The Tensor Spherical Harmonics

Home , Spherical basis, Spherical harmonics

Physics 214 Winter 2017

1 The Clebsch-Gordon coeﬃcients

Consider a system with orbital angular momentum L~ and spin angular momentum S~ . The total angular momentum of the system is denoted by J~ = L~ + S~ . Clebsch Gordon coeﬃcients allow us to express the total angular momentum basis j m ; ℓs in terms of the direct product | i basis, ℓ m ; s m ℓ m s m , | ℓ si≡| ℓi⊗| si ℓ s j m ; ℓs = ℓ m ; s m j m ; ℓs ℓ m ; s m . (1) | i h ℓ s | i| ℓ si =− m =−s mXℓ ℓ Xs The Clebsch-Gordon coeﬃcient is often denoted by (cf. pp. 412–415 of Ref. [1]):

ℓ m ; s m j m ℓ m ; s m j m ; ℓs , h ℓ s | i≡h ℓ s | i since including ℓs in j m ; ℓs on the right hand side above is redundant information. | i One important property of the Clebsch-Gordon coeﬃcients is

ℓ m ; s m j m = δ + ℓ m ; s m j m + m , (2) h ℓ s | i m , mℓ ms h ℓ s | ℓ si which implies that if m = m + m then the corresponding Clebsch-Gordon coefficient must 6 ℓ s vanish. This is simply a consequence of Jz = Lz + Sz. Likewise, ℓ s j ℓ + s (where 2j, ℓ and 2s are non-negative integers), otherwise the corresponding| Clebsc− | ≤h-Gordon≤ coefficients vanish. Recall that in the coordinate representation, the angular moment operator is a differential operator given by L~ = i~ x~ × ∇~ . − ~ 2 The spherical harmonics, Yℓmℓ (θ,φ) are simultaneous eigenstates of L and Lz, ~ 2 ~2 ~ L Yℓmℓ (θ,φ)= ℓ(ℓ + 1) Yℓmℓ (θ,φ) , Lz Yℓmℓ (θ,φ)= mℓ Yℓmℓ (θ,φ) .

We can generalize these results to systems with non-zero spin. First, we deﬁne χs ms to be the 2 simultaneous eigenstates of S~ and Sz , ~ 2 ~2 ~ S χs ms = s(s + 1)χs ms , Szχs ms = msχs ms .

The direct product basis in the coordinate representation is given by Yℓmℓ (θ,φ)χs ms .

1 2 Deﬁnition of the tensor spherical harmonics

In the coordinate representation, the total angular momentum basis consists of simultaneous 2 2 2 eigenstates of J~ , Jz , L~ , S~ . These are the tensor spherical harmonics, which satisfy, J~ 2 ℓs (θ,φ)= ~2j(j + 1) ℓs (θ,φ) , J ℓs (θ,φ)= ~m ℓs (θ,φ) , Yjm Yjm z Yjm Yjm L~ 2 ℓs (θ,φ)= ~2ℓ(ℓ + 1) ℓs (θ,φ) , S~ 2 ℓs (θ,φ)= ~2s(s + 1) ℓs (θ,φ) . Yjm Yjm Yjm Yjm As a consequence of eq. (1), the tensor spherical harmonics are deﬁned by

ℓ s ℓs (θ,φ)= ℓ m ; s m j m Y (θ,φ)χ Yjm h ℓ s | i ℓmℓ s ms =− m =−s mXℓ ℓ Xs s = ℓ , m m ; s m j m Y − (θ,φ)χ , (3) h − s s | i ℓ , m ms s ms m =−s Xs where the second line follows from the first line above since the Clebsch-Gordon coefficient above vanishes unless m = mℓ + ms. The general expressions for the Clebsch-Gordon coefficients in terms of j, mℓ, ℓ, s and ms are very complicated to write down. Nevertheless, the explicit expressions in the simplest cases of s =1/2 and s = 1 are manageable. Thus, we shall exhibit these two special cases below.

