Weight factor in Path Integral Formalism

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?

• Please define what you mean by "intuitive". There is no requirement for facts to have explanations of any particular kind. – ACuriousMind Sep 5 '16 at 11:48

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.
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.