A finite potential energy function U(x) allows the solution of the time-independent Schrödinger equation. to penetrate the classically forbidden region. Without assuming any particular function for U(x) show that b(x) must have an inflection point at any value of x where it enters a classically forbidden region.
The answer of given problem is
The wave function
By definition, the inflection point is a point where the total energy is equal to zero. Total energy E is equal to potential energy Uo
Traditionally authorised and classically banned territories are separated by turning points. A turning point is a point at which the second derivative vanishes.
So, the answer is
Simple models are very useful. Consider the twin finite wells shown in the figure, at First with a tiny separation. Then with increasingly distant separations, In all case, the four lowest allowed wave functions are planned on axes proportional to their energies. We see that they pass through the classically forbidden region between the wells, and we also see a trend. When the wells are very close, the four functions and energies are what we might expect of a single finite well, but as they move apart, pairs of functions converge to intermediate energies.
(a) The energies of the second and fourth states decrease. Based on changing wavelength alone, argue that is reasonable.
(b) The energies of the first and third states increase. Why? (Hint: Study bow the behaviour required in the classically forbidden region affects these two relative to the others.)
(c) The distant wells case might represent two distant atoms. If each atom had one electron, what advantage is there in bringing the atoms closer to form a molecule? (Note: Two electrons can have the same wave function.)
It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant . First, use the continuity/smoothness conditions to eliminate , , and in favor of in (21). Then make the change of variables and use the trigonometric relations
functions in region I, . The change of variables shifts the problem so that it is symmetric about , which requires that the probability density be symmetric and thus that be either an odd or even function of . By comparing the region II and region III functions, argue that this in turn demands that must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as and . Finally, plug these separately back into the region I solutions and show that
Note that is now a standard multiplicative normalization constant. Setting the integral of over all space to 1 would give it in terms of and , but because we can’t solve (22) exactly for k (or E), neither can we obtain an exact value for .
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