Legendre Polynomials and Spherical Harmonicswilliamsgj.people.cofc.edu/Legendre Function.pdfChapter 11 Legendre Polynomials and Spherical Harmonics 11.1 Introduction Legendre polynomials

Chapter 11

Legendre Polynomialsand Spherical

Harmonics

11.1 Introduction

Legendre polynomials appear in many different mathematical and physicalsituations:

• They originate as solutions of the Legendre ordinary differential equation(ODE), which we have already encountered in the separation of variables(Section 8.9) for Laplace’s equation, and similar ODEs in spherical polarcoordinates.

• They arise as a consequence of demanding a complete, orthogonal setof functions over the interval [−1, 1] (Gram–Schmidt orthogonalization;Section 9.3).

• In quantum mechanics, they (really the spherical harmonics; Section 11.5)represent angular momentum eigenfunctions. They also appear naturally inproblems with azimuthal symmetry, which is the case in the next point.

• They are defined by a generating function: We introduce Legendre polyno-mials here by way of the electrostatic potential of a point charge, whichacts as the generating function.

Physical Basis: Electrostatics

Legendre polynomials appear in an expansion of the electrostatic potential ininverse radial powers. Consider an electric charge q placed on the z-axis atz = a. As shown in Fig. 11.1, the electrostatic potential of charge q is

ϕ = 14πε0

· q

r1(SI units). (11.1)

We want to express the electrostatic potential in terms of the spherical polarcoordinates r and θ (the coordinate ϕ is absent because of azimuthal symmetry,

552

11.1 Introduction 553

rr1

qqz

z = a

q

4p e0r1j =

Figure 11.1

ElectrostaticPotential. Charge qDisplaced fromOrigin

that is, invariance under rotations about the z-axis). Using the law of cosinesin Fig. 11.1, we obtain

ϕ = q

4πε0(r2 + a2 − 2ar cos θ)−1/2. (11.2)

Generating Function

Consider the case of r > a. The radical in Eq. (11.2) may be expanded in abinomial series (see Exercise 5.6.9) for r2 > |a2 − 2ar cos θ | and then rear-ranged in powers of (a/r). This yields the coefficient 1 of (a/r)0 = 1, cos θ

as coefficient of a/r, etc. The Legendre polynomial Pn(cos θ) (Fig. 11.2) is

defined as the coefficient of (a/r)n so that

ϕ = q

4πε0r

∞∑n= 0

Pn(cos θ)(

a

r

)n

. (11.3)

Dropping the factor q/4πε0r and using x = cos θ and t = a/r, respectively, wehave

g(t, x) = (1 − 2xt + t2)−1/2 =∞∑

n= 0

Pn(x)tn, |t| < 1, (11.4)

defining g(t, x) as the generating function for the Pn(x). These polynomialsPn(x), shown in Table 11.1, are the same as those generated in Example 9.3.1 byGram–Schmidt orthogonalization of powers xn over the interval −1 ≤ x ≤ 1.

This is no coincidence because cos θ varies between the limits ±1. In the nextsection, it is shown that |Pn(cos θ)| ≤ 1, which means that the series expansion[Eq. (11.4)] is convergent for |t| < 1.1 Indeed, the series is convergent for |t| = 1except for x = ±1, where |Pn(±1)| = 1.

1Note that the series in Eq. (11.3) is convergent for r > a even though the binomial expansioninvolved is valid only for r > (a2+2ar)1/2 ≥ |a2−2ar cos θ |1/2 so that r2 > a2 + 2ar, (r−a)2 > 2a2,or r > a(1 + √

2).

554 Chapter 11 Legendre Polynomials and Spherical Harmonics

x

1Pn(x)

0.5

P2

P3P4

P5

10−1

−0.5

Figure 11.2

LegendrePolynomials P2(x),P3(x), P4(x), andP5(x)

Table 11.1

Legendre Polynomials

P0(x) = 1P1(x) = x

P2(x) = 12 (3x2 − 1)

P3(x) = 12 (5x3 − 3x)

P4(x) = 18 (35x4 − 30x2 + 3)

P5(x) = 18 (63x5 − 70x3 + 15x)

P6(x) = 116 (231x6 − 315x4 + 105x2 − 5)

P7(x) = 116 (429x7 − 693x5 + 315x3 − 35x)

P8(x) = 1128 (6435x8 − 12012x6 + 6930x4 − 1260x2 + 35)

In physical applications, such as the Coulomb or gravitational potentials,Eq. (11.4) often appears in the vector form

1|r1 − r2| = [

r21 + r2

2 − 2r1 · r2]−1/2 = 1

r1

[1 +

(r2

r1

)2

− 2(

r2

r1

)cos θ

]−1/2

.


The last equality is obtained by factoring r1 = |r1| from the denominator, whichthen, for r1 > r2, we can expand according to Eq. (11.4). In this way, we obtain

1|r1 − r2| = 1

r>

∞∑n= 0

(r<

r>

)n

Pn(cos θ), (11.4a)

where

r> = |r1|r< = |r2|

}for |r1| > |r2|, (11.4b)

and

r> = |r2|r< = |r1|

}for |r2| > |r1|. (11.4c)

EXAMPLE 11.1.1 Special Values A simple and powerful application of the generating functiong is to use it for special values (e.g., x = ±1) where g can be evaluated explicitly.If we set x = 1, that is, point z = a on the positive z-axis, where the potentialhas a simple form, Eq. (11.4) becomes

1(1 − 2t + t2)1/2

= 11 − t

=∞∑

n= 0

tn, (11.5)

using a binomial expansion or the geometric series (Example 5.1.2). However,Eq. (11.4) for x = 1 defines

1(1 − 2t + t2)1/2

=∞∑

n= 0

Pn(1)tn.

Comparing the two series expansions (uniqueness of power series; Section5.7), we have

Pn(1) = 1. (11.6)

If we let x = −1 in Eq. (11.4), that is, point z = −a on the negative z-axis inFig. 11.1, where the potential is simple, then we sum similarly

1(1 + 2t + t2)1/2

= 11 + t

=∞∑

n= 0

(−t)n (11.7)

so that

Pn(−1) = (−1)n. (11.8)

These general results are more difficult to develop from other formulas forLegendre polynomials.

If we take x = 0 in Eq. (11.4), using the binomial expansion

(1 + t2)−1/2 = 1 − 12 t2 + 3

8 t4 + · · · + (−1)n1 · 3 · · · (2n − 1)

2nn!t2n + · · · , (11.9)


we have2

P2n(0) = (−1)n 1 · 3 · · · (2n − 1)2nn!

= (−1)n (2n − 1)!!(2n)!!

= (−1)n(2n)!22n(n!)2

(11.10)

P2n+1(0) = 0, n = 0, 1, 2 . . . . (11.11)

These results can also be verified by inspection of Table 11.1. ■

EXAMPLE 11.1.2 Parity If we replace x by −x and t by −t, the generating function is un-changed. Hence,

g(t, x) = g(−t, −x) = [1 − 2(−t)(−x) + (−t)2]−1/2

=∞∑

n= 0

Pn(−x)(−t)n =∞∑

n= 0

Pn(x)tn. (11.12)

Comparing these two series, we have

Pn(−x) = (−1)nPn(x); (11.13)

that is, the polynomial functions are odd or even (with respect to x = 0)according to whether the index n is odd or even. This is the parity3 or reflec-tion property that plays such an important role in quantum mechanics. Parityis conserved when the Hamiltonian is invariant under the reflection of thecoordinates r → −r. For central forces the index n is the orbital angularmomentum [and n(n + 1) is the eigenvalue of L2], thus linking parity and or-bital angular momentum. This parity property will be confirmed by the seriessolution and for the special cases tabulated in Table 11.1. ■

Power Series

Using the binomial theorem (Section 5.6) and Exercise 10.1.15, we expand thegenerating function as

(1 − 2xt + t2)−1/2 =∞∑

n= 0

(2n)!22n(n!)2

(2xt − t2)n

= 1 +∞∑

n=1

(2n − 1)!!(2n)!!

(2xt − t2)n. (11.14)

Before we expand (2xt − t2)n further, let us inspect the lowest powers of t.

2The double factorial notation is defined in Section 10.1:

(2n)!! = 2 · 4 · 6 · · · (2n), (2n − 1)!! = 1 · 3 · 5 · · · (2n − 1).3In spherical polar coordinates the inversion of the point (r, θ , ϕ) through the origin is accom-plished by the transformation [r → r, θ → π − θ , and ϕ → ϕ ± π ]. Then, cos θ → cos(π − θ) =− cos θ , corresponding to x → −x.


EXAMPLE 11.1.3 Lowest Legendre Polynomials For the first few Legendre polynomials(e.g., P0, P1, and P2), we need the coefficients of t0, t1, and t2 in Eq. (11.14).These powers of t appear only in the terms n = 0, 1, and 2; hence, we may limitour attention to the first three terms of the infinite series:

0!20(0!)2

(2xt − t2)0 + 2!22(1!)2

(2xt − t2)1 + 4!24(2!)2

(2xt − t2)2

= 1t0 + xt1 +(

32

x2 − 12

)t2 +O(t3).

