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I have been reading a wikipedia entry on abstract Wiener space where they give as a motivation the necessity to define $$ \frac{1}{Z}\int_H f(v) e^{-\frac{1}{2} \Vert v\Vert^2} Dv, \tag{1} $$ which they say is an integral often met in physics. Here $H$ is some Hilbert space. The problem apparently is that it is hard to introduce a suirable measure $Dv$ on such space. For example, if one follows a construction of cylinder measures, one can get a nice behaving Gaussian set function, but only on an algebra, and it does not extend well to the $\sigma$-algebra generated by those cylinder sets. Please correct me if my understanding is wrong here.

Instead an idea is to consider a Banach space such that $i:H \to B$ makes $i(H)$ dense in $B$ and yet $B$ is sufficiently large so that the very same (?) cylinder set procedure with measure actually yeilds a nice behaving Gaussian measure $\gamma$ on the whole Borel $\sigma$-algebra of $B$. This is apparently a procedure authored by Leonard Grass, who also mentions something like: it is acutally $H$ that $\gamma$ very much depends on, not $B$. This space $H$ is referred to as the Cameron-Martin space.

I am very much lost here and hope that you could help my clarify understanding of this fact. My familiarity is with the classical Wiener space, let's say a Brownian motion law $\gamma_W$ on the Banach space $B_W = C_0(\Bbb R_+,\Bbb R)$ of all continuous function from $\Bbb R_+$ to $\Bbb R$ that start at $0$. Such construction was done using finite-dimensional distributions, Kolomogorov's extensions theorem and Kolmogorov's continuity theorem. An in particular it so happened that $\gamma_W(C_0^1) = 0$, i.e. a probability of getting a function with continuous derivative is $0$. Yet, in the wikipedia article one uses $H_W = L^{2,1}_0(\Bbb R_+, \Bbb R)$ which is the Hilbert space of all continuously differentiable square integrable functions starting at $0$ with the inner product being $$ \langle \sigma_1, \sigma_2 \rangle_{L_0^{2,1}} := \int_0^\infty \dot{\sigma}_1 (t) \dot{\sigma}_2 (t) \, dt. $$

Now what I don't understand at all is:

  1. How exactly does $H_W$ help in constructing $\gamma_W$? I mean, in the end we'll even end up having $\gamma_W(H_W) = 0$. How can it matter at all then?

  2. In the wikipedia article they use intergal $(1)$ as a motivation to formally define measures on Hilbert spaces. In the end the defined measure gives $0$ weight to $H$, casting all integral to be $0$ as well. How was this useful then for the original problem?

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  • $\begingroup$ 1. To construct a Wiener measure $\gamma$ on $B=C_0(\mathbb R_+,\mathbb R)$ one does not need a Cameron-Martin space $H$ (as we know by Kolmogorov's theorem). 2. You made a good observation. To this date there is no satisfactory rigorous definition of the Feynman path integral. 3. When you read a bit about the Cameron-Martin theorem it should become clearer what they are trying to do: functional calculus on an infinite dimensional space. They even proved an integration-by-parts formula which allows ... $\endgroup$
    – Kurt G.
    May 1, 2023 at 19:48
  • $\begingroup$ to define a divergence for functions on $B$. $\endgroup$
    – Kurt G.
    May 1, 2023 at 19:48
  • $\begingroup$ @KurtG. why don't you post an answer? you've addressed almost all of my questions, so perhaps you could also cover the remaining part :) $\endgroup$
    – SBF
    May 1, 2023 at 19:50
  • $\begingroup$ Maybe I'll do that in the coming days. Please read a bit more. My knowledge of these matters is quite rusty and there might come in better answers. $\endgroup$
    – Kurt G.
    May 1, 2023 at 19:51
  • 1
    $\begingroup$ To rephrase: 1. $H_W$ is important to do geometry on the infinite dimensional Wiener space. ($H_W$ is not important (and not even useful) to introduce only the Wiener probability measure that defines Brownian motion). The geometry stuff is sound and would not even require the motiation by (1). 2. I found some material to elaborate a bit why (1) is a nice motivation but falls short of being rigorous, precisely for the reason you mentioned. Will come up with an answer. $\endgroup$
    – Kurt G.
    May 2, 2023 at 9:44

1 Answer 1

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The Wikipedia article mentions that path integrals like (1)

"arise, notably, in the context of Euclidean path integral formulation of quantum field theory."

