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Nathanael A. Fortune edited Appendix.tex
over 8 years ago
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Derivation of Eq. \ref{eq:lowestlevel}:\newline
We have:
$$E_{n}=n(E_{a}+\delta_{n})$$
THIS IS WRONG, ELSE EQUATION (*) is WRONG.
SHOULD READ $$E_{n}=n E_{a}+\delta_{n}$$
$$\delta_{n}=n\frac{\lambda}{L} E_{a}$$
Plugging into:
$$\Delta E_{n}=E_{n}-E_{n-1}$$
$$\Delta E_{n}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})$$
$$\Delta E_{n}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$
$$\Delta E_{n}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$
(*)
Therefore, when n=$\frac{1}{2}$:
$$\Delta E_{n}=E_{a}$$
In other words: