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I have come across a limit of Gaussian integrals in the literature and am wondering if this is a well known result.

The background for this problem comes from the composition of Brownian motion and studying the densities of the composed process. So if we have a two sided Brownian motion $B_1(t)$ we replace t by an independent Brownian motion $B_2(t)$ and study the density of $B_1(B_2(t))$. If we iterate this composition n times we get the iterated interal in (**) below as an expression for the density of the n times iterated Brownian motion. The result I am interested in is derived in the following paper:

The original reference is "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin http://arxiv.org/abs/1102.4729

Line (3.14) of Orsingher and Beghins paper reads

(**) $\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|} $

  1. How do you prove this result without using probability?

  2. Is this a type of path integral (functional integral)? Or is this integrand some kind of kinetic plus potential term arsing in quantum mechanics? Do expressions like (**) ever come up in physics literature?

(I tried using the change of variable theorem for Wiener measure to transform (**) into a Wiener integral with respect a specific integrand and have had some success with this.. I think this shows how to compute a Wiener integral with respect to a function depending on a path and not just a finite number of variables but did not see how to take this any further - The change of variable theorem for Wiener Measure was taken from "The Feynman Integral and Feynman's Operational Calculus" by G. W. Johnson and M. L. Lapidus.)

  1. I have been studying a slight generalization of ** from the probability side of things and have been trying to use dominated convergence to show the LHS of ** is finite but I am having problems finding a dominating function over the interval $[1,\infty)^n$. Is dominated convergence the best way to just show the LHS of (**) is finite?
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I have posted this at mathoverflow as well: mathoverflow.net/questions/59513/… –  user7980 Mar 25 '11 at 7:02
    
This does not look at first glance like a path integral. It looks like an n dimensional integral perhaps solved by change of variable z->rcos $\theta$. Then take the limit for n. –  Roy Simpson Mar 25 '11 at 9:22
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@Roy: do you know about any other definition of path integral than a $n \to \infty$ limit of $n$-dimensional integrals? Or to put it more mathematically, as a limit of cylindrical integrals. That being said, I am also not sure whether OP's integral is a path integral or not. My guess wouldn't be that it isn't. –  Marek Mar 25 '11 at 9:32
    
@user2757 : Looking at the paper I see that this is a neat simplifying result which comes about after a large integral. I see that the integrand is solved by Gamma terms there. As far as links with Path Integration are concerned, there is a link between Stochastic processes and PI via e.g. the Feynman-Kac formula. Your formula might be useful in that context, but I dont know whether it is "known" there. –  Roy Simpson Mar 25 '11 at 10:54
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The most important step is doing the integral: $\int e^{x^2/2z}e^{z^2/2t}/(\pi\sqrt{tz})dz$. After that, you could use induction to find a general formula for your integral. Also note the RH is the characteristic function of a Cauchy distribution. Maybe Fourier transforming can help? –  Raskolnikov Mar 25 '11 at 16:58
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1 Answer

up vote 7 down vote accepted

Our plan of an answer is as follows. Firstly, we will introduce Planck's constant $\hbar$ so that the particular value $\hbar=1$ corresponds to the original problem. Secondly, we mention a connection to (what physicists often calls) the group property of Feynman path integrals. Thirdly, we will show that the sought-for formula happens to be the classical "instanton" contribution in a saddle point/steepest descent asymptotic expansion, which becomes valid as $\hbar\to 0$. We are currently unaware if semi-classical localization methods can be applied to justify the saddle point/steepest descent expansion, and we shall not attempt to make a justification here.

