Proposition 3.1. Consider a system of particles with indices n = 1, 2, . . . starting from positions y
1
y
2
. . .. Denote by x
n
t the position of particle with index n at time t. Then the joint distribution of particle positions is given by the Fredholm determinant
P
m
\
k=1
{x
n
k
t
k
≥ s
k
} = det
1
− ˜ χ
s
K ˜ χ
s ℓ
2
{n
1
,t
1
,...,n
m
,t
m
}×
Z
3.2 with n
1
, t
1
, . . . , n
m
, t
m
∈ S , and ˜ χ
s
n
k
, t
k
x =
1
x s
k
. The kernel K is given by Kn
1
, t
1
, x
1
; n
2
, t
2
, x
2
= −φ
n
1
,t
1
,n
2
,t
2
x
1
, x
2
+
n
2
X
k=1
Ψ
n
1
,t
1
n
1
−k
x
1
Φ
n
2
,t
2
n
2
−k
x
2
3.3 where
Ψ
n,t n
−l
x = 1
2 πi
I
Γ
dzz
x − y
l
−1
e
at z+btz
1 − v
1
z · · · 1 − v
n
z 1 − v
1
z · · · 1 − v
l
z ,
l = 1, 2, . . . , 3.4
the functions {Φ
n,t n
−k
}
n k=1
are uniquely determined by the orthogonality relations X
x ∈
Z
Ψ
n,t n
−l
xΦ
n,t n
−k
x = δ
k,l
, 1
≤ k, l ≤ n, 3.5
and by the requirement span {Φ
n,t n
−l
x, l = 1, . . . , n} = V
n
. The first term in 3.3 is given by φ
n
1
,t
1
,n
2
,t
2
x, y = 1
2 πi
I
Γ
dz z
y −x+1
e
at
1
−at
2
z
e
bt
1
−bt
2
z
1 − v
n
1
+1
z · · · 1 − v
n
2
z
1
[n
1
,t
1
≺n
2
,t
2
]
. 3.6
The notation Γ stands for any anticlockwise oriented simple loop including only the pole at 0.
3.2 Kernel for step initial condition
We set all the jump rates to 1: v
1
= v
2
= · · · = 1. The transition function 3.6 does not depend on initial conditions. It is useful to rewrite it in a slightly different form.
Lemma 3.2. The transition function can be rewritten as
φ
n
1
,t
1
,n
2
,t
2
x, y 3.7
= 1
2 πi
I
Γ
0,1
dw 1
w
x − y+1
w w
− 1
n
2
−n
1
e
at
1
w+bt
1
w
e
at
2
w+bt
2
w
1
[n
1
,t
1
≺n
2
,t
2
]
. Proof of Lemma 3.2. The proof follows by the change of variable z = 1
w in 3.6.
Lemma 3.3. Let y
i
= −i, i ≥ 1. Then, the functions Φ and Ψ are given by Ψ
n,t k
x = 1
2 πi
I
Γ
0,1
dw w − 1
k
w
x+n+1
e
atw+bt w
, Φ
n,t j
x = 1
2 πi
I
Γ
1
dz z
x+n
z − 1
j+1
e
−atz−btz
. 3.8
1389
Proof of Lemma 3.3. Ψ
n,t k
x comes from the change of variable z = 1w in 3.4. For k ≥ 0, the pole at w = 1 is irrelevant, but in the kernel Ψ
n,t k
enters also for negative values of k. We have to verify that the function Φ
n,t j
x satisfy the orthogonal condition 3.5 and span the space V
n
given in 3.1. For v
1
= · · · = v
n
= 1, V
n
= span1, x, . . . , x
n −1
. By the residue’s theorem, the function Φ
n,t j
x is a polynomial of degree j in x. Thus, spanΦ
n,t j
x, j = 0, . . . , n − 1 = V
n
. The second step is to compute
P
x ∈
Z
Φ
n,t j
xΨ
n,t k
x for 0 ≤ j, k ≤ n − 1. We divide it into the sum over x
≥ 0 and the one over x 0. We have X
x ≥0
Φ
n,t j
xΨ
n,t k
x = X
x ≥0
1 2πi
2
I
Γ
1
dz I
Γ
dw e
atw+bt w
e
atz+bt z
w − 1
k
z − 1
j+1
z
x+n
w
x+n+1
. 3.9
We choose the paths Γ and Γ
1
satisfying |z| |w|, so that we can take the sum inside the integrals
and use X
x ≥0
z
x
w
x+1
= 1
w − z
, 3.10
to get 3.9 =
1 2πi
2
I
Γ
1
dz I
Γ
