PHY3063 R. D. Field Department of Physics Chapter6_10.doc University of Florida The Infinite Square Well (4) Particle in a One-Dimensional Box: Consider the solution of )()()()(2222xExxVdxxdmψ=+−h, where h/)(),(iEtextx−=Ψ, for the case V(x) = ∞if x ≤-L/2and V(x) = ∞if x ≥L/2and V(x) = 0 for -L/2 < x < L/2 (Note that now V(-x) = V(x)). As before we have 0)(=xfor x ≤-L/2 and x ≥L/2For -L/2 < x < L/2we have )()(2)(2222xkxmEdxxd−=−=hwhere 22hmEk=. The most general solution is )sin(')cos(')(kxBkxABeAexikxikx+=+=−where A' and B' are constants. Boundary Conditions:We require that ψ(x) be “square-integrable” and that it be continuous and “single valued”. Thus at x = L/2(1)0)2/sin(')2/cos(')2/(=+==kLBkLALx. At x = -L/2we have (2)0)2/sin(')2/cos(')2/(=−
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