System of Complex Brownian Motions Associated with the O’Connell Process

Abstract

The O’Connell process is a softened version (a geometric lifting with a parameter a>0) of the noncolliding Brownian motion such that neighboring particles can change the order of positions in one dimension within the characteristic length a. This process is not determinantal. Under a special entrance law, however, Borodin and Corwin gave a Fredholm determinant expression for the expectation of an observable, which is a softening of an indicator of a particle position. We rewrite their integral kernel to a form similar to the correlation kernels of determinantal processes and show, if the number of particles is N, the rank of the matrix of the Fredholm determinant is N. Then we give a representation for the quantity by using an N-particle system of complex Brownian motions (CBMs). The complex function, which gives the determinantal expression to the weight of CBM paths, is not entire, but in the combinatorial limit a→0 it becomes an entire function providing conformal martingales and the CBM representation for the noncolliding Brownian motion is recovered.

Appendix: Sketch of a Non-rigorous Derivation of (3.1)

Here we provide a sketch of a non-rigorous approach to deriving Borodin and Corwin’s Theorem 4.1.40 [6]. We work with the Whittaker functions and use their orthogonality (2.3) and recurrence relations (2.6) to compute moments $\mathbb{E}^{\nu, a}[(e^{-X^{a}_{1}(t)/a})^{\kappa}]$, κ∈ℕ₀≡{0,1,2,…}. This computation can be done rigorously as Sect. 4.1.4 of [6]. We then take a power series of these moments in an attempt to recover the Laplace transform of the distribution of $e^{-X^{a}_{1}(t)/a}$, $\mathbb{E}^{\nu, a}[\exp(-u e^{-X^{a}_{1}(t)/a})]$, ℜu>0. This is the place where the derivation becomes non-rigorous because the power series is divergent for all values of u and the moments do not identify the Laplace transform of the distribution. Nevertheless, proceeding formally and working with the divergent series we can recover the formula from Theorem 4.1.40 of [6]. Borodin and Corwin work at the higher level of q-Whittaker measures where the analogues of the moments are bounded by one and can be used to rigorously compute the q-deformed version of the Laplace transform of the distribution, which can be written as a Fredholm determinant. They then proved that the q-Whittaker measure converges weakly to the Whittaker measure, the q-deformed Laplace transform converges to the Laplace transform, and the Fredholm determinant has a limit which yields Theorem 4.1.40 of [6]. The fact that the formal calculations we describe actually recover the correct answer can be attributed to the fact that these are limits of the rigorous calculations done one level higher.

The expectation at a single time t>0 given by (2.20) can be written as

(A.1)

First we consider the case with $f(\boldsymbol{x})=e^{-x_{1}/a}$. By (2.6),

(A.2)

and

where we used the orthogonality relation (2.3) extended to complex indices as in Sect. 4.1.4 of [6]. Then (A.1) gives

We can see

$$\frac{1}{N!} \sum_{\sigma\in\mathfrak{S}_N} \exp \Biggl\{ - \frac{t}{2 a^2} \sum_{p=1}^N \bigl(-i \boldsymbol{\nu}_{\sigma(p)}+i (\boldsymbol{e}_{\{j\} })_{\sigma(p)} \bigr)^2 \Biggr\} =e^{t|\boldsymbol{\nu}|^2/2a^2-t \nu_j/a^2+t/2a^2} $$

for ν∈ℝ^N. Then, if we set

$$ f_N^{\nu, t, a}(v)=e^{t v/a^2} \prod _{\ell=1}^N \frac{1}{v+\nu_{\ell}}, $$

(A.3)

we have the expression

(A.4)

Next we consider (A.1) in the case $f(x)=e^{-2 x_{1}/a}$. By (A.2),

Applying the recurrence relation (2.6), it becomes

Moreover, it is rewritten as

Then, by using the orthogonality relation (2.3) and following the similar procedure to the first case, we have

