130_notes.dvi

The total momentum in the (transverse) radiation field can also be computed (from the classical
formula for the Poynting vector).

P~=^1

c

∫

E~×B d~^3 x=

∑

k,α

̄h~k

(

Nk,α+

1

2

)

This time the^12 can really be dropped since the sum is over positive and negative~k, so it sums to
zero.
P~=

∑

k,α

̄h~k Nk,α

We can compute the energy and momentum of a single photon state by operating on that state with
the Hamiltonian and with the total momentum operator. The state for a single photon with a given
momentum and polarization can be written asa†k,α| 0 〉.

Ha†k,α| 0 〉=

(

a†k,αH+ [H,a†k,α]

)

| 0 〉= 0 + ̄hωa†k,α| 0 〉= ̄hωa†k,α| 0 〉

The energy of single photon state is ̄hω.

Pa†k,α| 0 〉=

(

a†k,αP+ [P,a†k,α]

)

| 0 〉= 0 + ̄h~ka†k,α| 0 〉= ̄h~ka†k,α| 0 〉

The momentum of the single photon state is ̄h~k. The mass of the photon can be computed.

E^2 = p^2 c^2 + (mc^2 )^2

mc^2 =

√

( ̄hω)^2 −( ̄hk)^2 c^2 = ̄h

√

ω^2 −ω^2 = 0

So the energy, momentum, and mass of a single photon state are aswe would expect.

The vector potential has been given two transverse polarizationsas expected from classical Electricity
and Magnetism. The result is two possible transverse polarization vectors in our quantized field. The
photon states are also labeled by one of two polarizations, that we have so far assumed were linear
polarizations. The polarization vector, and therefore the vectorpotential, transform like a Lorentz
vector. We know that the matrix element of vector operators (See section 29.9) is associated with an
angular momentum of one. When a photon is emitted, selection rules indicate it is carrying away an
angular momentum of one, so we deduce that the photon has spin one. We need not add anything to
our theory though; the vector properties of the field are alreadyincluded in our assumptions about
polarization.

Of course we could equally well use circular polarizations which are related to the linear set we have
been using by

ˆǫ(±)=∓

1

√

2

(ˆǫ(1)±iˆǫ(2)).

The polarization ˆǫ(±)is associated with them=±1 component of the photon’s spin. These are
the transverse mode of the photon,~k·ˆǫ(±)= 0. We have separated the field into transverse and
longitudinal parts. The longitudinal part is partially responsible for static E and B fields, while the
transverse part makes up radiation. Them= 0 component of the photon is not present in radiation
but is important in understanding static fields.

By assuming the canonical coordinates and momenta in the Hamiltonian have commutators like
those of the position and momentum of a particle, led to an understanding that radiation is made
up of spin-1 particles with mass zero. All fields correspond to a particle of definite mass and spin.
We now have a pretty good idea how to quantize the field for any particle.