# What type of non-differentiable continuous paths contribute for the path integral in quantum mechanics

Consider the path integral for a 1D particle subjected to a potential $V(x)$ in imaginary time

$$\int_{x(0)=x_0}^{x(T)=x_T} [dx] \, e^{- \int_0^T d\tau \left[\frac{1}{2}\dot{x}^2 + V(x(\tau))\right]} \ .$$

I want to understand what subset of continuous paths $\mathcal{C}[0,T]$ contribute in the above integral.

It seems clear that discontinuous paths will not contribute to the path integral because of the structure of the regularized kinetic term

$$\dot{x}^2 \simeq \frac{\left[x(\tau+\epsilon)-x(t)\right]^2}{\epsilon^2} \ .$$

If there is a jump a $\tau$ then as $\epsilon \to 0$ the above quantity diverges as $1/\epsilon^2$. If we take into account the extra factor of $\epsilon$ from the measure ($\Delta \tau = \epsilon$), then the Euclidean action for this path diverges as $1/\epsilon$ so the corresponding Boltzmann factor $e^{-S_E}$ vanishes and the path does not contribute.

In mathematical parlance I guess one would say that discontinuous paths have Wiener measure zero where the Wiener measure is heuristically defined by

$$d\mu_W(x) \simeq [dx] \, e^{-\frac{1}{2} \int_0^T d\tau \, \dot{x}^2} \ .$$

Hence any path which contributes must be continuous.

R. J. Rivers states (without justification) the following on p. 119 of "Path integral methods in quantum field theory":

"The continuous path $x(\tau), \, \tau \geq 0$ is said to have Holder continuity index $\alpha$ if $\exists K > 0$ such that $$|x(\tau_1) -x(\tau_2)| < K |\tau_1-\tau_2|^\alpha, \, \forall \tau_1,\tau_2 > 0.$$ It can be shown that the set of Holder continuous paths $a(\tau)$ of index $\alpha$ has $\mu_W$ measure zero if $\alpha \geq 1/2$. In particular, the differentiable paths for which, approximately, $\alpha = 1$ have $\mu_W$ measure zero. From this point of view finite action means zero measure because finite action paths are too smooth."

Is this statement really true? If so does anyone know of a reference that proves it? One thing that seems suspicious is that Brownian motion has $\alpha = 1/2$ which would suggest it doesn't contribute in the Wiener measure. (Correction: After reading the reference suggested by Abdelmalek Abdesselam, I realize that the Brownian paths have Holder continuity index $\alpha < 1/2$).

Additional comment: If we look at the structure of the action then it becomes apparent that we are really interested in

$$\epsilon \dot{x}^2 = \left[\frac{|x(t+\epsilon)-x(t)|}{\sqrt{\epsilon}}\right]^2 \sim \epsilon^{2(\alpha - 1/2)}$$

If $\alpha < 1/2$ then we get a divergence as $\epsilon \to 0$ so, following the same logic I used for the discontinuous paths, I would have thought that these are precisely the paths which don't contribute to the path integral because they have infinite Euclidean action. The Rivers books seems to be saying the opposite so I remain confused.*

Additional comment (part 2): I am trying to wrap my head around Abdelmalek Abdesselam's claim that (at least when we restrict to the space of continuous paths) that only those paths with infinite Euclidean action contribute.

It seems that it might be conceivable that the $S = \infty$ argument for discarding paths breaks down for the following reason. Let's consider a single a path which has $\alpha < 1/2$, then its corresponding Boltzmann factor presumably scales like $$Z \supset \exp ( - K \epsilon^{2(\alpha-1/2)})$$ where $K > 0$ is some positive constant. This exponential is indeed zero in the limit $\epsilon \to 0$. If for some reason, however, the paths with $\alpha < 1/2$ are sufficiently numerous then we should also taking into account the multiplicity factor $\mathcal{N}_\alpha$ for such paths

$$Z \supset \mathcal{N}_\alpha \exp ( - K \epsilon^{2(\alpha-1/2)}) = \exp \left[ S(\alpha)- K \epsilon^{2(\alpha-1/2)}\right]$$ where I have written the multiplicity factor in terms of an 'entropy' $S(\alpha)$ defined by $\mathcal{N}_\alpha = e^{S(\alpha)}$. Thus provided that the entropy for $\alpha<1/2$ paths grows at least as fast as $\epsilon^{2(\alpha-1/2) }$, then these paths have a chance to dominate the path integral in the $\epsilon\to 0$ limit. Is it possible to see this mathematically?

I suppose for the discontinuous paths the $S=\infty$ argument really works because for some reason these paths are not sufficiently numerous to overcome the $\exp(-K/\epsilon)$ damping factor. Again it would be nice to see a mathematical justification for this

A good discussion of regularity properties of the Wiener measure is in section II.5 of the book "Functional Integration and Quantum Physics" by Barry Simon. He gives the proofs of most of the relevant theorems except the borderline case $$\alpha=\frac{1}{2}$$, namely Levy's Theorem showing that $$\frac{|x(\tau_1)-x(\tau_2)|}{\sqrt|\tau_1-\tau_2|}$$ diverges logarithmically everywhere with probability one.

