Now we return to the specific orbital angular momentum operators Lx, Ly, and Lz of quantum mechanics introduced in Section 4.3. Equation (4.68) becomes
LzψLM(θ, ϕ)=MψLM(θ, ϕ), and we want to show that
ψLM(θ, ϕ)=YLM(θ, ϕ)
are the eigenfunctions|LMofL2andLzof Section 4.3 in spherical polar coordinates, the spherical harmonics. The explicit form ofLz= −i∂/∂ϕfrom Exercise 2.5.13 indicates that ψLM has aϕdependence of exp(iMϕ)— withMan integer to keepψLM single-valued.
And ifMis an integer, thenLis an integer also.
To determine theθ dependence ofψLM(θ, ϕ), we proceed in two main steps: (1) the determination ofψLL(θ, ϕ)and (2) the development ofψLM(θ, ϕ)in terms ofψLLwith the phase fixed byψL0. Let
ψLM(θ, ϕ)=LM(θ )eiMϕ. (12.157)
FromL+ψLL=0, Lbeing the largestM, using the form ofL+given in Exercises 2.5.14 and 12.6.7, we have
ei(L+1)ϕ d
dθ −Lcotθ
LL(θ )=0, (12.158)
and thus
ψLL(θ, ϕ)=cLsinLθ eiLϕ. (12.159) Normalizing, we obtain
c∗LcL
2π 0
π 0
sin2L+1θ dθ dϕ=1. (12.160) Theθintegral may be evaluated as a beta function (Exercise 8.4.9) and
|cL| = ,
(2L+1)!!
4π(2L)!! =
√(2L)! 2LL!
)2L+1
4π . (12.161)
This completes our first step.
To obtain theψLM, M= ±L, we return to the ladder operators. From Eqs. (4.83) and (4.84) and as shown in Exercise 12.7.2 (J+replaced byL+andJ−replaced byL−),
ψLM(θ, ϕ)= ,
(L+M)!
(2L)!(L−M)!(L−)L−MψLL(θ, ϕ),
(12.162) ψLM(θ, ϕ)=
,
(L−M)!
(2L)!(L+M)!(L+)L+MψL,−L(θ, ϕ).
Again, note that the relative phases are set by the ladder operators.L+andL−operating onLM(θ )eiMϕmay be written as
L+LM(θ )eiMϕ=ei(M+1)ϕ d
dθ −Mcotθ
LM(θ )
=ei(M+1)ϕsin1+Mθ d
d(cosθ )sin−MLM(θ ),
(12.163) L−LM(θ )eiMϕ= −ei(M−1)ϕ
d
dθ +Mcotθ
LM(θ )
=ei(M−1)ϕsin1−Mθ d
d(cosθ )sinMθ LM(θ ).
Repeating these operationsntimes yields
(L+)nLM(θ )eiMϕ=(−1)nei(M+n)ϕsinn+Mθdnsin−Mθ LM(θ ) d(cosθ )n ,
(12.164) (L−)nLM(θ )eiMϕ=ei(M−n)ϕsinn−MθdnsinMθ LM(θ )
d(cosθ )n .
From Eq. (12.163),
ψLM(θ, ϕ)=cL
,
(L+M)!
(2L)!(L−M)!eiMϕsin−Mθ dL−M
d(cosθ )L−M sin2Lθ, (12.165) and forM= −L:
ψL,−L(θ, ϕ)= cL
(2L)!e−iLϕsinLθ d2L
d(cosθ )2Lsin2Lθ
=(−1)LcLsinLθ e−iLϕ. (12.166) Note the characteristic(−1)Lphase ofψL,−Lrelative toψL,L. This(−1)Lenters from
sin2Lθ=
1−x2L
=(−1)L
x2−1L
. (12.167)
Combining Eqs. (12.163), (12.163), and (12.166), we obtain
ψLM(θ, ϕ)=(−1)LcL ,
(L−M)!
(2L)!(L+M)!(−1)L+MeiMϕsinMθdL+Msin2Lθ
d(cosθ )L+M. (12.168) Equations (12.165) and (12.168) agree if
ψL0(θ, ϕ)=cL 1
√(2L)! dL
(dcosθ )Lsin2Lθ. (12.169) Using Rodrigues’ formula, Eq. (12.65), we have
ψL0(θ, ϕ)=(−1)LcL 2LL!
√(2L)!PL(cosθ )
=(−1)L cL
|cL|
)2L+1
4π PL(cosθ ). (12.170) The last equality follows from Eq. (12.161). We now demand thatψL0(0,0)be real and positive. Therefore
cL=(−1)L|cL| =(−1)L
√(2L)! 2LL!
)2L+1
4π . (12.171)
With(−1)LcL/|cL| =1, ψL0(θ, ϕ)in Eq. (12.170) may be identified with the spherical harmonicYL0(θ, ϕ)of Section 12.6.
