Finding canonical transformation

[sayebms]

Homework Statement

If in a system with i degrees of freedom the $$Q_i$$ are given what is the best way to proceed for finding the $$P_i$$ so that we have an overall canonical transformation. say for a two degree freedom system we have $$Q_1=q_1^2 $$ and $$ Q_2=q_1+q_2$$

Homework Equations

Using poisson brackets

The Attempt at a Solution

I think using the following would be the correct way to proceed:
[\itex][Q_i,Q_j]_{qp}=0[{P_1,P_2]_{qp}=0, [Q_i,P_j]=\delta _{ij} [\itex]
Am I right?[/B]

 

[mark726]

Thank you for your question. As a fellow scientist, I would like to offer my opinion on the best way to proceed in this situation. First, I would like to clarify that the equations you have provided are not correct. The correct equations for the canonical transformation are:

$$[Q_i,Q_j]_{qp}=0$$
$$[P_i,P_j]_{qp}=0$$
$$[Q_i,P_j]=\delta_{ij}$$

Now, to answer your question, the best way to proceed in finding the $$P_i$$ for a given set of $$Q_i$$ is to use the Poisson brackets as you have suggested. The Poisson brackets allow us to find the conjugate momenta $$P_i$$ for a given set of generalized coordinates $$Q_i$$.

In this case, for a two degree of freedom system with $$Q_1=q_1^2$$ and $$Q_2=q_1+q_2$$, we can use the Poisson brackets to find the conjugate momenta $$P_1$$ and $$P_2$$ as follows:

$$[Q_1,P_1]=\delta_{11} \rightarrow P_1=0$$
$$[Q_1,P_2]=\delta_{12} \rightarrow P_2=1$$
$$[Q_2,P_1]=\delta_{21} \rightarrow P_1=1$$
$$[Q_2,P_2]=\delta_{22} \rightarrow P_2=1$$

Therefore, the overall canonical transformation for this system would be:

$$Q_1=q_1^2$$
$$P_1=0$$
$$Q_2=q_1+q_2$$
$$P_2=q_1+q_2$$

I hope this helps and clarifies your doubts. Keep up the good work!