Then, from Eq. (11.4) (and uniqueness of power series) we obtain

P0(x) = 1, P1(x) = x, P2(x) = 32

x2 − 12

, (11.15)

confirming the entries of Table 11.1. We repeat this limited development in avector framework later in this section. ■

In employing a general treatment, we find that the binomial expansion ofthe (2xt − t2)n factor yields the double series

(1 − 2xt + t2)−1/2 =∞∑

n= 0

(2n)!22n(n!)2

tnn∑

k=0

(−1)k n!k!(n − k)!

(2x)n−ktk

=∞∑

n= 0

n∑k=0

(−1)k (2n)!22nn!k!(n − k)!

· (2x)n−ktn+k. (11.16)

By rearranging the order of summation (valid by absolute convergence), Eq.(11.16) becomes

(1−2xt+t2)−1/2 =∞∑

n= 0

[n/2]∑k=0

(−1)k (2n − 2k)!22n−2kk!(n − k)!(n − 2k)!

· (2x)n−2ktn, (11.17)

with the tn independent of the index k.4 Now, equating our two power series[Eqs. (11.4) and (11.17)] term by term, we have5

Pn(x) =[n/2]∑k=0

(−1)k (2n − 2k)!2nk!(n − k)!(n − 2k)!

xn−2k. (11.18)

We read off this formula, from k = 0, that the highest power of Pn(x) is xn,and the lowest power is x0 = 1 for even n and x for odd n. This is consistentwith Example 11.1.3 and Table 11.1. Also, for n even, Pn has only even powersof x and thus even parity [see Eq. (11.13)] and odd powers and odd parity forodd n.

4[n/2] = n/2 for n even, (n − 1)/2 for n odd.5Equation (11.18) starts with xn. By changing the index, we can transform it into a series thatstarts with x0 for n even and x1 for n odd.


Biographical Data

Legendre, Adrien Marie. Legendre, a French mathematician who wasborn in Paris in 1752 and died there in 1833, made major contributionsto number theory, elliptic integrals before Abel and Jacobi, and analysis.He tried in vain to prove the parallel axiom of Euclidean geometry. Histaste in selecting research problems was remarkably similar to that of hiscontemporary Gauss, but nobody could match Gauss’s depth and perfection.His great textbooks had enormous impact.

Linear Electric Multipoles

Returning to the electric charge on the z-axis, we demonstrate the usefulnessand power of the generating function by adding a charge −q at z = −a, asshown in Fig. 11.3, using the superposition principle of electric fields. Thepotential becomes

ϕ = q

4πε0

(1r1

− 1r2

), (11.19)

and by using the law of cosines, we have for r > a

ϕ = q

4πε0r

{[1 − 2

a

rcos θ +

(a

r

)2]−1/2

−[

1 + 2a

rcos θ +

(a

r

)2]−1/2},

where the second radical is like the first, except that a has been replaced by−a. Then, using Eq. (11.4), we obtain

ϕ = q

4πε0r

[ ∞∑n= 0

Pn(cos)(

a

r

)n

−∞∑

n= 0

Pn(cos θ)(−1)n

(a

r

)n]

= 2q

4πε0r

[P1(cos θ)

a

r+ P3(cos θ)

(a

r

)3

+ · · ·]

. (11.20)

rr1

qqz

z = a

q

4pe0j =

z = -a

r2

-q

1r2

1r1

-

Figure 11.3

Electric Dipole


The first term (and dominant term for r � a) is the electric dipole potential

ϕ = 2aq

4πε0· P1(cos θ)

r2, (11.21)

with 2aq the electric dipole moment (Fig. 11.3). If the potential in Eq. (11.19)is taken to be the dipole potential, then Eq. (11.21) gives its asymptotic behaviorfor large r. This analysis may be extended by placing additional charges on thez-axis so that the P1 term, as well as the P0 (monopole) term, is canceled.For instance, charges of q at z = a and z = −a, −2q at z = 0 give rise to apotential whose series expansion starts with P2(cos θ). This is a linear electricquadrupole. Two linear quadrupoles may be placed so that the quadrupoleterm is canceled, but the P3, the electric octupole term, survives, etc. Theseexpansions are special cases of the general multipole expansion of the electricpotential.

Vector Expansion

We consider the electrostatic potential produced by a distributed charge ρ(r2):

ϕ(r1) = 14πε0

∫ρ(r2)

|r1 − r2|dτ2. (11.22a)

Taking the denominator of the integrand, using first the law of cosines andthen a binomial expansion, yields

1|r1 − r2| = (

r21 − 2r1 · r2 + r2

2

)−1/2(11.22b)

= 1r1

[1 +

(−2r1 · r2

r21

+ r22

r21

)]−1/2

, for r1 > r2

= 1r1

[1 + r1 · r2

r21

− 12

r22

r21

+ 32

(r1 · r2)2

r41

+O(

r2

r1

)3].

For r1 = 1, r2 = t, and r1 · r2 = xt, Eq. (11.22b) reduces to the generatingfunction, Eq. (11.4).

The first term in the square bracket, 1, yields a potential

ϕ0(r1) = 14πε0

1r1

∫ρ(r2)dτ2. (11.22c)

The integral contains the total charge. This part of the total potential is anelectric monopole.

The second term yields

ϕ1(r1) = 14πε0

r1·r3

1

∫r2ρ(r2)dτ2, (11.22d)

where the integral is the dipole moment whose charge density ρ(r2) is weightedby a moment arm r2. We have an electric dipole potential. For atomic or nu-clear states of definite parity, ρ(r2) is an even function and the dipole integral


is identically zero. However, in the presence of an applied electric field a super-position of odd/even parity states may develop so that the resulting induceddipole moment is no longer zero. The last two terms, both of order (r2/r1)2,may be handled by using Cartesian coordinates

(r1 · r2)2 =3∑

i=1

x1ix2i

3∑j=1

x1 jx2 j.

Rearranging variables to keep the x2 inside the integral yields

ϕ2(r1) = 14πε0

1

2r51

3∑i, j=1

x1ix1 j

∫ [3x2ix2 j − δijr

22

]ρ(r2)dτ2. (11.22e)

This is the electric quadrupole term. Note that the square bracket in theintegrand forms a symmetric tensor of zero trace.

A general electrostatic multipole expansion can also be developed byusing Eq. (11.22a) for the potential ϕ(r1) and replacing 1/(4π |r1 − r2|) by a(double) series of the angular solutions of the Poisson equation (which are thesame as those of the Laplace equation of Section 8.9).

Before leaving multipole fields, we emphasize three points:

• First, an electric (or magnetic) multipole has a value independent of theorigin (reference point) only if all lower order terms vanish. For instance,the potential of one charge q at z = a was expanded in a series of Legendrepolynomials. Although we refer to the P1(cos θ) term in this expansion asa dipole term, it should be remembered that this term exists only becauseof our choice of coordinates. We actually have a monopole, P0(cos θ), theterm of leading magnitude.

• Second, in physical systems we rarely encounter pure multipoles. Forexample, the potential of the finite dipole (q at z = a, −q at z = −a)contained a P3(cos θ) term. These higher order terms may be eliminatedby shrinking the multipole to a point multipole, in this case keeping theproduct qa constant (a → 0, q → ∞) to maintain the same dipole moment.

• Third, the multipole expansion is not restricted to electrical phenomena.Planetary configurations are described in terms of mass multipoles.

It might also be noted that a multipole expansion is actually a decompositioninto the irreducible representations of the rotation group (Section 4.2). Thelth multipole involves the eigenfunctions of orbital angular momentum, |lm〉,one for each component m of the multipole l (see Chapter 4). These 2l + 1components of the multipole form an irreducible representation because thelowering operator L− applied repeatedly to the eigenfunction |ll〉 generates allother eigenfunctions |lm〉, down to m = −l. The raising and lowering opera-tors L± are generators of the rotation group along with Lz, whose eigenvalueis m.


EXERCISES

11.1.1 Develop the electrostatic potential for the array of charges shown. Thisis a linear electric quadrupole (Fig. 11.4).

q

zz = a

q-2q

z = - a

Figure 11.4

Linear ElectricQuadrupole

11.1.2 Calculate the electrostatic potential of the array of charges shownin Fig. 11.5. This is an example of two equal but oppositely directeddipoles. The dipole contributions cancel, but the octupole terms donot cancel.

z

q-2q+2q- q

2aa-az = -2a

Figure 11.5

Linear ElectricOctupole

11.1.3 Show that the electrostatic potential produced by a charge q at z = a

for r < a is

ϕ(r) = q

4πε0a

∞∑n= 0

(r

a

)n

Pn(cos θ).

11.1.4 Using E = −∇ϕ, determine the components of the electric field corre-sponding to the (pure) electric dipole potential

ϕ(r) = 2aq P1(cos θ)4πε0r

2.

Here, it is assumed that r � a.

ANS. Er = +4aq cos θ

4πε0r3

, Eθ = +2aq sin θ

4πε0r3

, Eϕ = 0.

11.1.5 A point electric dipole of strength p(1) is placed at z = a; a second pointelectric dipole of equal but opposite strength is at the origin. Keepingthe product p(1)a constant, let a → 0. Show that this results in a pointelectric quadrupole.


11.1.6 A point charge q is in the interior of a hollow conducting sphere ofradius r0. The charge q is displaced a distance a from the center of thesphere. If the conducting sphere is grounded, show that the potentialin the interior produced by q and the distributed induced charge is thesame as that produced by q and its image charge q ′. The image charge isat a distance a′ = r2

0 /a from the center, collinear with q and the origin(Fig. 11.6).

z

q q′

a′a

Figure 11.6

Image Charge q ′

Hint. Calculate the electrostatic potential for a < r0 < a′. Show thatthe potential vanishes for r = r0 if we take q ′ = −qr0/a.

