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Today’s Outline - October 24, 2019 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

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Page 1: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 2: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 3: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 4: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 5: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 6: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 7: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 8: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Today’s Outline - October 24, 2019

• Aharonov-Bohm effect theory

• Aharonov-Bohm experiments

• Identical particles

• Fermions and Bosons

• Exchange forces

Homework Assignment #09:Chapter 4: 23,26,30,34,52,64due Tuesday, October 29, 2019

Midterm Exam 2: Tuesday, November 05, 2019covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 1 / 22

Page 9: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 10: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 11: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 12: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 13: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 14: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 15: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 16: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 17: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 18: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 19: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Review of 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 405 - Fall 2019 October 24, 2019 2 / 22

Page 20: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 21: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 22: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 23: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 24: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 25: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6

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, using the Coulombgauge

∇ · ~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 405 - Fall 2019 October 24, 2019 3 / 22

Page 26: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]

the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[

− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

Page 27: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]

the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[

− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

Page 28: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]

the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[

− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

Page 29: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[

− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[

− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2

+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

= Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

]ψ = Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

]ψ = Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

~A =Φ

2πrφ

H =1

2m

[−~2∇2 + q2A2 + 2i~q~A · ∇

]the Schrodinger equation thus becomes

but the wave function onlydepends on the polar angle,φ, so

∇ → 1

b

d

dφφ

1

2m

[− ~2

b2

d2

dφ2+

(qΦ

2πb

)2

+ i~qΦ

πb2

d

]ψ = Eψ

− ~2

b2

d2ψ

dφ2+ i

~qΦ

πb2

dφ+

[(qΦ

2πb

)2

− 2mE

]ψ = 0

d2ψ

dφ2− i

π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ = 0

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 4 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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

)Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

0 =d2ψ

dφ2− 2i

2π~dψ

dφ+

[2mb2E

~2−(

2π~

)2]ψ

0 =d2ψ

dφ2− 2iβ

dφ+

[2mb2E

~2− β2

0 =d2ψ

dφ2− 2iβ

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)���Ae iλφ

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 405 - Fall 2019 October 24, 2019 5 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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Example 4.6 (cont.)

The solenoid lifts the two-fold degeneracy of the bead on a ring system

a

b

q

I

B

E+n =~2

2mb2(|n| − β)2

a

b

q

I

B

E−n =~2

2mb2(|n|+ β)2

The energies depend on the field inside the solenoid even though the fieldat the particle is zero.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 6 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

Page 61: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′

→ ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′

→ ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′

→ ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′

→ ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′

→ ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 potential, ~A,the Schrodinger equation be-comes

this can be simplified by sub-stituting

Ψ = e igΨ′

i~∂Ψ

∂t=

[1

2m

(~i∇− q~A

)2

+ V

g(~r) ≡ q

~

∫ ~r

O~A(~r ′) · d~r ′ → ∇g = (q/~)~A

∇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 405 - Fall 2019 October 24, 2019 7 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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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 a longsolenoid before being recombined

the two beams should arrive with differentphases g± = ±(qΦ/2~)

I

B

A

Beamsplit

Beamrecombined

solenoid

“Significance of electromagnetic potentials in the quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491(1959).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 8 / 22

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Aharonov-Bohm effect

QJf jonrnal of experimental and theoretical physics established by E. L ¹chols i» l899

SEcoND SERIEs, VoL. 115, NO. 3 AUGUST 1, 1959

Significance of Electromagnetic Potentials in the Quantum Theory

Y. AHARONOV AND D. BOHIN

H. II. Wills Physics Laboratory,

University

of Bristol, Bristol, England

(Received May 28, 1959; revised manuscript received June 16, 1959)

In this paper, we discuss some interesting properties of the electromagnetic potentials in the quantumdomain. We shall show that, contrary to the conclusions of classical mechanics, there exist effects of poten-tials on charged particles, even in the region where all the fields (and therefore the forces on the particles)vanish. Ke shall then discuss possible experiments to test these conclusions; and, finally, we shall suggestfurther possible developments in the interpretation of the potentials.

1. INTRODUCTION

N classical electrodynamics, the vector and scalar&- potentials were first introduced as a convenientmathematical aid for calculating the fields. It is truethat in order to obtain a classical canonical formalism,the potentials are needed. Nevertheless, the funda-mental equations of motion can always be expresseddirectly in terms of the fields alone.

In the quantum mechanics, however, the canonicalformalism is necessary, and as a result, the potentialscannot be eliminated from the basic equations. Never-theless, these equations, as well as the physical quan-tities, are all gauge invariant; so that it may seem thateven in quantum mechanics, the potentials themselveshave no independent significance.

In this paper, we shall show that the above conclu-sions are not correct and that a further interpretationof the potentials is needed in the quantum mechanics.

2. POSSIBLE EXPERIMENTS DEMONSTRATINGTHE ROLE OF POTENTIALS IN THE

QUANTUM THEORY

In this section, we shall discuss several possible ex-periments which demonstrate the significance of poten-tials in the quantum theory. We shall begin with asimple example.

Suppose we have a charged particle inside a "Faradaycage" connected to an external generator which causesthe potential on the cage to alternate in time. This willadd to the Hamiltonian of the particle a term V(x,t)which is, for the region inside the cage, a function oftime only. In the nonrelativistic limit (and we shall

assume this almost everywhere in the following dis-cussions) we have, for the region inside the cage,H=Hp+V(t) where Hp is the Hamiltonian when thegenerator is not functioning, and V(t) =ep(t). Ifleap(s, t) is a solution of the Hamiltonian Hp, then thesolution for H will be

&=/pe '«&, S= —V(t)dt,

which follows from

8$ ( Bfp 85)i jt =

Ii jt —+pp—1

e 'sty= 1tH p+ V (t)]Q= HQ.a~ E a~ at &

The new solution differs from the old one just by aphase factor and this corresponds, of course, to nochange in any physical result.

Now consider a more complex experiment in which asingle coherent electron beam is split into two parts andeach part is then allowed to enter a long cylindricalmetal tube, as shown in Fig. 1.

After the beams pass through the tubes, they arecombined to interfere coherently at F. By means oftime-determining electrical "shutters" the beam ischopped into wave packets that are long comparedwith the wavelength ), but short compared with thelength of the tubes. The potential in each tube is deter-mined by a time delay mechanism in such a way thatthe potential is zero in region I (until each packet iswell inside its tube). The potential then grows as afunction of time, but differently in each tube. Finally,it falls back to zero, before the electron comes near the

85

Y. AHARONOV AND D. 13OHM

14 I Mif7: ~~ l~~ re~' I

Orbit of

Electron Bea

Electron

beam

fnter ferenceregion

%At M/l

re'QK')

Fn. 1. Schematic experiment to demonstrate interference withtime-dependent scalar potential. A, 8, C, D, E: suitable devicesto separate and divert beams. 5'I, W2. wave packets. Mi, M'g.'

cylindrical metal tubes. F: interference region.

other edge of the tube. Thus the potential is nonzeroonly while the electrons are well inside the tube (regionII). When the electron is in region III, there is again nopotential. The purpose of this arrangement is to ensurethat the electron is in a time-varying potential withoutever being in a fmld (because the 6eid does not penetratefar from the edges of the tubes, and is nonzero only attimes when the electron is far from these edges).