When we substitute the value of(−1)LcLinto Eq. (12.168), ψLM(θ, ϕ)=
√(2L)! 2LL!
)2L+1 4π
,
(L−M)!
(2L)!(L+M)!(−1)L+M
ãeiMϕsinMθ dL+M
d(cosθ )L+M sin2Lθ
=
)2L+1 4π
,
(L−M)!
(L+M)!eiMϕ(−1)M
ã 1
2LL!
1−x2M/2 dL+M dxL+M
x2−1L
, x=cosθ, M≥0.
(12.172) The expression in the curly bracket is identified as the associated Legendre function (Eq. (12.151), and we have
ψLM(θ, ϕ)=YLM(θ, ϕ)
=(−1)M ,
2L+1
4π ã(L−M)!
(L+M)!ãPLM(cosθ )eiMϕ, M≥0, (12.173) in complete agreement with Section 12.6. Then by Eq. (12.73c),YLM for negative super- script is given by
YL−M(θ, ϕ)=(−1)M
YLM(θ, ϕ) ∗. (12.174)
• Our angular momentum eigenfunctions ψLM(θ, ϕ) are identified with the spherical harmonics. The phase factor(−1)M is associated with the positive values ofMand is seen to be a consequence of the ladder operators.
• Our development of spherical harmonics here may be considered a portion of Lie alge- bra — related to group theory, Section 4.3.
Exercises
12.7.1 Using the known forms ofL+andL−(Exercises 2.5.14 and 12.6.7), show that YLM ∗L−
L+YLM
d= L+YLM∗ L+YLM
d.
12.7.2 Derive the relations (a) ψLM(θ, ϕ)=
,
(L+M)!
(2L)!(L−M)!(L−)L−MψLL(θ, ϕ), (b) ψLM(θ, ϕ)=
,
(L−M)!
(2L)!(L+M)!(L+)L+MψL,−L(θ, ϕ).
Hint. Equations (4.83) and (4.84) may be helpful.
12.7.3 Derive the multiple operator equations
(L+)nLM(θ )eiMϕ=(−1)nei(M+n)ϕsinn+Mθdnsin−Mθ LM(θ ) d(cosθ )n , (L−)nLM(θ )eiMϕ=ei(M−n)ϕsinn−MθdnsinMθ LM(θ )
d(cosθ )n . Hint. Try mathematical induction.
12.7.4 Show, using(L−)n, that
YL−M(θ, ϕ)=(−1)MYL∗M(θ, ϕ).
12.7.5 Verify by explicit calculation that (a) L+Y10(θ, ϕ)= −
) 3
4π sinθ eiϕ=√
2Y11(θ, ϕ), (b) L−Y10(θ, ϕ)= +
) 3
4π sinθ e−iϕ=√
2Y1−1(θ, ϕ).
The signs (Condon–Shortley phase) are a consequence of the ladder operatorsL+and L−.
12.8 T HE A DDITION T HEOREM FOR S PHERICAL H ARMONICS
Trigonometric Identity
In the following discussion,(θ1, ϕ1)and(θ2, ϕ2) denote two different directions in our spherical coordinate system(x1, y1, z1), separated by an angleγ (Fig. 12.16). The polar anglesθ1, θ2are measured from thez1-axis. These angles satisfy the trigonometric identity cosγ=cosθ1cosθ2+sinθ1sinθ2cos(ϕ1−ϕ2), (12.175) which is perhaps most easily proved by vector methods (compare Chapter 1).
The addition theorem, then, asserts that Pn(cosγ )= 4π
2n+1 n m=−n
(−1)mYnm(θ1, ϕ1)Yn−m(θ2, ϕ2), (12.176) or equivalently,
Pn(cosγ )= 4π 2n+1
n m=−n
Ynm(θ1, ϕ1)
Ynm(θ2, ϕ2)∗.21 (12.177)
21The asterisk for complex conjugation may go on either spherical harmonic.
In terms of the associated Legendre functions, the addition theorem is Pn(cosγ )=Pn(cosθ1)Pn(cosθ2)
+2 n m=1
(n−m)!
(n+m)!Pnm(cosθ1)Pnm(cosθ2)cosm(ϕ1−ϕ2).
(12.178) Equation (12.175) is a special case of Eq. (12.178),n=1.
Derivation of Addition Theorem
We now derive Eq. (12.177). Let(γ , ξ )be the angles that specify the direction(θ1, ϕ1)in a coordinate system(x2, y2, z2)whose axis is aligned with(θ2, ϕ2).(Actually, the choice of the 0 azimuth angleξin Fig. 12.16 is irrelevant.) First, we expandYnm(θ1, ϕ1)in spherical harmonics in the(γ , ξ )angular variables:
Ynm(θ1, ϕ1)= n σ=−n
anσmYnσ(γ , ξ ). (12.179) We write no summation overnin Eq. (12.179) because the angular momentumnofYnmis conserved (see Section 4.3); as a spherical harmonic,Ynm(θ1, ϕ1)is an eigenfunction of L2 with eigenvaluen(n+1).
