Question

2. A particle is confined to the interval (-L/2. L/2) by infinite potentials for rs -L/2 and * 1/2 - Votol - ܘܐ V(x) - [+ o 0 (+ for Is-L/2 for -L/2<x<L/2 for r 1/2 ܝ 02 This is the same as the "particle-in-one-dimensional-box" model of Problem 1, except the origin of the coordinate is taken at the midpoint of the interval. With this choice of the ori gin, potential energy function V () of the particle-in-one-dimensional-box" model becomes 2. When V(x) is even, eigenfunctions of the Hamiltonian (ie, energy eigenfunctions) are either even or odd. Use this fact to find the even and odd eigenfunctions of energy and the corresponding energy levels

Answer 1

Pauiele in one Dimensional Box Snside Sn tuis problema pautick ot mass placud in One dimensional box ot length pauticle fs tree to Move Box has Intinite wadl assumed that the potential energ pautiele s Rero 3t is ereuy wheu Snsidu the box (x)0 Thu time Sndependant onu dimensional Schrodinger equation toill be Brm tE-v)0 - 0) + Outsida tu box equation ) taukes tu torm + 81 m 1E-00) 0 clx Negleching is (empauison to co, uwe have OT dx

That is eo Outsida the box This means that the pauticle cannot exist out side th vegion With in th box th Schrondigeu equation tor th motion ot tha pauticle takes th form +20 dx2 8m2me tor equation (2) is A guneual Solution =A Sin(ax) + B Cos(x) (-80 Wheue A and B au Constants a- k- Constant Take ka h2 k2h2 8r2 m

Condi tions Boundauy d For first 0, x- 1 -Asink () + 6 Cos K ) sin(-0)--Sino and Cos(-0)- tCoso 0F ASink+ BCos Gi) 2 i and Gi) 2 BC0O Caskf=o c-(ii) 2A SinkoA Sinko A o Or Sinki o Possible Solubions A- 0 (oske -o Kl nn 2(1,2.5)

B Cos ( nn n)r (n = 153,5) 8EO SinkKe - o k 2 2 Knn (n a,4,6) Pin) Asin nH2 for odd x)=B (os n 1 (1,3, for even PCn) Asin(n (ns2,4,0) Eneugy Ra enugy 2 K2h2 8n2m n272h2 n2h 2 8mez