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PHOTON SCATTERING   75

watching the signs carefully, we discover that terms proportional to ω canceltoo (use

 I 

I I  = 1) leaving terms proportional to  ω2. We obtain

d σ 

d = r2e  m

2ω

4

I 

Aε · x

I I ε∗ · x

A+Aε∗ · x

I I ε · x

A

EI − EA

2

(2.102)The units of the right and left hand side of Equation (2.102) are lengthsquared, so we do not need to insert powers of    or   c. Note that we mustsum over all possible intermediate states  I  of the system  A and we must dothat before we square. Note too that the matrix elements in the terms areproportional to the (transition) electric dipole moments of the system  A, aconsequence of our dipole (or long wavelength) approximation. The crosssection is proportional to  ω4 and thus inversely proportional to  λ4 due tothe cancellation of the terms proportional to  ω  discussed earlier. Had thiscancellation not happened, the cross section would have been proportionalto   ω2 and thus inversely proportional to   λ2. For atoms in colorless gasesthe energy difference   |EI − EA| corresponds to the ultraviolet (wavelengthsless than blue light) region so the approximation   ω |EI − EA|   is valid.The   λ   dependence of the elastic scattering of light was discovered byRayleigh using classical electromagnetism with wavelengths in the opticalregion. The absence of      in Equation (2.102) is the reason that Rayleighwas able to derive the formula without the use of Quantum Physics. Thestrong dependence upon wavelength is responsible for the blue sky and redsunsets. Short wavelength (blue) light gets scattered much more than longwavelength (red) light. So if you look at a part of the sky away from thesun, light must scatter off molecules in the atmosphere to reach your eyesand mostly blue light will reach them. Looking at the sunset the oppositehappens. Scattering will preferentially remove blue light leaving the red toreach your eyes.

2.4.4 Thomson Scattering

In Thomson scattering we are dealing with elastic scattering with photonenergies much larger than typical energy differences between levels in thetarget. Thus we have   ω = ω |EI − EA|. Because  EI − EA  appear in thedenominators of the second and third terms of Equation (2.94), the firstterm with  δAB  dominates and we have

d σ 

d = r2e

ελ(k) · ε∗λ (k)

2 (2.103)

Note that elastic scattering requires   |k| = |k|   but   not   λ = λ. The crosssection depends upon the angle between   ελ(k) and   ε

∗λ

(k) and not directlyupon the angle between   k   and   k (the scattering angle). This makes for