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It is known that the path integral in quantum mechanics means the summation of all probable classical trajectories between first and last measurement of the quantum state. In QFT this formalism leads to generating function. Generating function can be changed to 1PI effective action, $\Gamma$, be Legendre Transformation, Which ends to $ e^{i \Gamma} \propto \int D\phi e^{iS} $ Now, I have a problem with the classical interpretation of this equation. Does this equation have a same classical interpretation as classical trajectory?

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closed as unclear what you're asking by AccidentalFourierTransform, Qmechanic May 4 '18 at 17:13

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You might know from Q.M. that momentum and energy are generators of space and time translation respectively:

$$e^{iP\delta x/\hbar}\psi(x,t)= \psi(x+\delta x,t)$$

$$e^{-iH\delta t/\hbar}\psi(x,t)=\psi(x, t+\delta t)$$

If you were to imagine a sin wave, the momentum operator would take you across the spatial extent of the the wave for a fixed time, while the Hamiltonian would fix your point in space and translate you to the value of the wave at different times.

The Lagrangian, however is a combination of these two generators such that you follow a fixed point on the wave. If you find a wave-peak at $(x,t)$, the Lagrangian as an operator will take you to the same peak at some other time and place $(x+\delta x, t+\delta t)$. The Lagrangian is the generator of paths.

A complication that arises is that generally $P$ and $H$ don't commute. However, for small $\delta x$ and $\delta t$:

$$e^{-iH\delta t/\hbar}e^{iP\delta x/\hbar}\approx e^{-iH\delta t/\hbar+iP\delta x/\hbar}=e^{i(P\delta x/{\delta t}-H)\delta t/\hbar}=e^{iL \delta t/\hbar}$$

Since this is only valid as an infitesimal generator, the trick of the path integral is to introduce a resolution of identity between every one of the above operators over the "states":

$$|x+\delta x, t+\delta t>$$

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