I've recently been reading Path Integrals and Quantum Processes by Mark Swanson; it's an excellent and pedagogical introduction to the Path Integral formulation. He derives the path integral and shows it to be: $$\int_{q_a}^{q_b} \mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$$

This is clear to me. He then likens it to a discrete sum $$\sum_\limits{\text{paths}}\exp\left(\frac{iS}{\hbar}\right)$$ where $S$ is the action functional of a particular path.

Now, this is where I get confused. He claims that, because some of these paths are discontinuous or non-differentiable and that these "un-mathematical"1 paths cannot be disregarded, the sum is not mathematically rigorous, and, thus, that the transition amplitude described by the path integral is not rigorous either. Please correct me if I am incorrect here.

Furthermore, he claims that this can be alleviated through the development of a suitable measure. There are two things that I don't understand about this. First, why isn't the integral rigorous? Though some of the paths might be difficult to handle mathematically, they aren't explicitly mentioned at all in the integral. Why isn't the answer that it spits out rigorous? And, second, why would a measure fix this problem?

1 Note: this is not the term he uses

  • 5
    $\begingroup$ Short answer: To define an integral rigorously, it's not enough to just say "and now take the limit $N \to \infty$". You need to prove that your discrete sum converges to something, and that it doesn't matter how you take the limit. $\endgroup$
    – Javier
    Mar 11, 2015 at 2:44
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    $\begingroup$ @Javier Badia does this have to do with non-differentiable paths or is it a separate issue? $\endgroup$
    – jimmy
    Mar 11, 2015 at 4:00
  • $\begingroup$ Can't one make everything work with proper regularization, and doesn't this regularization allow all the cases actually relevant to physics (as opposed to all possible mathematical corner cases)? $\endgroup$
    – DanielSank
    Mar 11, 2015 at 7:26
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    $\begingroup$ It is indeed folklore that path integral is not rigorous mathematically, or more precisely, the rigorous maths has not yet been rigorously developed. This is typical in physics. But the real problem is that, many people do not know they are doing handing waving when they are doing it. $\endgroup$ Mar 11, 2015 at 9:14
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    $\begingroup$ @NikolajK just in general. it is my first introduction to path integrals, but I am finding that the book is not too difficult to follow $\endgroup$
    – jimmy
    Mar 11, 2015 at 15:04

2 Answers 2


There are several points:

  • The first is that for usual self-adjoint Hamiltonians of the form $H=-\Delta +V(x)$, with a common densely defined domain (and I am being very pedantic here mathematically, you may just ignore that remark) the limit process is well defined and it gives a meaning to the formal expression

    $\int_{q_a}^{q_b} \mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$

    by means of trotter product formula and the corresponding limit of discrete sums. So the object has most of the time meaning, as long as we see it as a limit. Nevertheless, it would be suitable to give a more direct mathematical interpretation as a true integral on paths. This would allow for generalizations and flexibility in its utilization.

  • It turns out that a suitable notion of measure on the space of paths can be given, using stochastic processes such as brownian motion (there is a whole branch of probability theory that deals with such stochastic integration, called Itô integral). To relate this notion with our situation at hand there is however a necessary modification to make: the factor $-it$ in the quantum evolution has to be replaced by $-\tau$ (i.e. it is necessary to pass to "imaginary time"). This enables to single out the correct gaussian factors that come now from the free part of the Hamiltonian, and to recognize the correct Wiener measure on the space of paths. On a mathematical standpoint, the rotation back to real time is possible only in few special situations, nevertheless this procedure gives a satisfying way to mathematically define euclidean time path integrals of quantum mechanics and field theory (at least the free ones, and also in some interacting case). There are recent works of very renowned mathematicians on this context, most notably the work of the fields medal Martin Hairer (see e.g this paper and this one, or the recent work by A. Jaffe that gives an interesting overview; a more physical approach is given by Lorinczi, Gubinelli and Hiroshima among others).

  • The precise mathematical formulation of path integral in QM is called Feynman-Kac formula, and the precise statement is the following:

    Let $V$ be a real-valued function in $L^2(\mathbb{R}^3)+L^\infty(\mathbb{R}^3)$, $H=H_0+V$ where $H_0=-\Delta$ (the Laplacian). Then for any $f\in L^2(\mathbb{R}^3)$, for any $t\geq 0$: $$(e^{-tH}f)(x)=\int_\Omega f(\omega(t))e^{-\int_0^t V(\omega(s))ds}d\mu_x(\omega)\; ;$$ where $\Omega$ is the set of paths (with suitable endpoints, I don't want to give a rigorous definition), and $\mu_x$ is the corresponding Wiener measure w.r.t. $x\in\mathbb{R}^3$.


...He derives the path integral and shows it to be: $$\int_{q_a}^{q_b}\mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$$

This is clear to me. He then likens it to a discrete sum $$\sum_\limits{\text{paths}}\exp\left(\frac{iS}{\hbar}\right)$$ where $S$ is the action functional of a particular path.

Now, this is where I get confused.

At this point I think it will be helpful to make an analogy with an ordinary Riemann integral (which gives the area under a curve).

The area A under a curve f(x) from x="a" to x="b" is approximately proportional to the sum $$ A\sim\sum_i f(x_i)\;, $$ where the $x_i$ are chosen to be spaced out from a to b, say in intervals of "h". The greater the number of $x_i$ we choose the better an approximation we get. However, we have to introduce a "measure" to make the sum converge sensibly. In the case of the Riemann integral that measure is just "h" itself. $$ A=\lim_{h\to 0}h\sum_i f(x_i)\;, $$

In analogy, in the path integral theory of quantum mechanics, we have the kernel "K" to go from "a" to "b" being proportional to the sum of paths $$ K\sim\sum_\limits{\text{paths}}\exp\left(\frac{iS_{\tt path}}{\hbar}\right) $$

In this case too, it makes no sense to just consider the sum alone, since it does not have a sensible limit as more and more paths are added. We need to introduce some measure to make the sum approach a sensible limit. We did this for the Riemann integral simply by multiplying by "h". But there is no such simple process in general for the path integral which involves a rather higher order of infinity of number of paths to contend with...

To quote Feynman and Hibbs: "Unfortunately, to define such a normalizating factor seems to be a very difficult problem and we do not know how to do it in general terms." --Path Integrals and Quantum Mechanics, p. 33

In the case of a free particle in one-dimension Feynman and Hibbs show that the normalization factor is $$ {({\frac{m}{2\pi i\hbar\epsilon}})}^{N/2};\, $$ where there are N steps of size $\epsilon$ from $t_a$ to $t_b$, and N-1 integrations over the intermediate points between $x_a$ and $x_b$.

Again, quoting from Feynman and Hibbs regarding these normalization measures: "...we do know how to give the definition for all situations which so far seem to have practical value."

So, that should make you feel better...


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