1
$\begingroup$

My question regards a comment D. Gross makes in his unpublished lecture notes about quantum field theory (the one with no chapter 1).

In chapter 8 (path integrals) pag. 136, he reaches at the following expression for the non-relativistic QM version of the Lagrangian path integral in the Euclidean formulation ( $\epsilon$ is a small time step)

$$ \langle q_f \lvert e^{-\frac{1}{\hbar}H(\hat p , \hat q) T} \rvert q_i \rangle = \mathcal{N} \int^{q(T) = q_f}_{q(0)=q_i} \mathcal{D} q(t) e^{-\frac{1}{\hbar} \int_0^T dt L_E[q(t)] }, $$

where $L_E$ is written as formal limit $\epsilon \to 0$ of the discretized formula

$$ m \frac{(q_i - q_{i-1})^2}{2\epsilon} + \epsilon V(q_i). $$

He then comments that as $\epsilon\to0$, the measure is dominated by the kinetic term of order $O(\epsilon^{-1})$ and concludes that the determinant paths are paths for which $q_i - q_{i-1}\sim \sqrt{\epsilon}$ (because they are not suppressed by the exponential weight). Finally he says these paths typical of Brownian motion, which continuous but nowhere differentiable.

Until here, I believe the argument seems reasonable (at least for a physicist), even if it is a totally heuristic and intuitive argument (and because we know the answer).

The problem is when he generalizes to the Lagrangian path integral for QFT in (Euclidean) $d$ dimensions (real scalar field). He then writes the discretized version of the kinetic term (with lattice spacing $a$)

$$ a^d \left( \frac{\phi(x+an_{\nu}) - \phi(x)}{a}\right)^2, $$

where he then concludes $\phi(x+an_{\nu}) - \phi(x) \sim a^{1-d/2}$, exactly in the same way as before. I cite here his last two sentences:

"For $d\ge2$ this means that the fields that contribute to the path integral will not even be continuous. It is no surprise that the continuum limit might not even exist."

I am aware (see for example the first two pages of this paper (and its references) and this post ) that even in QFT the accounted paths (configurations) are still continuous but nowhere differentiable, even if the mathematical machinery needed is much more involved and not yet properly defined for, say, gauge theories in 4 dimensions.

So, is this quite right? Do the discontinuous paths contribute? What does mean "to contribute" in this context (maybe having non-zero measure)? In other words, I would like to know to what extent this heuristic argument holds and in what sense the (dis)continuous nowhere differentiable are "typical" of the path integral measure, both in QFT and QM.

$\endgroup$
  • $\begingroup$ "I am aware (books, and discussions in the internet) that even in quantum field theory the paths (configurations) are still continuous but nowhere differentiable" [citation needed]. $\endgroup$ – AccidentalFourierTransform Nov 30 '18 at 1:17
1
$\begingroup$

Let $\alpha$ be a number in the interval $(0,1)$. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is Hölder continuous of exponent $\alpha$, or belongs to the space $C^{\alpha}(\mathbb{R})$ if, roughly speaking, it satisfies a bound of the form $$ |f(x)-f(y)|\le O(1)\ |x-y|^{\alpha}\ . $$ One can generalize that to higher values of $\alpha$ by letting $C^{\alpha}(\mathbb{R})$ be the space of functions that are $n$ times continuously differentiable and whose $n$-th derivative is in $C^{\alpha-n}(\mathbb{R})$ where $n=\lfloor\alpha\rfloor$ is the integer part of $\alpha$. One can generalize this further to $d$ dimensions and define $C^{\alpha}(\mathbb{R}^d)$. More importantly, one can generalize such Hölder-Zygmund spaces to arbitrary $\alpha\in\mathbb{R}$, i.e., to negative exponents as well. These are also particular cases of Besov spaces $B_{p,q}^{\alpha}(\mathbb{R}^d)$ when $p=q=\infty$. They are all subspaces of the $S'(\mathbb{R}^d)$, the space of temperate Schwartz distributions. If I remember correctly the delta function $\delta^d(x)$ in $\mathbb{R}^d$ is in $C^{\alpha}(\mathbb{R})$ if and only if $\alpha\le -d$.

Now regarding the OP's questions:

The sentence "For d≥2 this means that the fields that contribute to the path integral will not even be continuous. It is no surprise that the continuum limit might not even exist." from the book may lead to some misunderstanding. Not only the fields that contribute will not be continuous functions, they will not be functions at all. They will be Schwartz distributions a.k.a. generalized functions. "contribute" means that the set of fields with the mentioned regularity measured by $\alpha$ has full measure (probability one). In general for a field of scaling dimension $\Delta$, i.e., satisfying $$ \langle \phi(x)\phi(y) \rangle\sim \frac{1}{|x-y|^{2\Delta}} $$ at short distances, one can show that almost surely a sample $\phi$ belongs to $C^{\alpha}(\mathbb{R})$ for $\alpha<-\Delta$. The other problem with the sentence is that the reason a continuum limit may or may not exist is more subtle than suggested. I tried to explain some of that in my MO post https://mathoverflow.net/questions/260854/a-roadmap-to-hairers-theory-for-taming-infinities

Also, for precise mathematical theorems about almost sure regularity of sample fields you can look at this article by Furlan and Mourrat and references therein.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.