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differential equation applications 213Although we say we are solving for the energy E, in practice we solve for the wavevector κ. The energy is negative for bound states, and so we relate the two byThe Schrödinger equation then takes the formκ 2 = − 2m¯h 2 E = 2m2|E|. (9.47)¯hd 2 ψ(x)dx 2− 2m¯h 2 V (x)ψ(x)=κ2 ψ(x). (9.48)When our problem tells us that the particle is bound, we are being told that it isconfined to some finite region of space. The only way to have a ψ(x) with a finiteintegral is to have it decay exponentially as x →±∞(where the potential vanishes):{e −κx , for x → +∞,ψ(x) →e +κx , for x →−∞.(9.49)In summary, although it is straightforward to solve the ODE (9.46) with thetechniques we have learned so far, we must also require that the solution ψ(x)simultaneously satisfies the boundary conditions (9.49). This extra condition turnsthe ODE problem into an eigenvalue problem that has solutions (eigenvalues) for onlycertain values of the energy E. The ground-state energy corresponds to the smallest(most negative) eigenvalue. The ground-state wave function (eigenfunction),which we must determine in order to find its energy, must be nodeless and even(symmetric) about x =0. The excited states have higher (less negative) energiesand wave functions that may be odd (antisymmetric).9.10.1 Model: Nucleon in a BoxThe numerical methods we describe are capable of handling the most realisticpotential shapes. Yet to make a connection with the standard textbook case and topermit some analytic checking, we will use a simple model in which the potentialV (x) in (9.46) is a finite square well (Figure 9.7):V (x)={−V0 = −83 MeV, for |x|≤a =2fm,0, for |x| >a=2fm,(9.50)where values of 83 MeV for the depth and 2 fm for the radius are typical fornuclei (these are the units in which we solve the problem). With this potential−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 213