Question 8.4: Prove that the transformation Q1 = p²1 ,Q2 = p²2 + q2 , P1 =......
Prove that the transformation
Q_{1} = p^{2}_{1} ,Q_{2} = p^{2}_{2} + q_{2} , P_{1}=- \frac{q_{1}}{2p_{1}} , P_{2} = p_{2} (8.54)
is canonical.
We have
\left[q_{1}, q_{2}\right]_{\left(Q,P\right) } =\left(\frac{\partial Q_{1} }{\partial q_{1}} \frac{\partial P_{1} }{\partial q_{2}}- \frac{\partial Q_{1} }{\partial q_{2}} \frac{\partial P_{1} }{\partial q_{1}}\right)+ \left(\frac{\partial Q_{2} }{\partial q_{1}} \frac{\partial P_{2} }{\partial q_{2}}- \frac{\partial Q_{2} }{\partial q_{2}} \frac{\partial P_{2} }{\partial q_{1}}\right) =0 (8.55)
and also
\left[p_{1}, p_{2}\right]_{\left(Q,P\right) } =\left(\frac{\partial Q_{1} }{\partial p_{1}} \frac{\partial P_{1} }{\partial p_{2}}- \frac{\partial Q_{1} }{\partial p_{2}} \frac{\partial P_{1} }{\partial p_{1}}\right)+ \left(\frac{\partial Q_{2} }{\partial p_{1}} \frac{\partial P_{2} }{\partial p_{2}}- \frac{\partial Q_{2} }{\partial p_{2}} \frac{\partial P_{2} }{\partial p_{1}}\right) =0, (8.56)
with similar computations showing that \left[q_{1}, p_{2}\right]_{\left(Q,P\right) } =\left[q_{2}, p_{1}\right]_{\left(Q,P\right) }=0. Finally,
\left[q_{1}, p_{1}\right]_{\left(Q,P\right) } = \left(\frac{\partial Q_{1} }{\partial q_{1}} \frac{\partial P_{1} }{\partial p_{1}}- \frac{\partial Q_{1} }{\partial p_{1}} \frac{\partial P_{1} }{\partial q_{1}}\right)+ \left(\frac{\partial Q_{2} }{\partial q_{1}} \frac{\partial P_{2} }{\partial p_{1}}- \frac{\partial Q_{2} }{\partial p_{1}} \frac{\partial P_{2} }{\partial q_{1}}\right)= −2p_{1} \left(- \frac{1}{2p_{1}}\right)=1 (8.57)
and
\left[q_{2}, p_{2}\right]_{\left(Q,P\right) } = \left(\frac{\partial Q_{1} }{\partial q_{2}} \frac{\partial P_{1} }{\partial p_{2}}- \frac{\partial Q_{1} }{\partial p_{2}} \frac{\partial P_{1} }{\partial q_{2}}\right)+ \left(\frac{\partial Q_{2} }{\partial q_{2}} \frac{\partial P_{2} }{\partial p_{2}}- \frac{\partial Q_{2} }{\partial p_{2}} \frac{\partial P_{2} }{\partial q_{2}}\right)=1. (8.58)
There is no need to compute the remaining Lagrange brackets because of their antisymmetry: \left[u, v\right]_{\left(\eta , \xi\right) } =- \left[v, u\right]_{\left(\eta , \xi\right) } . Since all conditions (8.44) are satisfied, the transformation is canonical.
\left[q_{i}, q_{j}\right]_{\left(Q,P\right) } =0 , \left[p_{i}, p_{j}\right]_{\left(Q,P\right) } =0 , \left[q_{i}, p_{j}\right]_{\left(Q,P\right) } =\delta_{ij}. (8.44)