Consider $H(q,p) = \frac{1}{2} \sum\limits_{j=1}^{n+1} {(p_j^2 + (q_{j}-q_{j-1})^2)}$
$H(q,p) $ is the Hamiltonian considered in the FermiPasta-Ulam problem. Consider canonical transformation
$Q = \sqrt{\frac{2}{n+1}} \sum\limits_{k=1}^{n} {q_{k}}$ $sin(\frac{kjn\pi}{n+1})$, $P = \sqrt{\frac{2}{n+1}} \sum\limits_{k=1}^{n} {p_{k}}$ $sin(\frac{kjn\pi}{n+1})$
Show that $H(Q,P) = \frac{1}{2} \sum\limits_{j=1}^{n} {(P_j^2 + \omega^2 Q_{j}^2)}$. I could not obtain $\sum\limits_{j=1}^{n} {(Q_{j}-Q_{j-1})^2 }= \sum\limits_{j=1}^{n} {\omega^2 Q_{j}^2} $. I used trigonometric identities to simplify the computation, but I could not get $H(Q,P) = \frac{1}{2} \sum\limits_{j=1}^{n} {(P_j^2 + \omega^2 Q_{j}^2)}$, so could you guys help me to do this problem?
Thanks
Let us first see how the $p$ part of the Hamiltonian is transformed, before looking at the $q$ part. Thus, forgetting some irrelevant factors, first consider $$ h_1(p)=\sum_kp_k^2. $$ The crucial remark is that, for some suitable family of parameters $s$, $$ h_1(p)=\sum_jP_j^2,\quad\text{where}\quad P_j=\sum_kp_ks(kj). $$ When does this happen? Well, note that, for every $j$, $$ P_j^2=\sum_k\sum_\ell p_kp_\ell s(kj)s(\ell j), $$ hence $$ \sum_jP_j^2=\sum_k\left(\sum_js(kj)^2\right)p_k^2+\sum_{k\ne\ell}\left(\sum_js(kj)s(\ell j)\right)p_kp_\ell, $$ and the trick works if $$ \sum_js(kj)s(\ell j)=\left\{\begin{array}{lcl}1&\text{if}&k=\ell\\ 0&\text{if}&k\ne\ell\end{array}\right. $$ As is well known, these conditions are met when $s$ is based on the sine function in the sense that, for some suitable angle $\theta$ and factor $\kappa$, $$ s(m)=\kappa\sin(m\theta), $$ hence the formulas for $P_j$ in your post.
Now we are ready to consider the second type of Hamiltonian in the question, namely, $$ h_2(q)=\sum_k(q_k-q_{k-1})^2=2\sum_kq_k^2-2\sum_kq_kq_{k-1}. $$ The first part of the RHS is liable to the same analysis as before hence we are left with $$ h_3(q)=-2\sum_kq_kq_{k-1}. $$ Of course, one cannot have $$ h_3(q)=\sigma\sum_jQ_j^2,\quad\text{where}\quad Q_j=\sum_kq_ks(kj), $$ for the same family $s$ as above and some parameter $\sigma$, but it happens that one can obtain $$ h_3(q)=\sum_j\sigma_jQ_j^2, $$ for some specific family $(\sigma_j)$. The same computations as before show that this happens when $$ \sum_j\sigma_js(kj)s(\ell j)=\left\{\begin{array}{lcl}0&\text{if}&k=\ell\\ -1&\text{if}&k=\ell\pm1\\ 0&\text{if}&|k-\ell|\geqslant2\end{array}\right. $$ that is, using the addition formula for cosines, when $$ \sum_j\sigma_j=0,\quad\sum_j\sigma_j\cos(j\theta)=1,\quad\sum_j\sigma_j\cos(rj\theta)=0\quad(r\geqslant2), $$ hence $\sigma_j=\sigma_0\cos(j\theta)$ solves this, for some suitable positive $\sigma_0$.
Finally, once one understands that the parameter $\omega^2$ in your question ought to be replaced by a family $(\omega_j^2)$, one gets $$ \sum_k(q_k-q_{k-1})^2=\sum_j\omega_j^2Q_j^2,\quad\text{where}\quad\omega_j^2=2+2\sigma_j, $$ hence $$ 2H(q,p)=\sum_kp_k^2+(q_k-q_{k-1})^2=\sum_jP_j^2+\omega_j^2Q_j^2. $$