Now let P(x, t) =iti'(x, t)+Ps'(x, t) be the wave func-tion when the potential is absent (Pi' and its' repre-senting the parts that pass through tubes 1 and 2,respectively). But since U is a function only of twherever it is appreciable, the problem for each tubeis essentially the same as that of the Faraday cage. Thesolution is then

f—$ oe isi///+p oe —/ss/s—

Fzt".. 2. Schematic experiment to demonstrate interferencewith time-independent vector potential.

suggests that the associated phase shift of the electronwave function ought to be

eAS//'t= —— A dx,

CS

where gA dx= J'H ds=p (the total magnetic fluxinside the circuit).

This corresponds to another experimental situation.By means of a current Qowing through a very closelywound cylindrical solenoid of radius E., center at theorigin and axis in the s direction, we create a magneticfield, H, which is essentially confined within the sole-noid. However, the vector potential, A, evidently,cannot be zero everywhere outside the solenoid, becausethe total Qux through every circuit containing theorigin is equal to a constant

where

Sy=e ~ pydt, 52=8 yqdt.yp —— H ds=)IA dx.

It is evident that the interference of the two parts atE will depend on the phase difference (Si—Ss)/It. Thus,there is a physical eGect of the potentials even thoughno force is ever actually exerted on the electron. TheeGect is evidently essentially quantum-mechanical innature because it comes in the phenomenon of inter-ference. We are therefore not surprised that it does notappear in classical mechanics.

From relativistic considerations, it is easily seen thatthe covariance of the above conclusion demands thatthere should be similar results involving the vectorpotential, A.

The phase difference, (Si—Ss)/5, can also be ex-pressed as the integral (e/It)gpdt around a closedcircuit in space-time, where q is evaluated at the placeof the center of the wave packet. The relativistic gener-alization of the above integral is

e ] Arpdt dx- —

/s E c i '

where the path of integration now goes over any closedcircuit in space-time.

As another special case, let us now consider a pathin space only (t=constant). The above argument

To demonstrate the eGects of the total Qux, we begin,as before, with a coherent beam of electrons. (But nowthere is no need to make wave packets. ) The beam issplit into two parts, each going on opposite sides of thesolenoid, but avoiding it. (The solenoid can be shieldedfrom the electron beam by a thin plate which casts ashadow. ) As in the former example, the beams arebrought together at J (Fig. 2).

The Hamiltonian for this case is

$P—(e/c) Aj'

In singly connected regions, where H= V&(A=O, wecan always obtain a solution for the above Hamiltonianby taking it =itse 'e/", where ps is the solution whenA=O and where V'S/5= (e/c)A. But, in the experimentdiscussed above, in which we have a multiply connectedregion (the region outside the solenoid), fee 'e/" is anon-single-valued function' and therefore, in general,not a permissible solution of Schrodinger's equation.Nevertheless, in our problem it is still possible to usesuch solutions because the wave function splits intotwo parts P =Pi+fs, where Pi represents the beam on

' Vniess go ——eigc/e, where e is an integer.

“Significance of electromagnetic potentials in quantum theory,” Y. Aharonov and D. Bohm, Phys. Rev. 115, 485-491 (1959).

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Aharonov-Bohm experiment

VQLUME 5) +UMBER 1 PHYSICAL REVIEW LETTERS JUx, v 1, 1960

the mass spectrographic analyses and Dr. M. E.Norberg of the Corning Glass Company for pro-viding us with porous Vycor glass.

This work was supported by the Atomic EnergyCommission and, in the case of one of the authors(F. R.$, also by the Alfred P. Sloan Foundation.

~L. Meyer and F. Reif, Phys. Rev. 110, 279 (1958).2F. Reif and L. Meyer, Phys. Rev. (to be published).3J. L. Yarnell, G. P. Arnold, P. J. Bendt, and

E. C. Kerr, Phys. Rev. 113, 1379 (1959).4K. R. Atkins, H. Seki, and E. U. Condon, Phys.

Rev. 102, 582 (1956). The onset temperature forsuper fluidity was about l.35'K in the Vycor used.

SK. R. Atkins, Phys. Rev. 116, 1339 (1959).'More exactly, Df& "(M+M /M&) "4, where M3 is

the effective mass of a Hes atom. A reasonable esti-mate is M3=2MHe.

YThis is deduced from the temperature range fromabout 0.6 to 0.5 K where the mobility differences aresufficiently large for the subtraction analysis to befeasible.

I. M. Khalatnikov and U. N. Kharkov, J. Exptl.Theoret. Phys. U. S.S.R. 32, 1108 (1957) ttranslation:Soviet Phys. -JETP 5, 905 (1957}].

SHIFT OF AN ELECTRON INTERFERENCE PATTERN BY ENCLOSED MAGNETIC FLUX

R. G. ChambersH. H. Wills Physics Laboratory, University of Bristol, Bristol, England

(Received May 27, 1960)

Aharonov and Bohm' have recently drawnattention to a remarkable prediction from quan-tum theory. According to this, the fringe patternin an electron interference experiment shouldbe shifted by altering the amount of magneticflux passing between the two beams (e.g. , in

region a of Fig. 1), even though the beamsthemselves pass only through field-free regions.Theory predicts a shift of n fringes for an en-closed flux 4 of nkc/e; it is convenient to referto a natural "flux unit, "hc/e =4.135X10 ' gausscm'. It has since been pointed out that the sameconclusion had previously been reached byEhrenberg and Siday, ' using semiclassical argu-ments, but these authors perhaps did not suf-ficiently stress the remarkable nature of theresult, and their work appears to have attractedlittle attention.

