Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT)...
Transcript of Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT)...
![Page 1: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/1.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 2: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/2.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 3: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/3.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 4: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/4.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 5: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/5.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 6: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/6.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 7: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/7.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 8: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/8.jpg)
Today’s Outline - March 28, 2016
• Gauge transformation review
• Aharonov-Bohm Effect
• Aharonov-Bohm experiment
• Changing double well
Homework Assignment #09:Chapter 10:1,2,4,5,6,7due Wednesday, March 30, 2016
Homework Assignment #10:Chapter 10:8,9,10; 11:2,4,5due Wednesday, April 06, 2016
Midterm Exam #2:Wednesday, April 11 or 13 – will be announced next class
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 1 / 11
![Page 9: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/9.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 10: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/10.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t,
~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 11: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/11.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 12: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/12.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 13: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/13.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t
~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 14: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/14.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 15: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/15.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 16: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/16.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 17: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/17.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 18: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/18.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 19: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/19.jpg)
Fields and gauge transformations
The potentials in classical electro-dynamics are not measurable butare used to describe the fields whichare
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
potentials can be arbitrarilychanged via a gauge transforma-tion, without altering the resultingfields
ϕ→ ϕ′ = ϕ− ∂Λ
∂t~A→ ~A′ = ~A +∇Λ
In quantum mechanics, potentialsare an integral part of the Hamilto-nian while fields are not
H =1
2m
(~i∇− q~A
)2
+ qϕ
even so, the Hamiltonian is invari-ant under gauge transformations
it was presumed that that therecould be no electromagnetic effectsunless there were non-zero fields
but the vector potential can affectthe quantum behavior of a chargedparticle even if the field is zero
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11
![Page 20: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/20.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 21: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/21.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 22: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/22.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 23: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/23.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t
= −∇ϕ+���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 24: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/24.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+∇∂Λ
∂t− ∂~A
∂t− ∂
∂t∇Λ
= −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 25: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/25.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ
= −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 26: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/26.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 27: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/27.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′
= ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 28: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/28.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +∇×∇Λ
= ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 29: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/29.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ
= ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 30: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/30.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 31: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/31.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 32: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/32.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 33: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/33.jpg)
Invariance proof (aka Problem 4.61)
~E = −∇ϕ− ∂~A
∂t, ~B = ∇× ~A
ϕ→ ϕ′ = ϕ− ∂Λ
∂t, ~A→ ~A′ = ~A +∇Λ
We can easily show the gauge invariance of the electric and magnetic fields
~E = −∇ϕ′ − ∂ ~A′
∂t= −∇ϕ+
���∇∂Λ
∂t− ∂~A
∂t−����∂
∂t∇Λ = −∇ϕ− ∂~A
∂t
~B = ∇× ~A′ = ∇× ~A +�����∇×∇Λ = ∇× ~A
It is also possible to show that the solution to the Schrodinger equation isalso invariant under a gauge transformation with the addition of only aphase factor
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′ Ψ′ ≡ e iqΛ/~Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 3 / 11
![Page 34: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/34.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′
=
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 35: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/35.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 36: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/36.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
= q(∇Λ)e iqΛΨ +~ie iqΛ∇Ψ
− q~Ae iqΛΨ−������q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 37: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/37.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
= q(∇Λ)e iqΛΨ +~ie iqΛ∇Ψ− q~Ae iqΛΨ
−������q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 38: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/38.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
= q(∇Λ)e iqΛΨ +~ie iqΛ∇Ψ− q~Ae iqΛΨ− q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 39: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/39.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 40: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/40.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ
[~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)
= −((((((((
((iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ
− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ
− q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)
− q~ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
![Page 46: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/46.jpg)
Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ
−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ− q~ie iqΛ(∇Λ · ∇Ψ)
+((((((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
− q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ− q~ie iqΛ(∇Λ · ∇Ψ) + q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~q
i(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ− q~ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)[~i∇− q~A′
]Ψ′ =
[~i∇− q~A− q∇Λ
]e iqΛΨ
=������q(∇Λ)e iqΛΨ +
~ie iqΛ∇Ψ− q~Ae iqΛΨ−���
���q(∇~A)e iqΛΨ
=~ie iqΛ∇Ψ− q~Ae iqΛΨ[
~i∇− q~A′
]2
Ψ′ =
[~i∇− q~A− q∇Λ
](~ie iqΛ∇Ψ− q~Ae iqΛΨ
)= −(((((
(((((
iq~(∇Λ · ∇Ψ)e iqΛΨ− ~2e iqΛ∇2Ψ− ~qi
(∇ · ~A)e iqΛΨ
−((((((((
q2(~A · ∇Λ)e iqΛ − q~ie iqΛ(~A · ∇Ψ)− q~
ie iqΛ(~A · ∇Ψ)
+ q2A2e iqΛΨ−���
������q~
ie iqΛ(∇Λ · ∇Ψ) +((((
((((q2(~A · ∇Λ)e iqΛΨ
= e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 4 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]
but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation
[~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ
=~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ
[~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ
=
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)
= −~2∇2Ψ− ~iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ
− ~iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ
− 2~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ)
+ q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have
[~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
[~i∇− q~A′
]2
Ψ′ = e iqΛ[−~2∇2Ψ + iq~(∇ · ~A)Ψ + 2iq~(~A · ∇Ψ) + q2A2Ψ
]but the term in square brackets is identical to the same portion of theuntransformed Schrodinger equation[
~i∇− q~A
]Ψ =
~i∇Ψ− q~AΨ[
~i∇− q~A
]2
Ψ =
[~i∇− q~A
](~i∇Ψ− q~AΨ
)= −~2∇2Ψ− ~
iq(∇ · ~A)Ψ− 2
~iq(~A · ∇Ψ) + q2A2Ψ
thus we have [~i∇− q~A′
]2
Ψ′ = e iqΛ
[~i∇− q~A
]2
Ψ
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 5 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t−���q∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t−���q∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t− q
∂Λ
∂tΨ
)e iqΛ
=���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t− q
∂Λ
∂tΨ
)e iqΛ = e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ− q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t− q
∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ− q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t−���q∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t−���q∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Invariance proof (cont.)
