F.2 Gaussian Unitary Ensemble (GUE)
479
(C N )
−1
=
∞
−∞
dx 1 ×. . .×
∞
−∞
dx N
1≤i
|x j − x i |
2
exp
−
N
j =1
x
2
j
= (2π)
N/2 2
−N 2 /2
[(2)]
−N
N
j =1
(1 + j).
(F.63)
Equations (F.62) and (F.63) give us the probability density of energy eigenvalues of
complex Hermitian Hamiltonian matrices.
Example GUE-3 (Probability Density for a 2×2 Complex Hermitian Matrix)
A 2×2 complex Hermitian matrix can be written
¯
H C =
h 11 h 12
h ∗
12 h 22
(F.64)
so that
Tr[ ¯
H C · ¯
H
†
C ] = h
2
11 + h
2
22 + 2(h
R
12 )
2 + 2(h
I
12 )
2 .
(F.65)
Then, using the definition of dd H C in Eq. (F.39), we obtain
dd H C P ( ¯
H C ) = 2
∞
−∞
dh 11
∞
−∞
dh 22
∞
−∞
dh
R
12
∞
−∞
dh
I
12
×
1
2πδ 2
2
exp
1
2δ 2 (h
2
11 + h
2
22 + 2(h
R
12 )
2 + 2(h
I
12 )
2 )
= 1.
(F.66)
F.2.4 Cluster Functions for GUE
The probability density for the Gaussian unitary ensemble (GUE) is
P C (x 1 , . . ., x N ) =
2 N(N−1)/2 π −N/2
N
n=1 (n + 1)
⎡
⎣
1≤i
(x j − x i )
2
⎤
⎦ exp
−
N
i=1
x
2
i
,
(F.67)
In Eqs. (F.8) and (F.9), we found that
exp
−
1
2
N
j =1
x
2
j
1≤i
(x j − x i )
=
N −1
n=0
(
√
π 2
n n!)
1
2
2
−N(N−1)/2 Det[ ¯
G N ],
(F.68)
479
(C N )
−1
=
∞
−∞
dx 1 ×. . .×
∞
−∞
dx N
1≤i
2
exp
−
N
j =1
x
2
j
= (2π)
N/2 2
−N 2 /2
[(2)]
−N
N
j =1
(1 + j).
(F.63)
Equations (F.62) and (F.63) give us the probability density of energy eigenvalues of
complex Hermitian Hamiltonian matrices.
Example GUE-3 (Probability Density for a 2×2 Complex Hermitian Matrix)
A 2×2 complex Hermitian matrix can be written
¯
H C =
h 11 h 12
h ∗
12 h 22
(F.64)
so that
Tr[ ¯
H C · ¯
H
†
C ] = h
2
11 + h
2
22 + 2(h
R
12 )
2 + 2(h
I
12 )
2 .
(F.65)
Then, using the definition of dd H C in Eq. (F.39), we obtain
dd H C P ( ¯
H C ) = 2
∞
−∞
dh 11
∞
−∞
dh 22
∞
−∞
dh
R
12
∞
−∞
dh
I
12
×
1
2πδ 2
2
exp
1
2δ 2 (h
2
11 + h
2
22 + 2(h
R
12 )
2 + 2(h
I
12 )
2 )
= 1.
(F.66)
F.2.4 Cluster Functions for GUE
The probability density for the Gaussian unitary ensemble (GUE) is
P C (x 1 , . . ., x N ) =
2 N(N−1)/2 π −N/2
N
n=1 (n + 1)
⎡
⎣
1≤i
2
⎤
⎦ exp
−
N
i=1
x
2
i
,
(F.67)
In Eqs. (F.8) and (F.9), we found that
exp
−
1
2
N
j =1
x
2
j
1≤i
=
N −1
n=0
(
√
π 2
n n!)
1
2
2
−N(N−1)/2 Det[ ¯
G N ],
(F.68)
