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.

  • $\begingroup$ The first identity can also be derived using canonical transformation as you can find here. $\endgroup$ – Victor Palea Sep 23 '17 at 23:57
  • $\begingroup$ The source you cited relies on Hamilton's equations, and therefore it seems to me that this is a classical result. Can you provide a derivation that does not rely explicitly or implicitly on the principle of least action? $\endgroup$ – Ian Sep 24 '17 at 0:19
  • $\begingroup$ This post (v1) could need some clarification. E.g.: What is $t$ in the two equations? The final time of a path? What variables does $S$ and $H$ depend on? $\endgroup$ – Qmechanic Sep 24 '17 at 11:41
  • $\begingroup$ @Qmechanic, does that edit answer your question? Here the Hamiltonian is defined as the usual Legendre transform of the Lagrangian. $\endgroup$ – Ian Sep 24 '17 at 19:33
  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.


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

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.

  • $\begingroup$ This answer isn't quite clear because it tries to use statements about the principle of least action to draw conclusions about the more general quantum path integral. Can you make your answer more clear? $\endgroup$ – Ian Sep 24 '17 at 5:11
  • $\begingroup$ Furthermore, I'd like to make two points about your answer. I believe that since we are considering the partial time derivative that we need not consider the time evolution of the fields or the path integral measure, contrary to what you have stated. It seems to me that these have implicit time dependence that would show up in the total time derivative via a \partial / \partial q term. Do you agree? $\endgroup$ – Ian Sep 24 '17 at 5:18

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