3 The spinor spherical harmonics

1 1 For spin s = 1/2, the possible values of j are j = ℓ + 2 and ℓ 2 , for ℓ = 1, 2, 3,... . If ℓ = 0 1 − then only j = 2 is possible (and the last row of Table 1 should be omitted). The corresponding table of Clebsch-Gordon coeﬃcients is exhibited in Table 1. Table 1: the Clebsch-Gordon coeﬃcients, ℓ m m ; 1 m jm . h − s 2 s | i j m = 1 m = 1 s 2 s − 2 1 2 1 2 ℓ + m + 1 / ℓ m + 1 / ℓ + 1 2 − 2 2 2ℓ +1 2ℓ +1 1 2 1 2 ℓ m + 1 / ℓ + m + 1 / ℓ 1 − 2 2 − 2 − 2ℓ +1 2ℓ +1

Comparing with eq. (3), the entries in Table 1 are equivalent to the following result: 1 j = ℓ 1 m = ℓ + 1 m ℓ m 1 ; 1 1 + ℓ + 1 m ℓ m + 1 ; 1 1 . ± 2 √2ℓ +1 ± 2 ± − 2 2 2 2 ∓ 2 2 − 2 E q E q E

2 1 1 1 1 1 0 We can represent 2 2 = ( 0 ) and 2 2 = ( 1 ). Then in the coordinate representation, the spin spherical harmonics| i are given| by− i

1 − 1 1 ℓ m + 2 Yℓ,m 2 (θ,φ) ℓ 2 1 1 ± ± ± 1 (θ,φ) θφ j = ℓ , m = . (4) j=ℓ 2 , m 2  q  Y ≡h | ± i √2ℓ +1 1 ℓ m + Yℓ,m+ 1 (θ,φ)  ∓ 2 2   q  If ℓ = 0, there is only one spin spherical harmonic,

1 − 1 1 2 + m Y0,m 2 (θ,φ) 0 2 1 1 1 (θ,φ) θφ j = , m = . (5) j= 2 , m 2 q  Y ≡h | i √2ℓ +1 1 m Y0,m+ 1 (θ,φ)  2 − 2  q  1 1 Note that when m = 2 the lower component of eq. (5) vanishes and when m = 2 the upper component of eq. (5) vanishes. In both cases, the non-vanishing component is proportional− to Y00(θ,φ)=1/√4π.

4 The vector spherical harmonics

For spin s = 1, the possible values of j are j = ℓ +1 ,ℓ,ℓ 1 for ℓ =1, 2, 3,.... If ℓ = 0 then − only j = 1 is possible (and the last two rows exhibited in Table 2 should be omitted). The corresponding table of Clebsch-Gordon coeﬃcients is exhibited in Table 2.

Table 2: the Clebsch-Gordon coeﬃcients, ℓ m m ; 1 m jm . h − s s | i j m =1 m =0 m = 1 s s s −

1 2 1 2 1 2 (ℓ + m)(ℓ + m + 1) / (ℓ m + 1)(ℓ + m + 1) / (ℓ m)(ℓ m + 1) / ℓ +1 − − − (2ℓ + 1)(2ℓ + 2) (ℓ + 1)(2ℓ + 1) (2ℓ + 1)(2ℓ + 2) 1 2 1 2 (ℓ m + 1)(ℓ + m) / m (ℓ m)(ℓ + m + 1) / ℓ − − − 2ℓ(ℓ + 1) 2ℓ(ℓ + 1) ℓ(ℓ + 1)

1 2 p 1 2 1 2 (ℓ m)(ℓ m + 1) / (ℓ m)(ℓ + m) / (ℓ + m)(ℓ + m + 1) / ℓ 1 − − − − 2ℓ(2ℓ + 1) − ℓ(2ℓ + 1) 2ℓ(2ℓ + 1)

3 1 0 0 Using a spherical basis, we can represent 11 = 0 , 10 = 1 and 1 1 = 0 . | i 0 | i 0 | − i 1 With respect to this basis, we can explicitly write out the three vector spherical harmonics , ℓ 1 ℓ 1 ± (θ,φ) and (θ,φ). For example, if ℓ = 0 then, Yj=ℓ 1 , m Yj=ℓ , m 6 1 2 (ℓ m + 1)(ℓ + m) / − Y −1(θ,φ) − 2ℓ(ℓ + 1) ℓ,m   m ℓ 1 (θ,φ)=  Yℓm(θ,φ)  . Yj=ℓ , m  ℓ(ℓ + 1)     1/2   (ℓ + mp+ 1)(ℓ m)   − Y −1(θ,φ)  2ℓ(ℓ + 1) ℓ,m    The other two vector spherical harmonics can be written out in a similar fashion. If ℓ = 0 then ℓ 1 (θ,φ) is the only surviving vector spherical harmonic. Yj=ℓ+1 , m It is instructive to work in a Cartesian basis, where the χ1 are eigenvectors of 3, and ,ms S the spin-1 spin matrices are given by ~S~, where ( ) = iǫ . In particular, Sk ij − ijk 0 i 0 − 3 = i 0 0 .   S 0 0 0 and 3χ1 = m χ1 . This yields the orthonormal eigenvectors, S ,ms s ,ms 1 0 1 ∓ χ1 ±1 = i , χ1 0 = 0 . (6) , √ −  ,   2 0 1 where the arbitrary overall phase factors are conventionally chosen to be unity. As an example, in the Cartesian basis, 1/2 1/2 [(ℓ m + 1)(ℓ + m)] Y −1(θ,φ) + [(ℓ + m + 1)(ℓ m)] Y +1(θ,φ) − ℓ,m − ℓ,m 1 ℓ 1  1/2 1/2  j=ℓ , m(θ,φ)= i [(ℓ m + 1)(ℓ + m)] Y −1(θ,φ) i [(ℓ + m + 1)(ℓ m)] Y +1(θ,φ) . Y 2 ℓ(ℓ + 1) − ℓ,m − − ℓ,m    2mY (θ,φ)  p  ℓm   (7)  This is a vector with respect to the basis xˆ , yˆ , zˆ . It is convenient to rewrite eq. (7) in terms of the basis rˆ, θˆ, φˆ using { } xˆ = rˆsin θ cos φ + θˆcos θ cos φ φˆ sin φ , − yˆ = rˆsin θ sin φ + θˆcos θ sin φ + φˆ cos φ , zˆ = rˆcos θ θˆsin θ . − ℓ 1 We can then greatly simplify the resulting expression for j=ℓ , m(θ,φ) by employing the recursion relation, Y

