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In quantum mechanics, we are used to thinking of states represented by vectors in a (possibly infinite dimensional, projective) Hilbert space, equipped with an inner product (e.g. the $L_2$ norm). Physical observables are represented by Hermitian linear operators, such as the Hamiltonian. In principle, I suppose one could define the Lagrangian as being an operator, but I don't know how to quantify that statement. Moreso, however, I am interested in what sort of operator one can think of the action, $S = \int dt \,L$, as being. The Schrodinger equation tells us that the Hamiltonian is the generator of unitary time evolution, and it seems that we could then say that the information content of a minimal quantum mechanics is given by "QM $\sim$ Hilbert Space + Schrodinger Equation". With this rough basis, one can go on to define the path integral, from which we see that the action rather naturally arises, and thereafter one can take the "classical limit" $\hbar \rightarrow 0$ to obtain the classical action principle by a saddle point argument. But I am wondering how one should think of the action in the context of QM. For example, we can define a partition function describing a thermal state according to $$Z \sim \mathrm{Tr}(e^{-\hat{H}/T})$$ and perhaps by analogy one can think of the path integral as being $$Z \sim \mathrm{Tr}(e^{i \hat{S}/\hbar})$$ but $\hat{H}$ is a linear operator in a vector space while "$\hat{S}$" is an "operator" in a space of "paths" (continuous sequences of states in a Hilbert space?).

I am wondering how one can make this more concrete. It seems to me that the fundamental difference lies in the fact that the action involves an integral over time while the Hamiltonian is the generator of time evolution. This suggests that one might think of $\hat{H}$ as being in some sort of "Lie algebra" while $S$ is in its "Lie group", if I can abuse some terminology and abstract the concepts somewhat (My understanding of these topics is still a bit vague and incomplete, if I understand correctly, the time evolution operators $U(t) = e^{-i\hat{H}t/\hbar}$ would be in the Lie group generated by $\hat{H}$. In that case, what is the space in which the action lives/acts?)

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  • $\begingroup$ It moves the system forward in time in configuration space. $\endgroup$ – Cinaed Simson Nov 29 '19 at 21:34
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When going from the operator formalism to the path integral formalism (by inserting infinitely many completeness relations$^1$) operators are replaced by number-valued variables/fields. The action arises as a functional (rather than an operator) of those.

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$^1$ For the derivation, see any good textbook on QM/QFT.

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In the path integral formulation of quantum mechanics there are no operators left - all the information has been recast as a functional integral over trajectories over which space the action takes ordinary real number values.

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