“Significance of Electromagnetic Potentials in the Quantum Theory”* The Aharanov-Bohm Effect Chad A. Middleton MSC Physics Seminar February 17, 2011 *Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959). D. J. Griffiths, Introduction to Quantum Mechanics, 2 nd ed., pps. 384- 391. D.J. Griffiths, Introduction to Electrodynamics, 3 rd ed.
“Significance of Electromagnetic Potentials in the Quantum Theory”* The Aharanov-Bohm Effect Chad A. Middleton MSC Physics Seminar February 17, 2011 *Y.
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“Significance of Electromagnetic Potentials
in the Quantum Theory”* The Aharanov-Bohm Effect
Chad A. MiddletonMSC Physics Seminar
February 17, 2011
*Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959).D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., pps. 384-391.
D.J. Griffiths, Introduction to Electrodynamics, 3rd ed.
Yakir Aharonov receives a 2009 National Medal of Science for his work in quantum physics which ranges from the Aharonov-Bohm effect to the notion of weak measurement.
http://www.nsf.gov/od/oia/activities/medals/2009/laureatephotos.jsp
Outline… Maxwell’s equations in terms of E & B fields
Scalar and vector potentials in E&M
Maxwell’s equations in terms of the potentials
Schrödinger equation
Schrödinger equation with E&M
Aharonov-Bohm Effect
A simple example
Maxwell’s equations in differential form (in vacuum)
€
r∇ ⋅
rE =
ρ
ε0
€
r∇ ×
rE +
∂r B
∂t= 0
Gauss’ law for E-field
Gauss’ law for B-field
Faraday’s law
Ampere’s law with Maxwell’s correction
€
r∇ ⋅
rB = 0
€
r∇ ×
rB − μ0ε0
∂r E
∂t= μ0
r J
€
rF = q
r E +
r v ×
r B ( )
these plusthe Lorentz force completely describe
classical Electromagnetic Theory
Taking the curl of the 3rd & 4th eqns (in free space when = J = 0) yield..
€
∇2 −1
c 2
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r E = 0
The wave equations for theE-, B-fields with
predicted wave speed
€
∇2 −1
c 2
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r B = 0
Light = EM wave!
€
c =1
μ0ε0
≅ 3.0 ×108 m /s
Maxwell’s equations…
€
r∇ ⋅
rE =
ρ
ε0
€
r∇ ×
rE +
∂r B
∂t= 0
Gauss’ law for E-field
Gauss’ law for B-field
Faraday’s law
Ampere’s law with Maxwell’s correction
€
r∇ ⋅
rB = 0
€
r∇ ×
rB − μ0ε0
∂r E
∂t= μ0
r J
Q: Can we write the Maxwell eqns in terms of potentials?
E, B in terms of A, Φ…
€
rB =
r ∇ ×
r A
€
rE = −
r ∇φ −
∂r A
∂t Φ is the scalar potential is the vector potential
€
rA
Write the (2 remaining) Maxwell equations in terms of the potentials.
Maxwell’s equations in terms of the scalar & vector potentials
€
∇2φ +∂
∂t
r ∇ ⋅
r A ( ) = −
ρ
ε0
∇ 2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r A −
r ∇r
∇ ⋅r A + μ0ε0
∂φ
∂t
⎡ ⎣ ⎢
⎤ ⎦ ⎥= −μ0
r J
Gauss’ Law
Ampere’s Law
Gauge invariance of A, Φ..
€
rB =
r ∇ ×
r A
Notice:E & B fields are invariant under the transformations:
€
φ→ ′ φ =φ−∂Λ∂t
for any function
€
Λ=Λ(r r , t)
€
rE = −
r ∇φ −
∂r A
∂t
€
rA →
r ′ A =
r A +
r ∇Λ
Show gauge invariance of E & B.
Maxwell’s equations in terms of the scalar & vector potentials
€
∇2φ +∂
∂t
r ∇ ⋅
r A ( ) = −
ρ
ε0
∇ 2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r A −
r ∇r
∇ ⋅r A + μ0ε0
∂φ
∂t
⎡ ⎣ ⎢
⎤ ⎦ ⎥= −μ0
r J
Gauss’ Law
Ampere’s Law
Coulomb gauge:
€
∇2φ = −ρ
ε0
∇ 2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r A = μ0ε0
r ∇∂φ
∂t− μ0
r J
Maxwell’s equations become..
€
r∇ ⋅
rA = 0
Easy
Hard
Gauss’ law is easy to solve for , Ampere’s law is hard to solve for
€
rA
€
φ
Maxwell’s equations in terms of the scalar & vector potentials
€
∇2φ +∂
∂t
r ∇ ⋅
r A ( ) = −
ρ
ε0
∇ 2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r A −
r ∇r
∇ ⋅r A + μ0ε0
∂φ
∂t
⎡ ⎣ ⎢
⎤ ⎦ ⎥= −μ0
r J
Gauss’ Law
Ampere’s Law
Lorentz gauge:
€
∇2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥φ = −
ρ
ε0
∇ 2 − μ0ε0
∂ 2
∂t 2
⎡
⎣ ⎢
⎤
⎦ ⎥r A = −μ0
r J
Maxwell’s equations become..