1/2 −iφ 2m cos θY (θ,φ) = sin θ [(ℓ + m + 1)(ℓ m)] e Y +1(θ,φ) − ℓm − ℓ,m

1/2 iφ + [(ℓ m + 1)(ℓ + m)] e Y −1(θ,φ) , − ℓ,m 4 and the following two diﬀerential relations, ∂ Y (θ,φ)= imY (θ,φ) , ∂φ ℓm ℓm

∂ 1 1/2 −iφ Y (θ,φ)= [(ℓ + m + 1)(ℓ m)] e Y +1(θ,φ) ∂θ ℓm 2 − ℓ,m 1 1/2 iφ [(ℓ m + 1)(ℓ + m)] e Y −1(θ,φ) . − 2 − ℓ,m Following a straightforward but tedious computation, the end result is: ˆ ℓ 1 i θ ∂ ˆ ∂ j=ℓ , m(θ,φ)= Yℓm(θ,φ) φ Yℓm(θ,φ) . Y ℓ(ℓ + 1) "sin θ ∂φ − ∂θ # p At this point, one should recognize the diﬀerential operator L~ expressed in the rˆ, θˆ, φˆ basis,

θˆ ∂ ∂ L~ = i~ x~ × ∇~ = i~ φˆ . − "sin θ ∂φ − ∂θ # Hence, we end up with

ℓ 1 1 = (θ,φ)= L~ Yℓm(θ,φ) , for ℓ =0 . (8) Yj ℓ , m ~2ℓ(ℓ + 1) 6 This is the vector spherical harmonic,p

i ~ X~ ℓm(θ,φ)= − x~ × ∇ Yℓm(θ,φ) , ℓ(ℓ + 1) employed by J.D. Jackson, Classical Electrodynamicsp , 3rd Edition (John Wiley & Sons, Inc., New York, 1999). Using the same methods, one can derive the following expressions for the other two vector spherical harmonics,

ℓ 1 1 ~ = −1 (θ,φ)= − (j + 1)nˆ r∇ Yjm(θ,φ) , for ℓ =0 , (9) Yj ℓ , m (j + 1)(2j + 1) − 6 h i p ℓ 1 1 ~ = +1 (θ,φ)= jnˆ + r∇ Yjm(θ,φ) , (10) Yj ℓ , m j(2j + 1) h i where x~ = rn~ and nˆ rˆ.p That is, the three independent normalized vector spherical harmonics ≡ can be chosen as:

ir ~ r ~ − nˆ × ∇ Yjm(θ,φ) , ∇ Yjm(θ,φ) , nˆ Yjm(θ,φ) . (11) j(j + 1) j(j + 1) It is often convenientp to rewrite p

r ∇~ Y (θ,φ)= r[nˆ(nˆ · ∇~ ) ∇~ ]Y (θ,φ)= r nˆ × (nˆ × ∇~ )Y (θ,φ) , jm − − jm − jm 5 after noting that ∂Y (θ,φ) nˆ · ∇~ Y (θ,φ)= jm =0 . jm ∂r Then, the list of the three independent normalized vector spherical harmonics takes the following form:

ir ~ r ~ − nˆ × ∇ Yjm(θ,φ) , − nˆ × (nˆ × ∇)Yjm(θ,φ) , nˆ Yjm(θ,φ) . (12) j(j + 1) j(j + 1) In particular,p the first two vector sphericalp harmonics listed in eq. (12) are transverse (i.e., perpendicular to nˆ), whereas the third vector spherical harmonic in eq. (12) is longitudinal (i.e., parallel to nˆ). This is convenient for the multipole expansion of the transverse electric and magnetic radiation fields, where only the first two vector spherical harmonics of eq. (11) appear. However, the second and third vector spherical harmonics listed in eq. (11) [or eq. (12)], ~ 2 r∇ Yjm(θ,φ) and nˆ Yjm(θ,φ), are not eigenstates of L~ since they consist of linear combinations of states with ℓ = j 1 [which can be explicitly derived by inverting eqs. (9) and (10)]. The algebraic steps± involved in establishing eqs. (8)–(10) are straightforward but tedious. A more streamlined approach to the derivation of these results is given in the next section.

5 The vector spherical harmonics revisited

Since Y (nˆ) is a spherical tensor of rank-ℓ, and nˆ x~ /r, L~ i~x~ × ∇~ and r∇~ are vector ℓm ≡ ≡ − operators, it is not surprising that the vector spherical harmonics are linear combinations of the quantities given in eq. (11). It is instructive to derive this result directly. For convenience, we denote the vector spherical harmonics in this section by

Y~ (nˆ) ℓ 1 (θ,φ) , for j = ℓ +1 ,ℓ,ℓ 1 , (13) jℓm ≡Yj m − where nˆ is a unit vector with polar angle θ and azimuthal angle φ. First, we recall that (see, e.g., eq. (12.5.20) of Ref. [1]):

1/2 L± ℓ m = ~ [(ℓ m)(ℓ m + 1)] j m 1 , L ℓ m = ~m ℓ m , (14) | i ∓ ± | ± i z| i | i where L± L iL . The spherical components of L~ are L (q = +1, 0, 1) where ≡ x ± y q − L± 1 L±1 = (Lx iLy) , L0 Lz . ≡ ±√2 √2 ± ≡ Using the Clebsch-Gordon coeﬃcients given in Table 2, it follows that

L ℓ m = ~( 1)q ℓ(ℓ + 1) ℓ , m + q ; 1 , q ℓ m ℓ , m + q . (15) q| i − h − | i| i In the coordinate representation,p eq. (15) is equivalent to

q L Y (nˆ)= ~( 1) ℓ(ℓ + 1) ℓ , m + q ; 1 , q ℓ m Y + (nˆ) . (16) q ℓm − h − | i ℓ , m q p 6 It is convenient to introduce a set of spherical basis vectors, 1 eˆ±1 (xˆ iyˆ) , eˆ0 zˆ. (17) ≡ ∓√2 ± ≡

It is not surprising that eˆq = χ1,q [cf. eq. (6)]. One can check that

q L~ = L xˆ + L yˆ + L zˆ = ( 1) L eˆ− , (18) x y z − q q q X where the sum over q runs over q = 1, 0, +1.1 Hence, eqs. (16) and (18) yield − L~ Y (nˆ)= ~ ℓ(ℓ + 1) eˆ− ℓ , m + q ; 1 , q ℓ m Y + (nˆ) . ℓm qh − | i ℓ , m q q p X Since the sum is taken over q = 1, 0, 1, we are free to relabel q q. Writing eˆ = χ1 , we − → − q ,q end up with

L~ Y (nˆ)= ~ ℓ(ℓ + 1) ℓ , m q ; 1 , q ℓ m Y − (nˆ)χ1 . ℓm h − | i ℓ , m q q q p X Comparing with eq. (3) for s = 1, it follows that [in the notation of eq. (13)]:

L~ Yℓm(nˆ)= ~ ℓ(ℓ + 1) Y~ ℓℓm(nˆ) (19) in agreement with eq. (8). p Next, we examine nˆYℓm(nˆ). It is convenient to expand nˆ x~ /r in a spherical basis. Using eq. (17), the following expression is an identity, ≡

4π q nˆ = xˆ sin θ cos φ + yˆ sin θ sin φ + zˆcos θ = ( 1) Y1 (nˆ) eˆ− . (20) 3 − q q r q X Hence,

4π q nˆY (nˆ)= ( 1) Y1 (nˆ)Y (nˆ) eˆ− . (21) ℓm 3 − q ℓm q r q X Using eq. (40) given in the Appendix, it follows that

