Does e^{iS} Equal Zero When S Approaches Infinity in Feynman Path Integrals?

[kakarukeys]

Hello there,

please tell me whether e^{iS} = 0 when S = +\infty

S is the action of a particle = \int L dt
e.g. when the particle goes to infinity and comes back, kinetic energy blows up,
then S blows up.

 

[LeonhardEuler]

Hello there,

please tell me whether e^{iS} = 0 when S = +\infty

It does'nt:
e^{ix}=\cos{x}+i\sin{x}
\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}
But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.

 

[kakarukeys]

That's what my math teacher told me,
but my physics teacher told me entirely different thing.

 

[Hurkyl]

I would suspect both your math teacher and physics teacher are right!

The map x → e^(ix) is, indeed, discontinuous at +∞, and thus one cannot continuously extend this map to have a value at +∞. (continuous extension is the process that is generally used to justify statements like arctan +∞ = π / 2 or x/x = 1).


However...


As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.

 

[kakarukeys]

Do you mean, for computation purpose, we can define the limit to be zero
it can't be proven?

 

[kakarukeys]

this is no college level!
Feymann Path Integral!

 

[inha]

this is no college level!
Feymann Path Integral!

College=university. Don't get freaked out.

 

[kakarukeys]

Feynmann Path Integral is at graduate level