Since there is no problem with classical or abstract Wiener spaces I will only concentrate on the Euclidean path integral which Wikipedia writes as $$\tag{1} \psi(x,t)=\frac{1}{Z}\int_{x(0)=x}e^{-S_{\text{Euclidean}}(\mathbf{x},\dot{\mathbf{x}})/\hbar}\,\psi_0(\mathbf{x}(t))\,{\cal D}\mathbf{x}\, $$ where the Euclidean action is the time integral of kinetic energy plus potential energy: $$\tag{2} S_{\text{Euclidean}}(\mathbf{x},\dot{\mathbf{x}})=\int_0^t\frac{m}{2}|\dot{\mathbf{x}}(s)|^2+V(\mathbf{x}(s))\,ds\,. $$ These expressions arise formally from Wick rotating Feynman's original path integral $$\tag{3} \psi(x,t)=\frac{1}{Z}\int_{x(0)=x}e^{\mathrm{i}S(\mathbf{x},\dot{\mathbf{x}})/\hbar}\,\psi_0(\mathbf{x}(t))\,{\cal D}\mathbf{x}\,. $$ (Feynman's Lagrangian $S$ in (3) has kinetic minus potential energy but otherwise is the same as (3). Wick rotation is nothing else than replacing time $t$ by imaginary time $it\,.$ After this $S$ becomes a Hamiltonian in (1) (kinetic plus potential energy). For details see [1] section 1.11.)

Wikipedia says now that this makes (1) rigorous but it does not. What makes (1) rigorous is the observation made by Mark Kac that (1) should be the expectation of a Wiener measure: $$\tag{4} \psi(x,t)=\frac{1}{Z}\mathbb E_x\Big[ e^{-\int_0^tV(\mathbf{x}(s))\,ds/\hbar}\psi_0(\mathbf{x}(t))\Big] $$ where $t\mapsto\mathbf{x}(t)$ is a Wiener process with $\mathbf{x}(0)=x$ and paths in the Banach space of continuous paths but not in a Hilbert space of differentiable paths.

  • The non differentiability of paths is the reason why kinetic term $\frac{m}{2}|\dot{\mathbf{x}}(s)|^2$ must disappear from (2) in favour of $\mathbb E_x\,.$

  • Revisiting the Kolmogorov construction of the Wiener measure shows that one effectively puts increments of the paths into the Gaussian PDF, $$\tag{5} \frac{1}{\sqrt{2\pi\Delta s}}\exp\Big(-\textstyle\frac{(\mathbf{x}(s+\Delta)-\mathbf{x}(s))^2}{2\Delta s}\Big) $$ takes a product of those probabilities and goes to the limit $\Delta s\to 0\,.$ In almost every introduction of the Path integral its motivation is taken from the purely formal convergence of that product to the kinetic term $$\tag{6} \exp\Big(-\int_0^t \frac{|\dot{\mathbf{x}}(s)|^2}{2}\,ds\Big)\,. $$

  • As one can read in [1],

Feynman's theory is not easily made rigorous. [...] Feynman objected to Kac's integral because it "spoils the physical unification of kinetic and potential parts of the action". The kinetic contribution is hidden from sight.

  • By Feynman's theory the authors of [1] mean (3). They give references [2-4] where rigorous approaches have been proposed.

[1] P. Cartier, C. DeWitt-Morette, Functional Integration - Action and Symmetries.

[2] C. DeWitt-Morette (1972), Feynman's path integral; definition without limiting procedure. Commun. Math. Phys. 28, 47-67.

[3] C. DeWitt-Morette (1974), Feynman path integrals; I. Linear and affine techniques. II. The Feynman Green function. Commun. Math. Phys. 37, 63-81.

[4] S.A. Albeverio, R.J. Hoegh-Krohn (1976), Mathematical Theory of Feynman Path Integrals.

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