Question

The Finite Square Wel A more realistic version of the infinite square well potential has a finite well depth: -a V(x)--V for -a<x <a for x <-a,'r > a =0 This assignment will consider the bound states of a particle (of mass m) in this potential (i.e. total energy E <0). (1) Determine the general solutions to the time-independent Schrödinger equation for the three regions x <-a, -a<x <a, and > a. Write these solutions in terms of k and j. given by 2m(E+%) h2 -2mE 22mV Since-V。< E < 0, both k and j are real. Also, k, + j.--- Note: although the general solution for - a can be written as a combination oy complex exponential functions, it is easiest in this example to write it as a combination of sine and cosine functions. 12 marks (2) Use the condition that the wavefunction must remain finite to eliminate any non- physical terms from the general solutions in part (1), and thus write the general form of the wavefunction.

Answer 1

"Given V(x) = -V_0 for -a < x < a = 0 for x < -a, x > a we have the following observations The potential energy is zero when Vertical-bar X Vertical-bar > a So we can gauss plane wave solutions in These regions when E > 0 The potential energy is negative ""-V_0"" So we are interested in the bound states, i.e. energy eigenstates which are normalizable the width of the potential is 2a So we are leaking at the following Scenario \r\n d^2 psi/dx^2 = -2m/h^2 (E + V_0)psi Vertical-bar X Vertical-bar < a j^2 = 2m/h^2 (V_0 + E) with E = - Vertical-bar E Vertical-bard^2 psi/dx^2 = -j^2 psi So V_0 + E = V_0 - Vertical-bar E Vertical-bar is the kinetic energy of the particle. So the only possible even solution to the problem isVertical-bar X Vertical-bar < a psi(x) = cos(jx) psi^'' = -2m(E - 0)/h^2 psi = 2m Vertical-bar E Vertical-bar /h^2 psi Vertical-bar X Vertical-bar > a psi^'' = k^2 psi The solution to this owing "