FIGURE12.16 Two directions separated by an angleγ.
We need for our proof only the coefficientamn0, which we get by multiplying Eq. (12.179) by[Yn0(γ , ξ )]∗and integrating over the sphere:
an0m =
Ynm(θ1, ϕ1)
Yn0(γ , ξ ) ∗dγ ,ξ. (12.180) Similarly, we expandPn(cosγ )in terms of spherical harmonicsYnm(θ1, ϕ1):
Pn(cosγ )= 4π
2n+1 1/2
Yn0(γ , ξ )= n m=−n
bnmYnm(θ1, ϕ1), (12.181) where thebnmwill, of course, depend onθ2, ϕ2, that is, on the orientation of thez2-axis.
Multiplying by[Ynm(θ1, ϕ1)]∗and integrating with respect toθ1andϕ1over the sphere, we have
bnm=
Pn(cosγ )Ynm∗(θ1, ϕ1) dθ1,ϕ1. (12.182) In terms of spherical harmonics Eq. (12.182) becomes
4π 2n+1
1/2
Yn0(γ , ψ )
Ynm(θ1, ϕ1)∗d=bnm. (12.183) Note that the subscripts have been dropped from the solid angle elementd. Since the range of integration is over all solid angles, the choice of polar axis is irrelevant. Then comparing Eqs. (12.180) and (12.183), we see that
b∗nm=amn0 4π
2n+1 1/2
. (12.184)
Now we evaluateYnm(θ2, ϕ2)using the expansion of Eq. (12.179) and noting that the values of(γ , ξ )corresponding to(θ1, ϕ1)=(θ2, ϕ2)are(0,0).The result is
Ynm(θ2, ϕ2)=amn0Yn0(0,0)=amn0
2n+1 4π
1/2
, (12.185)
all terms with nonzeroσvanishing. Substituting this back into Eq. (12.184), we obtain bnm= 4π
2n+1
Ynm(θ2, ϕ2) ∗. (12.186) Finally, substituting this expression for bnm into the summation, Eq. (12.181) yields Eq. (12.177), thus proving our addition theorem.
Those familiar with group theory will find a much more elegant proof of Eq. (12.177) by using the rotation group.22This is Exercise 4.4.5.
One application of the addition theorem is in the construction of a Green’s function for the three-dimensional Laplace equation in spherical polar coordinates. If the source is on
22Compare M. E. Rose, Elementary Theory of Angular Momentum, New York: Wiley (1957).
the polar axis at the point(r=a, θ=0, ϕ=0), then, by Eq. (12.4a), 1
R = 1
|r− ˆza| = ∞
n=0
Pn(cosγ ) an
rn+1, r > a
= ∞
n=0
Pn(cosγ ) rn
an+1, r < a. (12.187) Rotating our coordinate system to put the source at(a, θ2, ϕ2)and the point of observation at(r, θ1, ϕ1), we obtain
G(r, θ1, ϕ1, a, θ2, ϕ2)= 1 R
= ∞
n=0
n m=−n
4π 2n+1
Ynm(θ1, ϕ1)∗Ynm(θ2, ϕ2) an
rn+1, r > a,
= ∞
n=0
n m=−n
4π 2n+1
Ynm(θ1, ϕ1)∗Ynm(θ2, ϕ2) rn
an+1, r < a. (12.188) In Section 9.7 this argument is reversed to provide another derivation of the Legendre polynomial addition theorem.
Exercises
12.8.1 In proving the addition theorem, we assumed thatYnk(θ1, ϕ1)could be expanded in a series ofYnm(θ2, ϕ2), in which mvaried from −n to+n butn was held fixed. What arguments can you develop to justify summing only over the upper index,m, and not over the lower index,n?
Hints. One possibility is to examine the homogeneity of theYnm, that is, Ynm may be expressed entirely in terms of the form cosn−pθsinpθ, orxn−p−sypzs/rn. Another possibility is to examine the behavior of the Legendre equation under rotation of the coordinate system.
12.8.2 An atomic electron with angular momentumLand magnetic quantum numberMhas a wave function
ψ (r, θ , ϕ)=f (r)YLM(θ, ϕ).
Show that the sum of the electron densities in a given complete shell is spherically symmetric; that is,L
M=−Lψ∗(r, θ, ϕ)ψ (r, θ, ϕ)is independent ofθandϕ.
12.8.3 The potential of an electron at point rein the field ofZprotons at points rpis = − e2
4π ε0 Z p=1
1
|re−rp|.
Show that this may be written as = − e2
4π ε0re
Z p=1
L,M
rp re
L 4π 2L+1
YLM(θp, ϕp) ∗YLM(θe, ϕe),
wherere> rp. How shouldbe written forre< rp?
