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In Quantum Mechanics, transition amplitude between two states in given by (path integral approach): $$ \left\langle q';t'|q;t\right\rangle= \int[\mathrm dq] \exp \left(i \int L(q,\dot{q})~\mathrm d\tau\right) $$

This tells that contribution of the paths to the amplitude is given by the weight factor : "i times the action".

Can anybody explain "intuitively" why this should be the weight factor?

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  • $\begingroup$ Please define what you mean by "intuitive". There is no requirement for facts to have explanations of any particular kind. $\endgroup$ – ACuriousMind Sep 5 '16 at 11:48
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The intuitive explanation is really not there. I have a very fuzzy argument in the favor calling this "the weight factor". Let's see if that helps you at all.

The action is given by $S = \int L(q,\dot{q}) dt$. Now classically the path of the particle is determined by minimizing the action. i.e. the particle will follow the path for which the action is minimized(or in general extremized). This tells you that action somehow gives you the idea of how long the path is. Just like Fermat's Principle in optics.

Now come to the quantum mechanics using path integrals. Here we assume that the particle travels through all possible paths. But we know that the paths are not of same length(here length stands for the fact that action is different along different paths). So when the particle reaches to same final point and starts from same initial point but through two different paths, we expect the wave-function will catch a relative phase and its intuitively obvious that this phase must be proportional to the length of the path i.e. action. That's why you can call it "the weight factor" associated with the path.

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  • $\begingroup$ Thank you for your answer. Please tell me why this information of travelling through different path enters the wave function as a "phase"? $\endgroup$ – seeking_infinity Sep 19 '16 at 9:05
  • $\begingroup$ That is true even in optics. The length of two different path decides whether they are going to constructively interfere or destructively. Thus the length has a effect of "phase". $\endgroup$ – Ari Sep 20 '16 at 14:16
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Although you are looking for a more intuitive explanation i think the best way to see it is to simply derive it mathematically which is done in every QFT book.

Since $H=i \frac{\partial}{\partial t}$ the time evolution transition amplitude between two infinitesimally close states would be $\langle q_i|1-iH\Delta t|q_{i+1} \rangle = \langle q_i|e^{iH\Delta t}|q_{i+1} \rangle$. To obtain the full amplitude between states $|q \rangle$ and $ |q' \rangle$ one would need to sum over all infinitesimal time transitions resulting in some integrals. The definition of the Hamiltonian $H=\frac{p^2}{2m} - V(q)$ would allow for some gaussian integrations over $p$. Taking the analytical limit in the end the discretized sum over all intermediate states can be expressed as an continuous integral of the Lagrangian over time.

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