Clearly the first problem to consider, experi-mentally, is the effect on the fringe system ofstray fields not localized to region a but extend-ing, e.g. , over region a in Fig. 1. In addition

FIG. 1. Schematic diagram of interferometer, withsource s, observing plane o, biprism 8, f, and con-fined and extended field regions a and a'.

to the "quantum" fringe shift due to the enclosedflux, there will then be a shift due simply tocurvature of the electron trajectories by thefield. A straightforward calculation shows thatin a "biprism" experiment, such a field shouldproduce a fringe displacement which exactlykeeps pace with the deflection of the beams bythe field, so that the fringe system appears toremain undisplaced relative to the envelope ofthe pattern. A field of type a, on the other hand,should leave the envelope undisplaced, and pro-duce a fringe shift within it. In the Marton'interferometer, conditions are different, and afield of type a' should leave the fringes undis-placed in space. This explains how Marton et al. 'were able to observe fringes in the presence ofstray 60-cps fields probably large enough tohave destroyed them otherwise; this experimentthus constitutes an inadvertent check of theexistence of the "quantum" shift. '

To obtain a more direct check, a PhilipsEM100 electron microscope' has been modifiedso that it can be switched at will from normaloperation to operation as an interferometer.Fringes are produced by an electrostatic "biprism"consisting of an aluminized quartz fiber f (Fig. 1)flanked by two earthed metal plates e; alteringthe positive potential applied to f alters theeffective angle of the biprism. The distancess-f and f-o (Fig. 1) are about 6.7 cm and 13.4cm, respectively. With this microscope it wasnot possible to reduce the virtual source diameterbelow about 0.2 p. , so that it was necessary touse a fiber f only about 1.5 p, in diameter and a

VOLUME 5, NUMBER 1 P HY SI GAL REVI KW LETTERS JULY 1, 1960

'k .","":,, „*

~ m i

::'. '.4&A

(a) (b)

FIG. 2. (a) Fringe pattern due to biprisrn alone.(b) Pattern displaced by 2. 5 fringe widths by field oftype a'.

very small biprism angle, to produce a widepattern of fringes which would not be blurredout by the finite source size. The fringe yatternobtained is shown in Fig. 2(a); the fringe width

in the observing plane o is about 0.6 p, .We first examined the effect of a field of type

a', produced by a Helmholtz pair of single turns3 mm in diameter just behind the biprism. Fieldsup to 0.3 gauss were applied, sufficient to dis-place the pattern by uy to 30 fringe widths, andas predicted the appearance of the pattern wascompletely unchanged. Figure 2(b), for instance,shows the pattern in a field producing a dis-placement of about 2.5 fringe widths. In theabsence of the "quantum" shift due to the en-closed flux, this pattern would have had thelight and dark fringes interchanged. We alsoverified that with this interferometer, unlikeMarton's, a small ac field suffices to blur outthe fringe system completely. These resultsconfirm the presence of the quantum shift infields of type a'.

Of more interest is the effect predicted for afield of type a, where intuition might expect noeffect. Such a field was yroduced by an ironwhisker, ' about 1 p. in diameter and 0.5 mm long,placed in the shadow of the fiber f. Whiskersas thin as this are expected theoretically andfound experimentally to be single magneticdomains; moreover they are found to taper' witha slope of the order of 10 ', which is extremelyconvenient for the yresent purpose. An ironwhisker 1 p, in diameter will contain about 400flux units; if it tapers uniformly with a slope of10 ', the flux content will change along thelength at a rate d4/ds of about 1 flux unit per

(a) (b)

FIG. 3. (a) Tilted fringes produced by taperingwhisker in shadow of biprism fiber. (b) Fresnelfringes in the shadow of the whisker itself, just out-side shadow of fiber. (o) Same as (b), but from a dif-ferent part of the whisker, and with fiber out of thefield of view.

micron. Thus if such a whisker is placed inposition a (Fig. 1), we expect to see a patternin which the envelope is undisplaced, but thefringe system within the envelope is inclined atan angle of the order of one fringe width permicron. Since the fringe width in the observingplane is 0.6 p, , and there is a "pin-hole" magnifi-cation of x3 between the biprism-fiber assemblyand the observing plane, we thus expect thefringes to show a tilt of order 1 in 5 relative tothe envelope of the pattern. Precisely this isobserved experimentally, as shown in Fig. 3(a).It will be seen that the whisker taper is not uni-form, but in this example becomes very smallin the upper part of the picture.

In fact the biprism is an unnecessary refine-ment for this exyeriment: Fresnel diffractioninto the shadow of the whisker is strong enoughto produce a clear fringe pattern from the whiskeralone. Thus Fig. 3(b) shows the same section ofwhisker as Fig. 3(a), moved just out of the shadowof the biprism fiber. The biprism fringes arenow unperturbed; the Fresnel fringes in theshadow of the whisker show exactly the samepattern of fringe shifts along their length as in

Fig. 3(a). Figure 3(c) shows a further exampleof these fringes, from a different part of thesame whisker, with the biprism moved out ofthe way. The whisker here is tapering morerapidly.

These fringe shifts cannot be attributed to directinteraction between the electrons and the surfaceof the whisker, since in Fig. 3(a) the whisker

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 10 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope,

biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e),

Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

s: electron sourceo: observing planee, f : biprisma: confined field regiona′: extended field region

what about the effect of stray fieldsin region a′ which can curve theelectron beams electrostatically?

in the biprism, the quantum ef-fect exactly cancels the stray fieldleaving the interference pattern un-changed

a field solely in region a will leadto a quantum effect with the inter-ference fringes moving through theenvelope

modified electron microscope, biprism consists of an aluminized quartzfiber (f) and two grounded metal plates (e), Aharonov-Bohm effect willproduce a shift of n fringes for Φ = nhc/e

“Shift of an electron interference pattern by enclosed magnetic flux,”R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 11 / 22

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Aharonov-Bohm experiment

(a) is with no additional field ap-plied in extended region

(b) has 25mG, which alone wouldinvert the fringe, applied with novisible fringe shift

up to 300 mG applied in region a′

showed no shift

This calibration experiment shows that the Aharonov-Bhom effect ispresent and balances the electrostatic fringe shifts in a region where thereis both a flux AND a field

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 12 / 22

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Aharonov-Bohm experiment

(a) is with no additional field ap-plied in extended region

(b) has 25mG, which alone wouldinvert the fringe, applied with novisible fringe shift

up to 300 mG applied in region a′

showed no shift

This calibration experiment shows that the Aharonov-Bhom effect ispresent and balances the electrostatic fringe shifts in a region where thereis both a flux AND a field

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 12 / 22

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Aharonov-Bohm experiment

(a) is with no additional field ap-plied in extended region

(b) has 25mG, which alone wouldinvert the fringe, applied with novisible fringe shift

up to 300 mG applied in region a′

showed no shift

This calibration experiment shows that the Aharonov-Bhom effect ispresent and balances the electrostatic fringe shifts in a region where thereis both a flux AND a field

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 12 / 22

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Aharonov-Bohm experiment

(a) is with no additional field ap-plied in extended region

(b) has 25mG, which alone wouldinvert the fringe, applied with novisible fringe shift

up to 300 mG applied in region a′

showed no shift

This calibration experiment shows that the Aharonov-Bhom effect ispresent and balances the electrostatic fringe shifts in a region where thereis both a flux AND a field

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 12 / 22

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Aharonov-Bohm experiment

(a) is with no additional field ap-plied in extended region

(b) has 25mG, which alone wouldinvert the fringe, applied with novisible fringe shift

up to 300 mG applied in region a′

showed no shift

This calibration experiment shows that the Aharonov-Bhom effect ispresent and balances the electrostatic fringe shifts in a region where thereis both a flux AND a field

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 12 / 22

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Aharonov-Bohm experiment

(a) a tapered iron whisker producesa confined field and flux with a gra-dient along the z-axis manifested intilted fringes

(b) direct imaging, with the whiskeroutside the shadow of the biprismfiber, due to Fresnel diffractionin the shadow of the fiber showsbiprism fringes with tilted fringesjust to the side

(c) higher taper, again using Fresnel diffraction in fiber shadow but withbiprism removed shows more highly tilted fringes

Chambers says: “I am indebted for Mr. Aharonov and Dr. Bohm fortelling me of their work before publication..”