3(2ℓ + 1) 1 ′ ′ Y1q(nˆ)Yℓm(nˆ)= ℓ m ; 1 q ℓ , m + q ℓ 0;10 ℓ 0 Yℓ ′ , m+q(nˆ) , 4π √2ℓ ′ +1 h | ih | i r ℓ ′ X (22) Only two terms, corresponding to ℓ ′ = ℓ 1, can contribute to the sum over ℓ ′ since [cf. Table 2]: ± 1 2 ℓ +1 / , for ℓ ′ = ℓ +1 , 2ℓ +1  ′  0 , for ℓ ′ = ℓ 1 , ℓ 0;10 ℓ 0 =  (23) h | i  6 ±  1 2 ℓ / , for ℓ ′ = ℓ 1 . − 2ℓ +1 −  1  Henceforth, if left unspeciﬁed, sums overq will run over q = 1, 0, +1. − 7 Inserting eq. (22) on the right hand side of eq. (21) and employing eq. (23) then yields

1/2 q ℓ +1 nˆY (nˆ)= ( 1) eˆ− ℓ m ; 1 q ℓ +1 , m + q Y +1 + (nˆ) ℓm − q 2ℓ +3 h | i ℓ , m q q X 1 2 ℓ / ℓ m ; 1 q ℓ +1 , m + q Y −1 + (nˆ) . (24) − 2ℓ 1 h | i ℓ , m q − It is convenient to rewrite eq. (24) with the help of the following two relations, which can be obtained from Table 2,

1 2 2ℓ +3 / ℓ m ; 1 q ℓ +1 , m + q = ( 1)q ℓ +1 , m + q ; 1 , q ℓ m , (25) h | i − − 2ℓ +1 h − | i 1 2 2ℓ 1 / ℓ m ; 1 q ℓ 1 , m + q = ( 1)q − ℓ 1 , m + q ; 1 , q ℓ m . (26) h | − i − − 2ℓ +1 h − − | i The end result is:

1 2 ℓ +1 / nˆY (nˆ)= eˆ− ℓ +1 , m + q ; 1 , q ℓ m Y +1 + (nˆ) ℓm − q 2ℓ +1 h − | i ℓ , m q q X 1 2 ℓ / ℓ 1 , m + q ; 1 , q ℓ m Y −1 + (nˆ) . (27) − 2ℓ +1 h − − | i ℓ , m q

Using eq. (3) with s = 1 and χ1 q = eˆq and employing the notation of eq. (13), it follows that Y~ ±1 (nˆ)= eˆ− ℓ 1 , m + q ; 1 , q ℓ m Y ±1 + (nˆ) , (28) ℓ,ℓ , m q h ± − | i ℓ , m q q X after relabeling the summation index by q q. Hence, eq. (27) yields → − 1 2 1 2 ℓ +1 / ℓ / nˆY (nˆ)= Y~ +1 (nˆ)+ Y~ −1 (nˆ) (29) ℓm − 2ℓ +1 ℓ,ℓ , m 2ℓ +1 ℓ,ℓ , m ~ Finally, we examine r∇Yℓm(nˆ). First, we introduce the gradient operator in a spherical basis, =( +1 , 0 , −1), where ∇q ∇ ∇ ∇ 1 ∂ ∂ e±iφ ∂ cos θ ∂ i ∂ +1 = i = sin θ + , (30) ∇ ∓√2 ∂x ± ∂y ∓ √2 ∂r r ∂θ ± r sin θ ∂φ ∂ ∂ sin θ ∂ 0 = = cos θ . (31) ∇ ∂z ∂r − r ∂θ We can introduce a formal operator on the Hilbert space by deﬁning the coordinate space ∇q representation, x~ ℓ m = Y (nˆ) . h |∇q| i ∇q ℓm 8 Note that q is a vector operator. We shall employ the Wigner-Eckart theorem (see, e.g., pp. 240–241∇ of Ref. [2]), which states that

ℓ′ m′ ℓ m = ℓ m ; 1 q ℓ′ m′ ℓ′ ℓ , (32) h |∇q| i h | ih k∇k i where the reduced matrix element ℓ ℓ′ is independent of q, m and m′. To evaluate the h k∇k i ′ reduced matrix element, we consider the case of q = m = m = 0. Then,

′ ′ ′ ℓ 0 0 ℓ 0 = ℓ 0;10 ℓ 0 ℓ ℓ . h |∇ | i h | ih k∇k i Thus, ′ ′ ℓ 0 0 ℓ 0 ℓ ℓ = h |∇ | i . h k∇k i ℓ 0;10 ℓ′ 0 h | i Inserting this result into eq. (32) yields

′ ′ ′ ′ ℓ m ; 1 q ℓ m ′ ℓ m ℓ m = h | i ℓ 0 0 ℓ 0 . (33) h |∇q| i ℓ 0;10 ℓ′ 0 h |∇ | i h | i ′ We can evaluate ℓ 0 0 ℓ 0 explicitly in the coordinate representation using eq. (31), h |∇ | i

