PHY4604 R. D. Field Department of Physics Chapter2_3.doc University of Florida The Infinite Square Well (1) Particle in a One-Dimensional Box: Consider the solution of )()()()(2222xExxVdxxdmψ=+−h, where h/)(),(iEtextx−=Ψ, for the case V(x) = ∞if x ≤0and V(x) = ∞if x ≥Land V(x) = 0 for0 < x < L. For x ≤0and x ≥Lwe have 0)()()()()()(2222=→=+−∞→xxxVExdxxdxmVVh. For 0 < x < Lwe have )()(2)(2222xkxmEdxxd−=−=hwhere 22hmEk=. The most general solution is ikxikxBeAex−+=)(where A and B are constants. Bounday Conditions:We require that ψ(x) be “square-integrable” and that it be continuous and “single valued”. Thus at x = 00)0(=+==BAxand hence )sin(2)()(kxiAeeAxikxikx=−=−. At x = Lwe have 0)sin(2)(===kLiALxwhich implies that kL = nπwith n = 1, 2, 3,... Energy Levels:We see that only certain values of kare allowed which means that only the following energies are allowed: 022
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