Boundary Conditions in Quantum Mechanics: Solving Problems with Delta Function Potential -, Assignments of Quantum Physics

Download Boundary Conditions in Quantum Mechanics: Solving Problems with Delta Function Potential - and more Assignments Quantum Physics in PDF only on Docsity! Physics 486 Discussion Week 5 Solution See section 2.5.2 in Griffiths. Both problems are solved in Griffiths. Here are the important points: • Because V is not finite at x = 0, dψ/dx has a discontinuity there: d! dx + " d! dx + = " 2m# ! 2 ! (0) . This is obtained by integrating the SEQ from -ε to +ε (See Griffiths, eq. 2.123, p. 72). • ψ itself is continuous, because dψ/dx only suffers a finite discontinuity. 1) Consider a delta function potential, V(x) = αδ(x). (Note that α has dimensions of energy times length. Make sure you understand this.) Remember that ! (x) = 0, if x " 0 #, if x = 0 , and ! (x)dx =# # $ = 1. Consider α ≤ 0. Calculate the bound states of this potential. For x ≠ 0, the solution is simple, because E < 0: ψ(x) = Aeκx, for x < 0, and = Ae-κx, for x > 0, where κ = √(-2mE)/. Continuity requires the same A. A and κ are determined by the two boundary conditions: • Normalization: 2A/κ = 1 • Discontinuity of dψ/dx: -2Aκ = -(2mα/2)A So, κ = mα/2, or E = -mα2/22. There is exactly one bound state.