′ 1 ∗ ∂ ℓ 0 0 ℓ 0 = dΩ Y ′ (nˆ) sin θ Y 0(nˆ) . h |∇ | i −r ℓ 0 ∂θ ℓ Z 1/2 Using Y 0(nˆ) = [(2ℓ + 1)/(4π)] P (cos θ), and substituting x cos θ, ℓ ℓ ≡ ′ 1 ′ (2ℓ + 1)(2ℓ + 1) 2 ′ ′ ℓ 0 0 ℓ 0 = (1 x )Pℓ (x)Pℓ(x) dx , (34) h |∇ | i 2r − − p Z 1 ′ where Pℓ(x)= dPℓ(x)/dx. To evaluate eq. (34), we employ the recurrence relation,

2 ′ (1 x )P (x)= ℓP −1(x) ℓxP (x) , − ℓ ℓ − ℓ and the orthogonality relation of the Legendre polynomials,

1 2 Pℓ(x)Pℓ′ (x) dx = δℓℓ′ . − 2ℓ +1 Z 1 It follows that

′ 1 ′ (2ℓ + 1)(2ℓ + 1) 2ℓ ℓ 0 0 ℓ 0 = δℓ′,ℓ−1 ℓ xPℓ(x)Pℓ′ (x) dx . (35) h |∇ | i 2r 2ℓ 1 − −1 p − Z

To evaluate the remaining integral, we use x = P1(x) and the result of eq. (43) obtained in the Appendix to write:

1 1 2 ′ 2 ′ ′ xPℓ(x)Pℓ (x) dx = P1(x)Pℓ(x)Pℓ (x) dx = ′ 1 0 ; ℓ 0 ℓ 0 . − − 2ℓ +1h | i Z 1 Z 1

9 Using eq. (23), the above integral is equal to

1 2(ℓ + 1) 2ℓ xPℓ(x)Pℓ′ (x) dx = δℓ′,ℓ+1 + δℓ′,ℓ−1 . −1 (2ℓ + 1)(2ℓ + 3) (2ℓ 1)(2ℓ + 1) Z − Inserting this result back into eq. (35) yields

′ ′ (2ℓ + 1)(2ℓ + 1) 2ℓ(ℓ + 1) 2ℓ(ℓ + 1) ℓ 0 0 ℓ 0 = δ ′ −1 δ ′ +1 h |∇ | i 2r (2ℓ 1)(2ℓ + 1) ℓ ,ℓ − (2ℓ + 1)(2ℓ + 3) ℓ ,ℓ p ( − ) ℓ(ℓ + 1) 1 1 = δℓ′,ℓ−1 δℓ′,ℓ+1 . r√2ℓ +1 √2ℓ 1 − √2ℓ +3 − Using eq. (33), it follows that:

′ ′ ′ ′ ℓ m ; 1 q ℓ m ℓ(ℓ + 1) 1 1 ℓ m q ℓ m = h | i δℓ′,ℓ−1 δℓ′,ℓ+1 h |∇ | i ℓ 0;10 ℓ′ 0 r√2ℓ +1 √2ℓ 1 − √2ℓ +3 h | i −

1 ′ ′ ℓ ℓ +1 = ℓ m ; 1 q ℓ m (ℓ + 1) δ ′ −1 + ℓ δ ′ +1 , (36) −r h | i 2ℓ 1 ℓ ,ℓ 2ℓ +3 ℓ ,ℓ " r − r # after using eq. (23) to evaluate ℓ 0;10 ℓ′ 0 . h ~ | i We are now ready to evaluate r∇Yℓm(nˆ). First, we insert a complete set of states to obtain

′ ′ ′ ′ q ℓ m = ℓ m ℓ m q ℓ m ∇ | i ′ ′ | ih |∇ | i ℓX,m

1 ′ ′ ′ ′ ℓ ℓ +1 = ℓ m ℓ m ; 1 q ℓ m (ℓ + 1) δ ′ −1 + ℓ δ ′ +1 . −r | i h | i 2ℓ 1 ℓ ,ℓ 2ℓ +3 ℓ ,ℓ ℓ′,m′ ( " r r #) X − (37)

Note that in the sum over m′, only the terms corresponding to m′ = m + q survive, due to the presence of the Clebsch-Gordon coeﬃcient ℓ m ; 1 q ℓ′ m′ . Likewise, in the sum over ℓ′, only the terms corresponding to ℓ′ = ℓ 1 survive.h In the| coordinatei representation, eq. (37) is equivalent to ±

