410 Appendix: Exercises...
For circular polarized light with its electric field vector parallel toe=ex±iey,
the vector polarization is
〈L〉=±ez.Application ofLνonLμΦ 1 =−iεμκλφκeλyieldsLνLμΦ 1 =−εναβ̂rα∂∂̂rβεμκλφκ
eλ=−εναβεμβ λφαeλ=δνμφλeλ−φμeν. Now multiplication byΦ 1 ∗and subsequent
integration overd^2 ̂rleads to
〈LνLμ〉=δνμe∗λeλ−eμ∗eν.Notice thateλ∗eλ=1 and〈LνLν〉=2, in accord with(+ 1 )for=1. The
symmetric traceless part of〈LνLμ〉is the tensor polarization
〈LμLν〉=−eμ∗eν.Thus for the linear polarizationeν=exν, and for the circular polarization one finds
〈LμLν〉=−exμeνx,and
〈LμLν〉=−1
2
(
exμexν+eyμeyν)
=
1
2
ezμezν.Exercises Chapter 13
13.1 Verify the Normalization for the Spin 1 Matrices(p.241)
Compute explicitly sx^2 +s^2 y+sz^2 for the spin matrices(13.6)in order to check the
normalization relation(13.4),viz.s·s=s(s+ 1 ) 1.
Spin 1.Matrix multiplication yields
sxsx=1
2
⎛
⎝
101
020
101
⎞
⎠, sysy=^1
2⎛
⎝
10 − 1
020
−10 1
⎞
⎠, szsz=⎛
⎝
100
000
001
⎞
⎠, (A.14)
thus
sx^2 +sy^2 +sz^2 = 2⎛
⎝
100
010
001
⎞
⎠, (A.15)
in accord with (13.4), fors=1.