0,z
dw e
atw+bt w
e
atz+bt z
w − 1
k
z − 1
j+1
z
n
w
n
1 w
− z ,
3.11 where the subscript z in Γ
0,z
reminds that z is a pole for the integral over w. Next consider the sum over x
0. We have X
x
Φ
n,t j
xΨ
n,t k
x = X
x
1 2πi
2
I
Γ
dw I
Γ
1
dz e
atw+bt w
e
atz+bt z
w − 1
k
z − 1
j+1
z
x+n
w
x+n+1
. 3.12
This time we choose the paths Γ and Γ
1
satisfying |z| |w| and then take the sum inside the
integrals. Using X
x
z
x
w
x+1
= − 1
w − z
3.13 we obtain
3.12 = − 1
2πi
2
I
Γ
dw I
Γ
1,w
dz e
atw+bt w
e
atz+bt z
w − 1
k
z − 1
j+1
z
n
w
n
1 w
− z ,
3.14 where now w is a pole for the integral over z. Thus,
X
x ∈
Z
Φ
n,t j
xΨ
n,t k
x = 3.11 + 3.14. 3.15
We can deform the paths of integration in 3.14 so that they become as the integration paths of 3.11 up to correcting the contribution of the residue at z = w. Thus we finally get, for 0
≤ j, k ≤ n
− 1, X
x ∈
Z
Φ
n,t j
xΨ
n,t k
x = 1
2 πi
I
Γ
1
dzz − 1
k − j−1
= δ
j,k
. 3.16
As a side remark, the same computations would not hold for k, j 0, for which Φ
n,t j
x ≡ 0, because it is not possible to choose the paths with
|z| |w| without introducing an extra pole at z = 0. 1390
Proposition 3.4 Step initial condition, finite time kernel. The kernel for y
i
= −i, i ≥ 1, is given by Kn
1
, t
1
, x
1
; n
2
, t
2
, x
2
3.17 =
− 1
2 πi
I
Γ
dw 1
w
x
1
−x
2
+1
w 1
− w
n
2
−n
1
e
at
1
w+bt
1
w
e
at
2
w+bt
2
w
1
[n
1
,t
1
≺n
2
,t
2
]
+ 1
2πi
2
I
Γ
dw I
Γ
1
dz e
bt
1
w+at
1
w
e
bt
2
z+at
2
z
1 − w
n
1
w
x
1
+n
1
+1
z
x
2
+n
2
1 − z
n
2
1 w
− z .
The contours Γ and Γ
1
include the poles w = 0 and z = 1 and no other poles. This means in particular that Γ
and Γ
1
are disjoints, because of the term 1 w − z.
Proof of Proposition 3.4. Consider the main term of the kernel, namely
n
2
X
k=1
Ψ
n
1
,t
1
n
1
−k
x
1
Φ
n
2
,t
2
n
2
−k
x
2
=
n
2
X
k=1
1 2
πi I
Γ
0,1
dw w − 1
n
1
−k
w
x
1
+n
1
+1
e
at
1
w+bt
1
w
× 1
2 πi
I
Γ
1
dz z
x
2
+n
2
z − 1
n
2
−k+1
e
−at
2
z−bt
2
z
. 3.18 First we extend the sum to +
∞, since the second term is identically equal to zero for k n
2
. We choose the integration paths so that
|z − 1| |w − 1|. Then, we can take the sum inside the integral. The k-dependent terms are
X
k ≥1
z − 1
k −1
w − 1
k
= 1
w − z
. 3.19
Thus, we get 3.18
= 1
2πi
2
I
Γ
1
dz I
Γ
0,z
dw e
at
1
w+bt
1
w
e
at
2
z+bt
2
z
w − 1
n
1
w
x
1
+n
1
+1
z
x
2
+n
2
z − 1
n
2
1 w
− z . 3.20
Notice now we have a new pole at w = z, but the one at w = 1 vanished. The contribution of the pole at w = z is exactly equal to the contribution of the pole at z = 1 in the transition function 3.7.
Therefore in the final result the first term coming from 3.7 has the integral only around z = 0, and the second term is 3.20 but with the integral over w only around the pole at w = 0 and does not
contain z. Finally, a conjugation by a factor
−1
n
1
−n
2
gives the final result.
3.3 Kernel for flat initial condition
