E&M
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(x, t) = 4πρ(x, t)
∇ ·B(x, t) = 0
∇×E(x, t) = −1
c
∂
∂tB(x, t)
∇×B(x, t) =1
c
∂
∂tE(x, t) +
4π
cj(x, t)
E&M
Maxwell’s equations
Gaussian units
E = −∇Φ− 1
c
∂A
∂t
B = ∇×A
E&M
Scalar and Vector potentialQuantum applications require replacingelectric and magnetic fields! implies
From Faraday
so
or
in terms of vector and scalar potentials.Homogeneous equations are automatically solved.
∇ ·B = 0
∇×�E +
1
c
∂
∂tA
�= 0
E +1
c
∂
∂tA = −∇Φ
∇ ·A+1
c
∂Φ
∂t= 0
∇2Φ+1
c
∂
∂t(∇ ·A) = −4πρ
∇2A− 1
c2∂2A
∂t2−∇
�∇ ·A+
1
c
∂Φ
∂t
�= −4π
cj
E&M
Gauge freedom
Remaining equations using
To decouple one could choose (gauge freedom)
more later… first
∇× (∇×A) = ∇ (∇ ·A)−∇2A
E&M
Coupling to charged particles
Lorentz
Rewrite
Note
and
So that withF = −∇U +d
dt
∂U
∂vU = qΦ− q
cA · v
∂A
∂t+ (v ·∇)A =
dA
dt
F = q
�E +
1
cv ×B
�
v × (∇×A) = ∇ (v ·A)− (v ·∇)A
F = q
�−∇Φ− 1
c
∂A
∂t+
1
cv × (∇×A)
�
E&M
CheckYields Lorentz from
Equations of motion
Generalized momentum
Solve for v and substitute in Hamiltonian--> Hamiltonian for a charged particle
H = p · v − L =
�p− q
cA�2
2m+ qΦ
p =∂L
∂v= mv +
q
cA
d
dt
∂L
∂v− ∂L
∂x= 0
L = T − U =1
2mv2 − qΦ+
q
cA · v
E&M
Include external electromagnetic field in QM• Static electric field: nothing new (position --> operator)• Include static magnetic field with momentum and position
operators
• Note velocity operator
• Note Hamiltonian not “free” particle one
• Use
• to show that
• Gauge independent! So think in terms of
v =1
m
�p− q
cA�
H =
�p− q
cA(x)�2
2m
[pi, Aj ] =�i
∂Aj
∂xi
[vi, vj ] = iq�m2c
�ijkBk
H =1
2m|v|2
E&M
Include external electromagnetic field• Include uniform magnetic field• For example by
• Only nonvanishing commutator
• Write Hamiltonian as
• but now so this corresponds to free particle motion parallel to magnetic field (true classically too)
• Only consider
• Operators don’t commute but commutator is a complex number!
• So...
B(x) = Bz
[vx, vy] = iq�Bm2c
H =1
2m
�v2x + v
2y + v
2z
�
vz =pzm
H =1
2m
�v2x + v
2y
�
E&M
Harmonic oscillator again...• Motion perpendicular to magnetic field --> harmonic oscillator• Introduce
• with cyclotron frequency
• Straightforward to check
• So Hamiltonian becomes
• and consequently spectrum is (called Landau levels)
ωc =qB
mc
�a, a†
�= 1
H = �ωc
�a†a+ 1
2
�
En = �ωc (n+ 12 ) n = 0, 1, ....
a =
�m
2�ωc(vx + ivy)
a† =
�m
2�ωc(vx − ivy)
E&M
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
A =Br212r
φE&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
– Use Stokes theorem
– Only z-component of magnetic field so left-hand side becomes
– for any circular loop outside inner cylinder (and centered)
– Vector potential in the direction of and line integral -->
– Resulting in modifying the Hamiltonian and the spectrum!!
�
S(∇×A) · n da =
�
CA · d�
�
S(∇×A) · n da =
�
SBθ(r1 − ρ)da = Bπr21
φ 2πr
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