3. Chain Rule for Fourier-Space Derivatives

Since $x_j$ depends on all Fourier modes:

$$x_j = \frac{1}{\sqrt{N}} \sum_q \tilde{x}_q e^{i q j},$$ (35)

the chain rule gives

$$\frac{\partial H}{\partial \tilde{x}_k} = \sum_{j=0}^{N-1} \frac{... ... \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} \frac{\partial H}{\partial x_j} e^{i k j}.$$ (36)

Observation: For real $x_j$, the Fourier modes satisfy $\tilde{x}_{-k} = \tilde{x}_k^*$. Therefore, the correct derivative that appears in Hamilton's equations is with respect to $\tilde{x}_{-k}$:

$$\frac{\partial H}{\partial \tilde{x}_{-k}} = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} \frac{\partial H}{\partial x_j} e^{-i k j}.$$ (37)

Similarly, for momenta:

$$\frac{\partial H}{\partial \tilde{p}_{-k}} = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} \frac{\partial H}{\partial p_j} e^{-i k j}.$$ (38)