# path integrals: how/why can the phase be identified with the action?

In Peskin & Schroeder, chapter 9 introduces the functional methods.

The idea, to recall, is simply to sum over all the possible paths:

$U(x_a,x_b;T) = \sum_{\text{all paths}} e^{i . \text{phase}} = \int Dx(t) e^{i . \text{phase}}$

Then, it happens that the phase could be identified with the action. But I have not understood how this could be done ... Could anyone clarify that point?

Thanks in advance.

-
In the classical (small $\hbar$) limit you can find the dominant contribution to the path integral by the method of stationary phase. Because of that it is natural to identify the phase with something that is extremised in classical dynamics, the obvious candidate being the action. –  Michael Brown Jan 19 '13 at 10:14
Alternatively, you should follow a derivation of the equivalence of Feynman's path-integral approach to Hamiltonian-based approaches - to see how over the infinitesimal times, the Hamiltonian switches to the Lagrangian etc. This is obviously a topic for hours of study rather than three lines of screaming at this server. I am afraid that you haven't understood those issues because you have never studied them! Without an understanding of the equivalence of the path-integral approach to the operator approach (or without its classical limit), it's still a valid definition of a QM theory! –  Luboš Motl Jan 19 '13 at 11:42
Indeed, Feynman's original paper is still the most readable account I'm aware of. –  twistor59 Jan 19 '13 at 12:36