(84) Particle in Finite Square Potential Well

By symmetry the solution in the region to the right

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Unformatted text preview: ion in the region to the right of the well (i.e., ) is (1184) The solution inside the well (i.e., ) that satisfies the symmetry constraint is (1185) where (1186) and is a constant. The appropriate matching conditions at the edges of the well (i.e., and ) are that both be continuous [because a discontinuity in the wavefunction, or its first derivative, would generate a singular term in the time independent Schrödinger equation (i.e., the term involving ) that could not be balanced]. The matching conditions yield (1187) Let . It follows that (1188) where (1189) Moreover, Equation (1187) becomes (1190) with (1191) Here, must lie in the range , in order to ensure that Figure : The curves (solid) and lies in the range . Figure : The curves (solid) and (dashed), calculated for . The latter curve takes the value 0 when . The solutions of Equation (1190) correspond to the intersection of the curve . Figure 82 shows these two curves plotted for a particular value of with the curve . In this case, the curves intersect twice, indicating the existence of two totally symmetric bound states in the well. It is apparent, from the figure, that as increases (i.e., as the well becomes deeper) there are more and more bound states. However, it is also apparent that there is always at least one totally symmetric bound state, no matter how small no matter how shallow the well becomes). In the limit becomes (i.e., (i.e., the limit in which the...
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This note was uploaded on 08/25/2013 for the course PHY 315 taught by Professor Staff during the Fall '08 term at University of Texas at Austin.

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