(a) Consider
\frac{1}{N} \sum_{j=1}^{N} e^{i 2 \pi j r / N}=\frac{1}{N} \sum_{j=1}^{N}\left(e^{i 2 \pi r / N}\right)^{j}=\frac{1}{N} \frac{e^{2 i \pi r}-1}{1-e^{-2 i \pi r / N}} .
For any integer r the numerator vanishes and the sum must be zero unless the denominator also vanishes. The denominator vanishes for r = m N where m is any integer; when r = m N it is easy to see that the expression evaluates to 1. Now, to prove the first identity, k-k^{\prime}=m N only when m = 0 and hence k = k′. In the second case, k+k^{\prime}=m N only when m = 1 and k′ = N – k.
(b) We first consider
\left[a_{k-}, a_{k^{\prime}+}\right]=\frac{1}{N} \sum_{j, j^{\prime}} e^{i 2 \pi\left(j^{\prime} k^{\prime}-j k\right) / N} \frac{1}{2}\left\{-\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}\left[x_{j}, \frac{\partial}{\partial x_{j^{\prime}}}\right]+\sqrt{\frac{\omega_{k}^{\prime}}{\omega_{k}}}\left[\frac{\partial}{\partial x_{j}}, x_{j^{\prime}}\right]\right\} .
where I’ve dropped the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are -\delta_{j j^{\prime}} \text { and } \delta_{j j^{\prime}} respectively and we have
\left[a_{k-}, a_{k^{\prime}+}\right]=\left(\frac{1}{N} \sum_{j} e^{i 2 \pi j\left(k^{\prime}-k\right) / N}\right) \frac{1}{2}\left[\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}+\sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}}\right] .
We can now do the sum on j using the result from part (a), and that gives the first commutator. Next we take
\left[a_{k-}, a_{k^{\prime}-}\right]=\frac{1}{N} \sum_{j, j^{\prime}} e^{-i 2 \pi\left(j^{\prime} k^{\prime}+j k\right) / N} \frac{1}{2}\left\{\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}\left[x_{j}, \frac{\partial}{\partial x_{j^{\prime}}}\right]+\sqrt{\frac{\omega_{k}^{\prime}}{\omega_{k}}}\left[\frac{\partial}{\partial x_{j}}, x_{j^{\prime}}\right]\right\} .
Evaluating the two commutators we have
\left[a_{k-}, a_{k^{\prime}-}\right]=\left(\frac{1}{N} \sum_{j} e^{-i 2 \pi j\left(k^{\prime}+k\right) / N}\right) \frac{1}{2}\left[-\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}+\sqrt{\frac{\omega_{k}^{\prime}}{\omega_{k}}}\right] .
Doing the sum on j now gives the result that k′= N – k (or k′ = k if k = N) but since \omega_{N-k}=\omega_{k} this vanishes, proving the second result. The final commutator works out the same way.
(c)
\left(a_{k_{-}}+a_{N-k_{+}}\right)=\frac{1}{\sqrt{N}} \sum_{j=1}^{N}\left\{e^{-i 2 \pi j k / N}\left[\sqrt{\frac{m \omega_{k}}{2 \hbar}} x_{j}+\sqrt{\frac{\hbar}{2 m \omega_{k}}} \frac{\partial}{\partial x_{j}}\right]\right.
\left.+e^{i 2 \pi j(N-k) / N}\left[\sqrt{\frac{m \omega_{N-k}}{2 \hbar}} x_{j}-\sqrt{\frac{\hbar}{2 m \omega_{N-k}}} \frac{\partial}{\partial x_{j}}\right]\right\} .
\text { But } \omega_{N-k}=\omega_{k}, \text { and } e^{i 2 \pi j(N-k) / N}=e^{i 2 \pi j} e^{-i 2 \pi j k / N}=e^{-i 2 \pi j k / N} , so
\left(a_{k_{-}}+a_{N-k_{+}}\right)=\frac{2}{\sqrt{N}} \sum_{j=1}^{N} \sqrt{\frac{m \omega_{k}}{2 \hbar}} e^{-i 2 \pi j k / N} x_{j} .
