Tensors for Physics

160 9 Irreducible Tensors

The isomorphic Cartesian tensor has the same number of independent components,
though it is not obvious from its notation.
The expression (9.23) can also be applied to the irreducible tensoraμ 1 aμ 2 ···aμ
constructed from the components of the vectora. The resulting spherical components
a(m)are related to the-th order spherical harmonicY(m)(̂a)by

a(m)=a

√

4 π!

( 2 + 1 )!!

Y(m)(̂a). (9.24)

The scalar product, i.e. the total contraction of the two tensors in (9.10) can also be
expressed in terms of the pertaining spherical components, where the sum overm
corresponds to a scalar product, viz.

aμ 1 aμ 2 ...aμbμ 1 bμ 2 ...bμ =

∑

m=−

a(m)

(

b(m)

)∗

(9.25)

=ab

4 π!

( 2 + 1 )!!

∑

m=−

Y(m)(̂a)

(

Y(m)

)∗

(̂b).

Comparison of this equation with the right hand side of (9.10) yields the relation

P(̂a·̂b)=

4 π

( 2 + 1 )

∑

m=−

Y(m)(̂a)

(

Y(m)

)∗

(̂b), (9.26)

which expresses the Legendre polynomial with a scalar product of spherical harmon-
ics.
For ease of reference, the first few spherical harmonics, for= 0 , 1 ,2, are
listed here, where the Cartesian components ofrare denoted by{x,y,z}, and the
coefficientsc()=

√

( 2 + 1 )!!/ 4 πare used:

Y 0 (^0 )=c( 0 )= 1 /

√

4 π,

Y 1 (^0 )=c( 1 )r−^1 z=c( 1 )cosθ,

Y 1 (±^1 )=∓c( 1 )

1

√

2

r−^1 (x±iy)=∓c( 1 )

1

√

2

sinθexp(±iφ), (9.27)

Y 2 (^0 )=c( 2 )

√

3

2

r−^2

(

z^2 −

1

3

r^2

)

=c( 2 )

√

3

2

(

cos^2 θ−

1

3

)

,

Y 2 (±^1 )=∓c( 2 )

1

√

2

r−^2 z(x±iy)=∓c( 2 )

1

√

2

sinθcosθexp(±iφ),