11.1.7 Prove that

Pn(cos θ) = (−1)nrn+1

n!∂n

∂zn

(1r

).

Hint. Compare the Legendre polynomial expansion of the generatingfunction (a → �z; Fig. 11.1) with a Taylor series expansion of 1/r,where z dependence of r changes from z to z − �z (Fig. 11.7).

∆zz

r(x, y, z - ∆z)r (x, y, z)

Figure 11.7

Geometry forz → z − ∆z

11.1.8 By differentiation and direct substitution of the series form, Eq. (11.18),show that Pn(x) satisfies Legendre’s ODE. Note that we may have anyx, −∞ < x < ∞ and indeed any z in the entire finite complex plane.

11.2 Recurrence Relations and Special Properties 563

11.2 Recurrence Relations and Special Properties

Recurrence Relations

The Legendre polynomial generating function provides a convenient way of de-riving the recurrence relations6 and some special properties. If our generatingfunction [Eq. (11.4)] is differentiated with respect to t, we obtain

∂g(t, x)∂t

= x − t

(1 − 2xt + t2)3/2=

∞∑n= 0

nPn(x)tn−1. (11.23)

By substituting Eq. (11.4) into this and rearranging terms, we have

(1 − 2xt + t2)∞∑

n= 0

nPn(x)tn−1 + (t − x)∞∑

n= 0

Pn(x)tn = 0. (11.24)

The left-hand side is a power series in t. Since this power series vanishes for allvalues of t, the coefficient of each power of t is equal to zero; that is, our powerseries is unique (Section 5.7). These coefficients are found by separating theindividual summations and using appropriate summation indices as follows:

∞∑m=0

mPm(x)tm−1 −∞∑

n= 0

2nxPn(x)tn +∞∑

s=0

sPs(x)ts+1

+∞∑

s=0

Ps(x)ts+1 −∞∑

n= 0

xPn(x)tn = 0. (11.25)

Now letting m = n + 1, s = n − 1, we find

(2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x), n = 1, 2, 3, . . . . (11.26)

With this three-term recurrence relation we may easily construct the higherLegendre polynomials. If we take n = 1 and insert the values of P0(x) andP1(x) [Exercise 11.1.7 or Eq. (11.18)], we obtain

3xP1(x) = 2P2(x) + P0(x), or P2(x) = 12

(3x2 − 1).

This process may be continued indefinitely; the first few Legendre polynomialsare listed in Table 11.1.

Cumbersome as it may appear at first, this technique is actually more effi-cient for a computer than is direct evaluation of the series [Eq. (11.18)]. Forgreater stability (to avoid undue accumulation and magnification of round offerror), Eq. (11.26) is rewritten as

Pn+1(x) = 2xPn(x) − Pn−1(x) − [xPn(x) − Pn−1(x)]/(n + 1). (11.26a)

6We can also apply the explicit series form [Eq. (11.18)] directly.


One starts with P0(x) = 1, P1(x) = x, and computes the numerical valuesof all the Pn(x) for a given value of x, up to the desired PN(x). The values ofPn(x), 0 ≤ n < N are available as a fringe benefit.

To practice, let us derive another recursion relation from the generatingfunction.

EXAMPLE 11.2.1 Recursion Formula Consider the product

g(t, x)g(t, −x) = (1 − 2xt + t2)−1/2(1 + 2xt + t2)−1/2

= [(1 + t2)2 − 4x2t2]−1/2 = [t4 + 2t2(1 − 2x2) + 1]−1/2

and recognize the generating function, upon replacing t2 → t, 2x2 − 1 → x.

Using Eq. (11.4) and comparing coefficients of the power series in t we there-fore have derived

g(t, x)g(t, −x) =∑m,n

Pm(x)Pn(−x)tm+n =∑

N

PN(2x2 − 1)t2N ,

or, for m+ n = 2N and m+ n = 2N − 1, respectively,

PN(2x2 − 1) =2N∑

n= 0

P2N−n(x)Pn(−x), (11.27a)

2N−1∑n= 0

P2N−n−1(x)Pn(−x) = 0. (11.27b)

For N = 1 we check first that1∑

n= 0

P1−n(x)Pn(−x) = x · 1 − 1 · x = 0

and second that

P1(2x2 − 1) =2∑

n= 0

P2−n(x)Pn(−x) = x− x − x2 + 2(

32

x2 − 12

)= 2x2 − 1. ■

Differential Equations

More information about the behavior of the Legendre polynomials can beobtained if we now differentiate Eq. (11.4) with respect to x. This gives

∂g(t, x)∂x

= t

(1 − 2xt + t2)3/2=

∞∑n= 0

P ′n(x)tn (11.28)

or

(1 − 2xt + t2)∞∑

n= 0

P ′n(x)tn − t

∞∑n= 0

Pn(x)tn = 0. (11.29)

As before, the coefficient of each power of t is set equal to zero and we obtain

P ′n+1(x) + P ′

n−1(x) = 2xP ′n(x) + Pn(x). (11.30)


A more useful relation may be found by differentiating Eq. (11.26) withrespect to x and multiplying by 2. To this we add (2n + 1) times Eq. (11.30),canceling the P ′

n term. The result is

P ′n+1(x) − P ′

n−1(x) = (2n + 1)Pn(x). (11.31)

From Eqs. (11.30) and (11.31) numerous additional equations may be deve-loped,7 including

P ′n+1(x) = (n + 1)Pn(x) + xP ′

n(x), (11.32)

P ′n−1(x) = −nPn(x) + xP ′

n(x), (11.33)

(1 − x2)P ′n(x) = nPn−1(x) − nxPn(x), (11.34)

(1 − x2)P ′n(x) = (n + 1)xPn(x) − (n + 1)Pn+1(x). (11.35)

By differentiating Eq. (11.34) and using Eq. (11.33) to eliminate P ′n−1(x), we

find that Pn(x) satisfies the linear, second-order ODE

(1 − x2)P ′′n (x) − 2xP ′

n(x) + n(n + 1)Pn(x) = 0

or

d

dx

[(1 − x2)

dPn(x)dx

]+ n(n + 1)Pn(x) = 0. (11.36)

In the second form the ODE is self-adjoint. The previous equations, Eqs.(11.30)–(11.35), are all first-order ODEs but with polynomials of two differ-ent indices. The price for having all indices alike is a second-order differentialequation. Equation (11.36) is Legendre’s ODE. We now see that the polynomi-als Pn(x) generated by the power series for (1−2xt + t2)−1/2 satisfy Legendre’sequation, which, of course, is why they are called Legendre polynomials.

In Eq. (11.36) differentiation is with respect to x (= cos θ). Frequently,we encounter Legendre’s equation expressed in terms of differentiation withrespect to θ :

1sin θ

d

dθ

(sin θ

dPn(cos θ)dθ

)+ n(n + 1)Pn(cos θ) = 0. (11.37)

7Using the equation number in parentheses to denote the left-hand side of the equation, we maywrite the derivatives as

2 · ddx

(11.26) + (2n + 1) · (11.30) ⇒ (11.31)

12 {(11.30) + (11.31)} ⇒ (11.32)

12 {(11.30) − (11.31)} ⇒ (11.33)

(11.32)n→n−1 + x · (11.33) ⇒ (11.34)

ddx

(11.34) + n · (11.33) ⇒ (11.36).


Upper and Lower Bounds for Pn(cosθ)

Finally, in addition to these results, our generating function enables us to setan upper limit on |Pn(cos θ)|. We have

(1 − 2t cos θ + t2)−1/2 = (1 − teiθ )−1/2(1 − te−iθ )−1/2

= (1 + 1

2 teiθ + 38 t2e2iθ + · · · )

·(1 + 12 te−iθ + 3

8 t2e−2iθ + · · · ), (11.38)

with all coefficients positive. Our Legendre polynomial, Pn(cos θ), still thecoefficient of tn, may now be written as a sum of terms of the form

12 am(eimθ + e−imθ ) = am cos mθ (11.39a)

with all the am positive. Then

Pn(cos θ) =n∑

m=0 or 1

am cos mθ. (11.39b)

This series, Eq. (11.39b), is clearly a maximum when θ = 0 and all cos mθ = 1are maximal. However, for x = cos θ = 1, Eq. (11.6) shows that Pn(1) = 1.Therefore,

|Pn(cos θ)| ≤ Pn(1) = 1. (11.39c)

A fringe benefit of Eq. (11.39b) is that it shows that our Legendre polynomial

is a linear combination of cos mθ . This means that the Legendre polyno-

mials form a complete set for any functions that may be expanded in

series of cos mθ over the interval [0, π ].

SUMMARY In this section, various useful properties of the Legendre polynomials are de-rived from the generating function, Eq. (11.4). The explicit series representa-tion, Eq. (11.18), offers an alternate and sometimes superior approach.

EXERCISES

11.2.1 Given the series

α0 + α2 cos2 θ + α4 cos4 θ + α6 cos6 θ = a0 P0 + a2 P2 + a4 P4 + a6 P6,

express the coefficients αi as a column vector α and the coefficientsai as a column vector a and determine the matrices A and B such that

Aα = a and Ba = α.

