# Does Feynman path integral include discontinuous trajectories?

While reading this derivation of relation of Schrödinger equation to Feynman path integral, I noticed that $q_i$ can differ form $q_{i+1}$ very much, and when the limit of $N\to\infty$ is taken, there remain lots of paths, which are discontinuous (almost) everywhere — i.e. paths consisting of disconnected points.

Do I understand this wrongly? How do such discontinuous paths disappear on taking the limit? Or maybe they have zero contribution to integral?

• Related: Paths in the path integral where Slereah notes "In phase space, things get a bit more complicated, and only discontinuous paths in phase space contribute ('Feynman Path Integrals in a Phase Space' by Berezin)." – Alex Nelson May 3 '14 at 17:11

The discontinuous paths do 'disappear' when you take the continuum limit. They contribute nothing to the integral in the end. In the Euclidean picture, they are suppressed by the kinetic term in $e^{-S(\phi)}$, which looks like $\sum_t \frac{(\phi(t+a) - \phi(t))^2}{a}$.
One of the little details that makes QFT more difficult than QM is that the fluctuations of fields are not generally suppressed in the continuum limit. In scalar field theory in 4d, the kinetic terms looks like $\sum_x \sum_\mu a^2 (\phi(x + a e^\mu) - \phi(x))^2$, so you can get exponent 2 power law singularities in the correlation function of two field values.