# In the Feynman Path Integral, why must “contributing paths” be continuous? Or is this a false notion?

I have been studying the Feynman path integral and its various derivations, and I've run into a bit of a problem. The standard Feynman path integral appears as follows: $$\int \mathcal{D}[x(t)]\exp\left(\frac i \hbar \int_0^t d\tau\ \left( \frac m2 \dot{x}(\tau) - V(x(\tau) \right) \right)$$ Where the path integral is defined by passing the following into the continuum limit: $$\lim_{n \to \infty}\idotsint d^{n-1}x\ \mathcal{N} \exp\left( \frac i \hbar \sum_{j=1}^n \left( \frac m2 \left( \frac{x_{j+1} - x_j}{\Delta t} \right)^2 - V(x_j) \right) \Delta t \right)$$ where $$\Delta t = t/n$$. In this form, we independently integrate each $$x_j$$ from $$-\infty$$ to $$+\infty$$, and hence there is no innate notion that the paths considered must be continuous; $$x_{j+1}$$ and $$x_j$$ can be as far apart as we please.

In doing readings, I have found multiple sources that claim that the paths we consider are continuous, for example in this paper. In transforming to imaginary time ($$-t \to -i\tau$$), this claim makes sense, since discontinuities in the path will blow up the the sum and they will be exponentially suppressed. However, in the form presented above, I am not quite sure where and how we can make this claim.

I have seen the argument that paths that deviate greatly from the classical path will lead to extreme oscillatory behavior from our phase, and hence will be suppressed since they will on-average integrate to zero. However, I don't think I am satisfied with this argument to claim we must have continuity. This shows which paths contribute the most, but I don't think it rules out discontinuous paths. Does anyone have insight into this? Any help would be greatly appreciated.

• The measure of the discontinuous path is zero, from what I remember – Slereah Apr 8 at 19:59

The paths of a Feynman integral do not always have to be continuous, and the first example that comes to my mind is the Feynman integral in the Feynman-Vernon formalism, which is used to study the spin-boson model, among other things.

For the spin-boson problem, the $$x(t)$$ paths written in your question, are discontinuous. If the spin in your spin-boson problem is a spin-1/2 particle, then $$x(t)$$ will be either -1/2 or +1/2 at each time $$t$$, and the Feynman integral will be over all possible paths, of which I'll give four examples below:

• $$x(t)$$ = -1/2 for all $$t$$.
• $$x(t)$$ = +1/2 for all $$t$$.
• $$x(t)$$ = -1/2 from $$t=0$$ to $$t$$=5ps, and $$x(t)$$=+1/2 for the rest of the time.
• $$x(t)$$ = -1/2 from $$t=0$$ to $$t$$=5ps, $$x(t)$$=+1/2 from $$t$$=5ps to $$t$$=10ps, then $$x(t)$$=-1/2 for the rest of the time.

If you allow all possible flips between spin-up and spin-down between $$t=0$$ and the rest of your desired time for the Feynman integral, there will be an infinite number of such paths.

The software FeynDyn (Feynman Dynamics) can calculation such Feynman integrals for up to 16 spin-1/2 particles (or 16 qubits, if simulating a quantum computer in the presence of noise) and more information about this whole "Feynman-Vernon" formalism can be found in this paper (I have given a free link to the PDF rather than the original journal, in case COVID lockdown has caused you not to be on a campus where you can download papers easily).

My understanding is that most paths are inherently discontinuous, and the transition to functional integration is a symbolic one.

However, many applications of path integrals are centered on the stationary phase approximation, which in case of path integrals cited in the question corresponds to a quasi-classical trajectory and small fluctuations around it. In this limit the contribution of discontinuous paths is indeed negligeable.

• I don't think this is correct. The reason why only continuous paths contribute is that the kinetic term is $[(x_{j+1}-x_j)/\Delta t]^2$ and if paths are discontinuous that gives a highly oscillatory factor $e^{i/(\hbar\Delta t^2)}$ which sends the contribution to zero, regardless of whether or not we take the saddle-point limit $\hbar \to 0$ – MannyC Apr 8 at 20:20
• In fact, the paths that contribute the most to the path integral have $x_{j+1}-x_j \in O(\Delta t^{1/2})$. – MannyC Apr 8 at 20:22
• @MannyC is there any reference that shows how that is derived? – B. Sandoval Apr 8 at 20:28
• To the level of rigor of mathematical analysis I have no idea. But there is a more "physicist's proof" in Sec. 1.1 of Anselmi's book on Renormalization. – MannyC Apr 8 at 20:35
• @MannyC I don't see how what you say is different from my answer. Except that I am more careful in distinguishing a formal mathematical representation (where the continuity does not apply) and the approximation where on can consider continuous paths. – Vadim Apr 8 at 20:45