Here we consider a determinant of a matrix of size two

(A.5)

which is equal to the symmetrization of (v ₂−v ₁)/(v ₂−v ₁+1) with respect to indices j∈{1,2} of v _j ’s,

$$\frac{1}{2} \biggl[ \frac{v_2-v_1}{v_2-v_1+1} +\frac{v_1-v_2}{v_1-v_2+1} \biggr]. $$

Then we obtain the expression

(A.6)

By the similar calculation with the orthogonality relation (2.3) and the recurrence relation (2.6) of the Whittaker functions using the symmetrization identity

$$ \frac{1}{\kappa!} \sum_{\sigma\in\mathfrak{S}_{\kappa}}\, \prod _{1 \leq p < q \leq\kappa} \frac{v_{\sigma(q)}-v_{\sigma(p)}}{ v_{\sigma(q)}-v_{\sigma(p)}+1} = \det_{1 \leq j,\, k \leq\kappa} \biggl[ \frac{1}{v_j+1-v_{\ell}} \biggr], $$

(A.7)

one can prove the following. For any κ∈ℕ

(A.8)

where the summation is over all partitions

$$\lambda=(\lambda_1, \lambda_2, \dots)= 1^{m_1} 2^{m_2} \dots, \quad \lambda_1 \geq \lambda_2 \geq\cdots\geq0, \ m_j \in \mathbb{N}_0,\ j \geq1 $$

conditioned that |λ|≡∑ _j≥1 λ _j =κ. Here l(λ) denotes the length of λ. (Precisely speaking, by using the orthogonality relation (2.3) and the recurrence relation (2.6) of the Whittaker functions, Borodin and Corwin gave a multiple contour-integral representation for general moment, $\mathbb{E}^{\nu, a}[(e^{-X^{a}_{1}(t)/a})^{\kappa}]$, κ∈ℕ, in Lemma 4.1.29 in [6]. This integral formula involves nested contours. Then by deforming them to all be a small circle, denoted here by C(−ν), Borodin and Corwin derived the formula (A.8) (Proposition 6.2.7 in [6]), in which the identity (A.7) was used.)

One can rewrite (A.8) in a suggestive form as

For ℜu>0 one would like to recover the Laplace transform of the distribution of $e^{-X^{a}_{1}(t)/a}$ from the moments via

$$\sum_{\kappa=0}^{\infty} \frac{(-u)^{\kappa}}{\kappa!} \mathbb{E}^{\nu, a} \bigl[\bigl(e^{-X^a_1(t)/a}\bigr)^{\kappa} \bigr] =\mathbb{E}^{\nu, a} \bigl[ \exp \bigl(-u e^{-X^a_1(t)/a} \bigr) \bigr]. $$

One checks from (A.8) that the moments grow super-exponentially, so this interchange of expectation and summation is unjustifiable and constitutes the physics ‘replica trick’. Nevertheless we proceed now complete formally. By reordering terms in an unbounded manner, we arrive at the formula

(A.9)

which was given as (3.1) in the text, where

(A.10)

By (A.3),

$$f_N^{\nu, t, a}(v) f_N^{\nu, t, a}(v+1) \cdots f_N^{\nu, t, a}(v+n-1) = e^{tvn/a^2+tn^2/2a^2-tn/2a^2} \prod _{\ell=1}^N \frac{\Gamma(v+\nu_{\ell})}{\Gamma(n+v+\nu_{\ell})}. $$

Then (A.10) is equal to

$$ K_u\bigl(v,v'\bigr)=\sum _{n \in\mathbb{N}}(-1)^n \prod_{\ell=1}^N \frac{\Gamma(v+\nu_{\ell})}{\Gamma(n+v+\nu_{\ell})} \frac{u^n e^{tvn/a^2+tn^2/2a^2}}{v+n-v'}. $$

(A.11)

Since Γ(−s)Γ(1+s)=−π/sin(πs) by Euler’s reflection formula and −π/sin(πs) has simple poles at s=n∈ℤ with residues (−1)ⁿ, (A.11) can be reexpressed as (1.11).