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 13 / 22

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Aharonov-Bohm experiment

(a) a tapered iron whisker producesa confined field and flux with a gra-dient along the z-axis manifested intilted fringes

(b) direct imaging, with the whiskeroutside the shadow of the biprismfiber, due to Fresnel diffractionin the shadow of the fiber showsbiprism fringes with tilted fringesjust to the side

(c) higher taper, again using Fresnel diffraction in fiber shadow but withbiprism removed shows more highly tilted fringes

Chambers says: “I am indebted for Mr. Aharonov and Dr. Bohm fortelling me of their work before publication..”

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 13 / 22

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Aharonov-Bohm experiment

(a) a tapered iron whisker producesa confined field and flux with a gra-dient along the z-axis manifested intilted fringes

(b) direct imaging, with the whiskeroutside the shadow of the biprismfiber, due to Fresnel diffractionin the shadow of the fiber showsbiprism fringes with tilted fringesjust to the side

(c) higher taper, again using Fresnel diffraction in fiber shadow but withbiprism removed shows more highly tilted fringes

Chambers says: “I am indebted for Mr. Aharonov and Dr. Bohm fortelling me of their work before publication..”

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 13 / 22

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Aharonov-Bohm experiment

(a) a tapered iron whisker producesa confined field and flux with a gra-dient along the z-axis manifested intilted fringes

(b) direct imaging, with the whiskeroutside the shadow of the biprismfiber, due to Fresnel diffractionin the shadow of the fiber showsbiprism fringes with tilted fringesjust to the side

(c) higher taper, again using Fresnel diffraction in fiber shadow but withbiprism removed shows more highly tilted fringes

Chambers says: “I am indebted for Mr. Aharonov and Dr. Bohm fortelling me of their work before publication..”

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 13 / 22

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Aharonov-Bohm experiment

(a) a tapered iron whisker producesa confined field and flux with a gra-dient along the z-axis manifested intilted fringes

(b) direct imaging, with the whiskeroutside the shadow of the biprismfiber, due to Fresnel diffractionin the shadow of the fiber showsbiprism fringes with tilted fringesjust to the side

(c) higher taper, again using Fresnel diffraction in fiber shadow but withbiprism removed shows more highly tilted fringes

Chambers says: “I am indebted for Mr. Aharonov and Dr. Bohm fortelling me of their work before publication..”

“Shift of an electron interference pattern by enclosed magnetic flux,” R.G. Chambers, Phys. Rev. Lett. 5, 3-5 (1960).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 13 / 22

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A-B effect in a normal metal

VOLUME 54, NUMBER 25 PHYSICAL REVIEW LETTERS 24 JUNE 1985

Observation of h/e Aharonov-Bohm Oscillations in Normal-Metal Rings

R. A. Webb, S. Washburn, C. P. Umbach, and R. B. LaibowitzIBM Thomas J. 8'atson Research Center, Yorkto~n Heights, Xew Fork 20598

(Received 27 March 1985)

Magnetoresistance oscillations periodic with respect to the flux h/e have been observed insubmicron-diameter Au rings, along with weaker h/2e oscillations. The h/e oscillations persist tovery large magnetic fields. The background structure in the magnetoresistance was not symmetricabout zero field. The temperature dependence of both the amplitude of the oscillations and thebackground are consistent with the recent theory by Stone.

PACS numbers: 72.15.Gd, 72.90.+y, 73.60.Dt

Electron wave packets circling a magnetic fluxshould exhibit the phase shift introduced by the mag-netic vector potential A. ' In a metallic ring, smallenough so that the electron states are not randomizedby inelastic (or magnetic) scattering during the traver-sal of the arm of the ring, an interference patternshould be present in the magnetoresistance of the de-vice. Electrons traveling along one arm will acquire aphase change 5t, and electrons in the other arm will, ingeneral, suffer a different phase change 52. Changingthe magnetic flux encircled by the ring will tune thephase change along one arm of the ring by a well-defined amount Ba= (e/t) JA 11 and by —5& alongthe other arm. The phase tuning should appear as cy-cles of destructive and constructive interference of thewave packets, the period of the cycle being 40= h/e.This interference should be reflected in the transportproperties of the ring as described by Landauer's for-mula. 2 4 In this Letter, we describe the first experi-mental observation of the oscillations periodic withrespect to 40 in the magnetoresistance of a normal-metal ring.

Interference effects involving the flux h/e havebeen previously observed in a two-slit interference ex-periment involving coherent beams of electrons. 5

Magnetoresistance oscillations in single-crystal whisk-ers of bismuth periodic in h/e have been reported atlow fields for the case where the extremum of the Fer-mi surface is cut off by the sample diameter. 6 Resis-tance oscillations of period h/2e (flux quantization)have been seen in superconducting cylinders. 7 Fouryears ago, magnetoresistance oscillations of period—,4o were predicted on the basis of weak localizationin multiply connected devices. s This is the same fluxperiod as observed in superconductors, because of thesimilarity between the superconductor pairing and the"self-interference" described by the theory of weak lo-calization. 9 Since the first experiment by Sharvin andSharvin, to there have been several observations of thesuperconducting flux period —,

' 4o in normal-metalcylinders and networks of loops. " To date, there havebeen no observations of the one-electron flux period40, and its existence is controversial. Several recenttheoretical papers have argued that the h/e period will

be present in strictly one-dimensional rings, and evenin rings composed of wires with finite width. 4 Othershave claimed that only h/2e oscillations will be ob-served regardless of device size and topology. '

Theoretical work which relies upon ensemble-averaging techniques has uniformly predicted h/2e os-'cillationss'2; calculations of the conductance exclusiveof the averaging have predicted hje oscillations aswell. 2 4 The difference between a single ring and anetwork of rings or a long cylinder is, therefore, cru-cial. The network of many rings and the long cylinderextend much farther than the distance [L& = (Dr~) 'where D is the diffusion constant and r& is the timebetween phase-breaking co11isions] that the electrontravels before randomly changing its phase. For thisreason, it is believed that samples much longer than

L& physically incorporate the ensemble averaging.Each section (longer than L@) of a macroscopic sam-ple is quantum-mechanically independent because theelectron states are randomized between the sections.The single mesoscopic ring (diameter ( L&) does notaverage in this way because the entire sample isquantum-mechanically coherent. 4'3

There exists a further complication in normal met-als; the magnetic flux penetrates the wires composingthe device. Stone'4 has shown that the flux in the wireleads to an aperiodic fluctuation in the magnetoresis-tance. This fluctuation was the main complication in

interpreting the earlier experiments'5 where the diam-eter of the ring was not much larger than the widths ofthe wires. On the basis of the analysis, a predictionwas made that, in a ring having an area much largerthan the area covered by the wires, the oscillationswould be clearly observed, since the period would thenbe much smaller than the field scale of the fluctua-tions.

