13.6 Rotational Angular Momentum of Linear Molecules, Tensor Operators 253
are used. Thus the diagonal part(umuuν)diagofuμuνis given by
(uμuν)diag=−
1
2
(
J^2 −
3
4
)− 1
JμJν. (13.56)
The anisotropic part of the molecular polarizability tensorαμνis proportional to
uμuν, cf. Sect.5.3.3. The average of the diagonal part of this tensor is closely
related to thebirefringenceand thedepolarized Rayleigh light scatteringin gases of
rotating molecules, [22].
13.6.4 Diagonal Density Operator, Averages
Thepartofthedensityoperatorofagasofrotatinglinearmolecules,whichisdiagonal
with respect to the rotational quantum numbers, can be considered as a distribution
functionρ =ρ(J)depending on the operatorJ. The average〈Ψ〉of a function
Ψ=Ψ(J)is computed according to
〈Ψ〉=Tr{Ψ(J)ρ(J)}=
∑
j
tr{PjΨ(J)ρ(J)}, (13.57)
where it is understood thatρis normalized, viz. Tr{ρ(J)}=1.
In thermal equilibrium, and in the absence of orienting fields, the molecules
of a gas have a random orientation of their rotational angular momentaJ.The
square of the angular momentumJ^2 is distributed with the canonical weight factor
exp[−H/kBT]. The Hamiltonian of a linear rotator with the moment of inertiaθis
H=
1
2 θ
^2 J^2.
The equilibrium distribution operator is
ρeq=Z−^1 exp[−H/kBT]=Z−^1 exp
[
−
^2 J^2
2 θkBT
]
, (13.58)
with the ‘state sum’Zgiven by
Z=Tr exp[−H/kBT]=
∑
j
( 2 j+ 1 )exp
[
−
^2 j(j+ 1 )
2 θkBT