At the moment I'm working with the quantum action principle of J. Schwinger. For this I read the following paper: http://arxiv.org/abs/1503.08091. But I have some mathematical problems with this formalism. I don't really see why this is a variation principle since it is opartor-valued.

The action principle is defined by: $\delta S = G_1 -G_2$, where $S$ is the action operator $S = \int_{t_1}^{t_2} dt ~L$ and $G$ the generator operators of the transformations. What does it mean that the action principle is operator-valued? For the classical action where $S$ is real we search for a stationary point but for operators? I also have a problem that $\delta S \neq 0$. What interpretation has this additional term $G_1 - G_2$?


2 Answers 2


Schwinger's quantum action principle states that, \begin{equation} \delta\langle x_{2},t_{2}|x_{1},t_{1}\rangle=i\langle x_{2},t_{2}|\delta\hat{S}|x_{1},t_{1}\rangle \end{equation} where $|x,t\rangle$ is an eigenstate of the time dependent Heisenberg picture position operator and $\hat{S}$ is the action promoted to be an operator.

Schwinger argues that the variation of the amplitude $\langle x_{2},t_{2}|x_{1},t_{1}\rangle$ can only depend on the initial and final states since these are the only objects making the amplitude. The variation in the amplitude must be, \begin{equation} \delta\langle x_{2},t_{2}|x_{1},t_{1}\rangle=\delta (\langle x_{2},t_{2}|)|x_{1},t_{1}\rangle + \langle x_{2},t_{2}|\delta(|x_{1},t_{1}\rangle) \ . \end{equation} The variation of the eigenstates is assumed to be the result of a small unitary operator, \begin{equation} |x,t\rangle+\delta |x,t\rangle=e^{-i\hat{G}}|x,t\rangle =|x,t\rangle-i\hat{G}|x,t\rangle \end{equation} so that, \begin{equation} \delta |x,t\rangle=-i\hat{G}|x,t\rangle \ . \end{equation} The variation of the amplitude is now, \begin{equation} \delta\langle x_{2},t_{2}|x_{1},t_{1}\rangle=i\langle x_{2},t_{2}|\hat{G}_{2}|x_{1},t_{1}\rangle -i \langle x_{2},t_{2}|\hat{G}_{1}|x_{1},t_{1}\rangle =i\langle x_{2},t_{2}|(\hat{G}_{2}-\hat{G}_{1})|x_{1},t_{1}\rangle \ . \end{equation} Comparing this result with the statement of Schwinger's quantum action principle implies, \begin{equation} \delta\hat{S}=\delta\int_{t_{1}}^{t_{2}}\hat{L}dt=\hat{G}_{2}-\hat{G}_{1} \ . \end{equation} This relation is saying that the variation of the action operator only involves dynamical variables at the terminal times (Quantum Kinematics and Dynamics, page 78) and that over the interior of the duration the action operator is a constant operator.

As an example, suppose we vary the end time from $t_{2}$ to $t_{2}+\delta t_{2}$. In classical mechanics, the variation of the action as a function of the coordinates of the start and end of the path is, \begin{equation} dS=p_{2}dx_{2}-H_{2}dt_{2}-p_{1}dx_{1}+H_{1}dt_{1} \end{equation} so that varying the end time and promoting the result to an operator gives $\delta \hat{S}=-\hat{H}_{2}\delta t_{2}$. Schwinger's quantum action principle now gives Schrodinger's equation in the form, \begin{equation} \delta\langle x_{2},t_{2}|x_{1},t_{1}\rangle=-i\langle x_{2},t_{2}|\hat{H}_{2}|x_{1},t_{1}\rangle\delta t_{2} \ . \end{equation}

  • $\begingroup$ Thanks for your answer. But then the quantum "action principle" is not a variation principle in the mathematically sense cause we don't minimize some quantity? It is more an equation? $\endgroup$
    – Alpha001
    Oct 10, 2016 at 13:21
  • $\begingroup$ @Alpha001 : In Schwinger's quantum action principle, suppose we keep the terminal conditions fixed and vary the dynamical variables in the action operator. Since the terminal conditions are fixed, Schwinger's principle reduces to $\delta \hat{S}=0$ which says the action operator is stationary. So the quantum action principle is not minimizing some quantity because it makes no sense to say one operator is smaller than another, but it does make sense to say that the action operator is stationary in the sense that it is a constant to first order in the variations. $\endgroup$ Oct 12, 2016 at 20:10
  • $\begingroup$ Thanks for your answer. I have a request to this. Does this mean that $\delta S=0$ implies $DS=0$ (the total derivative vanish like in the classical version)? $\endgroup$
    – Alpha001
    Nov 21, 2016 at 16:17
  • $\begingroup$ @Alpha001 : Could you define your operator $D$ ? $\endgroup$ Dec 3, 2016 at 12:19
  • $\begingroup$ With $DS$ I mean the total derivative(en.wikipedia.org/wiki/…) more or less the generalized version of the Jacobian. It should also exists in this infinite dimensional case. $\endgroup$
    – Alpha001
    Dec 4, 2016 at 15:54

Schwinger's quantum action principle is not a variational principle in the sense of finding stationary points for a functional, cf. OP's title question (v3). Rather it gives a formula for how a quantum system (typically an overlap/transition amplitude $\langle A| B \rangle$) changes under a change of external parameters/sources in the action $S$.


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