With this in mind, we constructed several deviceseach containing a single loop or a lone wire. The sam-ples were drawn with a scanning transmission electronmicroscope (STEM) on a polycrystalline gold film 38nm thick having a resistivity p ——5 p, A cm at T= 4 K.The fabrication process has been described previous-ly. ' A photograph of the larger ring is shown in Fig.1. Here we will describe the results from two of the

2696 1985 The American Physical Society

VOLUME 54, NUMBER 25 PHYSICAL REVIEW LETTERS 24 JUNE 1985

0.05010x+

29.6—T= . 9 K

6 -'(b)

JI]I

«jII

~y V

29.4—0.2 0.0

H [T]

0.199 K

29.5

0.2

400)

cL 2

oO

0—0 100 200 30Q 40Q

1/bH [ 1/T ]

(b)

FIG. 2. (a) Magnetoresistance data from the ring in Fig. Iat several temperatures. (b) The Fourier transform of thedata in (a). The data at 0.199 and 0.698 K have been offsetfor clarity of display. The markers at the top of the figureindicate the bounds for the flux periods h/e and h/2e basedon the measured inside and outside diameters of the loop.

00 100

1/AH [ 1/T ]

200 300

FIG. 1. (a) Magnetoresistance of the ring measured atT=0.01 K. (b) Fourier power spectrum in arbitrary unitscontaining peaks at h/e and h/2e. The inset is a photographof the larger ring. The inside diameter of the loop is 784nm, and the width of the wires is 41 nm.

rings (average diameters 825 and 245 nm) and a lonewire (length 300 nm). The samples were cooled in themixing chamber of a dilution refrigerator, and theresistance was measured with a four-probe bridgeoperated at 205 Hz and 200 nA (rms).

Typical magnetoresistance data from the larger-diameter ring are displayed in Fig. 1(a). Periodic oscil-lations are clearly visible superimposed on a moreslowly varying background. The period of the high-frequency oscillations is AH = 0.007 59 T. This periodcorresponds to the addition of the flux 4p = h/e to thearea of the hole. From the average area (one half ofthe sum of the area from the inside diameter and thatfrom the outside diameter) measured with the STEM,4p = 0.007 80 T. The area measurement is accurate towithin = 10'/o. As a result of the large aspect ratio, wecan say unequivocally that the periodic oscillations arenot consistent with h/2e. They are certainly thesingle-electron process predicted recently. 2 4 In theFourier power spectrum [Fig. 1(b)] of these data, twopeaks are visible at I/AH=131 and 260 T ' corre-sponding respectively to h/e and h/2e. (Since the h/eoscillations are not strictly sinusoidal, we cannot becertain whether the h/2e peak is the self-interferenceprocess or harmonic content in the 4&p oscillations. )That the h/2e period is less significant than the h/eperiod is consistent with the theory for rings which aremoderately resistive. We note that the amplitude ofthe h/e oscillations at the lowest temperatures is about0.1% of the resistance at H= 0, at least a factor of 10

larger than the oscillations observed in normal-metalcylinders and networks of loops. s'p "

Figure 2(a) contains resistance data for three tem-peratures over a larger range of magnetic field.Surprisingly, the oscillations persist to rather highermagnetic field [H ) 8 T (our largest available field) orover 1000 periods] than expected from estimateswhich assumed that the phase difference between theinside edge of the ring and the outside edge shouldcompletely destroy the periodic effects. The argumentthat the flux in the metal should destroy the oscilla-tions relies on the simple assumption that the wireconsists of parallel but noninteracting conductionpaths. If instead the electron path in the wire is suffi-ciently erratic to "cover" the whole area of the wire,then no phase difference exists between the inside di-ameter and the outside diameter. '

Figure 2(b) contains the Fourier spectra of the datain Fig. 2(a). Again, the fundamental h/e period ap-pears as the large peak at I/b, H=131 T ', and nearI/AH=260 T ' there is a small feature in the spec-trum. There is also a peak near 5 T ' which is theaverage field scale of the aperiodic fluctuations. '4 Thedetailed structure of the h/e peak in the power spec-trum is probably the results of mixing of the fieldscales corresponding to the area of the hole in the ringand the area of the arms of the ring. ts (The simpledifference between inside and outside area implies asplitting of more than 20 T ', whereas the observedsplitting in the peak structure has never been morethan 7 T '.) A simple extension of the multichannelLandauer formula for a ring with flux piercing thearms implies that the Arharonov-Bohm oscillationswill be modulated by an aperiodic function. ' Roughlyspeaking, the field scale in which the aperiodic func-tion fluctuates is that for the addition of another fluxquantum to the arms of the ring. The field scale of themodulating function mixes with the Aharonov-Bohmperiod to give structure to the peak. As seen in Fig.

2697

“Observation of h/e Aharonov-Bohm oscillations in normal-metal rings,” R.A. Webb, et al., Phys. Rev. Lett. 54, 2696-2699(1985).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 14 / 22

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A-B effect in a normal metal

There is no reason why theAharonov-Bohm effect couldnot be observed in a conductingloop

The key would be to have theelectrons traveling on either sideof the “solenoid” maintain co-herence until they reunite

Make a small loop out of gold,cool to very low temperatures toincrease the mean free path ofthe electrons, and vary the field

“Observation of h/e Aharonov-Bohm oscillations in normal-metal rings,” R.A. Webb, et al., Phys. Rev. Lett. 54, 2696-2699(1985).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 15 / 22

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A-B effect in a normal metal

There is no reason why theAharonov-Bohm effect couldnot be observed in a conductingloop

The key would be to have theelectrons traveling on either sideof the “solenoid” maintain co-herence until they reunite

Make a small loop out of gold,cool to very low temperatures toincrease the mean free path ofthe electrons, and vary the field

