Writing $\mathbf{x}=\left(p_{1}, p_{2}, p_{3}, \ldots, p_{n}, q_{1}, q_{2}, q_{3}, \ldots, q_{n}\right)$, Hamilton's equations may be written in the form

$$\dot{\mathbf{x}}=\mathbf{J} \frac{\partial H}{\partial \mathbf{x}} \text {, }$$

where the $2 n \times 2 n$ matrix

$$\mathbf{J}=\left(\begin{array}{rr} 0 & -I \\ I & 0 \end{array}\right)$$

and $I$ and 0 denote the $n \times n$ unit and zero matrices respectively.

Explain what is meant by the statement that the transformation $\mathbf{x} \rightarrow \mathbf{y}$,

$$\left(p_{1}, p_{2}, p_{3}, \ldots, p_{n}, q_{1}, q_{2}, q_{3}, \ldots, q_{n}\right) \rightarrow\left(P_{1}, P_{2}, P_{3}, \ldots, P_{n}, Q_{1}, Q_{2}, Q_{3}, \ldots, Q_{n}\right)$$

is canonical, and show that the condition for this is that

$$\mathbf{J}=\mathcal{J} \mathbf{J} \mathcal{J}^{T}$$

where $\mathcal{J}$ is the Jacobian matrix with elements

$$\mathcal{J}_{i j}=\frac{\partial y_{i}}{\partial x_{j}}$$

Use this condition to show that for a system with $n=1$ the transformation given by

$$P=p+2 q, \quad Q=\frac{1}{2} q-\frac{1}{4} p$$

is canonical.