Recasting the entire Schrodinger equation
i~∂Ψ′
∂t=
[1
2m
(~i∇− q~A′
)2
+ qϕ′
]Ψ′
(i~∂Ψ
∂t−���q∂Λ
∂tΨ
)���e iqΛ =���e iqΛ
[1
2m
(~i∇− q~A
)2
+ qϕ−���q∂Λ
∂t
]Ψ
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ qϕ
]Ψ
and invariance under gauge transformation is shown
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 6 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid
a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid
a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid
a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid
a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
Consider a charged particle constrained to move on a ring of radius benclosing a long solenoid carrying current I
outside the solenoid the magnetic field iszero and the scalar potential is zero (nocharge)
however, the vector potential is not zero,and we can write
∇ · ~A = 0 −→ ~A =Φ
2πrφ (r > a)
where Φ = πa2B is the magnetic fluxthrough the solenoid a
b
q
I
B
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 7 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]
the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
H =1
2m
[−~2∇2 + q2A2 + 2i~q~A · ∇
]the Schrodinger equation thus becomes
but the wave function onlydepends on the azimuthalangle, φ, so
∇ → 1
b
d
dφφ
1
2m
[−~2
b2
d2
dφ2+
(qΦ
2πb
)2
+ i~qΦ
πb2
d
dφ
]ψ(φ) = Eψ(φ)
− ~2
b2
d2ψ
dφ2+ i
~qΦ
πb2
dψ
dφ+
[(qΦ
2πb
)2
− 2mE
]ψ = 0
d2ψ
dφ2− i
qΦ
π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ = 0
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 8 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~
and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~
and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
![Page 92: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/92.jpg)
Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε
= β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Effect of potential on Berry’s phase
0 =d2ψ
dφ2− 2i
qΦ
2π~dψ
dφ+
[2mb2E
~2−(
qΦ
2π~
)2]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+
[2mb2E
~2− β2
]ψ
0 =d2ψ
dφ2− 2iβ
dψ
dφ+ εψ
defining
β ≡ qΦ
2π~and
ε ≡ 2mb2E
~2− β2
this differential equation hasa solution
which, when substitutedback in gives
the boundary conditionψ(0) ≡ ψ(2π) requires thatλ be an integer
ψ = Ae iλφ
0 = −λ2 + 2βλ+ ε
λ = β ±√β2 + ε = β ± b
~√
2mE
n = β ± b
~√
2mE
En =~2
2mb2
(n − qΦ
2π~
)2
, n = 0,±1, . . .
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 9 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)
(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)
(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
= q~Ae igΨ′ +~ie ig (∇Ψ′)− q~Ae igΨ′
(~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′
(~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Consider the more general case where a particle moves through a regionwhere ~B = ∇× ~A = 0 but ~A 6= 0
for a static ~A the Schrodingerequation becomes
this can be simplified by sub-stituting
Ψ = e igΨ′
but ∇g = (q/~)~A so
i~∂Ψ
∂t=
[1
2m
(~i∇− q~A
)2
+ V
]Ψ
g(~r) ≡ q
~
∫ ~r
O~A(~r ′) · d~r ′
∇Ψ = e ig (i∇g)Ψ′ + e ig (∇Ψ′)(~i∇− q~A
)Ψ =
~ie ig (i∇g)Ψ′ +
~ie ig (∇Ψ′)− q~Ae igΨ′
=����
q~Ae igΨ′ +~ie ig (∇Ψ′)−���
�q~Ae igΨ′(
~i∇− q~A
)2
Ψ = −~2e ig∇2Ψ′
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 10 / 11
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Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
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Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
![Page 113: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/113.jpg)
Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
![Page 114: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/114.jpg)
Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
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Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
![Page 116: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/116.jpg)
Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11
![Page 117: Today’s Outline - March 28, 2016csrri.iit.edu/~segre/phys406/16S/lecture_18.pdf · C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 2 / 11. Fields and gauge transformations](https://reader033.fdocuments.in/reader033/viewer/2022050123/5f531f2ce218f20f5970f7ba/html5/thumbnails/117.jpg)
Aharonov-Bohm effect
Substituting into the Schro-dinger equation
Ψ′ satisfies the Schrodingerequation without ~A
i~e ig∂Ψ′
∂t= − 1
2m~2e ig∇2Ψ′ + Ve igΨ′
i~∂Ψ′
∂t= − ~2
2m∇2Ψ′ + VΨ′
thus the solution of a system where thereis a vector potential is trivial, just add ona phase factor e ig
Aharonov & Bohm proposed an experi-ment where an electron beam is split intwo and passed on either side of alongsolenoid before being recombined
the two beams should arrive with differentphases g± = ±(qΦ/2~)
I
B
A
Beamsplit
Beamrecombined
solenoid
C. Segre (IIT) PHYS 406 - Spring 2016 March 28, 2016 11 / 11