“Observation of h/e Aharonov-Bohm oscillations in normal-metal rings,” R.A. Webb, et al., Phys. Rev. Lett. 54, 2696-2699(1985).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 15 / 22

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A-B effect in a normal metal

There is no reason why theAharonov-Bohm effect couldnot be observed in a conductingloop

The key would be to have theelectrons traveling on either sideof the “solenoid” maintain co-herence until they reunite

Make a small loop out of gold,cool to very low temperatures toincrease the mean free path ofthe electrons, and vary the field

“Observation of h/e Aharonov-Bohm oscillations in normal-metal rings,” R.A. Webb, et al., Phys. Rev. Lett. 54, 2696-2699(1985).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 15 / 22

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A-B effect in a normal metal

There is no reason why theAharonov-Bohm effect couldnot be observed in a conductingloop

The key would be to have theelectrons traveling on either sideof the “solenoid” maintain co-herence until they reunite

Make a small loop out of gold,cool to very low temperatures toincrease the mean free path ofthe electrons, and vary the field

“Observation of h/e Aharonov-Bohm oscillations in normal-metal rings,” R.A. Webb, et al., Phys. Rev. Lett. 54, 2696-2699(1985).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 15 / 22

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Electrostatic A-B effect

Nature © Macmillan Publishers Ltd 1998

8

Magneto-electricAharonov–Bohmeffect inmetal ringsAlexander van Oudenaarden, Michel H. Devoret*,Yu. V. Nazarov & J. E. Mooij

Department of Applied Physics and DIMES, Delft Univesity of Technology,Lorentzweg 1, 2628 CJ Delft, The Netherlands* Permanent address: Service de Physique de l’Etat Condense, CEA-Saclay, F-91191, Gif-sur-Yvette, France.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The quantum-mechanical phase of the wavefunction of anelectron can be changed by electromagnetic potentials, as waspredicted by Aharonov and Bohm1 in 1959. Experiments onpropagating electron waves in vacuum have revealed both themagnetic2–4 and electrostatic5 Aharonov–Bohm effect. Surpris-ingly, the magnetic effect was also observed in micrometre-sizedmetal rings6–8, demonstrating that electrons keep their phasecoherence in such samples despite their diffusive motion. Thesearch for the electrostatic contribution to the electron phase in

these metal rings9,10 was hindered by the high conductivity ofmetal, which makes it difficult to apply a well defined voltagedifference across the ring. Here we report measurements ofquantum interference of electrons in metal rings that are inter-rupted by two small tunnel junctions. In these systems, a welldefined voltage difference between the two parts of the ring canbe applied. Using these rings we simultaneously explore theinfluence of magnetic and electrostatic potentials on theAharonov–Bohm quantum-interference effect, and we demon-strate that these two potentials play interchangeable roles.

To determine the combined influence of electrostatic and mag-netic potentials on the quantum interference of electron waves, weconsider the model shown in Fig. 1a and b. A metallic ring isinterrupted by two tunnel barriers denoted by the black regions inFig. 1b. Because the resistance of these tunnel barriers is much largerthan the resistance of the metallic part, a well defined potential Vcan be applied across the two halves of the ring. Electric transportoccurs because electrons can tunnel from an occupied state in theleft half to an empty state in the right half, as shown in Fig. 1a. Thisprocess is equivalent to the creation of an electron–hole pair duringa tunnel event. After tunnelling, the electron and the hole start todiffuse in the right and left halves of the ring, respectively. Becausethe transport is diffusive, the electron and hole have a finiteprobability of recombining at the other tunnel barrier as shownin Fig. 1b. Only those trajectories in which the electron and holetrajectories form a closed loop circling the ring are sensitive to themagnetic field through the ring. At the moment of recombination,the phase difference DfB between the electron and hole due to themagnetic field B is 2peBS/h (where e is the electron charge, h isPlanck’s constant, and S is the area of the ring). This results inalternating constructive and destructive interference between theelectron and hole wave as a function of B, with a period h/(eS). Thiseffect has analogies with the modulation of the Cooper pair currentwith period h/(2eS) in a superconducting ring interrupted by tunneljunctions (a superconducting quantum interference device,SQUID)11,12. But in the case of the SQUID the effect results fromcollective Cooper pair tunnelling (the Josephson effect), whereashere the interference takes place at the single electron level. Becauseof the voltage difference, the electron and hole have differentenergies with respect to the Fermi level in both halves of ourdevice. The sum of the energy of the hole eh and the electron ee

letters to nature

768 NATURE | VOL 391 | 19 FEBRUARY 1998

Figure 1 Quantum interference of electron–hole pair in mesoscopic ring. a,

Energy diagram modelling a tunnelling process. At the moment of tunnelling, an

electron (e)with energy ee is injected into the right part of the sample leavingahole

(h) with energy eh in the left part. b, Typical electron–hole trajectories which are

sensitive to an Aharonov–Bohm flux. At the moment of a tunnel event an

electron–hole pair is created. The hole and the electron diffuse separately to the

other tunnel barrier where they recombine and interfere. The interference of the

electron and hole wave is both sensitive to a magnetic field B piercing the ring

(magnetic Aharonov–Bohm effect) and a voltage V across the ring (electrostatic

Aharonov–Bohm effect). c, Scanning electron microscope image of the

Aharonov–Bohm ring which is interrupted by two small tunnel junctions. The

sample consists of a 0:9 mm 3 0:9 mm square loop. The width of the arms of the

loop is 60nm. The loop is interrupted by two small tunnel junctions of 60 3 60nm2.

The average distance L between the tunnel barriers is 1.6 mm. The mask of the

loop was defined by electron beam lithography. The junctions were fabricated in

three strips. First, a 25-nm-thick aluminium layer was deposited at an angle with

respect to the oxidized Si substrate. Second, the aluminium was oxidized to form

the insulating tunnel barriers. Finally, a second 40-nm-thick aluminium layer was

deposited at another angle. The tunnel junctions are denoted by the arrows. By

determining the overlap area of the different junctions, we conclude that the

tunnelling conductances do not differ by more than 10%. This is confirmed by

critical current measurements in the superconducting state, in which the sample

operates as a SQUID. The tunnel conductance GT of each junction is 55 6 5 mS,

which is about four orders of magnitude smaller than the conductance of the

metallic part of the ring.

Figure 2 Conductance G versus magnetic field B at T ¼ 20mK and V ¼ 500 mV.

Aharonov–Bohm oscillations with a period BAB ¼ 5mT areobserved as a function

of B. This period is in agreementwith the magnetic field of 5.3mTwhich is needed

to add a flux quantum h/e to the average area S ¼ 0:75 mm2 enclosed by the ring.

