Derivation of Eq. \ref{eq:lowestlevel}:
We have:
THIS IS WRONG, ELSE EQUATION (*) is WRONG. SHOULD READ
\begin{equation} E_{n}=nE_{a}+\delta_{n}\nonumber \\ \end{equation} \begin{equation} \delta_{n}=n\frac{\lambda}{L}E_{a}\nonumber \\ \end{equation}Plugging into:
\begin{equation} \Delta E_{n}=E_{n}-E_{n-1}\nonumber \\ \end{equation} \begin{equation} \Delta E_{n}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})\nonumber \\ \end{equation} \begin{equation} \Delta E_{n}=n(E_{a}+n\frac{\lambda}{L}E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L}E_{a})\nonumber \\ \end{equation} \begin{equation} \Delta E_{n}=E_{a}\frac{\lambda}{L}(2n-1)+E_{a}(*)\nonumber \\ \end{equation}Therefore, when n=\(\frac{1}{2}\):
\begin{equation} \Delta E_{n}=E_{a}\nonumber \\ \end{equation}In other words:
\begin{equation} {E_{a}=\Delta E_{(0.5)}}\nonumber \\ \end{equation}\cite{Rapior_2006}