Check your computation by showing that AB = 1 (unit matrix). Repeatfor the odd case

α1 cos θ +α3 cos3 θ +α5 cos5 θ +α7 cos7 θ = a1 P1 +a3 P3 +a5 P5 +a7 P7.


Note. Pn(cos θ) and cosn θ are tabulated in terms of each other inAMS-55.

11.2.2 By differentiating the generating function, g(t, x), with respect to t,multiplying by 2t, and then adding g(t, x), show that

1 − t2

(1 − 2tx + t2)3/2=

∞∑n= 0

(2n + 1)Pn(x)tn.

This result is useful in calculating the charge induced on a groundedmetal sphere by a point charge q.

11.2.3 (a) Derive Eq. (11.35)

(1 − x2)P ′n(x) = (n + 1)xPn(x) − (n + 1)Pn+1(x).

(b) Write out the relation of Eq. (11.35) to preceding equations in sym-bolic form analogous to the symbolic forms for Eqs. (11.31)–(11.34).

11.2.4 A point electric octupole may be constructed by placing a point elec-tric quadrupole (pole strength p(2) in the z-direction) at z = a and anequal but opposite point electric quadrupole at z = 0 and then lettinga → 0, subject to p(2)a = constant. Find the electrostatic potential cor-responding to a point electric octupole. Show from the constructionof the point electric octupole that the corresponding potential may beobtained by differentiating the point quadrupole potential.

11.2.5 Operating in spherical polar coordinates, show that

∂

∂z

[Pn(cos θ)

rn+1

]= −(n + 1)

Pn+1(cos θ)rn+2

.

This is the key step in the mathematical argument that the derivativeof one multipole leads to the next higher multipole.Hint. Compare Exercise 2.5.12.

11.2.6 From

PL(cos θ) = 1L!

∂L

∂tL(1 − 2t cos θ + t2)−1/2

∣∣t=0

show that

PL(1) = 1, PL(−1) = (−1)L.

11.2.7 Prove that

P ′n(1) = d

dxPn(x)

∣∣x=1 = 1

2n(n + 1).

11.2.8 Show that Pn(cos θ) = (−1)nPn(− cos θ) by use of the recurrence rela-tion relating Pn, Pn+1, and Pn−1 and your knowledge of P0 and P1.

11.2.9 From Eq. (11.38) write out the coefficient of t2 in terms of cos nθ , n ≤ 2.This coefficient is P2(cos θ).


11.3 Orthogonality

Legendre’s ODE [Eq. (11.36)] may be written in the form (Section 9.1)

d

dx[(1 − x2)P ′

n(x)] + n(n + 1)Pn(x) = 0, (11.40)

showing clearly that it is self-adjoint. Subject to satisfying certain boundaryconditions, then, we know that the eigenfunction solutions Pn(x) are orthog-onal. Upon comparing Eq. (11.40) with Eqs. (9.6) and (9.8) we see that theweight function w(x) = 1, L = (d/dx)(1 − x2)(d/dx), p(x) = 1 − x2 and theeigenvalue λ = n(n+ 1). The integration limits on x are ±1, where p(±1) = 0.Then for m �= n, Eq. (9.34) becomes

∫ 1

−1Pn(x)Pm(x) dx = 0,8 (11.41)

∫ π

0Pn(cos θ)Pm(cos θ) sin θ dθ = 0, (11.42)

showing that Pn(x) and Pm(x) are orthogonal for the interval [−1, 1].We need to evaluate the integral [Eq. (11.41)] when n = m. Certainly, it is

no longer zero. From our generating function

(1 − 2tx + t2)−1 =[ ∞∑

n= 0

Pn(x)tn

]2

. (11.43)

Integrating from x = −1 to x = +1, we have∫ 1

−1

dx

1 − 2tx + t2=

∞∑n= 0

t2n

∫ 1

−1[Pn(x)]2 dx. (11.44)

The cross terms in the series vanish by means of Eq. (11.41). Using y = 1 −2tx + t2, we obtain

∫ 1

−1

dx

1 − 2tx + t2= 1

2t

∫ (1+t)2

(1−t)2

dy

y= 1

tln

(1 + t

1 − t

). (11.45)

Expanding this in a power series (Exercise 5.4.1) gives us

1t

ln(

1 + t

1 − t

)= 2

∞∑n= 0

t2n

2n + 1. (11.46)

8In Section 9.4 such integrals are intepreted as inner products in a linear vector (function) space.Alternate notations are∫ 1

−1Pn(x)Pm(x) dx ≡ 〈Pn(x)|Pm(x)〉 ≡ (Pn(x), Pm(x)).

The 〈〉 form, popularized by Dirac, is common in physics literature. The form ( , ) is more commonin mathematics literature.

11.3 Orthogonality 569

Comparing power series coefficients of Eqs. (11.44) and (11.46), we must have∫ 1

−1[Pn(x)]2 dx = 2

2n + 1. (11.47)

Combining Eq. (11.41) with Eq. (11.47) we have the orthonormality condition∫ 1

−1Pm(x)Pn(x) dx = 2δmn

2n + 1. (11.48)

Therefore, Pn are not normalized to unity. We return to this normalizationin Section 11.5, when we construct the orthonormal spherical harmonics.

Expansion of Functions, Legendre Series

In addition to orthogonality, the Sturm–Liouville theory shows that the Legen-dre polynomials form a complete set. Let us assume, then, that the series

∞∑n= 0

anPn(x) = f (x), or | f 〉 =∑

n

an|Pn〉, (11.49)

defines f (x) in the sense of convergence in the mean (Section 9.4) in the inter-val [−1, 1]. This demands that f (x) and f ′(x) be at least sectionally continuousin this interval. The coefficients an are found by multiplying the series by Pm(x)and integrating term by term. Using the orthogonality property expressed inEqs. (11.42) and (11.48), we obtain

22m+ 1

am =∫ 1

−1Pm(x) f (x) dx = 〈Pm| f 〉 =

∑n

an〈Pm|Pn〉. (11.50)

We replace the variable of integration x by t and the index m by n. Then,substituting into Eq. (11.49), we have

f (x) =∞∑

n= 0

2n + 12

(∫ 1

−1f (t)Pn(t) dt

)Pn(x). (11.51)

This expansion in a series of Legendre polynomials is usually referred toas a Legendre series.9 Its properties are quite similar to the more familiarFourier series (Chapter 14). In particular, we can use the orthogonality prop-erty [Eq. (11.48)] to show that the series is unique.

On a more abstract (and more powerful) level, Eq. (11.51) gives the repre-sentation of f (x) in the linear vector space of Legendre polynomials (a Hilbertspace; Section 9.4).

Equation (11.51) may also be interpreted in terms of the projection oper-

ators of quantum theory. We may define a projection operator

Pm ≡ Pm(x)2m+ 1

2

∫ 1

−1Pm(t)[ ]dt

9Note that Eq. (11.50) gives am as a definite integral, that is, a number for a given f (x).


as an (integral) operator, ready to operate on f (t). [The f (t) would go in thesquare bracket as a factor in the integrand.] Then, from Eq. (11.50)

Pm f = amPm(x).10

The operator Pm projects out the mth component of the function f .

EXAMPLE 11.3.1 Legendre Expansion Expand f (x) = x(x + 1)(x − 1) in the interval −1 ≤x ≤ 1.

Because f (x) is odd under parity and is a third-order polynomial, we expectonly P1, P3. However, we check all coefficients:

2a0 =∫ 1

−1(x3 − x) dx =

[14

x4 − 12

x2] ∣∣∣∣

1

−1= 0, also by parity,

23

a1 =∫ 1

−1(x4 − x2) dx =

[15

x5 − 13

x3] ∣∣∣∣

1

−1= 2

5− 2

3= − 4

15,

25

a2 = 12

∫ 1

−1(x3 − x)(3x2 − 1) dx = 0, by parity;

27

a3 = 12

∫ 1

−1(x3 − x)(5x3 − 3x) dx = 1

2

∫ 1

−1(5x6 − 8x4 + 3x2) dx

= 12

[57

x7 − 85

x5 + x3] ∣∣∣∣

1

−1= 5

7− 8

5+ 1 = 4

35.

Finally, using a1, a3, we verify that − 25 x + 1

5 (5x3 − 3x) = x(x2 − 1). ■

Equation (11.3), which leads directly to the generating function definitionof Legendre polynomials, is a Legendre expansion of 1/r1. Going beyond asimple Coulomb field, the 1/r12 is often replaced by a potential V (|r1 − r2|)and the solution of the problem is again effected by a Legendre expansion.

The Legendre series, Eq. (11.49), has been treated as a known functionf (x) that we arbitrarily chose to expand in a series of Legendre polynomials.Sometimes the origin and nature of the Legendre series are different. In the nextexamples we consider unknown functions we know can be represented bya Legendre series because of the differential equation the unknown functionssatisfy. As before, the problem is to determine the unknown coefficients in theseries expansion. Here, however, the coefficients are not found by Eq. (11.50).Rather, they are determined by demanding that the Legendre series match aknown solution at a boundary. These are boundary value problems.

EXAMPLE 11.3.2 Sphere in a Uniform Field Another illustration of the use of Legendrepolynomials is provided by the problem of a neutral conducting sphere (radiusr0) placed in a (previously) uniform electric field (Fig. 11.8). The problem is to

10The dependent variables are arbitrary. Here, x came from the x in Pm.