In the inset, the Fourier power spectrum is shown as a function of the magnetic

andelectric ‘frequencies’ nB and nV. The lighter the grey scale the larger the Fourier

power. The magnetic and electric field modulation of GAB manifests itself as

peaks around 60.2mT−1 and 60.1 mV−1, respectively.

Nature © Macmillan Publishers Ltd 1998

8

equals eV (Fig. 1a). Owing to this energy difference, the electron–hole pair accumulates an electrostatic phase differenceDfV ¼ 2peVt=h, where t is the time the electron and hole spendin their respective parts of the ring before they recombine. Becauseelectron motion is diffusive there is not one unique time, but rathera distribution of times with an average value t0 ¼ L2=D where L isthe distance along the ring between the two tunnel barriers and D isthe diffusion coefficient. The electrostatic interference results in analternating constructive and destructive interference as a function ofV with a period h/(et0). Experimentally, one can explore both themagnetic and electrostatic quantum interference effect by measur-ing the Aharonov–Bohm flux-dependent part GAB of the conduc-tance. The magnetic field B and the bias voltage V have an equivalenteffect on GAB. Changing the magnetic field at fixed V leads to aperiodically oscillating GAB with a period which is solely defined bythe geometry of the ring. Measuring the conductance as a functionof Vat fixed B results in a periodically oscillating GAB with an averageperiod which is determined by t0. In other words, by exploring thevoltage dependence of GAB, t0 is measured experimentally. In theballistic regime related interference experiments13 have been per-formed in which one of the arms of the Aharonov–Bohm ring wasinterrupted by a quantum dot. But the relevant interference pro-cesses in such systems, where the number of electrons in the dot ischanging, differ significantly from the interference in the ring weconsider here.

To investigate the combined role of magnetic and electrostaticpotentials we designed the sample shown in Fig. 1c. The differentialconductance G at a voltage V and magnetic field B was measuredusing a lock-in technique. Details of the measurement set-up aregiven in ref. 14. All measurements were performed at B . 1 T todrive the aluminium loop into the normal state. At those fields time-reversal symmetry is broken, and effects related to this symmetrycan be neglected. In Fig. 2 the conductance G is plotted as a functionof B at a bias voltage V ¼ 500 mV and a temperature T ¼ 20 mK.Clear periodic oscillations of the conductance are observed with aperiod of 5 mT, which is in good agreement with the predictedperiod for h/e oscillations. The relative amplitude of the Aharonov–Bohm oscillations at V ¼ 500 mV is ,5%, which is considerablylarger than the results obtained for uniform rings6–10. The magneticfield not only pierces the hole of the loop, but also penetrates the

arms of the loop in contrast to the ideal geometry proposed byAharonov and Bohm1. This gives rise to aperiodic conductancefluctuations14, clearly visible in Fig. 2 as the slowly varying back-ground on top of the Aharonov–Bohm oscillations. Because of thelarge difference in magnetic field scales, it is possible to filter out theaperiodic conductance fluctuations using Fourier analysis. In thisway we can extract (from the conductance G) the Aharonov–Bohmconductance GAB, which we would measure if the field was appliedonly inside the ring. The Fourier power spectrum of GAB withrespect to B and V is shown in Fig. 2 inset.

Figure 3 shows how the Aharonov–Bohm conductance GAB

evolves when the bias voltage is changed. The voltage incrementbetween two successive traces is 0.48 mV. Trace a in Fig. 3 denotes aminimum of the Aharonov–Bohm conductance at V ¼ 519 mV.When the voltage is decreased by 2.5 mV (trace b) the Aharonov–Bohm oscillations have almost vanished. A further decrease ofthe voltage by 2.5 mV (trace c) leads again to an increase of theoscillations. However, the minimum near B ¼ 1:032 T for V ¼519 mV (trace a) turned into a maximum at V ¼ 514 mV (trace c).This observation unambiguously demonstrates the symmetry betweenthe magnetic and the electrostatic Aharonov–Bohm effect. A maxi-mum GAB (constructive electron–hole interference) is turned into aminimum GAB (destructive electron–hole interference) either byincreasing the magnetic field by 2.5 mT or by increasing the voltageby 5 mV. This symmetry is also reflected in the Fourier powerspectrum (Fig. 2 inset) by the peaks around 60.2 mT −1 and60.1 mV −1.

To obtain a more quantitative analysis, we calculated the correla-tion function CðDB;DVÞ ¼ 〈GABðB;VÞGABðB þ DB;V þ DVÞ〉,where the angle brackets denote an ensemble average. The quantityis shown in Fig. 4. The squares in Fig. 4 denote the experimentalCðDB;DV ¼ 0Þ. Its oscillatory behaviour is again the manifestationof the magnetic Aharonov–Bohm effect (we denote the period byBAB). The triangles in Fig. 4 denote the experimental cross-correlation

letters to nature

NATURE | VOL 391 | 19 FEBRUARY 1998 769

Figure 3 Aharonov–Bohm conductance as a function of magnetic field B at

T ¼ 20mK for bias voltages separated by 0.48 mV. The traces have been offset by

5 mS with respect to the conductance-axis for clarity. The Aharonov–Bohm

conductance GAB has been extracted from G by rejecting the Fourier components

of G for magnetic frequencies smaller than 0.1mT−1. The bold traces a–c in the

rectangular box show an example of the combined manifestation of the magnetic

and electrostatic Aharonov–Bohm effect.

Figure 4 Normalized correlation function versus DB/BAB at T ¼ 20mK. The

correlation function CðDB;DVÞ ¼ 〈GABðB;VÞGABðB þ DB;V þ DVÞ〉 has been nor-

malized to the variance Cð0; 0Þ ¼ 〈G2AB〉. Squares, triangles and circles correspond

to DV ¼ 0; 5 and 10 mV, respectively. The correlation functions are calculated from

a total data set consisting of 350 conductance traces as a function of B between

1.0 and 1.1Tor 2.0 and 2.1T. These traces are measured at different voltages in the

range 395–533 mV. The theoretical correlation function depends only on two

parameters: t0 and tf, which is the time during which an electron–hole pair

preserves its phase memory. A close agreement between the experimental

results and theory (full lines) is found for to ¼ 300 ps and tf ¼ 300 ps. These

values are consistent with previous experiments14.