1 ′ ℓ ℓ +1 Y (nˆ)= Y ′ + (nˆ) ℓ m ; 1 q ℓ , m+q (ℓ + 1) δ ′ −1 + ℓ δ ′ +1 . ∇q ℓm −r ℓ ,m q h | i 2ℓ 1 ℓ ,ℓ 2ℓ +3 ℓ ,ℓ ℓ′ ( " r r #) X − In analogy with eq. (18), we have

q ∇~ = ( 1) eˆ− . − q∇q q X

10 Hence, it follows that

q ℓ r∇~ Y (nˆ)= ( 1) eˆ− (ℓ + 1) ℓ m ; 1 q ℓ 1 , m + q Y −1 + (nˆ) − ℓm − q 2ℓ 1 h | − i ℓ ,m q q ( r X − ℓ +1 +ℓ ℓ m ; 1 q ℓ +1 , m + q Yℓ+1,m+q(nˆ) . r2ℓ +3 h | i ) It is convenient to employ eqs. (25) and (26) and rewrite the above result as

ℓ r∇~ Y (nˆ)= eˆ− (ℓ + 1) ℓ 1 , m + q ; 1 , q ℓ m Y −1 + (nˆ) ℓm q 2ℓ +1 h − − | i ℓ ,m q q ( r X ℓ +1 +ℓ ℓ +1 , m + q ; 1 , q ℓ m Yℓ+1,m+q(nˆ) . r2ℓ +1 h − | i ) Finally, using eq. (28), we end up with

~ ℓ ℓ +1 r∇Yℓm(nˆ)=(ℓ + 1) Y~ ℓ,ℓ−1,m(nˆ)+ ℓ Y~ ℓ,ℓ+1,m(nˆ) (38) r2ℓ +1 r2ℓ +1 which is known in the literature as the gradient formula. We can now use eqs. (29) and (38) to solve for Y~ ℓ,ℓ+1,m(nˆ) and Y~ ℓ,ℓ−1,m(nˆ) in terms of ~ nˆYℓm(nˆ) and r∇Yℓm(nˆ). Since these are linear equations, they are easily inverted, and we ﬁnd

1 ~ Y~ ℓ,ℓ+1,m(nˆ)= (ℓ + 1)nˆ + r∇ Yℓm(nˆ) , for ℓ =0, 1, 2, 3,..., (ℓ + 1)(2ℓ + 1) − h i p 1 ~ Y~ ℓ,ℓ−1,m(nˆ)= ℓnˆ + r∇ Yℓm(nˆ) , for ℓ =1, 2, 3,..., ℓ(2ℓ + 1) h i which are equivalentp to the results of eqs. (9) and (10) previously obtained. In addition, we also have eq. (19), which we can rewrite as

ir ~ Y~ ℓ,ℓ,m = − nˆ × ∇Yℓm(nˆ) , for ℓ =1, 2, 3, . . ., . ℓ(ℓ + 1)

Thus, we have identified thep three linearly independent vector spherical harmonics in terms of differential vector operators acting on Yℓm(nˆ). For the special case of ℓ = 0, only one vector spherical harmonic, Y~ 010(nˆ)=( nˆ + r∇~ )Y (nˆ), survives. − ℓm In books, one often encounters the vector spherical harmonic defined by nˆ × L~ Yℓm(nˆ). However, this is not independent of the vector spherical harmonics obtained above, since

∂ nˆ × L~ Y (nˆ)= i~r nˆ × (nˆ × ∇~ )Y (nˆ)= i~r nˆ ∇~ Y (nˆ)= i~r ∇~ Y (nˆ) . ℓm − ℓm − ∂r − ℓm ℓm 11 An alternative method for deriving the gradient formula [obtained in eq. (38)] is to evaluate nˆ × L~ Yℓm(nˆ) using the same technique employed in the computation of nˆYℓm(nˆ) given in this section. However, this calculation is much more involved and involves a product of four Clebsch- Gordon coeﬃcients. A certain sum involving a product of three Clebsch-Gordon coeﬃcients needs to be performed in closed form. This summation can be done (e.g., see Ref. [11] for the gory details), but the computation is much more involved than the simple analysis presented in this section based on the Wigner-Eckart theorem.

