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.

## REFERENCES

Awkwardly, the odd solutions $(\psi n(\u2212x)=\u2212\psi n(x))$ correspond to even indices and vice versa.

A particularly interesting example is $z0=\pi \u22154$ using $c1$. In this case $z1=\pi \u221523+\pi =0.633840$, and the exact answer is 0.633 159.

In this case $f1\u22152$ yields a quartic equation for the energies, while $f1$ leads to a cubic. Naturally, they provide more accurate results.

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 $z0=\pi \u22154$ the formula reduces to $zn=(\pi \u22152)(n\u22122+4\u22122n)$, but in this case there are only two solutions: $z1=(\pi \u22152)(2\u22121)=0.6506$ (compare Ref. 15) and a spurious $z2=0$.

Reference 1, Eq. (2.125).

The strategy is to nail down the two ends (0 and $\pi \u22152$). Equation (14) (using $s=1\u22152$ and $c1$) yields a more accurate approximation, $tanx\u2248x1\u2212(2\u2215\pi )4x2\u2215[1\u2212(2x\u2215\pi )2]$, but this approximation leads to a sixth-order polynomial equation for $xn$. A reasonable compromise is given by $tanx\u2248x(1\u2212cx2)\u2215[1\u2212(2x\u2215\pi )2]$, which gives a cubic equation for $xn$.

For $z>5$ we already know that $z1\u2248z2\u2248z0\u22152$.

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

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