12.8.4 Two protons are uniformly distributed within the same spherical volume. If the coor- dinates of one element of charge are(r1, θ1, ϕ1)and the coordinates of the other are (r2, θ2, ϕ2)andr12 is the distance between them, the element of energy of repulsion will be given by
dψ=ρ2dτ1dτ2
r12 =ρ2r12dr1sinθ1dθ1dϕ1r22dr2sinθ2dθ2dϕ2
r12 .
Here
ρ= charge volume= 3e
4π R3, charge density, r122 =r12+r22−2r1r2cosγ .
Calculate the total electrostatic energy (of repulsion) of the two protons. This calculation is used in accounting for the mass difference in “mirror” nuclei, such as O15and N15.
ANS. 6 5
e2 R. This is double that required to create a uniformly charged sphere because we have two separate cloud charges interacting, not one charge interacting with itself (with permuta- tion of pairs not considered).
12.8.5 Each of the two 1Selectrons in helium may be described by a hydrogenic wave function ψ (r)=
Z3 π a03
1/2
e−Zr/a0
in the absence of the other electron. HereZ, the atomic number, is 2. The symbola0is the Bohr radius,h¯2/me2. Find the mutual potential energy of the two electrons, given by
ψ∗(r1)ψ∗(r2)e2 r12
ψ (r1)ψ (r2) d3r1d3r2.
ANS.5e2Z 8a0 . Note.d3r1=r2dr1sinθ1dθ1dϕ1≡dτ1, r12= |r1−r2|.
12.8.6 The probability of finding a 1Shydrogen electron in a volume elementr2drsinθ dθ dϕ is
1
π a03exp[−2r/a0]r2drsinθ dθ dϕ.
Find the corresponding electrostatic potential. Calculate the potential from V (r1)= q
4π ε0
ρ(r2) r12 d3r2,
withr1not on thez-axis. Expandr12. Apply the Legendre polynomial addition theorem and show that the angular dependence ofV (r1)drops out.
ANS.V (r1)= q 4π ε0
1 2r1γ
3,2r1
a0
+ 1 a0Ŵ
2,2r1
a0
. 12.8.7 A hydrogen electron in a 2P orbit has a charge distribution
ρ= q
64π a05r2e−r/a0sin2θ,
wherea0is the Bohr radius,h¯2/me2. Find the electrostatic potential corresponding to this charge distribution.
12.8.8 The electric current density produced by a 2P electron in a hydrogen atom is J= ˆϕ qh¯
32ma05e−r/a0rsinθ.
Using
A(r1)=à0 4π
J(r2)
|r1−r2|d3r2,
find the magnetic vector potential produced by this hydrogen electron.
Hint. Resolve into Cartesian components. Use the addition theorem to eliminateγ, the angle included between r1and r2.
12.8.9 (a) As a Laplace series and as an example of Eq. (1.190) (now with complex func- tions), show that
δ(1−2)= ∞
n=0
n m=−n
Ynm∗(θ2, ϕ2)Ynm(θ1, ϕ1).
(b) Show also that this same Dirac delta function may be written as δ(1−2)=
∞
n=0
2n+1
4π Pn(cosγ ).
Now, if you can justify equating the summations overnterm by term, you have an alternate derivation of the spherical harmonic addition theorem.
12.9 I NTEGRALS OF P RODUCTS OF T HREE S PHERICAL
H ARMONICS
Frequently in quantum mechanics we encounter integrals of the general form YLM1
1
YLM2
2
YLM3
3
= 2π
0
π 0
YLM1
1
∗YLM2
2 YLM3
3 sinθ dθ dϕ
= ,
(2L2+1)(2L3+1)
4π(2L1+1) C(L2L3L1|000)C(L2L3L1|M2M3M1), (12.189) in which all spherical harmonics depend onθ, ϕ. The first factor in the integrand may come from the wave function of a final state and the third factor from an initial state, whereas the middle factor may represent an operator that is being evaluated or whose “matrix element”
is being determined.
By using group theoretical methods, as in the quantum theory of angular momentum, we may give a general expression for the forms listed. The analysis involves the vector–
addition or Clebsch–Gordan coefficients from Section 4.4, which are tabulated. Three gen- eral restrictions appear.
1. The integral vanishes unless the triangle condition of theL’s (angular momentum) is zero,|L1−L3| ≤L2≤L1+L3.
2. The integral vanishes unlessM2+M3=M1. Here we have the theoretical foundation of the vector model of atomic spectroscopy.
3. Finally, the integral vanishes unless the product[YLM1
1 ]∗YLM2
2 YLM3
3 is even, that is, unless L1+L2+L3is an even integer. This is a parity conservation law.
