Survival Probability of Mutually Killing Brownian Motions and the O’Connell Process

Abstract

Recently O’Connell introduced an interacting diffusive particle system in order to study a directed polymer model in 1+1 dimensions. The infinitesimal generator of the process is a harmonic transform of the quantum Toda-lattice Hamiltonian by the Whittaker function. As a physical interpretation of this construction, we show that the O’Connell process without drift is realized as a system of mutually killing Brownian motions conditioned that all particles survive forever. When the characteristic length of interaction killing other particles goes to zero, the process is reduced to the noncolliding Brownian motion (the Dyson model).

Appendix: One-Dimensional Killing Brownian Motion

By (1.12), we can put $K_{i \xi k}(e^{-x/\xi})=e^{i k x} \widehat{K}_{i \xi k}(e^{-x/\xi})$ with $\widehat{K}_{\nu}(z)=(2^{\nu} \Gamma(\nu+1/2)/\sqrt{\pi})\cdot\allowbreak \int_{0}^{\infty} du \cos(zu)/(1+u^{2})^{\nu+1/2}$. Changing the integral variable in (1.9) by $k \sqrt{t/2} -i(x-y)/\sqrt{2t}=\mu$, the transition probability density can be written as

with $\alpha(\mu) =\xi\sqrt{2/t} \{\mu+i(x-y)/\sqrt{2t}\}$. For |x|<∞, as $x/\sqrt{t} \to0$, $\alpha(\mu) \simeq\xi\sqrt{2/t}(\mu-iy/\sqrt{2t})\to0$, and $\Gamma(i \alpha(\mu))\simeq\{i \xi\sqrt{2/t} (\mu-iy/\sqrt{2t}) \}^{-1}$. Then

in $x/\sqrt{t} \to0$, since $\widehat{K}_{0}=K_{0}$ by definition. It gives through (1.10)

$$\mathcal{N}(t,x) = \xi^2 \sqrt{\frac{2}{\pi}} t^{-3/2}K_0\bigl(e^{-x/\xi}\bigr) I_t \times\biggl\{ 1+\mathcal{O} \biggl( \frac{x}{\sqrt{t}} \biggr) \biggr\} $$

(A.1)

with

By (1.13), for β>0, K ₀(e ^−βx) is written as

where we have set s=e ^βv. In the limit β→∞, e ^−β(x−v)+e ^−β(x+v)=0, if v<x and v>−x, and e ^−β(x−v)+e ^−β(x+v)=∞, otherwise. Therefore

$$\lim_{\beta\to\infty} \beta^{-1} K_0\bigl(e^{-\beta x}\bigr) = \frac{1}{2} \int_{-x}^{x} dv\mathbf{1}(x>0) = x \, \mathbf{1}(x>0). $$

(A.2)

Then $K_{0}(e^{-\sqrt{2t}u/\xi}) \simeq \sqrt{2t}u/\xi\mathbf{1}(u > 0)$ in t→∞, and

$$I_t = \frac{2t}{\xi} \int_{0}^{\infty}e^{-u^2} \bigl(1+2u^2\bigr) u \, du \times\bigl\{1+o(1)\bigr\} =\frac{3t}{\xi} \times\bigl\{1+o(1)\bigr\} \quad\mbox {in} \ t \to \infty.$$

Put this result into (A.1), then we obtain the asymptotics (1.14).