Appendix: An integral of a product of three spherical harmonics

In this Appendix, we state a number of results that are derived on pp. 231 of Ref. [2]. The product of two spherical harmonics can be obtained via an important expansion known as the Clebsch-Gordon series,

(2ℓ1 + 1)(2ℓ2 + 1) Y (nˆ) Y (nˆ)= ℓ1 m1 ; ℓ2 m2 ℓ m ℓ1 0 ; ℓ2 0 ℓ 0 Y (nˆ). (39) ℓ1m1 ℓ2m2 4π(2ℓ + 1) h | ih | i ℓm ℓ,m s X The sum over m can be performed using eq. (2). Only one term survives (corresponding to m = m1 + m2),

(2ℓ1 + 1)(2ℓ2 + 1) Y (nˆ) Y (nˆ)= ℓ1 m1 ; ℓ2 m2 ℓ m1+m2 ℓ1 0 ; ℓ2 0 ℓ 0 Y + (nˆ). ℓ1m1 ℓ2m2 4π(2ℓ + 1) h | ih | i ℓ , m1 m2 ℓ s X (40) Note that ℓ1 m1 ; ℓ2 m2 ℓ m1 + m2 = 0 unless the two conditions, ℓ1 ℓ2 ℓ ℓ1 + ℓ2 and h | i | − | ≤ ≤ m1 + m2 ℓ are both satisﬁed. This is simply a consequence of the rules for the addition of | | ≤ angular momentum in quantum mechanics. Consequently, the sum over ℓ in eq. (40) can be taken over the range of integer values that satisfy

max ℓ1 ℓ2 , m1 + m2 ℓ ℓ1 + ℓ2 . | − | ≤ ≤ ∗ If we multiply eq. (39) by Yℓ3m3 (nˆ) and then integrate over the solid angle using the orthogonality of the spherical harmonics,

∗ ′ ′ Yℓm(θ,φ) Yℓ′ m′ (θ,φ) dΩ= δℓℓ δmm , (41) Z one easily obtains the integral of a product of three spherical harmonics,

∗ (2ℓ1 + 1)(2ℓ2 + 1) Yℓ1m1 (θ,φ) Yℓ2m2 (θ,φ) Yℓ3m3 (θ,φ) dΩ= ℓ1 m1 ; ℓ2 m2 ℓ3 m3 ℓ1 0 ; ℓ2 0 ℓ3 0 . 4π(2ℓ3 + 1) h | ih | i Z s (42) 1/2 Using Yℓ0(nˆ) = [(2ℓ + 1)/(4π)] Pℓ(cos θ), the following special case of eq. (42) is then obtained, 1 2 2 Pℓ1 (x)Pℓ2 (x)Pℓ3 (x) dx = ℓ1 0 ; ℓ2 0 ℓ3 0 . (43) − 2ℓ3 +1 h | i Z 1

12 Bibliography

The basics of angular momentum theory, Clebsch-Gordon coeﬃcients, spherical tensors and the Wigner-Eckart theorem are treated in most quantum mechanics textbooks. Pedagogical treatments can be found in the following two well-known textbooks: 1. R. Shankar, Principles of Quantum Mechanics, 2nd Edition (Springer Science+Business Media, LLC, New York, NY, 1980). 2. J.J. Sakurai and Jim J. Napolitano, Modern Quantum Mechanics, 2nd edition (Addison- Wesley Publishing Company, Boston, MA, 2011).

Further details can be found in the following more specialized textbooks. In particular, Ref. 7 below is especially comprehensive in its treatment of tensor spherical harmonics. 3. M.E. Rose, Elementary Theory of Angular Momentum (John Wiley & Sons, Inc., New York, NY, 1957). 4. A.R. Edmonds, Angular Momentum in Quantum Mechanics, 2nd Edition (Princeton University Press, Princeton, NJ, 1960). 5. D.M. Brink and G.R. Sachler, Angular Momentum, 2nd Edition (Oxford University Press, Oxford, UK, 1968). 6. L.C. Biedenharn and J.D. Louck, Angular Momentum in Quantum Physics: Theory and Application (Cambridge University Press, Cambridge, UK, 1985). 7. D.A. Varshalovich, A.N. Moskalev and V.K. Khersonskii, Quantum Theory of Angular Momentum (World Scientiﬁc Publishing Co., Singapore, 1988). 8. Richard N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics (John Wiley & Sons, Inc., New York, NY, 1988). 9. William J. Thompson, Angular Momentum: An Illustrated Guide to Rotational Symme- tries for Physical Systems (John Wiley & Sons, Inc., New York, NY, 1994). 10. Masud Chaichian and Rolf Hagedorn, Symmetries in Quantum Mechanics: From An- gular Momentum to Supersymmetry (Institute of Physics Publishing, Ltd, Bristol, UK, 1998). 11. V. Devanathan, Angular Momentum Techniques in Quantum Mechanics (Kluwer Aca- demic Publishers, New York, NY, 2002).