91影视

which is zero unless both m and n are odd-in which case it is$卤2伪/a$So Eq. 6.15 says

(6.15).

But Eq. 2.27 says

$$\begin{gathered} E_n=\frac{{魔}^2{k}_n^2}{2m}=\frac{n^2{蟺}^2{魔}^2}{2m{a}^2} \\ {E}_n^0=\frac{{蟺}^2{魔}^2}{2m{a}^2}{n}^2,SO \end{gathered}$$ (2.27).

$E_n^2=\left\{\begin{array}{l}0,\mathrm{if}n\mathrm{is}\mathrm{even}\\ 2m{\left(\frac{2伪}{蟺魔}\right)}^2\underset{m鈮�n,o\mathrm{dd}}{鈭�}\frac{1}{\left(n^2-m^2\right)\text{'}}\mathrm{if}n\mathrm{is}\mathrm{odd}\end{array}\right.$

To sum the series, note that Thus,$\frac{1}{\left(n^2-m^2\right)}=\frac{1}{2n}\left(\frac{1}{m+n}-\frac{1}{m-n}\right).\mathrm{Thus}$

$$\begin{gathered} \mathrm{for}n=1:\underset{}{鈭�=}\frac{1}{2}\underset{3,5,7...}{鈭�}\left(\frac{1}{m+1}-\frac{1}{m-1}\right) \\ =\frac{1}{2}\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}...\right)=\frac{1}{2}\left(-\frac{1}{2}\right)=-\frac{1}{4} \\ \mathrm{for}n=3:\underset{}{鈭�=}\frac{1}{6}\underset{1,5,7...}{鈭�}\left(\frac{1}{m+3}-\frac{1}{m-3}\right), \\ =\frac{1}{6}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+...+\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}...\right)=\frac{1}{6}\left(-\frac{1}{6}\right)=-\frac{1}{36} \end{gathered}$$

In general, there is perfect cancellation except for the term 1/2n in the first sum, so the total is

$Therefore:E_n^2=\left\{\begin{array}{l}0,\mathrm{if}n\mathrm{is}\mathrm{even}\\ -2m{\left(伪/蟺魔n\right)}^2,\mathrm{if}n\mathrm{is}\mathrm{odd}.\end{array}\right.$

$$\begin{gathered} =\frac{\stackrel{藱}{{慰}^2魔蝇}}{16}\underset{m鈮�n}{鈭�}\frac{\left[\left(n+1\right)\left(n+1\right){未}_{m,n,+2}+n\left(n-1\right){未}_{m,n,-2}\right]}{\left(n-m\right)} \\ =\frac{\stackrel{藱}{{慰}^2魔蝇}}{32}\left(-n^2-3n-2+n^2-n\right)=\frac{\stackrel{藱}{{慰}^2魔蝇}}{32}\left(-4n-2\right)=-\stackrel{藱}{{慰}^2}\frac{1}{8}\stackrel{藱}{魔蝇}\left(n+\frac{1}{2}\right) \end{gathered}$$

(which agrees with the${蔚}^2$term in the exact solution Problem 6.2(a)).