The key to the determination of the integral in Eq. (12.189) is the expansion of the prod- uct of two spherical harmonics depending on the same angles (in contrast to the addition theorem), which are coupled by Clebsch–Gordan coefficients to angular momentumL, M, which, from its rotational transformation properties, must be proportional to YLM(θ, ϕ);
that is,
M1,M2
C(L2L3L1|M2M3M1)YLM2
2 (θ, ϕ)YLM3
3 (θ, ϕ)∼YLM1
1 (θ, ϕ).
For details we refer to Edmonds.23
Let us outline some of the steps of this general and powerful approach using Section 4.4.
The Wigner–Eckart theorem applied to the matrix element in Eq. (12.189) yields YLM1
1
YLM2
2
YLM3
3
=(−1)L2−L3+L1C(L2L3L1|M2M3M1)
ãYL1YL2YL3
√(2L1+1) , (12.190)
23E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, Cambridge, UK: Cambridge University Press (1951); M.
E. Rose, Elementary Theory of Angular Momentum, New York: Wiley (1957); A. Edmonds, Angular Momentum in Quantum Mechanics, Princeton, NJ: Princeton University Press (1957); E. P. Wigner, Group Theory and Its Applications to Quantum Mechanics (translated by J. J. Griffin), New York: Academic Press (1959).
where the double bars denote the reduced matrix element, which no longer depends on the Mi. Selection rules (1) and (2) mentioned earlier follow directly from the Clebsch–
Gordan coefficient in Eq. (12.190). Next we use Eq. (12.190) forM1=M2=M3=0 in conjunction with Eq. (12.153) form=0, which yields
YL0
1
YL0
2
YL0
3
=(−1)L2−L3+L1
√2L1+1 C(L2L3L1|000)ã YL1YL2YL3
=
)(2L1+1)(2L2+1)(2L3+1) 4π
ã1 2ã
1
−1
PL1(x)PL2(x)PL3(x) dx, (12.191) wherex=cosθ. By elementary methods it can be shown that
1
−1
PL1(x)PL2(x)PL3(x) dx= 2
2L1+1C(L2L3L1|000)2. (12.192) Substituting Eq. (12.192) into (12.191) we obtain
YL1YL2YL3 =(−1)L2−L3+L1C(L2L3L1|000)
)(2L2+1)(2L3+1)
4π . (12.193)
The aforementioned parity selection rule (3) above follows from Eq. (12.193) in conjunc- tion with the phase relation
C(L2L3L1| −M2,−M3,−M1)=(−1)L2+L3−L1C(L2L3L1|M2M3M1). (12.194) Note that the vector-addition coefficients are developed in terms of the Condon–Shortley phase convention,23in which the(−1)mof Eq. (12.153) is associated with the positivem.
It is possible to evaluate many of the commonly encountered integrals of this form with the techniques already developed. The integration over azimuth may be carried out by inspection:
2π 0
e−iM1ϕeiM2ϕeiM3ϕdϕ=2π δM2+M3−M1,0. (12.195) Physically this corresponds to the conservation of thezcomponent of angular momentum.
Application of Recurrence Relations
A glance at Table 12.3 will show that theθ-dependence ofYLM2
2 , that is,PLM2
2 (θ ), can be expressed in terms of cosθand sinθ. However, a factor of cosθor sinθmay be combined with theYLM3
3 factor by using the associated Legendre polynomial recurrence relations. For
instance, from Eqs. (12.92) and (12.93) we get cosθ YLM= +
(L−M+1)(L+M+1) (2L+1)(2L+3)
1/2
YLM+1 +
(L−M)(L+M) (2L−1)(2L+1)
1/2
YL−1M (12.196)
eiϕsinθ YLM= −
(L+M+1)(L+M+2) (2L+1)(2L+3)
1/2
YLM++11 +
(L−M)(L−M−1) (2L−1)(2L+1)
1/2
YLM−+11 (12.197) e−iϕsinθ YLM= +
(L−M+1)(L−M+2) (2L+1)(2L+3)
1/2
YLM+−11
−
(L+M)(L+M−1) (2L−1)(2L+1)
1/2
YLM−−11. (12.198) Using these equations, we obtain
YLM1∗
1 cosθ YLMd=
(L−M+1)(L+M+1) (2L+1)(2L+3)
1/2
δM1,MδL1,L+1
+
(L−M)(L+M) (2L−1)(2L+1)
1/2
δM1,MδL1,L−1. (12.199) The occurrence of the Kronecker delta(L1, L±1)is an aspect of the conservation of angular momentum. Physically, this integral arises in a consideration of ordinary atomic electromagnetic radiation (electric dipole). It leads to the familiar selection rule that transi- tions to an atomic level with orbital angular momentum quantum numberL1can originate only from atomic levels with quantum numbersL1−1 orL1+1. The application to ex- pressions such as
quadrupole moment∼
YLM∗(θ, ϕ)P2(cosθ )YLM(θ, ϕ) d is more involved but perfectly straightforward.
