130_notes.dvi

wave-vectors for the initial and final state. A look back at the calculation shows that we calculated
the transition rate from a state with one photon with wave-vector~kand polarizationǫ(α)to a final
state with polarizationǫ(α

′)

. We haveintegrated over the final state wave vector magnitude,
subject to the delta functiongiving energy conservation, but, we havenot integrated over
final state photon directionyet, as indicated by theddσΩ. There is no explicit angular dependence
but there is somehidden in the dot product between initial and final polarization vectors,
both of which must be transverseto the direction of propagation. We are ready to compute
four different differential cross sections corresponding to two initial polarizations times two final
state photon polarizations. Alternatively, we average and/or sum, if we so choose.

In the high energy approximation we have made, there is no dependence on the state of the atoms,
so we arefree to choose our coordinate systemany way we want. Set thez-axis to be along
the direction of the initial photonand set the x-axis so that thescattered photon is in the
x-z plane(φ= 0). The scattered photon is at an angleθto the initial photon direction and at
φ= 0. A reasonable set ofinitial state polarization vectorsis

ˆǫ(1) = ˆx

ˆǫ(2) = ˆy

Pick ˆǫ(1)

′
to be in thescattering plane(x-z) defined as the plane containing both~kand~k′and
ˆǫ(2)

′

to be perpendicular to the scattering plane. ˆǫ(1)

′

is then at an angleθto the x-axis. ˆǫ(2)

′
is
along the y-axis. We can compute all the dot products.

ˆǫ(1)·ǫˆ(1)

′

= cosθ

ˆǫ(1)·ǫˆ(2)

′

= 0

ˆǫ(2)·ǫˆ(1)

′

= 0

ˆǫ(2)·ǫˆ(2)

′

= 1

From these, we can compute any cross section we want. For example,averaging over initial state
polarization and summing over final is just half the sum of thesquares of the above.

dσ

dΩ

=

(

e^2

4 πmc^2

) 2

1

2

(1 + cos^2 θ)

Even if the initial state is unpolarized, thefinal state can be polarized. For example, forθ=π 2 ,

all of the above dot products are zero except ˆǫ(2)·ˆǫ(2)
′
= 1. That means only the initial photons
polarized along the y direction will scatter and that the scattered photon is100% polarized
transverse to the scattering plane(really just the same polarization as the initial state). The
angular distribution could also be used to deduce the polarization of the initial state if a large
ensemble of initial state photons were available.

For adefinite initial state polarization(at an angleφto the scattering plane, the compo-
nent along ˆǫ(1)is cosφand along ˆǫ(2)is sinφ. If we don’t observe final state polarization we sum
(cosθcosφ)^2 + (sinφ)^2 and have

dσ

dΩ

=

(

e^2

4 πmc^2

) 2

1

2

(cos^2 θcos^2 φ+ sin^2 φ)

Foratoms with more than one electron, this cross section will grow asZ^4.