Now let us get to business. Define endpoints $x_0\equiv x>0$ and $x_{n+1}\equiv t>0$. We start by introducing Planck's constant $\hbar$ into the $u_n$ function in eq. (1.9) of arXiv:1102.4729,

$$u_n(x,t,\hbar) :=\left[\prod_{j=1}^n 2 \int_0^{\infty} \frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right] \prod_{i=1}^{n+1}\frac{e^{-\frac{x_{i-1}^2}{2\hbar x_i}}}{\sqrt{2 \pi x_i}}. \qquad (1.9\hbar) $$ In particular, for $n=0$, we have

$$ u_{n=0}(x,t,\hbar)=\frac{e^{-\frac{x^2}{2\hbar t}}}{\sqrt{2 \pi t}}. $$

The $u_n$ function enjoys various scaling/homogeneity properties,

$$ u_n( x,t,\hbar) = \sqrt{\lambda} u_n(\lambda x,\lambda t,\lambda\hbar) = \lambda u_n(\lambda x,\lambda^{2^{n+1}} t,\hbar) $$ $$= \sqrt{\lambda} u_n( x,\lambda^{2^n} t,\frac{\hbar}{\lambda}), \qquad \lambda>0. \qquad (H) $$

With the help of the first homogeneity property in eq. ($H$), we can immediately deduce the corresponding $\hbar$ generalization of eq. (3.14) in arXiv:1102.4729,

$$ \lim_{n\to \infty}u_n(x,t,\hbar) =\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}}. \qquad (3.14\hbar) $$

So the question is basically how do we derive, understand, motivate, etc., eq. (3.14$\hbar$) physically? To get to a path integral interpretation, we note that the $u_n$ function has (what physicists often call) a group property,

$$ u_{n+1+m}(x,z,\hbar) = 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{n}(x,y,\hbar)u_{m}(y,z,\hbar), \qquad (G) $$

in close analogy with the Feynman propagator $K(x_f,t_f;x_i,t_i)$ with

$$ K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty}\mathrm{d}x_2 \ K(x_3,t_3;x_2,t_2) K(x_2,t_2;x_1,t_1).$$

So the "sum of histories" from $x$ to $z$ can be calculated by integrating over an intermediate point $y$. The $n$ in the $u_n$ function plays the role of a discretized time variable. As a consistency check, it is easy to see (by performing some elementary integrals) that the right-hand side of eq. (3.14$\hbar$),

$$u_{n=\infty}(x,t,\hbar) =\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}} \qquad \left(\to \sqrt{2\pi x} \delta(x) \quad \mathrm{for} \quad\hbar \to 0\right), $$

does indeed solve the group equation $(G)$ in the particular cases $n,m=0,\infty$,

$$ u_{\infty}(x,z,\hbar) = 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{\infty}(y,z,\hbar) $$ $$= 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{0}(y,z,\hbar)$$ $$= 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{0}(x,y,\hbar) u_{\infty}(y,z,\hbar). $$

Next introduce Gaussian momenta $p_1, \ldots,p_{n+1},$ with

$$\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}e^{-\frac{1}{2\hbar}x_ip_i^2} = \frac{1}{\sqrt{2 \pi x_i}}, \qquad x_i>0.$$

Then the $u_n$ function becomes

$$u_n(x,t,\hbar) = \left[\prod_{j=1}^n 2\int_0^{\infty}\frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right]\left[\prod_{i=1}^{n+1}\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}\right]e^{-\frac{S}{\hbar}}, $$

with Euclidean phase space action

$$S:=\frac{1}{2}\sum_{i=1}^{n+1}\left(\frac{x_{i-1}^2}{x_i}+x_i p_i^2\right).$$

Now let us turn to the saddle point/steepest descent asymptotic expansion. The classical equations of motion are

$$ 0 \approx \frac{\partial S}{\partial p_i} = x_i p_i, $$

$$ 0 \approx \frac{\partial S}{\partial x_i} = \frac{x_i}{x_{i+1}}-\frac{x_{i-1}^2}{2x_i^2} + \frac{p_i^2}{2}, $$

where we use $\approx$ sign instead of $=$ sign to emphasize when classical equations of motion have been applied. The classical solution is