\frac{1}{\sqrt{N}} \sum_{k=1}^{N-1} \sqrt{\frac{\hbar}{2 m \omega_{k}}}\left(a_{k_{-}}+a_{N-k_{+}}\right) e^{i 2 \pi j k / N}=\frac{1}{\sqrt{N}} \sum_{k=1}^{N-1} \sqrt{\frac{\hbar}{2 m \omega_{k}}} \frac{2}{\sqrt{N}} \sum_{j^{\prime}=1}^{N} \sqrt{\frac{m \omega_{k}}{2 \hbar}} e^{-i 2 \pi j^{\prime} k / N} x_{j^{\prime}} e^{i 2 \pi j k / N} .
=\frac{1}{N} \sum_{j^{\prime}=1}^{N} \sum_{k=1}^{N-1} e^{i 2 \pi\left(j-j^{\prime}\right) k / N} x_{j^{\prime}}=\frac{1}{N} \sum_{j^{\prime}=1}^{N}\left(\sum_{k=1}^{N} e^{i 2 \pi\left(j-j^{\prime}\right) k / N}-e^{i 2 \pi\left(j-j^{\prime}\right)}\right) x_{j^{\prime}} .
=\sum_{j^{\prime}=1}^{N}\left(\frac{1}{N} \sum_{k=1}^{N} e^{i 2 \pi\left(j-j^{\prime}\right) k / N}\right) x_{j^{\prime}}-\frac{1}{N} \sum_{j^{\prime}=1}^{N} x_{j^{\prime}}=\sum_{j^{\prime}=1}^{N} \delta_{j j^{\prime}} x_{j^{\prime}}-R=x_{j}-R .
That proves the first relation; to prove the second, we repeat these steps, starting with
\left(a_{k_{-}}-a_{N-k_{+}}\right)=\frac{1}{\sqrt{N}} \sum_{j=1}^{N}\left\{e^{-i 2 \pi j k / N}\left[\sqrt{\frac{m \omega_{k}}{2 \hbar}} x_{j}+\sqrt{\frac{\hbar}{2 m \omega_{k}}} \frac{\partial}{\partial x_{j}}\right]\right.
\left.-e^{i 2 \pi j(N-k) / N}\left[\sqrt{\frac{m \omega_{N-k}}{2 \hbar}} x_{j}-\sqrt{\frac{\hbar}{2 m \omega_{N-k}}} \frac{\partial}{\partial x_{j}}\right]\right\} .
=\frac{2}{\sqrt{N}} \sum_{j=1}^{N} \sqrt{\frac{\hbar}{2 m \omega_{k}}} e^{-i 2 \pi j k / N} \frac{\partial}{\partial x_{j}} .
\frac{1}{\sqrt{N}} \sum_{k=1}^{N-1} \sqrt{\frac{m \omega_{k}}{2 \hbar}}\left(a_{k_{-}}-a_{N-k_{+}}\right) e^{i 2 \pi j k / N}=\frac{1}{\sqrt{N}} \sum_{k=1}^{N-1} \sqrt{\frac{m \omega_{k}}{2 \hbar}} \frac{2}{\sqrt{N}} \sum_{j^{\prime}=1}^{N} \sqrt{\frac{\hbar}{2 m \omega_{k}}} e^{-i 2 \pi j^{\prime} k / N} \frac{\partial}{\partial x_{j^{\prime}}} e^{i 2 \pi j k / N} .
=\frac{1}{N} \sum_{j^{\prime}=1}^{N} \sum_{k=1}^{N-1} e^{i 2 \pi\left(j-j^{\prime}\right) k / N} \frac{\partial}{\partial x_{j^{\prime}}}=\frac{1}{N} \sum_{j^{\prime}=1}^{N}\left(\sum_{k=1}^{N} e^{i 2 \pi\left(j-j^{\prime}\right) k / N}-e^{i 2 \pi\left(j-j^{\prime}\right)}\right) \frac{\partial}{\partial x_{j^{\prime}}} .
