Inclusion of the electromagnetic field in
Quantum Mechanics
similar to Classical Mechanics
but with interesting consequences
• Maxwell’s equations
• Scalar and vector potentials
• Lorentz force
• Transform to Lagrangian
• Then Hamiltonian
• Minimal coupling to charged particles
E&M
Maxwell’s equations
Gaussian units
∇ · E(x, t)
=
4πρ(x, t)
∇ · B(x, t)
=
0
∇ × E(x, t)
=
∇ × B(x, t)
=
1 ∂
−
B(x, t)
c ∂t
1 ∂
4π
E(x, t) +
j(x, t)
c ∂t
c
E&M
Scalar and Vector potential
Quantum applications require replacing
electric and magnetic fields!
implies ∇ · B = 0
�
1 ∂
From Faraday ∇ × E +
A =0
c ∂t
1 ∂
A = −∇Φ
so E +
c ∂t
B
or
E
=
=
∇×A
�
1 ∂A
−∇Φ −
c ∂t
in terms of vector and scalar potentials.
Homogeneous equations are automatically solved.
E&M
Gauge freedom
Remaining equations using ∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A
1 ∂
∇ Φ+
(∇ · A)
c ∂t
�
�
2
1 ∂ A
1 ∂Φ
2
∇ A− 2 2 −∇ ∇·A+
c ∂t
c ∂t
2
=
−4πρ
=
4π
− j
c
To decouple one could choose (gauge freedom)
more later… first
1 ∂Φ
∇·A+
=0
c ∂t
E&M
Coupling to charged particles
Lorentz
Rewrite
Note
and
�
�
1
F =q E+ v×B
c
�
�
1 ∂A 1
F = q −∇Φ −
+ v × (∇ × A)
c ∂t
c
v × (∇ × A) = ∇ (v · A) − (v · ∇) A
∂A
dA
+ (v · ∇) A =
∂t
dt
d ∂U
So that F = −∇U +
dt ∂v
q
with U = qΦ − A · v
c
E&M
Check
Yields Lorentz from
1
q
2
L = T − U = mv − qΦ + A · v
2
c
Equations of motion
Generalized momentum
d ∂L ∂L
−
=0
dt ∂v
∂x
∂L
q
p=
= mv + A
∂v
c
Solve for v and substitute in Hamiltonian
--> Hamiltonian for a charged particle
�
�2
q
cA
p−
H =p·v−L=
2m
+ qΦ
E&M
Include external electromagnetic field in QM
• Static electric field: nothing new (position --> operator)
• Include static magnetic field with momentum and position
�
�2
operators
q
p − c A(x)
H=
2m
• Note velocity operator
1 �
q �
v=
p− A
m
c
• Note Hamiltonian not “free” particle one
• Use
• to show that
� ∂Aj
[pi , Aj ] =
i ∂xi
q�
[vi , vj ] = i 2 �ijk Bk
m c
1
• Gauge independent! So think in terms of H = m|v|2
2
E&M
Include external electromagnetic field
• Include uniform magnetic field
• For example by
B(x) = B ẑ
q�B
• Only nonvanishing commutator [vx , vy ] = i 2
m c
• Write Hamiltonian as
�
1 � 2
2
2
H = m vx + vy + vz
2
pz
• but now vz =
so this corresponds to free particle motion
m
parallel to magnetic field (true classically too)
• Only consider
�
1 � 2
2
H = m vx + vy
2
• Operators don’t commute but commutator is a complex number!
• So...
E&M
Harmonic oscillator again...
• Motion perpendicular to
� magnetic field --> harmonic oscillator
m
• Introduce
a =
(vx + ivy )
2�ωc
�
m
†
a =
(vx − ivy )
2�ωc
qB
ωc =
mc
• with cyclotron frequency
• Straightforward to check
• So Hamiltonian becomes
�
�
a, a
†
�
†
H = �ωc a a +
=1
1
2
�
• and consequently spectrum is (called Landau levels)
En = �ωc (n + 12 )
n = 0, 1, ....
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Aharanov-Bohm effect
• Consider hollow cylindrical shell
• Magnetic field inside inner cylinder either on or off
• Charged particle confined between inner and outer radius as well
as top and bottom
E&M
Discussion
• Without field:
– Wave function vanishes at the radii of the cylinders as well as top and
bottom --> discrete energies
• With field (think of solenoid)
– No magnetic field where the particle moves inside in z-direction and
constant
– Spectrum changes because the vector potential is needed in the Hamitonian
�
�
(∇ × A) · n̂ da =
A · d�
– Use Stokes theorem
S
C
– Only z-component of magnetic
field so left-hand
side becomes
�
�
S
(∇ × A) · n̂ da =
S
Bθ(r1 − ρ)da = Bπr12
– for any circular loop outside inner cylinder (and centered)
– Vector potential in the direction of φ̂ and line integral -->
– Resulting in
2πr
modifying the Hamiltonian and the spectrum!!
Br12
A=
φ̂
E&M
2r
Example
• No field
• Example of radial wave function
• Problem solved in cylindrical coordinates
•
Also with field -->
E&M
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Inclusion of the electromagnetic field in Quantum Mechanics similar