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214 chapter 9the Schrödinger equation (9.48) becomesd 2 ( )ψ(x) 2mdx 2 +¯h 2 V 0 − κ 2 ψ(x)=0, for |x|≤a, (9.51)d 2 ψ(x)dx 2 − κ 2 ψ(x)=0, for |x| >a. (9.52)To evaluate the ratio of constants here, we insert c 2 , the speed of light squared, intoboth the numerator and the denominator [L 96, Appendix A.1]:2m¯h 2= 2mc2(¯hc) 2 ≃ 2 × 940 MeV(197.32 MeV fm) 2 =0.0483 MeV−1 fm −2 . (9.53)9.11 Combined Algorithms: Eigenvalues viaODE Solver Plus SearchThe solution of the eigenvalue problem combines the numerical solution of the ordinarydifferential equation (9.48) with a trial-and-error search for a wave functionthat satisfies the boundary conditions (9.49). This is done in several steps:1. Start on the very far left at x = −X max ≃−∞, where X max ≫ a. Assume thatthe wave function there satisfies the left-hand boundary condition:ψ L (x = −X max )=e +κx = e −κXmax .2. Use your favorite ODE solver to step ψ L (x) in toward the origin (to theright) from x = −X max until you reach the matching radius x match . The exactvalue of this matching radius is not important, and our final solution shouldbe independent of it. On the left in Figure 9.7, we show a sample solutionwith x match = −a; that is, we match at the left edge of the potential well. Inthe middle and on the right in Figure 9.7 we see some guesses that do notmatch.3. Start on the very far right, that is, at x =+X max ≃ +∞, with a wave functionthat satisfies the right-hand boundary condition:ψ R (x =+κX max )=e −κx = e −κXmax .4. Use your favorite ODE solver (e.g., rk4) to step ψ R (x) in toward the origin(to the left) from x =+X max until you reach the matching radius x match . Thismeans that we have stepped through the potential well (Figure 9.7).5. In order for probability and current to be continuous at x = x match , ψ(x) andψ ′ (x) must be continuous there. Requiring the ratio ψ ′ (x)/ψ(x), called the−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 214