1 Physics 116C Fall 2012. The spherical harmonics 1. Solution to laplace 's equation in spherical coordinates In spherical coordinates, the Laplacian is given by 2.. ~ 2 1 2 1 1. = 2 r + 2 2 sin + 2 2 . (1). r r r r sin r sin 2. We shall solve laplace 's equation, ~ 2 T (r, , ) = 0 , (2). using the method of separation of variables, by writing T (r, , ) = R(r) ( ) ( ) . Inserting this decomposition into the laplace equation and multiplying through by r 2 /R . yields 1 1 d2 .. 1 d 2 dR 1 1 d d . r + sin + = 0. R dr dr sin d d sin2 d 2. Hence, 1 d2 sin2 d . 2 dR sin d d . 2. = r sin = m2 , (3). d R dr dr d d . where m2 is the separation constant, which is chosen negative so that the solutions for ( ) are periodic in , (. eim . ( ) = for m = 0, 1, 2, 3, .. e im . Note that m must be an integer since is a periodic variable and ( + 2 ) = ( ). In the case of m = 0, the general solution is ( ) = a + b, but we must choose a = 0 to be consistent with ( + 2 ) = ( ).)
2 Hence in the case of m = 0, only one solution is allowed. One can now recast eq. (3) in the following form, m2.. 1 d 2 dR 1 1 d d . r = sin + = ( + 1) , (4). R dr dr sin d d sin2 . where the separation variable at this step is denoted by ( + 1) for reasons that will shortly become clear. The resulting radial equation, d2 R dR. r2 2. + 2r ( + 1)R = 0 , (5). dr dr 1. is recognized to be an Euler equation. This means that the solution is of the form R = r s . To determine the exponent s, we insert this solution back into eq. (5). The end result is s(s + 1) = ( + 1) = s = or s = 1 . That is, (. r . R(r) =. r 1. Eq. (4) also yields m2.. 1 d d . sin + ( + 1) = 0. sin d d sin2 . Changing variables to x = cos and y = ( ), the above differential equation reduces to 2. m2.. 2 d y dy (1 x ) 2 2x + ( + 1) y = 0. (6). dx dx sin2 . This is the differential equation for associated Legendre polynomials, y(x) = P m (x) , for = 0, 1, 2, 3.)
3 And m = , + 1, .. , 1, . The restrictions of and |m| to non-negative integers with |m| is a consequence of the requirement that P m (x) should be non-singular at cos = 1. On p. 583, Boas gives the following result in eq. ( ), +m 1 2 m/2 d . P m (x) = (1 x ) x2. 1 , for m = , + 1, .. , 1, . (7). 2 ! dx +m The differential equation for the associated Legendre polynomials, given in eq. (6), depends on m2 and is therefore not sensitive to the sign of m. Consequently, P m (x) and P m (x). must be equivalent solutions and hence proportional to each other. Using eq. (7), it is straightforward to prove that ( m)! m P m (cos ) = ( 1)m P (cos ) . (8). ( + m)! . Combining all the results obtained above, we have found that the general solution to laplace 's equation is of the form im . r m e T (r, , ) = 1 P (cos ) im . r e where = 0, 1, 2, 3, .. and m = , + 1, .. , 1, . 2. 2. The spherical harmonics In obtaining the solutions to laplace 's equation in spherical coordinates, it is traditional to introduce the spherical harmonics , Y m ( , ), s (.)
4 (2 + 1) ( m)! = 0, 1, 2, 3, .. , Y m ( , ) = ( 1)m P m (cos ) eim , for 4 ( + m)! m = , + 1, .. , 1, . (9). m The phase factor ( 1) , introduced originally by Condon and Shortley, is convenient for applications in quantum mechanics. Note that eq. (8) implies that Y m ( , ) = ( 1)m Y m ( , ) , (10). where the star means complex conjugation. The normalization factor in eq. (9) has been chosen such that the spherical harmonics are normalized to one. In particular, these func- tions are orthonormal and complete. The orthonormality relation is given by: Z.. Y m ( , ) Y m ( , ) d = mm , (11). where d = sin d d is the differential solid angle in spherical coordinates. The complete- ness of the spherical harmonics means that these functions are linearly independent and there does not exist any function of and that is orthogonal to all the Y m ( , ) where . and m range over all possible values as indicated above.
5 The completeness property of the spherical harmonics implies that any well-behaved function of and can be written as . X. X. f ( , ) = a m Y m ( , ) . (12). =0 m= . for some choice of coefficients a m . For convenience, we list the spherical harmonics for = 0, 1, 2 and non-negative values of m. 1. = 0, Y00 ( , ) = . 4 . r 1 3. Y ( , ) = sin ei .. 1.. 8 .. = 1, r .. Y 0 ( , ) = 3. cos .. 1. 4 . r 1 15. Y22 ( , ) = sin2 e2i .. 4 2 .. r 15.. = 2, Y21 ( , ) = sin cos ei .. 8 .. r 1 5.. 0. Y2 ( , ) = (3 cos2 1).. 2 4 . The corresponding spherical harmonics for negative values of m are obtained using eq. (10). 3. In addition, eq. (9) yields the following useful relation, r 0 2 + 1. Y ( , ) = P (cos ) , for = 0, 1, 2, 3, .. , (13). 4 . which relates the Legendre polynomials to the spherical harmonics with m = 0. In terms of the spherical harmonics , the general solution to laplace 's equation can be written as: X.
