4.6: Square Potential Well
( \newcommand{\kernel}{\mathrm{null}\,}\)
Consider a particle of mass m and energy E interacting with the simple square potential well
V(x)={−V0 for −a/2≤x≤a/20 otherwise
Now, if E>0 then the particle is unbounded. Thus, when the particle encounters the well it is either reflected or transmitted. As is easily demonstrated, the reflection and transmission probabilities are given by Equations ([e5.28]) and ([e5.29]), respectively, where k2=2mEℏ2,q2=2m(E+V0)ℏ2.
Suppose, however, that E<0. In this case, the particle is bounded (i.e., |ψ|2→0 as |x|→∞). Is is possible to find bounded solutions of Schrödinger’s equation in the finite square potential well ([e5.71])?
Now, it is easily seen that independent solutions of Schrödinger’s equation ([e5.2]) in the symmetric [i.e., V(−x)=V(x)] potential ([e5.71]) must be either totally symmetric [i.e., ψ(−x)=ψ(x)], or totally anti-symmetric [i.e., ψ(−x)=−ψ(x)]. Moreover, the solutions must satisfy the boundary condition ψ→0 as |x|→∞
Let us, first of all, search for a totally symmetric solution. In the region to the left of the well (i.e. x<−a/2), the solution of Schrödinger’s equation which satisfies the boundary condition ψ→0 and x→−∞ is ψ(x)=Aekx,
Let y=qa/2. It follows that E=E0y2−V0,
Now, the solutions to Equation ([e5.84]) correspond to the intersection of the curve √λ−y2/y with the curve tany. Figure [well] shows these two curves plotted for a particular value of λ. In this case, the curves intersect twice, indicating the existence of two totally symmetric bound states in the well. Moreover, it is evident, from the figure, that as λ increases (i.e., as the well becomes deeper) there are more and more bound states. However, it is also evident that there is always at least one totally symmetric bound state, no matter how small λ becomes (i.e., no matter how shallow the well becomes). In the limit λ≫1 (i.e., the limit in which the well becomes very deep), the solutions to Equation ([e5.84]) asymptote to the roots of tany=∞. This gives y=(2j−1)π/2, where j is a positive integer, or q=(2j−1)πa.
For the case of a totally anti-symmetric bound state, similar analysis to the preceding yields −y√λ−y2=tany.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)