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`$$ \tan\left( \frac{L}{2}\sqrt{\frac{2m}{\hbar^2} \, W^+} \right) ~=~ \sqrt{\frac{V_0}{W^+} ~-~ 1} $$`

Here, the right and left sides of the equation were plotted (a bit rescaled) separately as a function of \(W^+\). This equation was obtained by solving the Schrödinger equation for an electron in a finite potential well described by a symmetric wave function. Und \(W^+\) entspricht der Energie des Elektrons. The intersections \(W^+_1 \) and \(W^+_2\) are allowed energies of the electron within the potential box. In this case the electron can occupy only two energies in the potential box if it has a symmetric state.