aula de hoje: potenciais e transformações de gaugemaplima/f789/2018/aula15.pdf · aula 15 aula de...
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![Page 1: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/1.jpg)
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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![Page 2: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/2.jpg)
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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![Page 3: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/3.jpg)
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
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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
~ .<latexit 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![Page 6: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/6.jpg)
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
![Page 8: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/8.jpg)
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�
![Page 9: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/9.jpg)
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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![Page 11: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/11.jpg)
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
![Page 12: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/12.jpg)
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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![Page 13: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/13.jpg)
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
.<latexit 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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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sha1_base64="4FVudDJFXhYdu0xTk9S4PtzTLj4=">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</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=">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</latexit><latexit sha1_base64="ROHTfGjc5B+yop5WIOtHUHYywEk=">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</latexit>
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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
![Page 17: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/17.jpg)
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 } .
![Page 18: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/18.jpg)
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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![Page 19: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/19.jpg)
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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![Page 22: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/22.jpg)
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.
![Page 23: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/23.jpg)
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�
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sha1_base64="VTkPyO8G6sVPEpNsT35l6LSEbjw=">AAAOVHicvVfNbxtFFH9paSkmNik99rIiorLVJFpHoFZFlvqhSIAq1ATSVsRpNLs7tlfer+7OlqSW/0g4IHHmwIkjcOHAmzcz6/V6bUethC2vZ968+b2P+c2bWScJ/EzY9m8bV65+cO36hzc+any82Wx9snXz0+dZnKcuP3bjIE5fOizjgR/xY+GLgL9MUs5CJ+AvnPETOf7iDU8zP45+EBcJPw3ZMPIHvssEiuKb17+EO9AHARzO8TmBR2Bh+zXkwOAP+B1i7HsoseB7cGEEKfyNfR8iGKI0RfkUGrCLGAPsMdSZYHsEDrZTeAX7OD7BZ0j/fZQPcE4f5zNsB/ickpaUJZAhsgU9/B0U/cacf9K7EJ8uehajBembxPLwl5HcR9Q/UebhqI99JR/o9gNsS5/7+L1D2GW/E/JaoC5DlHm/pjUacjTFaGMdqQV3K5hdPU9pTVfau6yt9XaMN+8f3RuSJIjqa9Sq5XfBfauxegvckUw50LPmWTTjR59WdojjgsZTjPgnWtcyUyRDQr3WamYb7e4UOdqpRNdBb45Qx4x34JB68zo/EkqnxJ+ZRUHcjNHWlEYc7A9pr0w0D3kxNovYo/gM7pQks/zswhjOwNY6y2NvFAg91JEeJYjUpthlFnyNo2xYa/yTWDZhydZTatkU87ydjGZbBfZXpGP6vQpPOK69YblixlN87pBkNlbe6xJd9n2KU2WO096e91h5tpjVuhVUGa5jdo+4GGHtU3MbK7Jdh7ws8wqzupsusw6HxTrI1r7OXKeI1+C+T87qxhq1/H5G3jPaU7NqVLU/2wUP9Omwvloc/S/VdZ0VaWF3wV/Dh7r6eqQ9cogl7UtWsvKeNnM7iGX2WDXvGZ2/guqZRDE1ZhVzzhFtXPir1nqs9+xrin7yDt4axi7jTP1uWdyXZi089NOcKovrel5ZUa80S85oo1715lFeLYNtcltlqjohpktyHiM3crwD/UXxKe0IY+N0xzD3DcWCHt6bvsUsKfvtUuY7xFoLHsN3S8b3Fvx6hAwN0Ia8hfHCRnnWGVo351VZ6uhKvVNCtPStqT7OGOcIWs8R3cVWcWpZjMqb1XEaj219QihGLkd0LoFool3GSGklJ7m8qTqaJRy1VsWxytruSn/XRW5WYH69v0FfM6qj6tYdUWV6W9TZ8QI/JCcY/UcU2ZAqmdI+QHvlszbUt+2czrfqjbD+nj6rZ2NCCwsEtXfvVirdspO9M1/hzra27T2bPtZio6sb26A/z862ful7sZuHPBJuwLLspGsn4nTCUuG7AZ82+nnGE+aO2ZCfYDNiIc9OJ/RSNLU+R4lnDeIUf5GwSFqeMWFhll2EDmqGTIyy6pgU1o2d5GJw/3TiR0kueOQqQ4M8sERsyTcsy/NT7orgAhvMTX301XJHLGWuwPewBiahWw15sfF8f69r73UPv9h+uK/TcQNuw2dU8u7BQ/gaD+NjcDd/3vynCc2N5q/Nf1tXW9eU6pUNPecWzH1arf8ACwjLkw==</latexit><latexit sha1_base64="VTkPyO8G6sVPEpNsT35l6LSEbjw=">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</latexit>
![Page 24: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/24.jpg)
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.
![Page 25: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/25.jpg)
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
![Page 26: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/26.jpg)
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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![Page 27: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/27.jpg)
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.
![Page 28: Aula de hoje: Potenciais e transformações de Gaugemaplima/f789/2018/aula15.pdf · Aula 15 Aula de hoje: Potenciais e transformações de Gauge Em Mecanica Classica o ponto zero](https://reader034.vdocuments.net/reader034/viewer/2022042805/5f66a8ddf96f741ada3b4164/html5/thumbnails/28.jpg)
28 MAPLima
F789 Aula 15
OefeitodeAharanov-Bohm