- #1

- 19

- 0

## Main Question or Discussion Point

In http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" [Broken], he mentions this:

I'm trying to get the same result, but I'm stuck. Has anyone done this before?...So I simply put them equal, taking the simplest example where the Lagrangian is [tex]\frac{m}{2}\dot{x}^2 - V(x)[/tex] but soon found I had to put a constant of proportionality A in, suitably adjusted. When I substituted [tex]Ae^{i \epsilon L / \hbar}[/tex] for K to get

[tex]\varphi(x',t+\epsilon) = \int A e^{\frac{i}{\hbar} L(\frac{x'-x}{\epsilon},x) \epsilon} \varphi(x,t) dx[/tex]

and just calculated things out by Taylor series expansion, out came the Schrödinger equation.

Last edited by a moderator: