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

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 $$(\alpha,\beta)$$?

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.

• The transformation is canonical with respect to $H_0$ only. Why does adding $\Delta H$ yields these equations ? We could have gotten for example that the new K requires an additional generating function, or anything else for it to be canonical. Wouldn't a starting point be the condition that a transformation is canonical if $p\dot{q} - H_0 = P\dot{Q} - H_0 + \dot{F}$ ? In other words why being canonical for $H_0$ implies it being canonical for $\Delta H$ but now with respect to the Kamiltonian $K = H_0 + \dot{S} + \Delta H$ ? Commented Aug 1 at 17:02
• Where is the answer to : why being canonical for $H_0$ implies it being canonical for $\Delta H$ but now with respect to the Kamiltonian $K = H_0 + \dot{S} + \Delta H$ ? Commented Aug 1 at 18:24
• Ah, could you elaborate on point 3. How can I note that (D) is a canonical transformation for K from this ? This would answer my question, thank you Commented Aug 1 at 19:12
• I updated the answer. Commented Aug 3 at 12:29