Example we consider the operator lx d dx 109 x d dx

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (θ) = λf (θ), or f (θ) − λf (θ) = 0 104 has the general solution f (θ) = ae( √ λθ ) + be−( √ λθ ) . Note that a = 0 ⇒ f (θ) → ±∞ as θ → ∞, while b = 0 ⇒ f (θ) → ±∞ as θ → −∞. Neither of these is consistent with the periodicity conditions f (θ + 2π ) = f (θ), so we conclude that a = b = 0, and we obtain no nontrivial solutions in this case. √ Case 3: λ < 0. In this case, we set ω = −λ, and rewrite the eigenvalue problem as f (θ) + ω 2 f (θ) = 0, f (θ + 2π ) = f (θ). We recognize once again our old friend, the differential equation of simple harmonic motion, which has the general solution f (θ) = a cos(ωθ) + b sin(ωθ) = A sin(ω (θ − θ0 )). The periodicity condition f (θ + 2π ) = f (θ) implies that ω = n, where n is an integer, which we can assume is positive, and f (θ) = an cos(nθ) + bn sin(nθ). Thus we see that the general solution to the eigenvalue problem (4.26) is λ=0 and f (θ) = a0 , 2 or λ = −n2 where n is a positive integer and f (θ) = an cos(nθ) + bn sin(nθ). Now we need to find the corres...
View Full Document

This document was uploaded on 01/12/2014.

Ask a homework question - tutors are online