Change of Variables and Besselian Path Integrals

Start with the effective Lagrangian from the Steigl/Hinterleitner orderings:

(1)
\begin{align} L_{eff} = \int dt \left[ \frac{1}{2} a \dot a^2 - \left( V \left( a \right) + \frac{\hbar^2}{2}p \left( i+k \right) a^{-3} \right) \right] \end{align}

where $V \left( a \right) = \frac{1}{2} \left( -a + \frac{\Lambda}{3}a^3 \right)$ and $p \left( i+k \right) = - \frac{7}{16} + \left( i+k \right) - \frac{ \left( i+k \right)^2}{4}$.

Make the change of variables $x = \frac{2}{3} a^{3/2}, \dot x = a^{1/2} \dot a$:

(2)
\begin{align} \implies L_{eff} = \int dt \left[ \frac{1}{2} \dot x^2 - \frac{1}{2} \left( - \left( \frac{3}{2} \right)^{2/3} x^{2/3} + \frac{3 \Lambda}{4} x^2 + \frac{4 \hbar^2}{9} p \left( i+k \right) x^{-2} \right) \right] \end{align}

(Note that also the path integral measure for the variable $a,$ $\mathcal{D} \left[ a^{1/2} \left( t \right) a \left( t \right) \left] = \prod_{n=1}^{N-1} a_{n}^{1/2} da_n$ should be transformed to $\mathcal{D} \left[ x \left( t \right) \right] = \prod_{n=1}^{N-1} dx_n$ since $dx_n = a_{n}^{1/2} da_n$.)

Now we can use the method in Fischer, Leschke, Muller to eliminate the inverse-square potential (Sections 3 and 4 of that paper, summarized below):

Let $I_{\nu}$ be the modified Bessel function of the first kind with index $\nu.$ Define the transition density

(3)
\begin{align} b_{t}^{ \left( \nu \right) } \left( x', x \right) = \frac{x'}{t} \left( \frac{x'}{x} \right)^{\nu} \exp \left\{ - \frac{\left( x' \right)^2 + x^2}{2t} \right\} I_{\nu} \left( \frac{x'x}{t} \right), \quad x, x' \in \mathbb{R}_{+} \end{align}

representing the Bessel process transition density for propagation from $x$ to $x'$. This satisfies the Markov semigroup properties with respect to Lebesgue measure, and hence induces a path space measure $dB_{\nu}$ on $\mathcal{C} \left( \mathbb{R}_+, x \right)$. The Bessel transition density satisfies

(4)
\begin{align} b_{t}^{\left( d/2 - 1 \right)} \left( x', x \right) = \left( x' \right)^{d-1} \int d \Omega_{x'} \ w_{t}^{ \left( d \right) } \left( x', x \right), \quad d \ge 2 \end{align}

where $\int d \Omega_{x'}$ denotes integration over the unit sphere in $\mathbb{R}^d$ ($x'$ space) and $w_{t}^{ \left( d \right) }$ is the transition density for the Weiner process:

(5)
\begin{align} w_{t}^{ \left( d \right) } \left( \mathbf{x'}, \mathbf{x} \right) = \left( 2 \pi t \right)^{-d/2} \exp \left\{ - \frac{\left( \mathbf{x' - x} \right)^2}{2t} \right\}, \quad \mathbf{x, x'} \in \mathbb{R}^d. \end{align}

This yields

(6)
\begin{align} \int_{\mathcal{C} \left( \mathbb{R}^d, x' \right) } dW^d \left[ x \right] F \left[ \left| x \right| \right] = \int_{\mathcal{C} \left( \mathbb{R}_+, x' \right) } d B_{d/2-1} \left[ x \right] F \left[ x \right]. \end{align}

Important result from Fischer/Leschke/Muller ("Statement 1"):

For

(7)
\begin{align} \mu, \nu \ge 0 \\ F_t: \mathcal{C} \left( \mathbb{R}_+, x' \right) \rightarrow \mathbb{R} \end{align}

where $F$ is a functional depending only on paths up through some time $t \ge 0$ (for Feynman-Kac functionals, $F_t \left[ x \right] = \delta \left( x \left( t \right) - x'' \right) \exp \left\{ - \int_{0}^{t} dt' \ V \left( x \left( t' \right) \right) \right\}$), we have

(8)
\begin{align} \int_{\mathcal{C} \left( \mathbb{R}_+, x' \right) } dB_{\nu} \left[ x \right] F_t \left[ x \right] = \int_{\mathcal{C} \left( \mathbb{R}_+, x' \right) } dB_{\mu} \left[ x \right] \left( \frac{x \left( t \right)}{x'} \right)^{\nu - \mu } \exp \left\{ \left( \mu^2 - \nu^2 \right) \int_{0}^{t} \frac{dt'}{2x^2 \left( t' \right)} \right\} F_t \left[ x \right] \end{align}

Now consider particle restricted to $\mathbb{R}_+$, under influence of potential $V \left( x \right)$ as well as an infinitely high barrier at the origin, to represent the restriction to$\mathbb{R}_+$. The Dirichlet boundary condition is that the transition density to $0$ should be $0$. So we get the transition density

(9)
\begin{align} w_{t}^{ \left( 1 \right) } \left( x', x \right) - w_{t}^{ \left( 1 \right) } \left( -x', x \right) \end{align}

since this satisfies the diffusion equation $\frac{\partial}{\partial t} G = \frac{1}{2} \frac{\partial^2}{\partial x^2} G$ and satisfies the Dirichlet boundary condition (i.e. vanishes at $x' = 0$).

