Ning Zhu edited Appendix.tex  over 8 years ago

Commit id: ac15af226908fe5e550bab69f8320a5dd989b247

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$$\Delta_{E_{n}}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})$$  $$\Delta_{E_{n}}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$  $$\Delta_{E_{n}}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$  Therefore, when n=\frac{1}{2}: n=$\frac{1}{2}$:  $$\Delta_{E_{n}}=E_{a}$$