Question

Consider a particle of mass in a 10 finite potential well of height V. the domain – a < x < a. a) Show that solutions for – a < x < a take the form on (x) = A cos(knx) for odd n, and on (x) = A sin(knx) for even n. . Show a) Match the boundary conditions at x = a to prove that cos(ka) = Bk where k is the wave vector for -a < x < a, and ß = - that for Vo → oo this reduces to the infinite square well solutions kn = "7 for all n = 1, 2, 3, 4, 5, .... 12ml b) Show that the relative amplitude of the ground state wave function at the boundary x = a is equal to the square root of the ratio of the ħ2 K2 kinetic energy inside the box to the potential Vo, that is, show cos(ka) = 2mV 22 c) The function cos(kx)can be expanded as a Taylor series in the parameter Ak about the positions kn = 27. Use this technique to show that the ground state wave vector becomes k' = 2a + Ak = 2078). Show that the ground state energy is therefore E1 = = 7., which 4m(a+b)2 "?" suggests that the parameter ß acts like an effective "growth distance" of the well width due to the finite height of the well. This effective stretch lowers the energy of the ground state. Is this effective "growth distance" B the same for E2, E3, ...? When is this Taylor series technique valid? d) Show that the kinetic energy p2 = h?k? of the particle for x > a is negative! 2m

Answer 1

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For odd So17 Y( 94) = DAin( Pa) (%) A e-T%2 - Ad e DxB Co( Pa2) eun cot ( P9) = this is carlled Treinscendeted ean. let 17 with sadiwg y Cirele Called eyn 3- - 35 boyn d sak one bound slake

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