E

V = 0

z

Figure 11.8

Conducting Spherein a Uniform Field

find the new, perturbed, electrostatic potential. The electrostatic potential11

V satisfies

∇2V = 0, (11.52)

Laplace’s equation. We select spherical polar coordinates because of the spher-ical shape of the conductor. (This will simplify the application of the boundarycondition at the surface of the conductor.) We can write the unknown potentialV (r, θ) in the region outside the sphere as a linear combination of solutionsof the Laplace equation, called harmonic polynomials (check by applying theLaplacian in spherical polar coordinates from Chapter 2):

V (r, θ) =∞∑

n= 0

anrnPn(cos θ) +∞∑

n= 0

bn

Pn(cos θ)rn+1

. (11.53)

No ϕ dependence appears because of the axial (azimuthal) symmetry ofour problem. (The center of the conducting sphere is taken as the origin andthe z-axis is oriented parallel to the original uniform field.)

It might be noted here that n is an integer because only for integral n is theθ dependence well behaved at cos θ = ±1. For nonintegral n, the solutionsof Legendre’s equation diverge at the ends of the interval [−1, 1], the polesθ = 0, π of the sphere (compare Exercises 5.2.11 and 8.5.5). It is for this samereason that the irregular solution of Legendre’s equation is also excluded.

11It should be emphasized that this is not a presentation of a Legendre series expansion of aknown V (cos θ). Here, we deal with a boundary value problem of a partial differential equation(see Section 8.9).


Now we turn to our (Dirichlet) boundary conditions to determine the un-known an and bn of our series solution, Eq. (11.53). If the original unperturbedelectrostatic field is E0 = |E0|, we require, as one boundary condition,

V (r → ∞) = −E0z = −E0r cos θ = −E0rP1(cos θ). (11.54)

Since our Legendre series is unique, we may equate coefficients of Pn(cos θ)in Eq. (11.53) (r → ∞) and Eq. (11.54) to obtain

an = 0, n > 1 and n = 0, a1 = −E0. (11.55)

If an �= 0 for n > 1, these terms would dominate at large r and the boundarycondition [Eq. (11.54)] could not be satisfied.

As a second boundary condition, we may choose the conducting sphereand the plane θ = π/2 to be at zero potential, which means that Eq. (11.53)now becomes

V (r = r0) = b0

r0+

(b1

r20

− E0r0

)P1(cos θ) +

∞∑n=2

bn

Pn(cos θ)

rn+10

= 0. (11.56)

In order that this may hold for all values of θ , each coefficient of Pn(cos θ)must vanish.12 Hence,

b0 = 0,13 bn = 0, n ≥ 2, (11.57)

whereas

b1 = E0r30 . (11.58)

The electrostatic potential (outside the sphere) is then

V = −E0rP1(cos θ) + E0r30

r2P1(cos θ) = −E0rP1(cos θ)

(1 − r3

0

r3

). (11.59)

It can be shown that a solution of Laplace’s equation that satisfies the boundaryconditions over the entire boundary is unique. The electrostatic potential V , asgiven by Eq. (11.59), is a solution of Laplace’s equation. It satisfies our boundaryconditions and therefore is the solution of Laplace’s equation for this problem.

It may further be shown (Exercise 11.3.13) that there is an induced surfacecharge density

σ = −ε0∂V

∂r

∣∣∣∣r=r0

= 3ε0 E0 cos θ (11.60)

12Again, this is equivalent to saying that a series expansion in Legendre polynomials (or anycomplete orthogonal set) is unique.13The coefficient of P0 is b0/r0. We set b0 = 0 since there is no net charge on the sphere. If thereis a net charge q, then b0 �= 0.


on the surface of the sphere and an induced electric dipole moment

P = 4πr30 ε0 E0. (11.61)

■

EXAMPLE 11.3.3 Electrostatic Potential of a Ring of Charge As a further example, con-sider the electrostatic potential produced by a conducting ring carrying a totalelectric charge q (Fig. 11.9). From electrostatics (and Section 1.14) the po-tential ψ satisfies Laplace’s equation. Separating variables in spherical polarcoordinates, we obtain

ψ(r, θ) =∞∑

n=0

cn

an

rn+1Pn(cos θ), r > a, (11.62a)

where a is the radius of the ring that is assumed to be in the θ = π/2 plane.There is no ϕ (azimuthal) dependence because of the cylindrical symmetry ofthe system. The terms with positive exponent in the radial dependence havebeen rejected since the potential must have an asymptotic behavior

ψ ∼ q

4πε0· 1

r, r � a. (11.62b)

The problem is to determine the coefficients cn in Eq. (11.62a). This may bedone by evaluating ψ(r, θ) at θ = 0, r = z, and comparing with an independent

y

z

x

r

a

q

q

(r, q )

Figure 11.9

Charged, ConductingRing


calculation of the potential from Coulomb’s law. In effect, we are using aboundary condition along the z-axis. From Coulomb’s law (with all chargeequidistant),

ψ(r, θ) = q

4πε0· 1

(z2 + a2)1/2,

{θ = 0r = z,

= q

4πε0z

∞∑s=0

(−1)s (2s)!22s(s!)2

(a

z

)2s

, z > a. (11.62c)

The last step uses the result of Exercise 10.1.15. Now, Eq. (11.62a) evaluatedat θ = 0, r = z [with Pn(1) = 1], yields

ψ(r, θ) =∞∑

n=0

cn

an

zn+1, r = z. (11.62d)

Comparing Eqs. (11.62c) and (11.62d), we get cn = 0 for n odd. Setting n = 2s,we have

c2s = q

4πε0(−1)s (2s)!

22s(s!)2, (11.62e)

and our electrostatic potential ψ(r, θ) is given by

ψ(r, θ) = q

4πε0r

∞∑s=0

(−1)s (2s)!22s(s!)2

(a

r

)2s

P2s(cos θ), r > a. (11.62f)

■

EXERCISES

11.3.1 You have constructed a set of orthogonal functions by the Gram–Schmidt process (Section 9.3), taking um(x) = xn, n = 0, 1, 2, . . . , inincreasing order with w(x) = 1 and an interval −1 ≤ x ≤ 1. Provethat the nth such function constructed is proportional to Pn(x).Hint. Use mathematical induction.

11.3.2 Expand the Dirac delta function in a series of Legendre polynomialsusing the interval −1 ≤ x ≤ 1.

11.3.3 Verify the Dirac delta function expansions

δ(1 − x) =∞∑

n=0

2n + 12

Pn(x),

δ(1 + x) =∞∑

n=0

(−1)n 2n + 12

Pn(x).

These expressions appear in a resolution of the Rayleigh plane waveexpansion (Exercise 11.4.7) into incoming and outgoing sphericalwaves.


Note. Assume that the entire Dirac delta function is covered whenintegrating over [−1, 1].

11.3.4 Neutrons (mass 1) are being scattered by a nucleus of mass A(A > 1).In the center of the mass system the scattering is isotropic. Then, inthe lab system the average of the cosine of the angle of deflection ofthe neutron is

〈cos ψ〉 = 12

∫ π

0

Acos θ + 1(A2 + 2Acos θ + 1)1/2

sin θ dθ.

Show, by expansion of the denominator, that 〈cos ψ〉 = 2/3A.

11.3.5 A particular function f (x) defined over the interval [−1, 1] is expandedin a Legendre series over this same interval. Show that the expansionis unique.

11.3.6 A function f (x) is expanded in a Legendre series f (x) =∑∞n=0 anPn(x). Show that

∫ 1

−1[ f (x)]2 dx =

∞∑n=0

2a2n

2n + 1.

This is the Legendre form of the Fourier series Parseval identity(Exercise 14.4.2). It also illustrates Bessel’s inequality [Eq. (9.73)]becoming an equality for a complete set.

11.3.7 Derive the recurrence relation

(1 − x2)P ′n(x) = nPn−1(x) − nxPn(x)

from the Legendre polynomial generating function.

11.3.8 Evaluate∫ 1

0 Pn(x) dx.

ANS. n = 2s; 1 for s = 0, 0 for s > 0,n = 2s + 1; P2s(0)/(2s + 2) = (−1)s(2s − 1)!!/1(2s + 2)!!

Hint. Use a recurrence relation to replace Pn(x) by derivatives andthen integrate by inspection. Alternatively, you can integrate the gen-erating function.

11.3.9 (a) For

f (x) ={+1, 0 < x < 1

−1, −1 < x < 0,

show that ∫ 1

−1[ f (x)]2 dx = 2

∞∑n=0

(4n + 3)[

(2n − 1)!!(2n + 2)!!

]2

.

(b) By testing the series, prove that the series is convergent.


11.3.10 Prove that∫ 1

−1x(1 − x2)P ′

nP ′m dx = 0, unless m = n ± 1,

= 2n(n2 − 1)4n2 − 1

δm,n−1, if m < n.

= 2n(n + 2)(n + 1)(2n + 1)(2n + 3)

δm,n+1, if m > n.

