# Prove that a transformation is canonical by using $\mathbb{M}^T\cdot \mathbb{J}\cdot \mathbb{M}$ [closed]

So, I was given the following problem to solve:

A system with two degrees of freedom is described by the following hamiltonian $$$$H=p_1^2+p_2^2+\frac{1}{2}(q_1-q_2)^2+\frac{1}{8}(q_1+q_2)^2$$$$ Show that the following transformation is canonical by using $$\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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – G. Smith, John Rennie, Kyle Kanos, GiorgioP, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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