Even after making the previous blog post, I realised that I still did not sufficiently understand the concept of an active canonical transformation, and how it can affect the value but not the functional form of some functions, so let me work out the classic example here - the canonical transformation corresponding to time evolution.
Let us, again, consider the Hamiltonian \[\mathcal{H}=\frac{p^2}{2m}+\frac 12k\left(x-vt\right)^2\]
We consider an infinitesimal canonical transformation generated by the Hamiltonian \(F_2=xP+\epsilon\mathcal{H}\). The momentum is given by \(p=P+\epsilon k(x-vt)\), and the coordinate is given by \(Q=x+\epsilon\frac{\partial\mathcal{H}}{\partial P}\). Recall that for infinitesimal canonical transformations, since \(p\) and \(P\) differ only by an infinitesimal amount, we can replace \(P\) by \(p\) in the derivative (up to 1st order). Then \(Q=x+\frac{\epsilon p}{m}\). At this point, note that when replacing \(\epsilon\) by \(dt\), we get the expected coordinates after an infinitesimal time evolution.
The Hamiltonian is given by \(K=\mathcal{H}+\epsilon k(x-vt)\). At this point it might be helpful to use this example to clarify the explanation given by Goldstein about the “change in the Hamiltonian”. He explains that the Hamiltonian is defined as the function that gives the canonical equations in the given phase space. Thus, the Hamiltonian goes not just from \(\mathcal{H}(\mathscr{A})\) to \(\mathcal{H}(\mathscr{A}’)\), but to \(K(\mathscr{A}’)). Let us explicitly write down what these three quantities are.
\[\mathcal{H}(\mathscr{A})=\frac{p^2}{2m}+\frac 12k\left(x-vt\right)^2\] \[\mathcal{H}(\mathscr{A}’)=\frac{[P+\epsilon k(x-vt)]^2}{2m}+\frac 12k\left(\left(Q-\frac{\epsilon p}{m}\right)-vt\right)^2\]
Note that there are some \(x\) and \(p\) in the expression for \(\mathcal{H}(\mathscr{A}’)\) that have not been replaced yet. Replacing them is left as an exercise to the reader. Also of importance is that \(\mathcal{H}(\mathscr{A}’)\) has the same numerical as the original Hamiltonian. However, it is not the Hamiltonian - it does not give the canonical equations in \(Q\), \(P\). In fact, the new Hamiltonian \(K(\mathscr{A}’)\) is given by
\[K(\mathscr{A}’)=\frac{[P+\epsilon k(x-vt)]^2}{2m}+\frac 12k\left(\left(Q-\frac{\epsilon p}{m}\right)-vt\right)^2+\epsilon k(x-vt)\]
Whereupon we can see that the difference is simply \(\partial\mathcal{H}=-\epsilon\frac{d\mathcal{H}}{dt}\)