⇢

@

@⇢+

1

⇢2@2

@'2+@2

@z2� a2

⇢2 � 2ai

⇢2@

@'= �2mE

~2 e = R(⇢)Q(')Z(z)

nos leva a:

8>>>>>><

>>>>>>:

d2Z(z)dz2 = �k20Z ! Z(z) = exp (±ik0z)

8><

>:

Z(0) = Z(L) = 0

Z(z) = sin k0z;

k0 = `⇡L , ` inteiro

d2Q(')d'2 = �⌫2Q ! Q(') = exp (±i⌫')

(Q(0) = Q(2⇡)

⌫ inteiro positivo

Para ⇢ agora temos:

@2R

@⇢2+

1

⇢

@R

@⇢� ⌫2

⇢2R� a2

⇢2R� 2ai

⇢2(±i⌫)R+

�2mE

~2 � k20�R = 0

@2R

@⇢2+

1

⇢

@R

@⇢� 1

⇢2(⌫ ± a)2R+

�2mE

~2 � k20�R = 0 se

(x = k⇢

k =q

2mE~2 � k20

! d2R

dx2+

1

x

dR

dx+ (1� (⌫ ± a)2

x2)R = 0 com R = AJ⌫0(k⇢) +BN⌫0(k⇢).

Note que ⌫0 = |⌫ ± a| nao e inteiro. Aplique R(k⇢a) = R(k⇢b) = 0,

!

8><

>:

AJ⌫0(k⇢a) +BN⌫0(k⇢a) = 0

AJ⌫0(k⇢b) +BN⌫0(k⇢b) = 0

J⌫0(k⇢a)N⌫0(k⇢b)� J⌫0(k⇢b)N⌫0(k⇢a) = 0| {z }<latexit 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27 MAPLima

F789 Aula 15

OefeitodeAharanov-BohmExperimentodecorrelação

CilindroB 6= 0

fonte de partıculasregiao de interferencia

Pacote de ondas antes de passar pelo cilindro impenetrável

Pacote dividido após passar pelo cilindro: Parte do pacote passou por cima. Parte passou por baixo.

B = 0Outro caminho

de Feynman

Um caminho de Feynman

FI

Queremos estudar como a probabilidade de encontrar a partıcula

na regiao de interferencia I depende do fluxo magnetico. O estudo

sera feito com integrais de Feynman. A partıcula nao pode penetrar

no cilindro, assim, de novo, ela percebe o potencial vetor, mas nao

vai onde B 6= 0. B = 0 fora do cilindro.

28 MAPLima

F789 Aula 15

OefeitodeAharanov-Bohm