8.02 - ESG Independent Study Spring 2005 Unit 5: Mathematical Appendix

Those with experience in 18.03 techniques or theory may recognize that since the separated equation for f_{n}(θ), given in Unit 5 as

0 = n(n + 1)f_{n} + \frac{1}{sin θ} \frac{d}{dθ} (sin θ \frac{d}{dθ} f_{n}) ,

is a second order linear ordinary differential equation, and should have two linearly independent solutions in any interval not including sin θ = 0, in our case 0 < θ < π. The form of the solution presented, that is, a polynomial of degree n for integer n ≥ 0, seems to disdain the other solution. To cut to the chase, the other solution can be shown to have an infinite power series solution which diverges as sin θ → 0, and so is unacceptable.

For more detailed sources on the following, consult any Differential Equations text; check the index under “Legendre’s Equation” or “Legendre Polynomials.” Another good source is Jackson’s “Classical Electrodynamics”; the page numbers change with edition, but either check the index or the back endpapers.

Before getting fancy, let’s “do the drill” in the simple case n = 0, in which case we find, more or less by inspection, a solution f_{0,1} ≡ 0 (the extra subscript will be 1 or 2, just to distinguish two indpendent solutions). Let’s do better, and integrate the above equation (with n = 0) to find

sin θ \frac{d}{dθ} f_{n} = −c_{2}
\frac{d}{dθ} f_{n} = −c_{2} csc θ
f_{n} = c_{2} ln (csc θ + cot θ) + c_{1}.

Note that the names given the constants of integration were seriously rigged so that c_{1} = 1 corresponds to f_{0,1}, and we can take c_{2} = 1 to correspond to f_{0,2}. Although it doesn’t matter right now, it will help later to note that

f_{0,2} = ln (csc θ + cot θ) = \frac{1}{2} ln (\frac{1 + cos θ}{1 − cos θ} ) .

This shows us two things: First, the argument of the square root, and hence the logarithm, is never negative, so we don’t have to worry about absolute value signs. Next, the solution clearly diverges at cos θ = ±1, and so is not part of our physical solution.

The above procedure gets nasty for integer n > 0, though. For instance, for n = 1, f_{1,1} = cos θ, use the standard “Reduction of Order” technique; that is, let f_{1,2} = v(θ) f_{1,1} = v(θ) cos θ, insert into equation, turn crank, cross fingers and arrive at

\frac{v′′(θ)}{v′(θ)} = \frac{d}{dθ} ln(v′(θ) = 2 tan θ − cot θ  ⟶  v′(θ) = sec^{2} θ csc θ

if done correctly (hey, it took me a few tries). The primes denote differentiation with respect to θ. Play with this last expression at your leisure; I know of no way to do the needed integral that does not involve integration by parts. The result is f_{1,2} = 1 − \frac{cos θ}{2} ln (\frac{1+cos θ}{1−cos θ} ), which has similar difficulties at cos θ = ±1.

Before going on, note that f_{0,1} is a (trivially) even function of cos θ, f_{0,2} is an odd function of cos θ, f_{1,2} is an odd function of cos θ and f_{1,2} is an even function of cos θ. Indeed, this pattern can be shown to be valid for all integer n ≥ 0.

To look at these functions further and a bit more rigorously, it helps to make the change of dependent variable x = cos θ; indeed, this substitution is almost the only way to do the above integral to find v from v′. The differential equation for f_{n} then becomes

n(n + 1) f_{n} + \frac{d}{dx} ((1 − x^{2}) \frac{d}{dx}f_{n}) = 0.

In this form, the usual methods of obtaining series solutions (“Frobenius Solutions”) may be employed. The technique won’t be reproduced in these notes, but it can be seen that near x = 0 (cos θ = 0, θ = π/2), two independent Frobenius Series solutions exist, one with lowest power x^{0} and one with lowest power x^{1}, regardless of the value of n. It is further seen, however, that if n is an integer, one series will truncate after the x^{n} term, and the other will have an infinite series expression which diverges logarithmically at x → ±1.

The reason for presenting this in this appendix is to emphasize that the two solutions are not found by considering the problem as an “exceptional case;” the roots of the indicial equation, 0 and 1, while differing by an integer, both lead to solutions valid for −1 < x < 1, but we discard the divergent solutions if we want results physically valid for the whole sphere.

Warning: Quantum Mechanics is, of course, different. Sometimes we want and even seek divergent solutions.