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So, I was given the following problem to solve:

A system with two degrees of freedom is described by the following hamiltonian \begin{equation} H=p_1^2+p_2^2+\frac{1}{2}(q_1-q_2)^2+\frac{1}{8}(q_1+q_2)^2 \end{equation} Show that the following transformation is canonical by using $\mathbb{M}^T\cdot \mathbb{J}\cdot\mathbb{M}$, where $\mathbb{M}$ is the jacobian matrix and $\mathbb{J}=\begin{pmatrix}\mathbb{O} & \mathbb{I}\\\ -\mathbb{I} & \mathbb{O}\end{pmatrix}$, where $\mathbb{O}$ is an $n\times n$ matrix with all null elements and $\mathbb{I}$ is an $n\times n$ identity matrix. $$q_1=\sqrt{Q_1}\cdot cos(P_1)+\sqrt{2Q_2}\cdot cos(P_2)$$ $$q_2=-\sqrt{Q_1}\cdot cos(P_1)+\sqrt{2Q_2}\cdot cos(P_2)$$ $$p_1=\sqrt{Q_1}\cdot sin(P_1)+\sqrt{Q_2/2}\cdot sin(P_2)$$ $$p_2=-\sqrt{Q_1}\cdot sin(P_1)+\sqrt{Q_2/2}\cdot sin(P_2)$$

Well, I've started the problem by organizing the matrix $\mathbb{M}$. $$\mathbb{M}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_1}{\partial P_1} & \frac{\partial q_1}{\partial P_2} & \\\ \frac{\partial q_2}{\partial Q_1} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial q_2}{\partial P_1} & \frac{\partial q_2}{\partial P_2} & \\\ \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_1}{\partial P_2} & \\\ \frac{\partial p_2}{\partial Q_2} & \frac{\partial p_2}{\partial Q_1} & \frac{\partial p_2}{\partial P_1} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix}$$ Then I found the transpose matrix of $\mathbb{M}$ $$\mathbb{M^T}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_2}{\partial Q_1} & \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_2}{\partial Q_1} & \\\ \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_2}{\partial Q_2} & \\\ \frac{\partial q_1}{\partial P_1} & \frac{\partial q_2}{\partial P_1} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_2}{\partial P_1} & \\\ \frac{\partial q_1}{\partial P_2} & \frac{\partial q_2}{\partial P_2} & \frac{\partial p_1}{\partial P_2} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix}$$

So $$\mathbb{M^T}\cdot \mathbb{J}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_2}{\partial Q_1} & \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_2}{\partial Q_1} & \\\ \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_2}{\partial Q_2} & \\\ \frac{\partial q_1}{\partial P_1} & \frac{\partial q_2}{\partial P_1} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_2}{\partial P_1} & \\\ \frac{\partial q_1}{\partial P_2} & \frac{\partial q_2}{\partial P_2} & \frac{\partial p_1}{\partial P_2} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 1 & \\\ 0 & 0 & 1 & 1 & \\\ -1 & -1 & 0 & 0 & \\\ -1 & -1 & 0 & 0 & \end{pmatrix} $$ gives me a $2n\times 2n$ null matrix. What I am doing wrong?

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closed as off-topic by G. Smith, John Rennie, Kyle Kanos, GiorgioP, stafusa Jun 19 at 22:04

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    $\begingroup$ $q_1=-q_2$ and $p_1=-p_2$?? $\endgroup$ – Qmechanic Jun 18 at 8:01
  • $\begingroup$ I got that from the transformations. $\endgroup$ – Lincon Ribeiro Jun 18 at 11:19
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    $\begingroup$ But that isn't true, given the definitions of $q_i$ and $p_i$, is it? $\endgroup$ – Kyle Kanos Jun 18 at 11:35
  • $\begingroup$ I used the following: $$q_2=-\sqrt{Q_1}cos(P_1)+\sqrt{2Q_2}cos(P_2)=-(q_1)$$ $\endgroup$ – Lincon Ribeiro Jun 18 at 12:27
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    $\begingroup$ Ops, I guess you're right. My mistake on that $\endgroup$ – Lincon Ribeiro Jun 18 at 12:32
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For Linear Canonical Transformation the equation $M^T\,J\,M=J$ must be fulfilled

with

$$M=\left[ \begin {array}{cccc} 1/2\,{\frac {\cos \left( {\it P1} \right) }{\sqrt {{\it Q1}}}}&1/2\,{\frac {\sqrt {2}\cos \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&-\sqrt {{\it Q1}}\sin \left( {\it P1} \right) &-\sqrt {2}\sqrt {{\it Q2}}\sin \left( {\it P2} \right) \\-1/2\,{\frac {\cos \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/2\,{\frac {\sqrt {2}\cos \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&\sqrt {{\it Q1}}\sin \left( {\it P1} \right) &- \sqrt {2}\sqrt {{\it Q2}}\sin \left( {\it P2} \right) \\1/2\,{\frac {\sin \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/4\,{\frac {\sqrt {2}\sin \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&\sqrt {{\it Q1}}\cos \left( {\it P1} \right) &1/2 \,\sqrt {2}\sqrt {{\it Q2}}\cos \left( {\it P2} \right) \\-1/2\,{\frac {\sin \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/4\,{\frac {\sqrt {2}\sin \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&-\sqrt {{\it Q1}}\cos \left( {\it P1} \right) &1/ 2\,\sqrt {2}\sqrt {{\it Q2}}\cos \left( {\it P2} \right) \end {array} \right] $$

and

$$J= \left[ \begin {array}{cccc} 0&0&1&0\\ 0&0&0&1 \\ -1&0&0&0\\ 0&-1&0&0\end {array} \right] $$ you get the write answer

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