=\sum_{j^{\prime}=1}^{N}\left(\frac{1}{N} \sum_{k=1}^{N} e^{i 2 \pi\left(j-j^{\prime}\right) k / N}\right) \frac{\partial}{\partial x_{j^{\prime}}}-\frac{1}{N} \sum_{j^{\prime}=1}^{N} \frac{\partial}{\partial x_{j^{\prime}}}=\sum_{j^{\prime}=1}^{N} \delta_{j j^{\prime}} \frac{\partial}{\partial x_{j^{\prime}}}-\frac{1}{N} \sum_{j^{\prime}=1}^{N} \frac{\partial}{\partial x_{j^{\prime}}}=\frac{\partial}{\partial x_{j}}-\frac{1}{N} \frac{\partial}{\partial R} .
In the last step I used \frac{\partial}{\partial R}=\sum_{j=1}^{N} \frac{\partial x_{j}}{\partial R} \frac{\partial}{\partial x_{j}}, \text { and } \frac{\partial x_{j}}{\partial R}=1 (for instance, if N = 2; R=\frac{1}{2}\left(x_{1}+x_{2}\right) , and r=\frac{1}{2}\left(x_{1}-x_{2}\right) \text {, so } x_{1}=R+r \text { and } x_{2}=R-r \Rightarrow \partial x_{i} / \partial R=1 \text { ) } .
(d)
\left(x_{j+1}-x_{j}\right)=\frac{1}{\sqrt{N}} \sum_{k=1}^{N-1} \sqrt{\frac{\hbar}{2 m \omega_{k}}}\left(a_{k_{-}}+a_{N-k_{+}}\right) e^{i 2 \pi j k / N}\left(e^{i 2 \pi k / N}-1\right) .
\left(x_{j+1}-x_{j}\right)^{2}=\frac{1}{N} \sum_{k, k^{\prime}}^{N-1} \frac{\hbar}{2 m \sqrt{\omega_{k} \omega_{k^{\prime}}}}\left(a_{k_{-}}+a_{N-k_{+}}\right)\left(a_{k_{-}^{\prime}}+a_{N-k^{\prime}+}\right) e^{i 2 \pi j\left(k+k^{\prime}\right) / N}
\times\left(e^{i 2 \pi k / N}-1\right)\left(e^{i 2 \pi k^{\prime} / N}-1\right) ,
\sum_{j=1}^{N} \frac{1}{2} m \omega^{2}\left(j+1-x_{j}\right)^{2}=\sum_{k, k^{\prime}}^{N-1} \frac{\hbar \omega^{2}}{4 \sqrt{\omega_{k} \omega_{k^{\prime}}}}\left(a_{k_{-}}+a_{N-k_{+}}\right)\left(a_{k_{-}^{\prime}}+a_{N-k^{\prime}+}\right) \delta_{k^{\prime}, N-k}
\times\left(e^{i 2 \pi k / N}-1\right)\left(e^{i 2 \pi k^{\prime} / N}-1\right) .
=\sum_{k=1}^{N-1} \frac{\hbar \omega^{2}}{4 \omega_{k}}\left(a_{k_{-}}+a_{N-k_{+}}\right)\left(a_{N-k_{-}}+a_{k_{+}}\right) \frac{\omega_{k}^{2}}{\omega^{2}} .
=\sum_{k=1}^{N-1} \frac{\hbar \omega_{k}}{4}\left(a_{k_{-}} a_{k+}+a_{N-k_{+}} a_{k_{+}}+a_{k_{-}} a_{N-k_{-}}+a_{N-k_{+}} a_{N-k_{-}}\right) .