“Magneto-electric Aharonov-Bohm effect in metal rings,” A. van Oudenaarden, et al., Nature 391, 768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 16 / 22

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Electrostatic A-B effect

tunnel barriers permit the application of apotential which is used to inject electronsand holes into opposite arms of the ring

electron and hole recombine at the othertunnel junction with a relative phase shiftdependent on the enclosed magnetic flux

sweeping the magnetic fieldresults in Aharonov-Bohmoscillations

“Magneto-electric Aharonov-Bohm effect in metalrings,” A. van Oudenaarden, et al., Nature 391,768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 17 / 22

Page 107: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Electrostatic A-B effect

tunnel barriers permit the application of apotential which is used to inject electronsand holes into opposite arms of the ring

electron and hole recombine at the othertunnel junction with a relative phase shiftdependent on the enclosed magnetic flux

sweeping the magnetic fieldresults in Aharonov-Bohmoscillations

“Magneto-electric Aharonov-Bohm effect in metalrings,” A. van Oudenaarden, et al., Nature 391,768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 17 / 22

Page 108: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Electrostatic A-B effect

tunnel barriers permit the application of apotential which is used to inject electronsand holes into opposite arms of the ring

electron and hole recombine at the othertunnel junction with a relative phase shiftdependent on the enclosed magnetic flux

sweeping the magnetic fieldresults in Aharonov-Bohmoscillations

“Magneto-electric Aharonov-Bohm effect in metalrings,” A. van Oudenaarden, et al., Nature 391,768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 17 / 22

Page 109: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Electrostatic A-B effect

tunnel barriers permit the application of apotential which is used to inject electronsand holes into opposite arms of the ring

electron and hole recombine at the othertunnel junction with a relative phase shiftdependent on the enclosed magnetic flux

sweeping the magnetic fieldresults in Aharonov-Bohmoscillations

“Magneto-electric Aharonov-Bohm effect in metalrings,” A. van Oudenaarden, et al., Nature 391,768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 17 / 22

Page 110: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Electrostatic A-B effect

tunnel barriers permit the application of apotential which is used to inject electronsand holes into opposite arms of the ring

electron and hole recombine at the othertunnel junction with a relative phase shiftdependent on the enclosed magnetic flux

sweeping the magnetic fieldresults in Aharonov-Bohmoscillations

“Magneto-electric Aharonov-Bohm effect in metalrings,” A. van Oudenaarden, et al., Nature 391,768-770 (1998).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 17 / 22

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Electrostatic A-B effect

Aharonov-Bohm oscillationsare seen also for constantmagnetic field as applied po-tential is varied

periodicity as seen in se-quence a→ b → c is definedby diffusion time of electrons& holes around the ring

Aharonov-Bohm effect continues to be an active area of research nearly 60years after it was first proposed!

“Magneto-electric Aharonov-Bohm effect in metal rings,” A. van Oudenaarden, et al., Nature 391, 768-770 (1998).

“The Aharonov-Bohm effects: Variations on a subtle theme,” H. Batelaan & A. Tonomura, Physics Today 62 (9), 38-43 (2009).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 18 / 22

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Electrostatic A-B effect

Aharonov-Bohm oscillationsare seen also for constantmagnetic field as applied po-tential is varied

periodicity as seen in se-quence a→ b → c is definedby diffusion time of electrons& holes around the ring

Aharonov-Bohm effect continues to be an active area of research nearly 60years after it was first proposed!

“Magneto-electric Aharonov-Bohm effect in metal rings,” A. van Oudenaarden, et al., Nature 391, 768-770 (1998).

“The Aharonov-Bohm effects: Variations on a subtle theme,” H. Batelaan & A. Tonomura, Physics Today 62 (9), 38-43 (2009).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 18 / 22

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Electrostatic A-B effect

Aharonov-Bohm oscillationsare seen also for constantmagnetic field as applied po-tential is varied

periodicity as seen in se-quence a→ b → c is definedby diffusion time of electrons& holes around the ring

Aharonov-Bohm effect continues to be an active area of research nearly 60years after it was first proposed!

“Magneto-electric Aharonov-Bohm effect in metal rings,” A. van Oudenaarden, et al., Nature 391, 768-770 (1998).

“The Aharonov-Bohm effects: Variations on a subtle theme,” H. Batelaan & A. Tonomura, Physics Today 62 (9), 38-43 (2009).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 18 / 22

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Electrostatic A-B effect

Aharonov-Bohm oscillationsare seen also for constantmagnetic field as applied po-tential is varied

periodicity as seen in se-quence a→ b → c is definedby diffusion time of electrons& holes around the ring

Aharonov-Bohm effect continues to be an active area of research nearly 60years after it was first proposed!

“Magneto-electric Aharonov-Bohm effect in metal rings,” A. van Oudenaarden, et al., Nature 391, 768-770 (1998).

“The Aharonov-Bohm effects: Variations on a subtle theme,” H. Batelaan & A. Tonomura, Physics Today 62 (9), 38-43 (2009).

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 18 / 22

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Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 116: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 117: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 118: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 119: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 120: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 121: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 19 / 22

Page 122: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

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Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 124: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 125: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 126: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles.

There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 127: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 128: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 129: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 130: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosons

particles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 131: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

If the particles are identical, the wavefunction must not choose betweenstates for the two particles. There are two such states:

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

particles with integer spin are called bosonsparticles with half-integer spin are called fermions

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 20 / 22

Page 132: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 133: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 134: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 135: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)]

= 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 136: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 137: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 138: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 139: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 140: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Fermion properties

ψ−(~r1,~r2) = A [ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)]

This wave function, for two identical fermions, describes a fundamentalproperty of this kind of particle.

Suppose that we have two identical fermions, both in the same state, ψa.

ψ−(~r1,~r2) = A [ψa(~r1)ψa(~r2)− ψa(~r1)ψa(~r2)] = 0

The form of the identical particle wavefunction mandates that no twoidentical fermion can occupy the same state!

This is called the Pauli exclusion principle and is a fundamental property ofall fermions (electrons included).

By contrast bosons have no such restriction and there can be an infinitenumber in the same state

What are some examples of bosons?

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 21 / 22

Page 141: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat

and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 142: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat

and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 143: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat

and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 144: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat

and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 145: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 146: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 147: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 148: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 149: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 150: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 151: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 152: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 153: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22

Page 154: Today’s Outline - October 24, 2019csrri.iit.edu/~segre/phys405/19F/lecture_19.pdf · covers through Chapter 4 C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 20191/22. Review of

Exchange symmetry

A more formal way of describing this isto define an exchange operator, P, suchthat and we see that P2 = 1 so that theeigenvalues of P are

Identical particles must be treated identi-cally in the Hamiltonian

and therefore, P and H commute andthere can be simultaneous eigenfunctionsof both

P f (~r1,~r2) = f (~r2,~r1)

= αf (~r1,~r2)

P2 f (~r1,~r2) = α2f (~r1,~r2)

α = ±1

m1 = m2

V (~r1,~r2) = V (~r2,~r1)

[P,H] = 0

The consequence is that there are solutions to the Schrodinger equationwhich are symmetric and antisymmetric under exchange

ψ(~r1,~r2) = ±ψ(~r2,~r1)

Exchange symmetry is time-independent and is a requirement for the twotypes of particles, fermions and bosons.

C. Segre (IIT) PHYS 405 - Fall 2019 October 24, 2019 22 / 22