€
r∇ ⋅
rA + μ0ε0
∂φ
∂t= 0
Lorentz gauge puts scalar and vector potentials on equal footing.
Schrödinger equation for a particle of mass m
where is the wave function with physical meaning given by:
€
−h2
2m∇ 2 + V
⎡
⎣ ⎢
⎤
⎦ ⎥Ψ = ih
∂
∂tΨ
€
Ψ
€
dP = Ψ*Ψd3x
How do we include E&M in QM?
In QM, the Hamiltonian is expressed in terms of and NOT .
Schrödinger equation for a particle of mass m and charge q in an electromagnetic field
€
1
2mih
r ∇ + q
r A ( )
2+ qφ
⎡ ⎣ ⎢
⎤ ⎦ ⎥Ψ = ih
∂
∂tΨ
€
rE ,
r B
is the scalar potential is the vector potential
€
rA
€
φ
€
φ,r A
Gauge invariance of A, Φ..Notice:E & B fields and the Schrödinger equation are invariant under the transformations:
€
φ→ ′ φ =φ−∂Λ∂t
for any function
€
Λ=Λ(r r , t)
€
rA →
r ′ A =
r A +
r ∇Λ
€
Ψ→ ′ Ψ =Ψe iqΛ / h
Since and differ only by a phase factor, they represent the same physical state.
€
Ψ
€
′ Ψ
The Aharonov-Bohm Effect
In 1959, Y. Aharonov and D. Bohm showed that the vector potential affects the behavior of a charged particle, even in a region where the E & B fields are zero!
http://physicaplus.org.il/zope/home/en/1224031001/Tonomura_en
A simple example:
Consider: A long solenoid of radius a A charged particle constrained to move
in a circle of radius b, with a < b
Magnetic field of solenoid:
€
rB = B0
ˆ k r < ar B = 0 r > a
Vector potential of solenoid? (in Coulomb gauge)
A simple example:
Consider: A long solenoid of radius a A charged particle constrained to move
in a circle of radius b, with a < b
Magnetic field of solenoid:
€
rB = B0
ˆ k r < ar B = 0 r > a
Vector potential of solenoid? (in Coulomb gauge)
€
rA =
ΦB
2πrˆ e ϕ , r > a
Notice:The wave function for a bead on a wire is only a
function of the azimuthal angle
€
ψ =ψ(ϕ ) as r = b, θ = π /2
∴r
∇ = ˆ e ϕ1
b
d
dϕ
Notice:The wave function for a bead on a wire is only a
function of the azimuthal angle
€
ψ =ψ(ϕ ) as r = b, θ = π /2
∴r
∇ = ˆ e ϕ1
b
d
dϕ
The time-independent Schrödinger eqn takes the form..
€
1
2m−
h2
b2
⎛
⎝ ⎜
⎞
⎠ ⎟d2
dϕ 2+ i
hqΦB
πb2
⎛
⎝ ⎜
⎞
⎠ ⎟d
dϕ+
q2ΦB2
4π 2b2
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥ψ = Eψ
The time-independent Schrödinger equation yields a solution of the form..
€
ψ(ϕ ) = Ae iλϕ
€
λ =qΦB
2πh±
b
h2mE
where
Notice:The wave function must satisfy the boundary condition
€
ψ(ϕ ) =ψ (ϕ + 2π ) ∴ e iλ 2π =1
this yields…
€
λ =qΦB
2πh±
b
h2mE = n where n = 0,±1,±2,...
Finally,solving for the energy…
€
En =h2
2mb2n −
qΦB
2πh
⎛
⎝ ⎜
⎞
⎠ ⎟2
where
€
n = 0,±1,±2,...
Notice: positive (negative) values of n represent particle
moving in the same (opposite) direction of I.
Finally,solving for the energy…
€
En =h2
2mb2n −
qΦB
2πh
⎛
⎝ ⎜
⎞
⎠ ⎟2
where
€
n = 0,±1,±2,...
Notice: positive (negative) values of n represent particle
moving in the same (opposite) direction of I. particle traveling in same direction as I has a lower
energy than a particle traveling in the opposite direction.
Finally,solving for the energy…
€
En =h2
2mb2n −
qΦB
2πh
⎛
⎝ ⎜
⎞
⎠ ⎟2
where
€
n = 0,±1,±2,...
Notice: positive (negative) values of n represent particle
moving in the same (opposite) direction of I. particle traveling in same direction as I has a lower
energy than a particle traveling in the opposite direction.
Allowed energies depend on the field inside the solenoid, even though the B-field at the location of the particle is zero!
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