The principle of least action says that a body moves in such a way that the action value $S=\int L dt$ is stationary (often minimal). The principle is written as $$\delta S =0 \ .$$

In contrast, Schwinger's quantum action principle between an in and an out state is $$ \langle{\rm out}| \delta S |{\rm in}\rangle = \frac{\hbar}{i} \delta \langle {\rm out}|{\rm in}\rangle \ .$$ I have a simple question: what is minimized in this case? Or, more generally: what does the principle state, if only words are used? That would be my question 1.

Question 2. Least action implies that the actual trajectory is special: it differs from all the others, because it has minimal/stationary action. In what sense does Schwinger's principle make the actual state evolution special from all the others?

Answering a similar, older question, Qmechanic had written:

Schwinger's quantum action principle is not a variational principle in the sense of finding stationary points for a functional. Rather it gives a formula for how a quantum system (typically an overlap/transition amplitude ⟨𝐴|𝐵⟩) changes under a change of external parameters/sources in the action 𝑆.

Can one add a few sentences to complete the answer to question 2? In what sense is the quantum motion special? How does it apply to free particles? Or: how can one express the quantum action principle for free particles in words?

Question 3. How does Schwinger's action principle become the least action principle in the classical limit? Ok, if $\hbar$ is zero, the two equations are very similar. Is that the way the transition is made? What can be said to make it clearer?

Question 4. The right hand side of Schwinger's principle is $-i\hbar$ times a complex number with a magnitude equal or smaller than one. Why does it enter the principle? Is it correct to say that for long times, the overlap is small, and thus the right hand side is zero?

  • $\begingroup$ Possible duplicate: Is the quantum action principle of Schwinger a variational principle? $\endgroup$
    – Qmechanic
    Jul 11, 2020 at 15:03
  • $\begingroup$ @Qmechanic I had read that question, including your answer. But it does not answer questions 1 and 3, and only partially question 2. Please leave the topic open - the answers are dear to me, and I would very much like to know them. $\endgroup$
    – user85598
    Jul 11, 2020 at 15:11
  • $\begingroup$ @Qmechanic Do you mean that the answer to question 1 is "nothing"? $\endgroup$
    – user85598
    Jul 11, 2020 at 15:13

1 Answer 1


In the quantum variational principle nothing is minimized: there is no variation that is set to zero.

Rather, the principle is the infinitesimal version of Feynman's path integral. Indeed, integrating the variational principle one obtains the usual formula for the functional integral, as very clearly emphasized by Schwinger's student Bryce DeWitt, cf. ref 1 §10. In fact, it is hard to find a better presentation of the variational principle than that of this reference.

As the variational principle is completely equivalent to the functional integral, the physical interpretation is the same: the amplitude for any given field configuration is a phase, determined by the classical action of that configuration. This is a postulate, there is no deeper explanation for where this comes from. One can derive all of quantum mechanics from this principle, but one cannot derive the principle itself. DeWitt has a very good motivation though. Go check it.

Moreover, and for the same reason, the classical limit is obtained in the same way: As usual, the classical configuration – being a critical point – gives a particularly large contribution to the total amplitude, at least as long as $S/\hbar$ is sufficiently large. In the limit $S/\hbar\to\infty$, the classical configuration is the only configuration – the rest all interfere destructively. One should keep in mind that, if $S/\hbar$ is not large, purely quantum effects may dominate, in which case the classical configuration is entirely irrelevant to the dynamics.

I don't understand subquestion 4. But no: the r.h.s. is not (typically) negligible for large times. It is negligible in the classical limit, and only in that limit, by definition. Sometimes, $t\to\infty$ might be equivalent to the classical limit (e.g. if we adiabatically turn off interactions or something like that). But not in general.

As a side comment, one can reformulate the quantum variational principle as a classical variational principle, by replacing the classical action by the effective (quantum) action. In fancier terms one can introduce the quantum BV bracket, etc., all of which replace classical objects. See ref. 1 §24 if interested. This is beyond the scope of this post.


  1. DeWitt B.S. - The global approach to quantum field theory Vol.1.
  • $\begingroup$ Thank you, I am getting it, slowly. You write "the amplitude for any given field configuration is a phase, determined by the classical action of that configuration." But in the postulate, there are "in" and "out" states. So should the statement be read as: "the amplitude for any given field transition is a phase, determined by the classical action of that transition"? $\endgroup$
    – user85598
    Jul 17, 2020 at 9:14
  • 1
    $\begingroup$ Ans "S/hbar" should probably be "large" instead of "small" everywhere? $\endgroup$
    – user85598
    Jul 17, 2020 at 9:18
  • $\begingroup$ If you have time to answer my comments I would release the bounty... $\endgroup$
    – user85598
    Jul 19, 2020 at 19:19
  • $\begingroup$ @Christian Hi. 1) The in/out labels refer to generic states. They typically depend functionally on the fields. The variation refers to their implicit dependence thereon, and only on that dependence (we do not allow variations that change the functional dependence on the fields). So: the amplitude ⟨in|out⟩ depends as a functional on the fields, and this dependence is postulated to be a phase, determined by the action evaluated on that configuration. This is explained in much more detail in the reference. 2) Yes! 3) No rush to award the bounty, it's best to wait in case a better answer appears! $\endgroup$ Jul 20, 2020 at 19:35
  • $\begingroup$ By the way, the book by DeWitt is terrible. A lot of formalism, little physics. A disaster book for learning QFT. $\endgroup$
    – user85598
    Jul 26, 2020 at 15:34

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