# Equivalence Picard-Lefschetz path integrals and “Feynman's” path integrals

I have just seen the Picard lefschetz method applied to path integrals in order to make these more convergent. I understand how we could modify the contour of integration for a real integral but what I don't understand is how can that be equivalent to the usual Feynman's integration over all paths since with this method we only integrate on a discrete number of Lefschetz thimble.

Even in one dimensions, we should integrate over all $$x(t)$$ having the right boundary conditions but with Picard-Lefschetz method, it apparently suffies to integrate over one lefschetz thimble for the free particle for instance ( see https://arxiv.org/abs/1406.2386 ).

What allows us to reduce the number of integration to make?

Let me clarify issues with your one-dimensional problem. Originally, you integrate all the real paths $$x(t)$$ which constitute an infinite dimensional space $$\{x(t)\}$$. To picture the use of complex analysis in this infinite dimensional space, let us formally think of it $$n$$-dimensional with $$n=\infty$$ kept in mind.
Now, we complexify this $$n$$-dimensional space to $$\{z(t)\}$$ which is $$2n$$-dimensional in real field. The Lefschetz thimble is an $$n$$-dimensional contour in this $$2n$$-dimensional space which passes some stationary points. So the Lefschetz thimble, quite different from what you thought, is an infinite dimensional space (remember $$n=\infty$$) and lose no information about your original real paths $$x(t)$$ provided the contour is equivalent to the original real plane via the Cauchy theorem.
• Is the fact that a lefschetz thimble is n-dimensional due to the fact that it is equivalent thanks to cauchy theorem to all real paths ( in the case you just discuss ) ? Because I thought the Lefschetz Thimble was just one path and integrating over all paths meant for me integrating x times ( x is the cardinality of $\left{x(t)\right}$ ) but that gives me an infinite since the integration over the thimble is a constant ( it does not change with the real paths ) and we integrate an infinite number of time... Is there kind of a "normalisation constant" ?And how does it arise in the integrals ? – thephysics17 Mar 18 '19 at 19:42