Tensors for Physics

Appendix: Exercises... 409

where the prime indicates the differentiation with respect tor. This corresponds to

g+=−γτrg′eq,

see (12.128).
The second order contribution is given by

δg(^2 )=−γτy

∂

∂x

δg(^1 )=(γ τ )^2

[

x^2 y^2 r−^1 (r−^1 g′eq)′+y^2 r−^1 g′eq

]

.

With the help of (9.6),

x^2 y^2 =exμexνe

y

λe

y

κrμrνrλrκ+

1

7

r^2 (x^2 +y^2 )−

1

35

r^4

is obtained, where the first term involves irreducible tensors of rank 4. The resulting
second rank contributions proportional tox^2 −y^2 andz^2 −r^2 /3, cf. (12.128), are

g−=−(γ τ )^2 rgeq′,

g 0 =−(γ τ )^2

[

1

2

rg′eq+

1

7

r^3 (r−^1 g′eq)′

]

=−(γ τ )^2

1

7

[

5

2

rg′eq+r^2 g′′eq

]

,

and one hasg−=γτg+.

12.5 Compute the Vector and Tensor Polarization for a= 1 State(p.237)
Hint: use the wave function(12.141)with eμ=exμand eμ=(exμ+ie
y
μ)/

√

2 for
the linear and circular polarized cases. For the angular momentum operator and its
properties see Sect.7.6.2.Furthermore, notice, the term polarization is used here with
two distinct, although related meanings. In connection with electric field, “polariza-
tion” indicates the direction of the field. In connection with angular momenta and
spins, this term refers to their average orientation.

Application of the angular momentum operatorLμ=−iLμon the wave function

Φ 1 =eνφνwithφν=

√

3 ̂rνyields

LμΦ 1 =−iεμκτ̂rκ

∂

∂̂rτ

φλeλ=−iεμκλφκeλ.

Multiplication byΦ 1 ∗and subsequent integration overd^2 ̂rleads to

〈Lμ〉=−iεμκλe∗κeλ.

Clearly, one has〈Lμ〉=0 whene∗κ=eκ, as in the case of linear polarization. On the

otherhand,forthecircularpolarization,theunitvectorisgivenbyeν=(exλ+ie
y
λ)/

√

2

ande∗κ=(exκ−ieκy)/

√

2. Thus one has〈Lμ〉=−iεμκλ(−ieyκexλ+iexκeyλ)/ 2 =
   εμκλexκeyλ=ezμ.