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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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