Edit as per the question in the new comment: finding out if paths contribute is a bit more complicated than just seeing if the action $$S=\infty$$ or not. One can see this in the simpler setting of a countable product of standard Gaussians, namely, the probability measure on the space of sequences $$x=(x_n)_{n\ge 0}$$ formally given by $$e^{-S(x)} \ \prod_{n\ge 0} \frac{dx_n}{\sqrt{2\pi}}$$ where the action is $$S(x)=\frac{1}{2}\sum_{n\ge 0} x_n^2\ .$$ In this case the product probability measure makes sense by the Kolmogorov Extension Theorem. However, with probability one the action will diverge. In fact if one allows different variances $$v_n$$ for the coordinates to generalize this a bit, i.e., one works with the measure $$e^{-S(x)} \ \prod_{n\ge 0} \frac{dx_n}{\sqrt{2\pi v_n}}$$ where the action is $$S(x)=\frac{1}{2}\sum_{n\ge 0} \frac{x_n^2}{v_n}\ ;$$ one has a clear cut dichotomy:

1. If $$\sum v_n<\infty$$ then $$l^2$$ has full measure, namely, the action (old one with no $$v$$'s) converges almost surely.
2. If $$\sum v_n=\infty$$ then $$l^2$$ has measure zero, namely, the action (old one with no $$v$$'s) diverges almost surely.

• Thanks for the reference. I have updated my question based on what I understood from there. It clarifies my question about the Brownian motion but I am still confused why the contributing paths have $\alpha < 1/2$. May 23, 2015 at 5:39
• I added more material which should shed some light on why for this kind of probability measures the paths which contribute must have $S=\infty$. May 26, 2015 at 13:12
• Thanks. I updated my question with some additional comments. Do you know of a good reference which discusses the countable product of standard gaussians you describe above? May 26, 2015 at 20:33
• it's not too hard to do by hand. The proof of 1 is trivial. For 2 you need to use fourth moment estimates in order to analyze the variance of $x_n^2$. Basically you need to use the divergence of the series of expectations (the hypothesis), a bound on the variance and most importantly the independence of the $x_n^2$. Otherwise you can look at any graduate textbook in probability. The key words are Kolmogorov's three series theorem. The convergence of the series of $x_n^2$ is a typical example of tail event involving what happens at $n\rightarrow\infty$. Kolmogorov's zero one law here says... May 30, 2015 at 10:38
• ...that the probability of convergence is either zero or one. Usually, it is nontrivial to figure out which of the alternatives holds, but here it's not too bad. May 30, 2015 at 10:39

Lubos Motl has some handwavy discussion of the $S = \infty$ puzzle towards the end of

http://motls.blogspot.com/2012/06/why-feynmans-path-integral-doesnt.html?m=1

Edit: It occurs to me that there is an analogous phenomenon taking place in the $d=2$ Ising model which leads to the second-order phase transition.

The idea is to consider spin configurations consisting of a droplet of "up" spins immersed in a sea of "down" spins. The Boltzmann factor for each such configuration is $e^{-2\beta L}$ where $\beta$ is the inverse temperature and $L$ is the perimeter of the droplets. So naively it looks like the droplets or large size $L$ will be strongly suppressed for all temperatures $1/\beta$ and the system will remain in a state of long-range order (this is precisely what happens in the $d=1$ Ising model).

On the other hand the number of droplets of perimeter $L$ goes like $C^L = e^{L \log C}$ where $C$ is a constant. So taking into account the entropy contribution we get

$$e^{L \log C - 2\beta L}$$

Thus if $\beta < (1/2)\log C$ the entropy contribution wins and there is a proliferation of droplets which disorders the system.

I wonder if this can help to understand the partition function for quantum mechanics.

• this kind of energy/entropy arguments, like the Peierls one for 2d Ising, occurs in lots of places in stat mech, and indeed something of this kind underlies the typical regularity of Brownian paths. However figuring out the Holder 1/2 behavior is quite subtle. For instance you can use a sine basis to decompose the Brownian bridge e.g. for $x_0=x_r=0$, then your Fourier modes $a_k$ have variance $1/k^2$, as in Lobos' post. The dichotomy theorem in my answer implies that $\sum k^{2\alpha}|a_k|$ converges a.s. provided that $\alpha<1/2$. Namely paths are in the Sobolev space $H^{\alpha}$... May 30, 2015 at 10:49
• ...unfortunately this is not enough to get you by a Sobolev imbedding theorem in the space of continuous functions, let alone those which are Holder 1/2. If fact if one organizes the momenta $k$ into dyadic shells $2^i<|k|<2^{i+1}$, then the corresponding lacunary Fourier series converges a.s. and this is one way to prove the Holder 1/2 property. This is essentially how Wiener constructed his measure. Another way to get the Holder 1/2 is to use the Kolmogorov-Censov continuity theorem. This is explained in the book by Simon, I mentioned earlier. Note that the shell decomposition... May 30, 2015 at 10:54
• ...is reminiscent of the renormalization group. It is interesting here that randomness gives an exponent 1/2 improvement in regularity versus a more deterministic approach via a Sobolev imbedding theorem. May 30, 2015 at 10:57
• apologies to Luboš for misspelling his name. Unfortunately I cannot edit my previous comment. May 30, 2015 at 10:59
• it's really annoying not to be able to edit comments...the series I meant above is $\sum k^{2\alpha}|a_k|^{2}$ and not $\sum k^{2\alpha}|a_k|$. Also, if I remember correctly the a.s. convergence of the lacunary series is uniform. May 30, 2015 at 11:01