The functions which are of interest in the theory of the hydrogen atom are fn(x) =

xy
= xy + y

y
= 3x2 y

xy
= y

y

= 4y

y

= y

y

2y
+ y = 0

x2 y

3xy
+ 3y = 0

(x2 + 2x)y

2(x + 1)y
+ 2y = 0

(x2 + 1)y

2xy
+ 2y = 0

y

4xy
+ (4x2 2)y = 0

x x3

7x4 3x + 1

x5

Show that any polynomial of degree n can be written as a linear combination of Legendre polynomials with l n.

Expand the potential V = K/d in (5.11) in the following way in order to see how the terms depend on the tensors mentioned above. In Figure 5.1 let R have the coordinates X, Y , Z and r have coordinates x, y, z. [Note: The coordinate x here is not the x in (5.14).] Then V = K d = K[(X x) 2 + (Y y) 2 + (Z z) 2 ] 1/2 . Consider X, Y , Z as constants and expand V (x, y, z) in a three variable power series about the origin. (See Chapter 4, Section 2, for discussion of two-variable power series and generalize the method.) You should find V =K R + K R2 X R x + + K R3 3 2 X2 R2 1 2 x2 + + 3 2 X R Y R 2xy + + and similar terms in y, z, y2, xz, and so on. Now letting r = ri and K = K
qi for a charge distribution as in (5.18), and summing (or integrating) over the charge distribution, show that: the first term is just (K
/R) total charge; the next group of terms (in x, y, z) involve the three components of the electric dipole moment; the sum of these terms is (K
/R2)component of the dipole moment in the R direction; the next group (quadratic terms) involve six quantities of the form ZZZ x2 d and similar y, z integrals, ZZZ 2xy d and similar xz, yz integrals. If we split the 2xy term into xy and yx (and similarly for the 2xz and 2yz terms), we have the nine components of a 2nd-rank tensor called the quadrupole moment. Use the direct product method of Chapter 10, Section 2 to show that it is a 2nd-rank

Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let f(x) be the given function to be approximated. Let the functions pl(x) be the normalized Legendre polynomials, that is, pl(x) = r2l + 1 2 Pl(x) so that Z 1 1 [pl(x)]2 dx = 1. Show that the Legendre series for f(x) as far as the p2(x) term is f(x) = c0p0(x) + c1p1(x) + c2p2(x) with cl = Z 1 1 f(x)pl(x) dx. Write the quadratic polynomial satisfying the least squares condition as b0p0(x) + b1p1(x) + b2p2(x) (by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find b0, b1, b2 so that I = Z 1 1 [f(x) (b0p0(x) + b1p1(x) + b2p2(x))]2 dx is a minimum. Square the bracket and write I as a sum of integrals of the individual terms. Show that some of the integrals are zero by orthogonality, some are 1 because the pls are normalized, and others are equal to the coefficients cl. Add and subtract c2 0 + c2 1 + c2 2 and show that I = Z 1 1 [f 2 (x)+(b0 c0) 2 + (b1 c1) 2 + (b2 c2) 2 c 2 0 c 2 1 c 2 2] dx. Now determine the values of the bs to make I as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree n.

xy

+ y
+ 9xy = 0

y2(x)

Computer plot on the same axes several Ip(x) functions together with their common asymptotic approximation. Then computer plot each function with its small x approximation.

As in Problem 19, study the Kp(x) functions. It is interesting to note (see Problem 17.4) that K1/2(x) is equal to the asymptotic approximation.

Verify that the polynomials Lk n(x) in (22.25) satisfy (22.26). Hint: Write (22.21) with n replaced by n + k and differentiate k times by Leibniz rule.

Verify that the polynomials given by (22.27) are the same as the Lk n(x) defined in (22.25). Hints: Show that the functions in (22.27) satisfy (22.26) as follows. Let v = exxn+k and show that xv
= (n+kx)v. (Compare Problem 14.) Differentiate this equation n + 1 times by Leibniz rule, and use dnv/dxn = n! exxkLk n(x) from (22.27). Also show that the coefficient of xn in both (22.25) and (22.27) is (1)n/n! [Thus, assuming that (22.26) for one k has only one polynomial solution of degree n (which can be shown by series solution), (22.27) gives the same polynomials as (22.25) for integral k.]

Verify the recursion relation relations (22.28) as follows: (a) In (22.24b), replace n by n + k and differentiate k times by Leibniz rule; in (22.24a), replace k by n + k and differentiate k 1 times. Subtract k times the second result from the first. (b) In (22.24c), replace n by n + k and differentiate k times

Show that the functions Lk n(x) are orthogonal on (0,) with respect to the weight function xkex. Hint: Write the differential equation (22.26) as xkex d dx(xk+1exy
) + ny = 0 and see Sections 7 and 19.

Evaluate the normalization integrals (22.29) and (22.30). Hints: Use (22.27) for one of the Lk n(x) factors in (22.29); integrate by parts n times. Use (22.25) and then (22.19) to evaluate dn/dxnLk n(x). Compare Problem 19. To evaluate (22.30), multiply (22.28a) by xkex and integrate; use (22.29) both for m = n and m = n.

Solve the following eigenvalue problem (see end of Section 2 and Problem 11): Given the differential equation y

+ x 1 4 l(l + 1) x2 y = 0 where l is an integer 0, find values of such that y 0 as x , and find the corresponding eigenfunctions. Hint: let y = xl+1ex/2v(x), and show that v(x) satisfies the differential equation xv

+ (2l + 2 x)v
+ ( l 1)v = 0. Compare (22.26) to show that if is an integer > l, there is a polynomial solution v(x) = L2l+1 l1(x).

The functions which are of interest in the theory of the hydrogen atom are fn(x) = xl+1ex/2nL2l+1 nl1 x n where n and l are integers with 0 l n 1. (Note that here k = 2l + 1, and we have replaced n by n l 1; in this problem L3 2 , say, means l = 1, n = 4.) For l = 1, show that f2(x) = x2 ex/4 , f3(x) = x2 ex/6 4 x 3 , f4(x) = x2 ex/8 10 5x 4 + x2 32 . Hint: Find the polynomials L3 0, L3 1, L3 2 as in Problem 20 (with k = 3) and then replace x by x/n. The functions fn(x) are very different from those in (22.29) since x/n changes from one function to the next. However, it can be shown (Problem 23.25) that for one fixed l, the set of functions fn(x), n l + 1, is an orthogonal set on (0,). Verify this for these three functions. Hint: The integrals are functionssee Chapter 11, Section 3.

Repeat Problem 27 for l = 0, n = 1, 2, 3.

Show that Rp = p x D and Lp = p x + D where D = d/dx, are raising and lowering operators for Bessel functions, that is, show that RpJp(x) = Jp+1(x) and LpJp(x) = Jp1(x). Hint: Use equations (15.5). Note that these operators depend on p as well as x, so they are not as simple as the Hermite function raising and lowering operators (22.7) and (22.8). If you want to operate, say, on Jp+1, you must change p in R or L to p+ 1, etc. Making this adjustment, show that the equations LRJp = Jp and RLJp = Jp both give Bessels equation.

Find raising and lowering operators (see Problem 29) for spherical Bessel functions. Hint: See problems 17.15 and 17.16.