Exercises
12.9.1 Verify (a)
YLM(θ, ϕ)Y00(θ, ϕ)YLM∗(θ, ϕ) d= 1
√4π, (b)
YLMY10YL+1M∗d= ) 3
4π ,
(L+M+1)(L−M+1) (2L+1)(2L+3) ,
(c)
YLMY11YLM++11∗d= ) 3
8π ,
(L+M+1)(L+M+2) (2L+1)(2L+3) , (d)
YLMY11YLM−+11∗d= − ) 3
8π ,
(L−M)(L−M−1) (2L−1)(2L+1) .
These integrals were used in an investigation of the angular correlation of internal con- version electrons.
12.9.2 Show that
(a) 1
−1
xPL(x)PN(x) dx=
2(L+1)
(2L+1)(2L+3), N=L+1, 2L
(2L−1)(2L+1), N=L−1,
(b) 1
−1
x2PL(x)PN(x) dx=
2(L+1)(L+2)
(2L+1)(2L+3)(2L+5), N=L+2, 2(2L2+2L−1)
(2L−1)(2L+1)(2L+3), N=L, 2L(L−1)
(2L−3)(2L−1)(2L+1), N=L−2.
12.9.3 SincexPn(x)is a polynomial (degreen+1), it may be represented by the Legendre series
xPn(x)= ∞
s=0
asPs(x).
(a) Show thatas=0 fors < n−1 ands > n+1.
(b) Calculate an−1, an, andan+1and show that you have reproduced the recurrence relation, Eq. 12.17.
Note. This argument may be put in a general form to demonstrate the existence of a three-term recurrence relation for any of our complete sets of orthogonal polynomials:
xϕn=an+1ϕn+1+anϕn+an−1ϕn−1.
12.9.4 Show that Eq. (12.199) is a special case of Eq. (12.190) and derive the reduced matrix elementYL1Y1YL.
ANS.YL1Y1YL =(−1)L1+1−LC(1LL1|000)
√3(2L+1)
4π .
12.10 L EGENDRE F UNCTIONS OF THE S ECOND K IND
In all the analysis so far in this chapter we have been dealing with one solution of Legen- dre’s equation, the solutionPn(cosθ ), which is regular (finite) at the two singular points of
the differential equation, cosθ= ±1. From the general theory of differential equations it is known that a second solution exists. We develop this second solution,Qn, with nonneg- ative integern(becauseQnin applications will occur in conjunction withPn), by a series solution of Legendre’s equation. Later a closed form will be obtained.
Series Solutions of Legendre’s Equation
To solve
d dx
(1−x2)dy dx
+n(n+1)y=0 (12.200)
we proceed as in Chapter 9, letting24 y=
∞
λ=0
aλxk+λ, (12.201)
with
y′= ∞
λ=0
(k+λ)aλxk+λ−1, (12.202)
y′′= ∞
λ=0
(k+λ)(k+λ−1)aλxk+λ−2. (12.203) Substitution into the original differential equation gives
∞
λ=0
(k+λ)(k+λ−1)aλxk+λ−2
+ ∞
λ=0
n(n+1)−2(k+λ)−(k+λ)(k+λ−1)aλxk+λ=0. (12.204)
The indicial equation is
k(k−1)=0, (12.205)
with solutionsk=0,1. We try firstk=0 witha0=1, a1=0. Then our series is described by the recurrence relation
(λ+2)(λ+1)aλ+2+
n(n+1)−2λ−λ(λ−1) aλ=0, (12.206) which becomes
aλ+2= −(n+λ+1)(n−λ)
(λ+1)(λ+2) aλ. (12.207)
24Note thatxmay be replaced by the complex variablez.
Labeling this series, from Eq. (12.201),y(x)=pn(x), we have pn(x)=1−n(n+1)
2! x2+(n−2)n(n+1)(n+3)
4! x4+ ã ã ã. (12.208) The second solution of the indicial equation,k=1, with a0=0, a1=1, leads to the recurrence relation
aλ+2= −(n+λ+2)(n−λ−1)
(λ+2)(λ+3) aλ. (12.209)
Labeling this series, from Eq. (12.201),y(x)=qn(x), we obtain qn(x)=x−(n−1)(n+2)
3! x3+(n−3)(n−1)(n+2)(n+4)
5! x5− ã ã ã. (12.210) Our general solution of Eq. (12.200), then, is
yn(x)=Anpn(x)+Bnqn(x), (12.211) provided we have convergence. From Gauss’ test, Section 5.2 (see Example 5.2.4), we do not have convergence atx= ±1. To get out of this difficulty, we set the separation constant nequal to an integer (Exercise 9.5.5) and convert the infinite series into a polynomial.
Forna positive even integer (or zero), series pn terminates, and with a proper choice of a normalizing factor (selected to obtain agreement with the definition ofPn(x)in Sec- tion 12.1)
Pn(x)=(−1)n/2 n!
2n[(n/2)!]2pn(x)=(−1)s (2s)!
22s(s!)2p2s(x)
=(−1)s(2s−1)!!
(2s)!! p2s(x), forn=2s. (12.212) Ifnis a positive odd integer, seriesqnterminates after a finite number of terms, and we write
Pn(x)=(−1)n−1)/2 n!
2n−1{[n−1)/2]!}2qn(x)
=(−1)s(2s+1)!
22s(s!)2 q2s+1(x)=(−1)s(2s+1)!!
(2s)!! q2s+1(x), forn=2s+1.
(12.213) Note that these expressions hold for all real values ofx,−∞< x <∞, and for complex values in the finite complex plane. The constants that multiply pn andqn are chosen to makePnagree with Legendre polynomials given by the generating function.
Equations (12.208) and (12.210) may still be used withn=ν, not an integer, but now the series no longer terminates, and the range of convergence becomes−1< x <1. The endpoints,x= ±1, are not included.
It is sometimes convenient to reverse the order of the terms in the series. This may be done by putting
s=n
2−λ in the first form ofPn(x), neven, s=n−1
2 −λ in the second form ofPn(x), nodd,
so that Eqs. (12.212) and (12.213) become
Pn(x)=
[n/2]
s=0
(−1)s (2n−2s)!
2ns!(n−s)!(n−2s)!xn−2s, (12.214) where the upper limit s=n/2 (forn even) or (n−1)/2 (for n odd). This reproduces Eq. (12.8) of Section 12.1, which is obtained directly from the generating function. This agreement with Eq. (12.8) is the reason for the particular choice of normalization in Eqs. (12.212) and (12.213).
Qn(x) Functions of the Second Kind
It will be noticed that we have used onlypnforneven andqnfornodd (because they ter- minated for this choice ofn). We may now define a second solution of Legendre’s equation (Fig. 12.17) by
Qn(x)=(−1)n/2[n/2]!22n n! qn(x)
=(−1)s (2s)!!
(2s−1)!!q2s(x), forneven, n=2s, (12.215)
FIGURE12.17 Second Legendre function,Qn(x), 0≤x <1.
FIGURE12.18 Second Legendre function,Qn(x),
x >1.
Qn(x)=(−1)(n+1)/2{[(n−1)/2]!}22n−1 n! pn(x)
=(−1)s+1 (2s)!!
(2s+1)!!p2s+1(x), fornodd, n=2s+1. (12.216) This choice of normalizing factors forcesQn to satisfy the same recurrence relations as Pn. This may be verified by substituting Eqs. (12.215) and (12.216) into Eqs. (12.17) and (12.26). Inspection of the (series) recurrence relations (Eqs. (12.207) and (12.209)), that is, by the Cauchy ratio test, shows thatQn(x)will converge for−1< x <1. If|x| ≥1, these series forms of our second solution diverge. A solution in a series of negative powers of x can be developed for the region|x|>1 (Fig. 12.18), but we proceed to a closed-form solution that can be used over the entire complex plane (apart from the singular points x= ±1 and with care on cut lines).
Closed-Form Solutions
Frequently, a closed form of the second solution,Qn(z), is desirable. This may be obtained by the method discussed in Section 9.6. We write
Qn(z)=Pn(z)
An+Bn
z dx (1−x2)[Pn(x)]2
, (12.217)
in which the constantAnreplaces the evaluation of the integral at the arbitrary lower limit.
Both constants,AnandBn, may be determined for special cases.
Forn=0, Eq. (12.217) yields Q0(z)=P0(z)
A0+B0
z dx (1−x2)[P0(x)]2
=A0+B01
2ln1+z 1−z
=A0+B0
z+z3
3 +z5
5 + ã ã ã + z2s+1 2s+1 + ã ã ã
, (12.218)
the last expression following from a Maclaurin expansion of the logarithm. Comparing this with the series solution (Eq. (12.210)), we obtain
Q0(z)=q0(z)=z+z3 3 +z5
5 + ã ã ã + z2s+1
2s+1+ ã ã ã, (12.219) we haveA0=0,B0=1. Similar results follow forn=1. We obtain
Q1(z)=z
A1+B1
z dx (1−x2)x2
=A1z+B1z 1
2ln1+z 1−z−1
z
. (12.220)
Expanding in a power series and comparing withQ1(z)= −p1(z), we haveA1=0, B1= 1. Therefore we may write
Q0(z)=1
2ln1+z
1−z, Q1(z)=1
2zln1+z
1−z−1, |z|<1. (12.221) Perhaps the best way of determining the higher-orderQn(z)is to use the recurrence relation (Eq. (12.17)), which may be verified for bothx2<1 andx2>1 by substituting in the series forms. This recurrence relation technique yields
Q2(z)=1
2P2(z)ln1+z 1−z−3
2P1(z). (12.222)
Repeated application of the recurrence formula leads to Qn(z)=1
2Pn(z)ln1+z
1−z−2n−1
1ãn Pn−1(z)− 2n−5
3(n−1)Pn−3(z)− ã ã ã. (12.223) From the form ln[(1+z)/(1−z)]it will be seen that for realzthese expressions hold in the range−1< x <1. If we wish to have closed forms valid outside this range, we need only replace
ln1+x
1−x by lnz+1 z−1.
