Field Theory exam II – Solutions

From (1) we obtain
∫
q lm =
∫
=
With (2) and e −0i = 1 we get
Y ∗
lm(θ, φ)r l ρ(r, θ, φ)d 3 x (8)
Y ∗
lm(θ, φ)r l q (δ(⃗x − ⃗x 1 ) − δ(⃗x − ⃗x 2 )) d 3 x (9)
= qYlm(θ ∗
1 , φ 1 )r1 l − qYlm(θ ∗
2 , φ 2 )r2 l (10)
= qr l (Ylm(θ ∗
1 , 0) − Ylm(θ ∗
2 , 0)) (11)
q lm = qr l N lm (P lm (cos(θ 1 )) − P lm (cos(θ 2 ))) (12)
We have
( ) 1
cos(θ 1 ) = cos
4 π = √ 1
(13)
2
Using (4) we finally arrive at
cos(θ 2 ) = cos(π − θ 1 ) = − cos(θ 1 ) (14)
q lm = qr l N lm P lm (cos(θ 1 )) ( 1 − (−1) l+m) (15)
Obviously, q lm = 0 if l + m is an even number. Otherwise,
q lm = 2qr l N lm P lm (cos(θ 1 )) (16)
(b) From (2) we get the transformation of the spherical harmonic functions
under rotation
Y lm (θ, φ) = N lm P lm (cos(θ))e imφ (17)
= N lm P lm (cos(θ))e im(φ−β) e imβ (18)
= Y lm (θ, φ − β)e imβ (19)
⇒ Y ∗
lm(θ, φ) = Y ∗
lm(θ, φ − β)e −imβ (20)
If ρ is an arbitrary charge distribution and ¯ρ the same distribution rotated
by an angle β around the z-axis, we have ¯ρ(r, θ, φ) = ρ(r, θ, φ − β). For
the multipole moments, we compute from (1)
∫
¯q lm = Ylm(θ, ∗ φ)r l ¯ρ(r, θ, φ)d 3 x (21)
∫
= Ylm(θ, ∗ φ − β)e −imβ r l ρ(r, θ, φ − β)d 3 x (22)
= e −imβ ∫
= e −imβ ∫
Y ∗
lm(θ, φ − β)r l ρ(r, θ, φ − β)d 3 x (23)
Y ∗
lm(θ, φ)r l ρ(r, θ, φ)d 3 x (24)
= e −imβ q lm (25)
where (23) and (24) are equal because the integral runs over all angles φ.
2