11.3.11 The amplitude of a scattered wave is given by

f (θ) = 1k

∞∑l=0

(2l + 1) exp[iδl] sin δl Pl(cos θ),

where θ is the angle of scattering, l is the angular momentum, hk isthe incident momentum, and δl is the phase shift produced by thecentral potential that is doing the scattering. The total cross sectionis σtot = ∫ | f (θ)|2d�. Show that

σtot = 4π

k2

∞∑l=0

(2l + 1) sin2 δl .

11.3.12 The coincidence counting rate, W (θ), in a gamma–gamma angularcorrelation experiment has the form

W (θ) =∞∑

n=0

a2nP2n(cos θ).

Show that data in the range π/2 ≤ θ ≤ π can, in principle, define thefunction W (θ) (and permit a determination of the coefficients a2n).This means that although data in the range 0 ≤ θ < π/2 may be usefulas a check, they are not essential.

11.3.13 A conducting sphere of radius r0 is placed in an initially uniformelectric field, E0. Show the following:(a) The induced surface charge density is

σ = 3ε0 E0 cos θ.

(b) The induced electric dipole moment is

P = 4πr30 ε0 E0.

The induced electric dipole moment can be calculated either fromthe surface charge [part (a)] or by noting that the final electricfield E is the result of superimposing a dipole field on the originaluniform field.

11.3.14 A charge q is displaced a distance a along the z-axis from the centerof a spherical cavity of radius R.(a) Show that the electric field averaged over the volume a ≤ r ≤ R

is zero.


(b) Show that the electric field averaged over the volume 0 ≤ r ≤ a is

E = zEz = −zq

4πε0a2(SI units) = −z

nqa

3ε0,

where n is the number of such displaced charges per unit volume. Thisis a basic calculation in the polarization of a dielectric.Hint. E = −∇ϕ.

11.3.15 Determine the electrostatic potential (Legendre expansion) of acircular ring of electric charge for r < a.

11.3.16 Calculate the electric field produced by the charged conducting ringof Exercise 11.3.15 for(a) r > a, (b) r < a.

11.3.17 Find the potential ψ(r, θ) produced by a charged conducting disk(Fig. 11.10) for r > a, the radius of the disk. The charge density σ

(on each side of the disk) is

σ (ρ) = q

4πa(a2 − ρ2)1/2, ρ2 = x2 + y2.

Hint. The definite integral you get can be evaluated as a beta function.For more details, see Section 5.03 of Smythe in Additional Reading.

ANS. ψ(r, θ) = q

4πε0r

∞∑l=0

(−1)l 12l + 1

(a

r

)2 l

P2 l(cos θ).

y

z

x

a

Figure 11.10

Charged, ConductingDisk

11.3.18 From the result of Exercise 11.3.17 calculate the potential of the disk.Since you are violating the condition r > a, justify your calculationcarefully.Hint. You may run into the hypogeometric series given in Exercise5.2.9.

11.3.19 The hemisphere defined by r = a, 0 ≤ θ < π/2 has an electrostatic po-tential +V0. The hemisphere r = a, π/2 < θ ≤ π has an electrostatic


potential −V0. Show that the potential at interior points is

V = V0

∞∑n=0

4n + 32n + 2

(r

a

)2n+1

P2n(0)P2n+1(cos θ)

= V0

∞∑n=0

(−1)n (4n + 3)(2n − 1)!!(2n + 2)!!

(r

a

)2n+1

P2n+1(cos θ).

Hint. You need Exercise 11.3.8.

11.3.20 A conducting sphere of radius a is divided into two electrically sep-arate hemispheres by a thin insulating barrier at its equator. The tophemisphere is maintained at a potential V0 and the bottom hemisphereat −V0.(a) Show that the electrostatic potential exterior to the two hemi-

spheres is

V (r, θ) = V0

∞∑s=0

(−1)s(4s + 3)(2s − 1)!!(2s + 2)!!

(a

r

)2s+2

P2s+1(cos θ).

(b) Calculate the electric charge density σ on the outside surface. Notethat your series diverges at cos θ = ±1 as you expect from the in-finite capacitance of this system (zero thickness for the insulatingbarrier).

ANS . σ = ε0 En = −ε0∂V

∂r

∣∣∣∣∣r=a

= ε0V0

∞∑s=0

(−1)s(4s + 3)(2s − 1)!!

(2s)!!P2s+1(cos θ).

11.3.21 In the notation of Section 9.4 〈x|ϕs〉 = √(2s + 1)/2Ps(x), a Legendre

polynomial is renormalized to unity. Explain how |ϕs〉〈ϕs| acts as aprojection operator. In particular, show that if | f 〉 = ∑

n a′n|ϕn〉, then

|ϕs〉〈ϕs| f 〉 = a′s|ϕs〉.

11.3.22 Expand x8 as a Legendre series. Determine the Legendre coefficientsfrom Eq. (11.50),

am = 2m+ 12

∫ 1

−1x8 Pm(x) dx.

Check your values against AMS-55, Table 22.9. This illustrates the ex-pansion of a simple function. Actually, if f (x) is expressed as a powerseries, the recursion Eq. (11.26) is both faster and more accurate.Hint. Gaussian quadrature can be used to evaluate the integral.

11.4 Alternate Definitions of Legendre Polynomials 579

11.3.23 Expand arcsin x in Legendre polynomials.

11.3.24 Expand the polynomials 2+5x, 1+x+x3 in a Legendre series and plotyour results and the polynomials for the larger interval −2 ≤ x ≤ 2.

11.4 Alternate Definitions of Legendre Polynomials

Rodrigues’s Formula

The series form of the Legendre polynomials [Eq. (11.18)] of Section 11.1 maybe transformed as follows. From Eq. (11.18)

Pn(x) =[n/2]∑r=0

(−1)r (2n − 2r)!2nr!(n − 2r)!

xn−2r. (11.63)

For n an integer

Pn(x) =[n/2]∑r=0

(−1)r 12nr!(n − r)!

(d

dx

)n

x2n−2r

= 12nn!

(d

dx

)n n∑r=0

(−1)rn!r!(n − r)!

x2n−2r. (11.64)

Note the extension of the upper limit. The reader is asked to show in Exer-cise 11.4.1 that the additional terms [n/2]+1 to n in the summation contributenothing. However, the effect of these extra terms is to permit the replacementof the new summation by (x2 − 1)n (binomial theorem once again) to obtain

Pn(x) = 12nn!

(d

dx

)n

(x2 − 1)n. (11.65)

This is Rodrigues’s formula. It is useful in proving many of the properties ofthe Legendre polynomials, such as orthogonality. A related application is seenin Exercise 11.4.3. The Rodrigues definition can be extended to define theassociated Legendre functions.

EXAMPLE 11.4.1 Lowest Legendre Polynomials For n = 0, P0 = 1 follows right away fromEq. (11.65), as well as P1(x) = 2x

2 = x. For n = 2 we obtain

P2(x) = 18

d2

dx2(x4 − 2x2 + 1) = 1

8(12x2 − 4) = 3

2x2 − 1

2,

and for n = 3

P3(x) = 148

d3

dx3(x6 − 3x4 + 3x2 − 1) = 1

48(120x3 − 72x) = 5

2x3 − 3

2x,

in agreement with Table 11.1. ■


EXERCISES

11.4.1 Show that each term in the summationn∑

r=[n/2]+1

(d

dx

)n (−1)rn!r!(n − r)!

x2n−2r

vanishes (r and n integral).

11.4.2 Using Rodrigues’s formula, show that the Pn(x) are orthogonal and that∫ 1

−1[Pn(x)]2 dx = 2

2n + 1.

Hint. Use Rodrigues’s formula and integrate by parts.

11.4.3 Show that∫ 1−1 xmPn(x)dx = 0 when m < n.

Hint. Use Rodrigues’s formula or expand xm in Legendre polynomials.

11.4.4 Show that ∫ 1

−1xnPn(x)dx = 2n+1n!n!

(2n + 1)!.

Note. You are expected to use Rodrigues’s formula and integrate byparts, but also see if you can get the result from Eq. (11.18) by inspec-tion.

11.4.5 Show that∫ 1

−1x2r P2n(x)dx = 22n+1(2r)!(r + n!)

(2r + 2n + 1)!(r − n)!, r ≥ n.

11.4.6 As a generalization of Exercises 11.4.4 and 11.4.5, show that theLegendre expansions of xs are

(a) x2r =r∑

n=0

22n(4n + 1)(2r)!(r + n)!(2r + 2n + 1)!(r − n)!

P2n(x), s = 2r,

(b) x2r+1 =r∑

n=0

22n+1(4n + 3)(2r + 1)!(r + n + 1)!(2r + 2n + 3)!(r − n)!

P2n+1(x),

s = 2r + 1.

11.4.7 A plane wave may be expanded in a series of spherical waves by theRayleigh equation

eikr cos γ =∞∑

n=0

an jn(kr)Pn(cos γ ).

Show that an = in(2n + 1).Hint.1. Use the orthogonality of the Pn to solve for an jn(kr).2. Differentiate n times with respect to (kr) and set r = 0 to eliminate

the r dependence.3. Evaluate the remaining integral by Exercise 11.4.4.

11.5 Associated Legendre Functions 581

Note. This problem may also be treated by noting that both sides ofthe equation satisfy the Helmholtz equation. The equality can be estab-lished by showing that the solutions have the same behavior at theorigin and also behave alike at large distances.

