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differential equation applications 215x match0V(x)x–V 0–a aHigh ELow E0x0Figure 9.7 Left: Computed wave function and the square-well potential (bold lines). Thewave function computed by integration in from the left is matched to the one computed byintegration in from the right (dashed curve) at a point near the left edge of the well. Notehow the wave function decays rapidly outside the well. Right: A first guess at a wave functionwith energy E that is 0.5% too low (dotted curve). We see that the left wave function does notvary rapidly enough to match the right one at x = 500. The solid curve shows a second guessat a wave function with energy E that is 0.5% too high. We see that now the left wavefunction varies too rapidly.logarithmic derivative, to be continuous encapsulates both continuity conditionsinto a single condition and is independent of ψ’s normalization.6. Even though we do not know ahead of time which energies E or κ values areeigenvalues, we still need a starting value for the energy in order to use ourODE solver. Such being the case, we start the solution with a guess for theenergy. A good guess for ground-state energy would be a value somewhat upfrom that at the bottom of the well, E>−V 0 .7. Because it is unlikely that any guess will be correct, the left- and right-wavefunctions will not quite match at x = x match (Figure 9.7). This is okay becausewe can use the amount of mismatch to improve the next guess. We measurehow well the right and left wave functions match by calculating the difference∆(E,x)= ψ′ L (x)/ψ L(x) − ψ R ′ (x)/ψ R(x)ψL ′ (x)/ψ L(x)+ψR ′ (x)/ψ R(x) ∣ , (9.54)x=xmatchwhere the denominator is used to avoid overly large or small numbers. Nextwe try a different energy, note how much ∆(E) has changed, and use this todeduce an intelligent guess at the next energy. The search continues until theleft- and right-wave ψ ′ /ψ match within some tolerance.−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 215