Preservation of exact equations of motion in time-dependent perturbation theory for the Hamilton-Jacobi equations

Question

From the Hamilton-Jacobi formalism the solution for the unperturbed hamiltonian $H_0$ has a generating function $S(q,\alpha,t)$ such that $$K_0 = H_0(q, \frac{\partial S}{\partial q},t) + \frac{\partial S}{\partial t}= 0.$$ For the dynamics generated by $H_0$, $(\alpha,\beta)$ are constants since $\dot{\alpha} = -\frac{\partial K_0}{\partial \beta } = 0$ and similarily for $\dot{\beta} = \frac{\partial K_0}{\partial \alpha}=0$. Futhermore this transformation $(q,p)$ -> $(\alpha,\beta)$ preserves the equations of motion.

For the new Hamiltonian $H = H_0 + \Delta H$, $(\alpha,\beta)$ are still valid coordinates such that the exact equations of motions for those variables are now given by $\dot{\alpha} = -\frac{\partial K}{\partial \beta }$ and $\dot{\beta} = \frac{\partial K}{\partial \alpha }$.

where $K = K_0 +\Delta H$. Why is this the case, how come these new equations for $(\dot{\alpha}, \dot{\beta})$ will give the exact equations of motion for $(\alpha,\beta)$?

Reference: page 2 of https://ocw.mit.edu/courses/8-09-classical-mechanics-iii-fall-2014/resources/mit8_09f14_chapter_5/

Answer 1

1. In a perhaps more logical notation, let $$H(q,p,t)=H_0(q,p,t) + \Delta H(q,p,t).\tag{A}$$ Let the Kamiltonian $$K(Q,P,t)~:=~\Delta H(q,p,t) \tag{B}$$ be designed to yield the interaction picture, i.e. the un-perturbed Kamiltonian $$K_0~\equiv~ 0\tag{C}$$ is trivial.

2. Now perform a canonical transformation (CT) $$(q,p,t)\quad\longrightarrow\quad(Q,P,t)\tag{D}$$ with type-2 generating function $S(q,P,t)$ satisfying the un-perturbed Hamilton-Jacobi (HJ) eq. $$\begin{align} \frac{\partial S}{\partial t} ~=~&K-H~\stackrel{(A)+(B)+(C)}{=}~K_0-H_0\cr ~=~&-H_0(q,\frac{\partial S}{\partial q},t).\end{align} \tag{E}$$

3. Note that the coordinate transformation (D) is both a type 2 CT wrt. the perturbed pair $(H,K)$ and the unperturbed pair $(H_0,K_0)$ because $K-H=K_0-H_0$.

4. The un-perturbed Kamilton's equations read $$ \frac{dQ^i}{dt}~=~\frac{\partial K_0}{\partial P_i} \quad\text{and}\quad \frac{dP_i}{dt}~=~-\frac{\partial K_0}{\partial Q^i}.\tag{F}$$ The perturbed Kamilton's equations read $$ \frac{dQ^i}{dt}~=~\frac{\partial K}{\partial P_i} \quad\text{and}\quad \frac{dP_i}{dt}~=~-\frac{\partial K}{\partial Q^i}.\tag{G}$$ That they have the exact same form as Hamilton's equations is the defining property of a CT, cf. OP's title question.

5. Note that even though the new coordinates $(Q,P)$ are exactly the same for the un-perturbed and the perturbed system, the solutions for the un-perturbed eq. (F) and the perturbed eq. (G) are not (necessarily) the same.