We investigate the problem of a quantum particle in a one-dimensional finite square well. In the standard approach the allowed energies are determined implicitly as the solutions to a transcendental equation. We obtain the spectrum analytically as the solution to a pair of parametric equations and algebraically using a remarkably accurate approximation to the cosine function. The approach is also applied to a variety of other quantum wells.

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16.
Reference 1, Sec. 2.6.
17.

Awkwardly, the odd solutions (ψn(x)=ψn(x)) correspond to even indices and vice versa.

18.
This technique was used in Ref. 2, pp.
209
210
.
19.

A particularly interesting example is z0=π4 using c1. In this case z1=π23+π=0.633840, and the exact answer is 0.633 159.

20.

A plot similar to Fig. 5 was used by Pitkanen4 to analyze the bound-state solutions graphically.

21.

In this case f12 yields a quartic equation for the energies, while f1 leads to a cubic. Naturally, they provide more accurate results.

22.

For z0=15 Eq. (22) yields 1.49 for n=1 and 14.44 for n=10; the correct values are 1.47 and 14.42, respectively (see Table I). For z0=π4 the formula reduces to zn=(π2)(n2+42n), but in this case there are only two solutions: z1=(π2)(21)=0.6506 (compare Ref. 15) and a spurious z2=0.

23.

Reference 1, Eq. (2.125).

24.

The strategy is to nail down the two ends (0 and π2). Equation (14) (using s=12 and c1) yields a more accurate approximation, tanxx1(2π)4x2[1(2xπ)2], but this approximation leads to a sixth-order polynomial equation for xn. A reasonable compromise is given by tanxx(1cx2)[1(2xπ)2], which gives a cubic equation for xn.

25.

For z>5 we already know that z1z2z02.

26.

Using the four-parameter fit, Eq. (45), leads to a cubic equation for z1.

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