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Ning Zhu edited Appendix.tex
over 8 years ago
Commit id: 7ae70c62c21241dfa4b34f738535d30dbd1898fb
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$$\Delta_{E_{n}}=E_{n}-E_{n-1}$$
$$\Delta_{E_{n}}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})$$
$$\Delta_{E_{n}}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$
$$\Delta_{E_{n}}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$
Therefore, when n=\frac{1}{2}:
$$\Delta_{E_{n}}=E_{a}$$