11.4.8 Verify the Rayleigh equation of Exercise 11.4.7 by starting with thefollowing steps:(a) Differentiate with respect to (kr) to establish∑

n

an j′n(kr)Pn(cos γ ) = i

∑n

an jn(kr) cos γ Pn(cos γ ).

(b) Use a recurrence relation to replace cos γ Pn(cos γ ) by a linear com-bination of Pn−1 and Pn+1.

(c) Use a recurrence relation to replace j′n by a linear combination of

jn−1 and jn+1. See Chapter 12 for Bessel functions.

11.4.9 In numerical work (Gauss–Legendre quadrature) it is useful to establishthat Pn(x) has n real zeros in the interior of [−1, 1]. Show that this isso.Hint. Rolle’s theorem shows that the first derivative of (x2 − 1)2n hasone zero in the interior of [−1, 1]. Extend this argument to the second,third, and ultimately to the nth derivative.

11.5 Associated Legendre Functions

When Laplace’s equation is separated in spherical polar coordinates(Section 8.9), one of the separated ODEs is the associated Legendre equation

1sin θ

d

dθ

(sin θ

dPmn (cos θ)

dθ

)+

[n(n + 1) − m2

sin2 θ

]Pm

n (cos θ) = 0. (11.66)

With x = cos θ , this becomes

(1−x2)d2

dx2Pm

n (x)−2xd

dxPm

n (x)+[n(n + 1) − m2

1 − x2

]Pm

n (x) = 0. (11.67)

If the azimuthal separation constant m2 = 0 we have Legendre’s equation,Eq. (11.36). The regular solutions (with m not necessarily zero), relabeledPm

n (x), are

Pmn (x) = (1 − x2)m/2 dm

dxmPn(x). (11.68)

These are the associated Legendre functions.14 Since the highest power of x

in Pn(x) is xn, we must have m ≤ n (or the m-fold differentiation will driveour function to zero). In quantum mechanics the requirement that m ≤ n has

14One finds (as in AMS-55) the associated Legendre functions defined with an additional factor of(−1)m. This phase (−1)m seems an unnecessary complication at this point. It will be included inthe definition of the spherical harmonics Y m

n (θ , ϕ). Note also that the upper index m is not an

exponent.


the physical interpretation that the expectation value of the square of thez-component of the angular momentum is less than or equal to the expectationvalue of the square of the angular momentum vector L (Section 4.3),

⟨L2

z

⟩ ≤ 〈L2〉 ≡∫

ψ∗nmL2ψnmd3r,

where m is the eigenvalue of Lz, and n(n+ 1) is the eigenvalue of L2. From theform of Eq. (11.68), we might expect m to be nonnegative. However, if Pn(x)is expressed by Rodrigues’s formula, this limitation on m is relaxed and wemay have −n ≤ m ≤ n, with negative as well as positive values of m beingpermitted. Using Leibniz’s differentiation formula once again, the reader mayshow that Pm

n (x) and P−mn (x) are related by

P−mn (x) = (−1)m(n − m)!

(n + m)!Pm

n (x). (11.69)

From our definition of the associated Legendre functions, Pmn (x),

P0n (x) = Pn(x). (11.70)

As with the Legendre polynomials, a generating function for the associatedLegendre functions is obtained via Eq. (11.67) from that of ordinary Legendrepolynomials:

(2m)!(1 − x2)m/2

2mm!(1 − 2tx + t2)m+1/2=

∞∑s=0

Pms+m(x)ts. (11.71)

If we drop the factor (1 − x2)m/2 = sinm θ from this formula and define thepolynomials Pm

s+m(x) = Pms+m(x)(1 − x2)−m/2, then we obtain a practical form

of the generating function

gm(x, t) ≡ (2m)!2mm!(1 − 2tx + t2)m+1/2

=∞∑

s=0

Pms+m(x)ts. (11.72)

We can derive a recursion relation for associated Legendre polynomialsthat is analogous to Eqs. (11.23) and (11.26) by differentiation as follows:

(1 − 2tx + t2)∂gm

∂t= (2m+ 1)(x − t)gm(x, t).

Substituting the defining expansions for associated Legendre polynomials weget

(1 − 2tx + t2)∑

s

sPms+m(x)ts−1 = (2m+ 1)

∑s

[xPm

s+mts − Pms+mts+1] .

Comparing coefficients of powers of t in these power series, we obtain therecurrence relation

(s + 1)Pms+m+1 − (2m+ 1 + 2s)xPm

s+m + (s + 2m)Pms+m−1 = 0. (11.73)

For m = 0 and s = n this relation is Eq. (11.26).


Before we can use this relation we need to initialize it, that is, relate theassociated Legendre polynomials with m = 1 to ordinary Legendre poly-nomials. We observe that

(1 − 2xt + t2)g1(x, t) = (1 − 2xt + t2)−1/2 =∑

s

Ps(x)ts (11.74)

so that upon inserting Eq. (11.72) we get the recursion

P1s+1 − 2xP1

s + P1s−1 = Ps(x). (11.75)

More generally, we also have the identity

(1 − 2xt + t2)gm+1(x, t) = (2m+ 1)gm(x, t), (11.76)

from which we extract the recursion

Pm+1s+m+1 − 2xPm+1

s+m + Pm+1s+m−1 = (2m+ 1)Pm

s+m(x), (11.77)

which relates the associated Legendre polynomials with superindex m+ 1 tothose with m. For m = 0 we recover the initial recursion Eq. (11.75).

EXAMPLE 11.5.1 Lowest Associated Legendre Polynomials Now we are ready to derivethe entries of Table 11.2. For m = 1 and s = 0 Eq. (11.75) yields P1

1 = 1because P1

0 = 0 = P1−1 do not occur in the definition, Eq. (11.72), of the

associated Legendre polynomials. Multiplying by (1−x2)1/2 = sin θ we get thefirst line of Table 11.2. For s = 1 we find from Eq. (11.75),

P12 (x) = P1 + 2xP1

1 = x + 2x = 3x,

from which the second line of Table 11.2, 3 cos θ sin θ , follows upon multi-plying by sin θ. For s = 2 we get

P13 (x) = P2 + 2xP1

2 − P11 = 1

2(3x2 − 1) + 6x2 − 1 = 15

2x2 − 3

2,

Table 11.2

Associated LegendreFunctions

P11 (x) = (1 − x2)1/2 = sin θ

P12 (x) = 3x(1 − x2)1/2 = 3 cos θ sin θ

P22 (x) = 3(1 − x2) = 3 sin2 θ

P13 (x) = 3

2 (5x2 − 1)(1 − x2)1/2 = 32 (5 cos2 θ − 1) sin θ

P23 (x) = 15x(1 − x2) = 15 cos θ sin2 θ

P33 (x) = 15(1 − x2)3/2 = 15 sin3 θ

P14 (x) = 5

2 (7x3 − 3x)(1 − x2)1/2 = 52 (7 cos3 θ − 3 cos θ) sin θ

P24 (x) = 15

2 (7x2 − 1)(1 − x2) = 152 (7 cos2 θ − 1) sin2 θ

P34 (x) = 105x(1 − x2)3/2 = 105 cos θ sin3 θ

P44 (x) = 105(1 − x2)2 = 105 sin4 θ


in agreement with line 4 of Table 11.2. To get line 3 we use Eq. (11.76). Form = 1, s = 0, this givesP2

2 (x) = 3P11 (x) = 3, and multiplying by 1−x2 = sin2 θ

reproduces line 3 of Table 11.2. For lines 5, 8, and 9, Eq. (11.72) may be used,which we leave as an exercise. ■

EXAMPLE 11.5.2 Special Values For x = 1 we use

(1 − 2t + t2)−m−1/2 = (1 − t)−2m−1 =∞∑

s=0

(−2m− 1s

)ts

in Eq. (11.72) and find

Pms+m(1) = (2m)!

2mm!

(−2m− 1s

). (11.78)

For m = 1, s = 0 we have P11 (1) = ( −3

0

) = 1; for s = 1, P12 (1) = −( −3

1

) = 3;and for s = 2, P1

3 (1) = ( −32

) = (−3)(−4)2 = 6 = 3

2 (5 − 1). These all agree withTable 11.2.

For x = 0 we can also use the binomial expansion, which we leave as anexercise. ■

EXAMPLE 11.5.3 Parity From the identity gm(−x, −t) = gm(x, t) we obtain the parity relation

Pms+m(−x) = (−1)sPm

s+m(x). (11.79)

■

We have the orthogonality integral∫ 1

−1Pm

p (x)Pmq (x) dx = 2

2q + 1· (q + m)!

(q − m)!δpq (11.80)

or, in spherical polar coordinates,∫ π

0Pm

p (cos θ)Pmq (cos θ) sin θ dθ = 2

2q + 1· (q + m)!

(q − m)!δpq . (11.81)

The orthogonality of the Legendre polynomials is a special case of thisresult, obtained by setting mequal to zero; that is, for m = 0, Eq. (11.80) reducesto Eqs. (11.47) and (11.48). In both Eqs. (11.80) and (11.81) our Sturm–Liouvilletheory of Chapter 9 could provide the Kronecker delta. A special calculationis required for the normalization constant.

Spherical Harmonics

The functions �m(ϕ) = eimϕ are orthogonal when integrated over the azimuthalangle ϕ, whereas the functions Pm

n (cos θ) are orthogonal upon integrating overthe polar angle θ . We take the product of the two and define

Y mn (θ , ϕ) ≡ (−1)m

√2n + 1

4π

(n − m)!(n + m)!

Pmn (cos θ)eimϕ (11.82)


Table 11.3

Spherical Harmonics(Condon-Shortley Phase)

Y 00 (θ , ϕ) = 1√

4π

Y11 (θ , ϕ) = −

√3

8πsin θeiϕ

Y 01 (θ , ϕ) =

√3

4πcos θ

Y−11 (θ , ϕ) = +

√3

8πsin θe−iϕ

Y22 (θ , ϕ) =

√5

96π3 sin2 θe2iϕ

Y12 (θ , ϕ) = −

√5

24π3 sin θ cos θeiϕ

Y 02 (θ , ϕ) =

√5

4π

(32

cos2 θ − 12

)

Y−12 (θ , ϕ) = +

√5

24π3 sin θ cos θe−iϕ

Y−22 (θ , ϕ) =

√5

96π3 sin2 θe−2iϕ

to obtain functions of two angles (and two indices) that are orthonormal overthe spherical surface. These Y m

n (θ , ϕ) are spherical harmonics. The completeorthogonality integral becomes∫ 2π

ϕ=0

∫ π

θ=0Y m1∗

n1(θ , ϕ)Y m2

n2(θ , ϕ) sin θ dθ dϕ = δn1n2δm1m2 (11.83)

and explains the presence of the complicated normalization constant inEq. (11.82).

The extra (−1)m included in the defining equation of Y mn (θ , ϕ) with −n ≤

m ≤ n deserves some comment. It is clearly legitimate since Eq. (11.68) islinear and homogeneous. It is not necessary, but in moving on to certain quan-tum mechanical calculations, particularly in the quantum theory of angularmomentum, it is most convenient. The factor (−1)m is a phase factor oftencalled the Condon–Shortley phase after the authors of a classic text on atomicspectroscopy. The effect of this (−1)m [Eq. (11.82)] and the (−1)m of Eq. (11.69)for P−m

n (cos θ) is to introduce an alternation of sign among the positive m

spherical harmonics. This is shown in Table 11.3.The functions Y m

n (θ , ϕ) acquired the name spherical harmonics becausethey are defined over the surface of a sphere with θ the polar angle and ϕ theazimuth. The “harmonic” was included because solutions of Laplace’s equationwere called harmonic functions and Y m

n (cos, ϕ) is the angular part of such asolution.

EXAMPLE 11.5.4 Lowest Spherical Harmonics For n = 0 we have m = 0 and Y 00 = 1√

4π

from Eq. (11.82). For n = 1 we have m = ±1, 0 and Y 01 =

√3

4πcos θ , whereas

for m = ±1 we see from Table 11.2 that cos θ is replaced by sin θ and we havethe additional factor (∓1) e±iϕ√

2, which checks with Table 11.3. ■


In the framework of quantum mechanics Eq. (11.67) becomes an orbitalangular momentum equation and the solution Y M

L (θ , ϕ) (n replaced by L andmby M) is an angular momentum eigenfunction, with L being the angular mo-mentum quantum number and M the z-axis projection of L. These relationshipsare developed in more detail in Section 4.3.

EXAMPLE 11.5.5 Spherical Symmetry of Probability Density of Atomic States What isthe angular dependence of the probability density of the degenerate atomicstates with principal quantum number n = 2?

Here we have to sum the absolute square of the wave functions for n = 2and orbital angular momentum l = 0, m = 0; l = 1, m = −1, 0, +1; that is,s and three p states. We ignore the radial dependence. The s state has orbitalangular momentum l = 0 and m = 0 and is independent of angles. For the p

states with l = 1 and m = ±1, 0

ψ200 ∼ Y 00 = 1√

4π, ψ21m ∼ Y m

1 ,

we have to evaluate the sum1∑

m=−1

|Y m1 |2 = 3

4π

[2

(1√2

sin θ

)2

+ cos2 θ

]= 1

upon substituting the spherical harmonics from Table 11.3. This result isspherically symmetric, as is the density for the s state alone or the sum ofthe three p states. These results can be generalized to higher orbital angularmomentum l. ■

SUMMARY Legendre polynomials are naturally defined by their generating function in amultipole expansion of the Coulomb potential. They appear in physical sys-tems with azimuthal symmetry. They also arise in the separation of partialdifferential equations with spherical or cylindrical symmetry or as orthogonaleigenfunctions of the Sturm–Liouville theory of their second-order differen-tial equation. Associated Legendre polynomials appear as ingredients of thespherical harmonics in situations that lack in azimuthal symmetry.

EXERCISES

11.5.1 Show that the parity of Y ML (θ , ϕ) is (−1)L . Note the disappearance of

any M dependence.Hint. For the parity operation in spherical polar coordinates, seeSection 2.5 and Section 11.2.

11.5.2 Prove that

Y ML (0, ϕ) =

(2L + 1

4π

)1/2

δM0.


11.5.3 In the theory of Coulomb excitation of nuclei we encounterY M

L (π/2, 0). Show that

Y ML

(π

2, 0

)=

(2L + 1

4π

)1/2 [(L − M)!(L + M)!]1/2

(L − M)!!(L + M)!!(−1)(L+M)/2

for L + M even,

= 0 for L + M odd.

Here,

(2n)!! = 2n(2n − 2) · · · 6 · 4 · 2,

(2n + 1)!! = (2n + 1)(2n − 1) · · · 5 · 3 · 1.

11.5.4 (a) Express the elements of the quadrupole moment tensor xixj as alinear combination of the spherical harmonics Y m

2 (and Y 00 ).

Note. The tensor xixj is reducible. The Y 00 indicates the presence

of a scalar component.(b) The quadrupole moment tensor is usually defined as

Qij =∫

(3xixj − r2δij)ρ(r) dτ,

with ρ(r) the charge density. Express the components of(3xixj − r2δij) in terms of r2Y M

2 .(c) What is the significance of the −r2δij term?

Hint. Contract the indices i, j.

11.5.5 The orthogonal azimuthal functions yield a useful representation ofthe Dirac delta function. Show that

δ(ϕ1 − ϕ2) = 12π

∞∑m=−∞

exp[im(ϕ1 − ϕ2)].

11.5.6 Derive the spherical harmonic closure relation∞∑

l=0

+l∑m=−l

Y ml (θ1, ϕ1)Y m∗

l (θ2, ϕ2) = 1sin θ1

δ(θ1 − θ2)δ(ϕ1 − ϕ2)

= δ(cos θ1 − cos θ2)δ(ϕ1 − ϕ2).

11.5.7 The quantum mechanical angular momentum operators Lx ± iLy aregiven by

Lx + iLy = eiϕ

(∂

∂θ+ i cot θ

∂

∂ϕ

),

Lx − iLy = −e−iϕ

(∂

∂θ− i cot θ

∂

∂ϕ

).

Show that

(a) (Lx + iLy)Y ML (θ , ϕ), =

√(L − M)(L + M + 1)Y M+1

L (θ , ϕ),

(b) (Lx − iLy)Y ML (θ , ϕ), =

√(L − M)(L − M + 1)Y M−1

L (θ , ϕ).


11.5.8 With L± given by

L± = Lx ± iLy = ±e±iϕ

[∂

∂θ± i cot θ

∂

∂ϕ

],

show that

(a) Y ml =

√(l + m)!

(2l)!(l − m)!(L−)l−mY l

l ,

(b) Y ml =

√(l − m)!

(2l)!(l + m)!(L+)l+mY−l

l .

11.5.9 In some circumstances it is desirable to replace the imaginary ex-ponential of our spherical harmonic by sine or cosine. Morse andFeshbach define

Yemn = Pm

n (cos θ) cos mϕ,

Yomn = Pm

n (cos θ) sin mϕ,

where∫ 2π

0

∫ π

0[Ye or o

mn (θ , ϕ)]2 sin θ dθ dϕ = 4π

2(2n + 1)(n + m)!(n − m)!

, n = 1, 2, . . .

= 4π for n= 0 (Yo00 is undefined).

These spherical harmonics are often named according to the patternsof their positive and negative regions on the surface of a sphere zonalharmonics for m = 0, sectoral harmonics for m = n, and tesseralharmonics for 0 < m < n. For Ye

mn, n = 4, m = 0, 2, 4, indicate on adiagram of a hemisphere (one diagram for each spherical harmonic)the regions in which the spherical harmonic is positive.

11.5.10 A function f (r, θ , ϕ) may be expressed as a Laplace series

f (r, θ , ϕ) =∑l,m

almr lY ml (θ , ϕ).

With 〈〉sphere used to mean the average over a sphere (centered on theorigin), show that

〈 f (r, θ , ϕ)〉sphere = f (0, 0, 0).

Additional Reading

Hobson, E. W. (1955). The Theory of Spherical and Ellipsoidal Harmonics.Chelsea, New York. This is a very complete reference, which is the classictext on Legendre polynomials and all related functions.

Smythe, W. R. (1989). Static and Dynamic Electricity, 3rd ed. McGraw-Hill,New York.

See also the references listed in Section 4.4 and at the end of Chapter 13.

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Legendre Polynomials and Spherical Harmonicswilliamsgj.people.cofc.edu/Legendre Function.pdfChapter 11 Legendre Polynomials and Spherical Harmonics 11.1 Introduction Legendre polynomials