Tensors for Physics

11.3 Generalized Cross Product,-Tensors 189

11.3.3 Action of the Differential Operator

13.3 Irreducible Spin Tensors

ApplicationofthedifferentialoperatorLλ=ελνκrν∇κ,cf.(7.80),onthecomponents
of the position vectorryields

Lλrμ=εμλνrν.

Likewise, the action ofLon the symmetric traceless tensors of rank, constructed
from the components ofr, can be expressed with the help of the-tensors:

Lλrμ 1 rμ 2 ···rμ =()μ 1 μ 2 ···μ,λ,μ′

1 μ′ 2 ···μ′

rμ′ 1 rμ′ 2 ···rμ′. (11.29)

Application of the second derivativeLλLλon the-th rank tensor yields

LλLλrμ 1 rμ 2 ···rμ =^2 ()μ 1 μ 2 ···μ,λ,ν 1 ν 2 ···ν()ν 1 ν 2 ···ν,λ,μ′

1 μ

′

2 ···μ

′



rμ′ 1 rμ′ 2 ···rμ′.

(11.30)

Use of (11.26) leads to

LλLλrμ 1 rμ 2 ···rμ =−(+ 1 )rμ 1 rμ 2 ···rμ. (11.31)

Thus the tensors rμ 1 rμ 2 ···rμ are eigenfunctions of the operatorL^2 with the
eigenvalues−(+ 1 ). The same applies to the corresponding tensors constructed
fromthecomponentsoftheunitvector̂r,aswellastothemultipoletensorsXμ 1 μ 2 ···μ,
cf. Sect.10.1, since the differential operatorLacts just on the angular part ofr,but
not on its magnitude.
The result (11.31) can be derived via an alternative route. The irreducible tensors
rμ 1 rμ 2 ···rμ are solutions of the Laplace equation, cf. Sect.10.2. On the other
hand, the Laplace operatorΔcan be split into its radial partΔrand the angular part
involvingL·L,cf.(7.90) with (7.91). Thus one has

(Δr+r−^2 LλLλ)rμ 1 rμ 2 ···rμ = 0.

Due to

Δr=r−^2

∂

∂r

(

r^2

∂

∂r

)

,

andrμ 1 rμ 2 ···rμ ∼r, the application of the radial part of the Laplace operator
on the tensor yields