Probability Amplitude phases

1. Nov 10, 2007

Frankww

Quesion in approaching to Path Integral

I've just read "Quantum Mechanics and Path Integral" book which was written by Feynman, he said the phase of probability amplitude is proportional to "the ACTION $$S$$ in units of Quantum of action $$\hbar$$. What is the reason to be that? Can any body explain it to me physically? Thank you for all replies.

Last edited: Nov 10, 2007
2. Nov 11, 2007

blechman

There might be a better answer than this, but one way to see it is to plug the wavefunction $e^{iS/\hbar}$ into the Schrodinger equation and get an equation for S. The equation you get (to leading order in an expansion in $\hbar$) is just the (classical) Hamilton-Jacobi equation, where S plays the role of the classical action. This is very strong evidence in favor of Feynman's arguments.

In the end, I don't think there's an explicit "proof" that you should use the classical action in this way, just like there's no "proof" that the Schrodinger equation is correct - at the end of the day, you just ask if the equations are consistent with experiment (and the other descriptions of QM). And in this case, they are. In particular:

1. You can derive the Schrodinger equation from the path integral formalism.
2. There is a nice classical limit as $\hbar\rightarrow 0$.

3. Nov 12, 2007

Frankww

4. Nov 12, 2007

quantumfireball

Read the first few pages of QFT in Nutshell by ZEE

The motivation for introducing path integral formulation of QM is startted as
a doubt that feynmann had regarding the inteference of photon as it goes through a doble slit...