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

E&M

Example• No field• Example of radial wave function

• Problem solved in cylindrical coordinates

• Also with field -->