Consider time-invariant canonical transformations to new coordinates \(Q_i(q_i,p_i)\), \(P_i(q_i,p_i)\), which preserve the Hamiltonian, \(H\). By definition, Hamilton’s canonical equations are satisfied:
\begin{align}
\dot Q_i&=\die{H}{P_i}=\sum_j\left( \die{H}{q_j}\die{q_j}{P_i}+\die{H}{p_j}\die{p_j}{P_i} \right)\\
\dot P_i&=-\die{H}{Q_i}=-\sum_j\left( \die{H}{q_j}\die{q_j}{Q_i}+\die{H}{p_j}\die{p_j}{Q_i} \right)\end{align}
Consider the time derivative of the \(Q_i\) and \(P_i\):
\begin{align}
\dot Q_i&=\sum_j\left(\die{Q_i}{q_j}\dot q_j+\die{Q_i}{p_j}\dot p_j\right)=\sum_j\left(\die{Q_i}{q_j}\die{H}{p_j}-\die{Q_i}{p_j}\die{H}{q_j}\right)\\
\dot P_i&=\sum_j \left(\die{P_i}{q_j}\die{H}{p_j}+\die{P_i}{p_j}\dot p_j\right)=\sum_j \left(\die{P_i}{q_j}\die{H}{p_j}-\die{P_i}{p_j}\die{H}{q_j}\right)
\tag{8.12}\end{align}
By comparing terms, we can get the “Direct Conditions” for the transformation to be canonical:
\begin{align}
\die{p_j}{P_i}&=\die{Q_i}{q_j}\\
\die{q_j}{P_i}&=-\die{Q_i}{p_j}\\
\die{q_j}{Q_i}&=\die{P_i}{p_j}\\
\die{p_j}{Q_i}&=-\die{P_i}{q_j}\end{align}
Here, we obtained these relations by assuming that the transformation is canonical (that is, that we could get the time derivative of \(Q\) and \(P\) from the Hamiltonian). It is also possible to show that the Direct Conditions also hold if the transformations are time-dependent.
Now, consider the Poisson Brackets of the \(q_i\) and \(p_i\) evaluated in the coordinates \(Q_i\) and \(P_i\), where we use the Direct conditions to simplify:
\begin{align}
\{q_i,q_j\}_{Q,P}=&\sum_k\left(\die{q_i}{Q_k}\die{q_j}{P_k}-\die{q_i}{P_k}\die{q_j}{Q_k}\right)=\sum_k\left(-\die{q_i}{Q_k}\die{Q_k}{p_j}-\die{q_i}{P_k}\die{P_k}{p_j}\right)\\
=&-\die{q_i}{p_j}=0\\
\{p_i,p_j\}_{Q,P}=&\sum_k\left(\die{p_i}{Q_k}\die{p_j}{P_k}-\die{p_i}{P_k}\die{p_j}{Q_k}\right) =\sum_k\left(\die{p_i}{Q_k}\die{Q_k}{q_j}+\die{p_i}{P_k}\die{P_k}{q_j}\right)\\
=&\die{p_i}{q_j}=0\\
\{q_i,p_j\}_{Q,P}=&\sum_k\left(\die{q_i}{Q_k}\die{p_j}{P_k}-\die{q_i}{P_k}\die{p_j}{Q_k}\right)
=\sum_k\left(\die{q_i}{Q_k}\die{Q_k}{q_j}+\die{q_i}{P_k}\die{P_k}{q_j}\right)\\
=&\die{q_i}{q_j}=\delta_{ij}
\tag{8.14}\end{align}
Thus it is clear that the Poisson Brackets between the coordinates is independent of the coordinate system. From the exact same algebra, it follows that:
\begin{align}
\{Q_i,P_j\}&=\delta_{ij}\\
\{Q_i,Q_j\}&=\{P_i,P_j\}=0
\tag{8.15}\end{align}
which are equivalent to the Direct Conditions for checking that a transformation is canonical.
Example 8-4
Show that the transformation \(Q=p\), \(P=-q\) is canonical
We can test this by checking the Poisson Brackets between \(Q\) and \(P\):
\begin{align*}
\{Q,Q\}&=\die{Q}{q}\die{Q}{p}-\die{Q}{p}\die{Q}{q}=0\nonumber\\
\{P,P\}&=\die{P}{q}\die{P}{p}-\die{P}{p}\die{P}{q}=0\nonumber\\
\{Q,P\}&=\die{Q}{q}\die{P}{p}-\die{Q}{p}\die{P}{q}=1
\end{align*}
thus the transformation is canonical. Note that checking the first two is un-necessary, since they are identically zero.
Now, consider two functions in the old coordinates, \(U(q_i,p_i)\) and \(V(q_i,p_i)\), with Poisson Bracket:
\begin{align}
\{U,V\}_{q,p}=\sum_i^n\left(\die{U}{q_i}\die{V}{p_i}-\die{U}{p_i}\die{V}{q_i}\right)
\tag{8.16}\end{align}
Consider the same Poisson Bracket, expressed in terms of the new coordinates, \(Q_i\), \(P_i\), where the functions are expressed as \(U(Q_i,P_i)\) and \(V(Q_i,P_i)\):
\begin{align}
\{U,V\}_{Q,P}=&\sum_i^n\left(\die{U}{Q_i}\die{V}{P_i}-\die{U}{P_i}\die{V}{Q_i}\right)\\
=&\sum_i^n\left[\left(\sum_j\die{U}{q_j}\die{q_j}{Q_i}+\die{U}{p_j}\die{p_j}{Q_i}\right)\left(\sum_j\die{V}{q_j}\die{q_j}{P_i}+\die{V}{p_j}\die{p_j}{P_i}\right) \right.\\
&\left. -\left(\sum_j\die{U}{q_j}\die{q_j}{P_i}+\die{U}{p_j}\die{p_j}{P_i}\right)\left(\sum_j\die{V}{q_j}\die{q_j}{Q_i}+\die{V}{p_j}\die{p_j}{Q_i}\right)\right]\\
=&\sum_{j}^n\left[\die{U}{q_j}\die{V}{q_j}\{ q_j,q_j\}_{Q,P}+\die{U}{q_j}\die{V}{p_j}\{ q_j,p_j\}_{Q,P}\right.\\
&\left. +\die{U}{p_j}\die{V}{p_j}\{ p_j,p_j\}_{Q,P}+\die{U}{p_j}\die{V}{q_j}\{p_j,q_j\}_{Q,P}\right ]\\
=&\sum_{j}^n\left[\die{U}{q_j}\die{V}{p_j}-\die{U}{p_j}\die{V}{q_j}\right ]\\
=&\{U,V\}_{q,p}
\tag{8.17}\end{align}
We thus find that the Poisson Bracket of two quantities is preserved in a canonical transformation. Poisson Brackets are canonical invariants.