Nathanael A. Fortune edited Appendix.tex  over 8 years ago

Commit id: abb6db976c164ee4487400d3602bf5f813637bc0

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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: