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.

  • $\begingroup$ 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 ? $\endgroup$ – thephysics17 Mar 18 '19 at 19:42
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    $\begingroup$ @thephysics17 For a recent reference see arxiv.org/abs/1905.04236 where some conceptual issues related to yours should be clarified. $\endgroup$ – Wein Eld Oct 18 '19 at 8:50

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