In the penultimate line I used
\left(e^{i 2 \pi k / N}-1\right)\left(e^{i 2 \pi(N-k) / N}-1\right)=e^{i \pi k / N}\left(e^{i \pi k / N}-e^{-i \pi k / N}\right) e^{-i \pi k / N}\left(e^{-i \pi k / N}-e^{i \pi k / N}\right) .
=(2 i \sin (\pi k / N))(-2 i \sin (\pi k / N))=[2 \sin (\pi k / N)]^{2}=\left(\frac{\omega_{k}}{\omega}\right)^{2} .
Meanwhile,
-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x_{j}^{2}}=\frac{1}{N} \sum_{k, k^{\prime}}^{N-1}\left(-\frac{\hbar^{2}}{2 m}\right) \frac{m}{2 \hbar} \sqrt{\omega_{k} \omega_{k^{\prime}}}\left(a_{k_{-}}-a_{N-k_{+}}\right)\left(a_{k_{-}^{\prime}}-a_{N-k^{\prime}+}\right) e^{i 2 \pi j\left(k+k^{\prime}\right) / N}
-\frac{\hbar^{2}}{2 m} \frac{1}{N^{3 / 2}}\left[\frac{\partial}{\partial R} \sum_{k=1}^{N-1} \sqrt{\frac{m \omega_{k}}{2 \hbar}}\left(a_{k_{-}}-a_{N-k_{+}}\right) e^{i 2 \pi j k / N}\right.
\left.+\sum_{k=1}^{N-1} \sqrt{\frac{m \omega_{k}}{2 \hbar}}\left(a_{k_{-}}-a_{N-k_{+}}\right) e^{i 2 \pi j k / N} \frac{\partial}{\partial R}\right]-\frac{\hbar^{2}}{2 m} \frac{1}{N^{2}} \frac{\partial^{2}}{\partial R^{2}} .
The middle terms vanish when summed over j \left(\sum_{j=1}^{N} e^{i 2 \pi j k / N}=N \delta_{k, 0}=0\right) , leaving
-\frac{\hbar^{2}}{2 m} \sum_{j=1}^{N} \frac{\partial^{2}}{\partial x_{j}^{2}}=\sum_{k=1}^{N-1}\left(-\frac{\hbar \omega_{k}}{4}\right)\left(a_{k_{-}}-a_{N-k_{+}}\right)\left(a_{N-k_{-}}-a_{k_{+}}\right)-\frac{\hbar^{2}}{2 N m} \frac{\partial^{2}}{\partial R^{2}} .
=\sum_{k=1}^{N-1} \frac{\hbar \omega_{k}}{4}\left(a_{k_{-}} a_{k_{+}}-a_{N-k_{+}} a_{k_{+}}-a_{k_{-}} a_{N-k_{-}}+a_{N-k_{+}} a_{N-k_{-}}\right)-\frac{\hbar^{2}}{2 N m} \frac{\partial^{2}}{\partial R^{2}} .
Adding the two results:
\hat{H}=-\frac{\hbar^{2}}{2 N m} \frac{\partial^{2}}{\partial R^{2}}+\sum_{k=1}^{N-1} \frac{\hbar \omega_{k}}{2}\left(a_{k_{-}} a_{k_{+}}+a_{N-k_{+}} a_{N-k_{-}}\right)=-\frac{\hbar^{2}}{2 N m} \frac{\partial^{2}}{\partial R^{2}}+\sum_{k=1}^{N-1} \frac{\hbar \omega_{k}}{2}\left(a_{k_{-}} a_{k_{+}}+a_{k_{+}} a_{k_{-}}\right) .
(the change in the final two subscripts just means adding these terms in reverse order, given that \omega_{N-k}=\omega_{k} ).
Thus
\hat{H}=-\frac{\hbar^{2}}{2(N m)} \frac{\partial^{2}}{\partial R^{2}}+\sum_{k=1}^{N-1} \hbar \omega_{k}\left(a_{k_{+}} a_{k_{-}}+\frac{1}{2}\right) .