Derivation of $\partial S / \partial t = -H$ for non-classical trajectories

Question

In classical mechanics, one can show that

$$\frac{\partial S}{\partial t} = -H,\tag{1} $$

where $$S=\int_0^t L(q, \dot{q}, t')dt'\tag{2} $$ is the action associated with a trajectory and $H$ is the Hamiltonian. $t$ is the final time of the trajectory under consideration. Using the path of least action to calculate the action $S$, this is proven in Landau and Lifschitz, Mechanics section 43.

Is this identity also true for general paths, not only limited to the path of least action? In particular, within a path integral formulation for quantum mechanics, is the following true?

$$\frac{\partial}{\partial t}\biggr[\int \mathcal{Dq}e^{iS[q]/\hbar}\biggr]=\int \mathcal{Dq}\frac{-i}{\hbar}H[q]e^{iS[q]/\hbar}\tag{3} $$

If possible, I am looking for a simple proof of the first identity shown above that does not rely on the principle of least action.

Answer 1

1. OP's eq. (1) can mean different things depending on context, e.g. the Hamilton-Jacobi equation. However, because OP is referring to $\S43$ in Ref. 1, it becomes clear that he is talking about the Dirichlet on-shell action function $$S(q_f,t_f; q_i, t_i),$$ see e.g. this Phys.SE post. Eq. (1) becomes $$ \frac{\partial S}{\partial t_f}~=~-h_f, \tag{1} $$ which is proven in a Lemma of my Phys.SE answer here. Here $h_f$ is the Lagrangian energy at the final time $t_f$. ($h_f$ is technically speaking not the Hamiltonian as this is purely a Lagrangian construction.)

2. One neat thing about eq. (1) for classical/on-shell paths is that it only depends on (final) boundary data. For off-shell/virtual paths, it would in general also receive bulk contributions, making eq. (1) no longer true. This gives a negative conclusion to OP's title question (v3).

3. On one hand, OP's eq. (3) follows from eq. (1) if $S$ is Dirichlet on-shell action function. However, that would render the path integral (3) meaningless. On the other hand, if $S$ is the off-shell action functional, then eq. (3) does not hold.

References:

1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S43$.

Answer 2

I believe that $$\frac{\partial}{\partial t}\biggr[\int \mathcal{Dq}e^{iS[q]/\hbar}\biggr]=\int \mathcal{Dq}\frac{-i}{\hbar}H[q]e^{iS[q]/\hbar}$$ does not hold in general. The principle of least action declares that the final and initial states are fixed and derives Hamiltonian/Lagrangian mechanics by considering the extremum of the action. The issue with doing this over the space of paths $\Gamma$ is that without fixed states or an extremum you would have to take into account time evolution of the fields and of the measure. Therefore you would have a $$\frac{\partial \mathcal{D}q(t)}{\partial t}.$$ Furthermore, even if I did have fixed states, I could always consider a really really wiggly path with a large length and that would not be the energy of the actual particle trajectory.