6 X . T (r, , ) = (a m r + b m r 1 )Y m ( , ) . =0 m= . In particular, ~ 2 (a m r + b m r 1)Y m ( , ) = 0 .. Making use of eq. (1) for the Laplacian and using . 2 . r f (r) = ( + 1)f (r) , for f (r) = ar + br 1 , r r it follows from eq. (2) that1. ~ 2 Y m ( , ) = ( + 1)Y m ( , ) . r 2 (15). 3. The laplace series As noted in the previous section, the completeness property of the spherical harmonics implies that any well-behaved function of and can be written as . X. X. f ( , ) = a m Y m ( , ) . (16). =0 m= . 1. Indeed, eq. (1) can be written as ~2. ~2= 1 . r 2 L. 2 , (14). r2 r r r ~ 2 is the differential operator, where L. 1 2.. ~2 1 . L sin , sin2 sin2 2. which depends only on the angular variables and . Then, eqs. (14) and (15) imply that: ~ 2 Y m ( , ) = ( + 1)Y m ( , ) . L. ~ 2 , with corresponding That is, the spherical harmonics are eigenfunctions of the differential operator L. eigenvalues ( + 1), for = 0, 1, 2, 3.
7 4. The coefficients a m can be determined from eq. (11). Namely, multiply both sides of . eq. (16) by Y m ( , ).. and then integrate over all solid angles. Using eq. (11), the right hand side of eq. (16) becomes . X. X. a m mm = a m . =0 m= . Dropping the primes, we conclude that Z. a m = f ( , )Y m ( , ) d . (17). Two special cases are notable. First, suppose that the function f ( , ) is independent of . In this case, we write f ( , ) = f ( ) and eq. (17) yields Z 2 Z 1. a m = N m e im . d f ( )P m (cos )d cos , 0 1. where we have written Y m ( , ) = N m P m (cos )eim , where N m is the normalization constant exhibited in eq. (9). However, the integral over is straightforward, Z 2 . e im d = 2 m0 , 0. where the Kronecker delta indicates that the above integral is nonzero only when m = 0. Using eq. (13), we end up with p Z 1. a 0 = (2 + 1) f ( )P (cos )d cos . 1. This means that the laplace series reduces to a sum over Legendre polynomials.
8 2 + 1 1. X Z. f ( ) = c P (cos ) , where c = f ( )P (cos )d cos . =0. 2 1. The coefficients c are related to the a 0 by r 2 + 1. c = a 0 . 4 . That is, for problems with azimuthal symmetry, the laplace series reduces to a sum over Legendre polynomials. The second special case of interest is one in which f ( , ) satisfies ~ 2 f ( , ) = ( + 1)f ( , ) . r 2 (18). In this case, we can conclude that . X. f ( , ) = bm Y m ( , ) . (19). m= . 5. The proof is simple. First, we expand f ( , ) in a laplace series, . X X . f ( , ) = a m Y m ( , ) . (20). =0 m = . If we operate on both sides of the above equation with the Laplacian and use eqs. (14) and (18), then it follows that: . X X . [ ( + 1) ( + 1)] a m Y m ( , ) = 0 . =0 m = . Since the spherical harmonics are linearly independent (and complete), the overall coeffi- . cient of Y m ( , ) in the above equation must vanish for any choice of and m . Thus, for 6= we conclude that a m = 0 , for 6= and m = , + 1.
9 , 1, . Inserting this result into eq. (20), only the term = survives in the sum over , and we end up with X .. f ( , ) = a m Y m ( , ) . m = . Dropping the primes on m and identifying a m bm , we end up with eq. (19), as was to be proven. 4. The addition formula for the spherical harmonics Suppose we have two vectors r = (r, , ) , ~ r = (r , , ) , ~. which are designated by their spherical coordinates, as shown in Figure 1 below. The angle between these two vectors, denoted by , is easily computed. Noting that the unit vectors are given by r = i sin cos + cos , j sin sin + k and similarly for r , it follows that cos = r r = cos cos + sin sin cos( ) . (21). The addition theorem for the spherical harmonics is . 4 X m P (cos ) = Y ( , )Y m ( , ) . 2 + 1 m= . 6. Figure 1: The two vectors ~. r and ~r are denoted in this figure by x and x , respectively. The angle between the two vectors is denoted by . Note that if one sets = 1 in the addition theorem and employs P1 (cos ) = cos , then the result coincides with eq.
10 (21) (check this!). That is, the addition theorem generalizes the geometric relation exhibited by eq. (21). To prove the addition theorem, we first note that ~ 2 P (cos ) = ( + 1)P (cos ) . r 2 (22). This result is justified by considering a coordinate system in which the z-axis is aligned along ~. r . In this coordinate system, is the polar angle of the vector ~. r . Noting that r 4 . P (cos ) = Y 0 ( , ) , 2 + 1 . where is the corresponding azimuthal angle in the new coordinate system, it then follows that eq. (22) is a consequence of eq. (15). However, ~ 2 is a scalar operator which is invariant with respect to rigid rotations of the coordinate system. The length r is also invariant with respect to rotations. Thus, eq. (22) must be true in the original coordinate system as well! By virtue of eq. (21), P (cos ) can be viewed as a function of and (where and . are held fixed). Thus, we can expand this function in a laplace series.