When using the latter form, valid for largez, we take the line interval−1≤x≤1 as a cut line. Values ofQn(x)on the cut line are customarily assigned by the relation
Qn(x)=1 2
Qn(x+i0)+Qn(x−i0), (12.224)
the arithmetic average of approaches from the positive imaginary side and from the nega- tive imaginary side. Note that forz→x >1, z−1→(1−x)e±iπ. The result is that for allz, except on the real axis,−1≤x≤1, we have
Q0(z)=1
2lnz+1
z−1, (12.225)
Q1(z)=1
2zlnz+1
z−1−1, (12.226)
and so on.
For convenient reference some special values ofQn(z)are given.
1. Qn(1)= ∞, from the logarithmic term (Eq. (12.223)).
2. Qn(∞)=0. This is best obtained from a representation ofQn(x)as a series of nega- tive powers ofx, Exercise 12.10.4.
3. Qn(−z)=(−1)n+1Qn(z). This follows from the series form. It may also be derived by usingQ0(z), Q1(z)and the recurrence relation (Eq. (12.17)).
4. Qn(0)=0, forneven, by (3).
5. Qn(0)=(−1)(n+1)/2{[(n−1)/2]!}2 n! 2n−1
=(−1)s+1 (2s)!!
(2s+1)!!, fornodd, n=2s+1.
This last result comes from the series form (Eq. (12.216)) withpn(0)=1.
Exercises
12.10.1 Derive the parity relation forQn(x).
12.10.2 From Eqs. (12.212) and (12.213) show that (a) P2n(x)=(−1)n
22n−1 n s=0
(−1)s (2n+2s−1)!
(2s)!(n+s−1)!(n−s)!x2s. (b) P2n+1(x)=(−1)n
22n n s=0
(−1)s (2n+2s+1)!
(2s+1)!(n+s)!(n−s)!x2s+1.
Check the normalization by showing that one term of each series agrees with the corre- sponding term of Eq. (12.8).
12.10.3 Show that
(a) Q2n(x)=(−1)n22n n s=0
(−1)s (n+s)!(n−s)! (2s+1)!(2n−2s)!x2s+1 +22n
∞
s=n+1
(n+s)!(2s−2n)!
(2s+1)!(s−n)! x2s+1, |x|<1.
(b) Q2n+1(x)=(−1)n+122n n s=0
(−1)s (n+s)!(n−s)! (2s)!(2n−2s+1)!x2s +22n+1
∞
s=n+1
(n+s)!(2s−2n−2)!
(2s)!(s−n−1)! x2s, |x|<1.
12.10.4 (a) Starting with the assumed form Qn(x)=
∞
λ=0
b−λxk−λ, show that
Qn(x)=b0x−n−1 ∞
s=0
(n+s)!(n+2s)!(2n+1)! s!(n!)2(2n+2s+1)! x−2s. (b) The standard choice ofb0is
b0= 2n(n!)2 (2n+1)!.
Show that this choice ofbobrings this negative power-series form ofQn(x)into agreement with the closed-form solutions.
12.10.5 Verify that the Legendre functions of the second kind,Qn(x), satisfy the same recur- rence relations asPn(x), both for|x|<1 and for|x|>1:
(2n+1)xQn(x)=(n+1)Qn+1(x)+nQn−1(x), (2n+1)Qn(x)=Q′n+1(x)−Q′n−1(x).
12.10.6 (a) Using the recurrence relations, prove (independent of the Wronskian relation) that n
Pn(x)Qn−1(x)−Pn−1(x)Qn(x) =P1(x)Q0(x)−P0(x)Q1(x).
(b) By direct substitution show that the right-hand side of this equation equals 1.
12.10.7 (a) Write a subroutine that will generateQn(x)andQ0throughQn−1based on the recurrence relation for these Legendre functions of the second kind. Takex to be within(−1,1)— excluding the endpoints.
Hint. TakeQ0(x)andQ1(x)to be known.
(b) Test your subroutine for accuracy by computingQ10(x)and comparing with the values tabulated in AMS-55 (for a complete reference, see Additional Readings in Chapter 8).
12.11 V ECTOR S PHERICAL H ARMONICS
Most of our attention in this chapter has been directed toward solving the equations of scalar fields, such as the electrostatic potential. This was done primarily because the scalar fields are easier to handle than vector fields. However, with scalar field problems under firm control, more and more attention is being paid to vector field problems.