$$p_i \approx 0, \qquad q_i \approx q_{i-1}^2,$$

where we have defined $q_i :=\frac{x_i}{2x_{i+1}}$. So $q_i \approx q_{i-1}^2 \approx q_{i-2}^4 \approx \ldots \approx q_0^{2^i}$. Now the telescopic product

$$\prod_{i=0}^n 2q_i =\prod_{i=0}^n\frac{x_i}{x_{i+1}}=\frac{x_0}{x_{n+1}}=\frac{x}{t},$$

is fixed by boundary conditions $x$ and $t$. So

$$q_0^{2^{n+1}-1}=q_0^{\sum_{i=0}^n 2^{i}}\approx\prod_{i=0}^n q_i =\frac{x}{2^{n+1}t},$$

and therefore the unique classical solution is

$$q_i \approx \left( \frac{x}{2^{n+1}t} \right)^{\frac{2^i}{2^{n+1}-1}} \to 1 \qquad \mathrm{for} \qquad n \to \infty. $$

Hence classically $x_i \approx 2^{-i}x$ for $n=\infty$. The classical value of the action is

$$ S_{\mathrm{cl}} \approx \sum_{i=0}^{n} x_i q_i \to x\sum_{i=0}^{\infty}2^{-i} =2x \qquad \mathrm{for} \qquad n \to \infty, $$

so the classical "instanton" contribution $e^{-\frac{S_{\mathrm{cl}}}{\hbar}}$ happens to be the right-hand side of eq. (3.14$\hbar$), up to a $\sqrt{\hbar}$ factor. This is our main observation.

A more complete treatment would now calculate the one-loop Van Vleck determinant $\det(\partial^2S)$ in the saddle point/steepest descent asymptotic expansion. Here we will only make a couple of further remarks. The Hessian $\partial^2 S$ of the action is

$$ \frac{\partial^2 S}{\partial x_i\partial x_j} = \delta_{i,j}\left(\frac{1}{x_{i+1}}+\frac{x_i^2}{x_{i+1}^3}\right)-\delta_{i+1,j}\frac{x_i}{x_j^2}-\delta_{i-1,j}\frac{x_j}{x_i^2}, $$

$$ \frac{\partial^2 S}{\partial p_i\partial x_j} = \frac{\partial^2 S}{\partial x_i\partial p_j} = \delta_{i,j}p_i\approx 0, \qquad \frac{\partial^2 S}{\partial p_i\partial p_j} = \delta_{i,j}x_i. $$

Since we have $n+1$ momenta $p_i$, but only $n$ positions $x_i$, we would naively expect the Van Vleck determinant $\det(\partial^2S) \sim x$ to be proportional to $x$ on-shell. This would mean a $1/\sqrt{x}$ factor in the expansion. It would be interesting to see a detailed calculation of the Van Vleck determinant $\det(\partial^2S)$.

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Thank you this is exactly what I needed help with! –  user7980 Mar 25 '11 at 18:01
    
I am a little confused on the use of $\approx$ and what you mean by the equations of motion have been applied. Did you have a choice of possible solutions besides from the classical one? Thank you again for all you help I have been loosing sleep struggling with applying Laplace method/Saddlepoint techniques .... –  user7980 Mar 25 '11 at 18:17
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The expression for the derivative of the Action you set equal to zero and I am wondering how to justify this rigorously. $$ 0 \approx \frac{\partial S}{\partial p_i} = x_i p_i, $$ $$ 0 \approx \frac{\partial S}{\partial x_i} $$ Does this follow from the Euler-Lagrange equations or something similar? –  user7980 Mar 25 '11 at 18:39
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Actually, he is applying the steepest descent method for integrals. He coated it in a quantum mechanical language, but that is not necessary. It's just important to understand that the largest contribution to the integral comes from the points that extremize the exponent. Hence the Euler-Lagrange equations. Very nice approach! +1 –  Raskolnikov Mar 26 '11 at 10:08
    
Thank you your the comment Raskolonikov. –  user7980 Mar 26 '11 at 16:43
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