# Notes9 - Lesson 9 I Particle In A Finite Well We return to...

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Lesson 9 I. Particle In A Finite Well We return to solving simple systems that can be used to model real physical systems like the nucleus. The first system considered is a particle of energy E in a finite well of width L and height U o as finite as shown below. ( 29 = < < L x 0 elsewhere 0 0 U x U A. Solution Method 1. Since the potential is not a function of time, we already know the time dependent part of the solution and can solve the Time Independent Schrodinger Equation to find the spatial part of the solution. Ψ - = Ψ 2 2 2 k dx d where ( 29 2 2 U E 2m k - = x 0 L U o E

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2. We can simplify the problem by breaking the problem into regions where the wave number is constant (i.e. regions of constant U). In this case, we can break the problem into following three regions: Region I: 0 x < < - Region II: L x 0 Region III: < < x L The advantage to this approach is that we know the solution to the problem in each region by inspection. The disadvantage is that we have to use properties of the wave function and its derivatives at the boundaries between the various regions to merge our individual solutions into the total wave function solution. B. Solution For Region I In this region, we have the differential equation I 2 2 I 2 dx d Ψ = Ψ α where ( 29 2 o 2 E U 2m - α As we have discussed previously in this course, the general solution to this differential equation is ( 29 x x I e B e A x α α + = Ψ - C. Solution For Region III In this region, we have the same differential equation as region I III 2 2 III 2 dx d Ψ = Ψ α where ( 29 2 o 2 E U 2m - α
Thus, the general solution for region III is of the same form as the solution for region I. ( 29 x x III e F e E x α α + = Ψ - D. Solution For Region II In this region, we have the differential equation II 2 2 II 2 k dx d Ψ - = Ψ where 2 2 E m 2 k As we have discussed previously in this course, the general solution to this differential equation is ( 29 ( 29 ( 29 kx cos D kx sin C x II + = Ψ E. Applying Boundary Conditions To Ψ and Ψ There are four boundaries in this problem. The boundaries are located at - = x , 0 x = , L x = , and + = x . To represent a quantum system, the wave function must have the following required properties: 1. The wave function must be continuous and finite. 2. The wave function must go to zero at ± = x . 3. The wave function must be normalizable. 1 dx = Ψ Ψ - 4. The spatial derivative of the wave function must be continuous at all points in space where the potential function is finite.

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