Note that

(10)
\begin{align} w_{t}^{ \left( 1 \right) } \left( x', x \right) - w_{t}^{ \left( 1 \right) } \left( -x', x \right) = \left( \frac{x}{x'} \right) b_{t}^{1/2} \left( x', x \right), \end{align}

since

(11)
\begin{align} I_{1/2} \left( \frac{x'x}{t} \right) = \frac{1}{\sqrt{2 \pi }} \left( \frac{x'x}{t} \right)^{-1/2} \sinh \left( \frac{x'x}{t} \right). \end{align}

Consider potentials of the form $V \left( x \right) = U \left( x \right) + \frac{g}{2x^2}$, where $U \left( x \right)$ is less singular than $x^{-2}$ at $x=0$ and $g \ge - \frac{1}{4}$.

We have $\hat{H}_D = \frac{1}{2} \hat{p}^2 + V \left( \hat x \right)$ (D denoting Dirichlet boundary condition), and

(12)
\begin{align} \left< x_2 \left| \exp \left\{ -t \hat{H}_D \right\} \right| x_1 \right> = \frac{x_1}{x_2} \int_{\mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{1/2} \left[ x \right] \ \delta \left( x \left( t \right) - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \ V \left( x \left( t' \right) \right) \right\}. \end{align}

Then use Statement 1 to get

(13)
\begin{align} \begin{align} \frac{x_1}{x_2} \int_{\mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{1/2} \left[ x \right] \delta \left( x \left( t \right) - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \ V \left( x \left( t' \right) \right) \right\} = \\ \frac{x_1}{x_2} \int_{\mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{\nu} \left[ x \right] \left( \frac{x \left( t \right)}{x_1} \right)^{\frac{1}{2}- \nu} \exp \left\{ \left( \nu^2 - \frac{1}{4} \right) \int_{0}^{t} \frac{dt'}{2x^2 \left( t' \right)} \right\} \times \\ \delta \left( x \left( t \right) - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \ V \left( x \left( t' \right) \right) \right\} \\ = \left( \frac{x_1}{x_2} \right)^{ \frac{1}{2} + \nu } \int_{ \mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{\nu} \left[ x \right] \ \delta \left( x \left( t \right) - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \left[ U \left( x \left( t' \right) \right) + \frac{g - \nu^2 + \frac{1}{4}}{2x^2 \left( t' \right)} \right] \right\} \end{align}. \end{align}

To eliminate the $x^{-2}$ part of the potential, take $\nu = \sqrt{ g + \frac{1}{4} }$.

So in our case,

(14)
\begin{align} U \left( x \right) = \frac{1}{2} \left( - \left( \frac{3}{2} \right)^{2/3} x^{2/3} + \frac{3 \Lambda}{4} x^2 \right) \\ \frac{g}{2x^2} = \frac{1}{2} \cdot \frac{4 \hbar^2}{9} p \left( i+k \right) x^{-2} \\ \implies g = \frac{4 \hbar^2}{9} p \left( i+k \right), \end{align}

and we should take $\nu = \sqrt{ \frac{1}{4} + \frac{4 \hbar^2}{9} p \left( i+k \right)$ as our change of Bessel index in order to eliminate the inverse-square potential (i.e. $\nu$ is always a Planck-scale deviation from $\frac{1}{2}$).

Note that in the Laplace-Beltrami case $p \left( i+k \right) = 0$, this means that we can use

(15)
\begin{align} \int_{\mathcal{C} \left( \mathbb{R}^3, x_1 \right) } dW^3 \left[ x \right] F \left[ \left| x \right| \right] = \int_{\mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{1/2} \left[ x \right] F \left[ x \right] \end{align}

to write

(16)
\begin{align} \left< x_2 \left| \exp \left\{ -t \hat{H}_D \right\} \right| x_1 \right> &= \frac{x_1}{x_2} \int_{\mathcal{C} \left( \mathbb{R}_+, x_1 \right) } dB_{1/2} \left[ x \right] \delta \left( x \left(t \right) - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \ U \left( x \left( t' \right) \right) \right\} \\ &= \frac{x_1}{x_2} \int_{\mathcal{C} \left( \mathbb{R}^3, x_1 \right) } dW^3 \left[ x \right] \delta \left( \left| x \left( t \right) \right| - x_2 \right) \exp \left\{ - \int_{0}^{t} dt' \ U \left( \left| x \left( t \right) \right| \right) \right\}, \\ \\ U \left( x \right) &= \frac{1}{2} \left( - \left( \frac{3}{2} \right)^{2/3} x^{2/3} \left( t \right) + \frac{3 \Lambda}{4} x^2 \left( t \right) \right). \end{align}

Notice that the path integral here is just that of an anharmonic oscillator where the anharmonic part of the potential goes like $x^{2/3}$. This might possibly be computable (I'm doing a literature search - the closest in Grosche/Steiner seems to be the path integral for a forced harmonic oscillator (formula 6.2.1.9) where the anharmonic part goes like $x$ rather than $x^{2/3}$). If we ignore the $x^{2/3}$ piece, we can explicitly perform the path integral, not just for the Laplace-Beltrami case but for any value of $\nu$. As shown in Fischer/Leschke/Muller, this is done by converting the Besselian path integral to an ordinary path integral a la the formula above for values of $\nu = d/2 - 1$, and then verifying that the resulting expression is in fact the correct propagator when we extend to all values of $\nu$. If we can actually perform the path integral with the anharmonic piece for values of $\nu = d/2 - 1$, we may similarly be able to verify that the result is in fact valid for all values of $\nu$.

page revision: 80, last edited: 09 Dec 2009 23:18