Introduction to Random Matrix Theorymath.as.uky.edu/sites/default/files/McLaughlin-1.pdf1. Random...
Transcript of Introduction to Random Matrix Theorymath.as.uky.edu/sites/default/files/McLaughlin-1.pdf1. Random...
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1. Random matrices: intro
2. Orthogonal polynomial connection
3. Density of eigenvalues as N grows
Ken McLaughlin, Univ. of Arizona
Introduction to Random Matrix Theory
Thursday, May 15, 14
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Homework:
1. Enjoy numerical simulations of random matrices using matlab
2. Understand the connection between Random Matrix Theory and Orthogonal Polynomials
3. Work out the OPs and mean density in a simple example
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The basic example of random matrices: N ⇥N Hermitian matrices:
M
jj
: Gaussian random variable: Prob {Mjj
< x} = 1p2⇡
R
x
�1 e
� 12 t
2
dt
M
jk
: Complex Gaussian random variable, Mjk
= M
(R)jk
+ iM
(I)jk
, with
Probn
M
(R)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
Probn
M
(I)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
0
B
B
B
B
@
M11 M
(R)12 + iM
(I)12 · · · M
(R)1N + iM
(I)1N
M
(R)12 � iM
(I)12 M22 · · · M
(R)2N + iM
(I)2N
.... . .
. . ....
M
(R)1N � iM
(I)1N M
(R)2N � iM
(I)2N · · · M
NN
1
C
C
C
C
A
This is referred to as the Gaussian Unitary Ensemble
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The basic example of random matrices: N ⇥N Hermitian matrices:
M
jj
: Gaussian random variable: Prob {Mjj
< x} = 1p2⇡
R
x
�1 e
� 12 t
2
dt
M
jk
: Complex Gaussian random variable, Mjk
= M
(R)jk
+ iM
(I)jk
, with
Probn
M
(R)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
Probn
M
(I)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
0
B
B
B
B
@
M11 M
(R)12 + iM
(I)12 · · · M
(R)1N + iM
(I)1N
M
(R)12 � iM
(I)12 M22 · · · M
(R)2N + iM
(I)2N
.... . .
. . ....
M
(R)1N � iM
(I)1N M
(R)2N � iM
(I)2N · · · M
NN
1
C
C
C
C
A
This is referred to as the Gaussian Unitary Ensemble
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The basic example of random matrices: N ⇥N Hermitian matrices:
M
jj
: Gaussian random variable: Prob {Mjj
< x} = 1p2⇡
R
x
�1 e
� 12 t
2
dt
M
jk
: Complex Gaussian random variable, Mjk
= M
(R)jk
+ iM
(I)jk
, with
Probn
M
(R)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
Probn
M
(I)jk
< x
o
= 1p⇡
R
x
�1 e
�t
2
dt
0
B
B
B
B
@
M11 M
(R)12 + iM
(I)12 · · · M
(R)1N + iM
(I)1N
M
(R)12 � iM
(I)12 M22 · · · M
(R)2N + iM
(I)2N
.... . .
. . ....
M
(R)1N � iM
(I)1N M
(R)2N � iM
(I)2N · · · M
NN
1
C
C
C
C
A
This is referred to as the Gaussian Unitary Ensemble
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Using Matlab, we can generate random matrices and compute their eigenvalues
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Using Matlab, we can generate random matrices and compute their eigenvalues
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Using Matlab, we can generate random matrices and compute their eigenvalues
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Using Matlab, we can generate random matrices and compute their eigenvalues
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Using Matlab, we can generate random matrices and compute their eigenvalues
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N=1
N=2
N=3
N=4
N=8
N=12
N=16
N=20
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N = 200
N = 400
N = 600
N = 800
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The basic example of random matrices: N ⇥N Hermitian matrices:
Mjj : Gaussian random variable, 1⇤2�
e�12 M2
jj dMjj
Mjk: Complex Gaussian random variable, 1� e�|Mjk|2dM (R)
jk dM (I)jk
Joint prob. measure:1
#Nexp
⇤�Tr
�12M2
⇥⌅dM,
dM =⇧
j<k
dM (R)jk dM (I)
jk
N⇧
j=1
dMjj
This is referred to as the Gaussian Unitary Ensemble
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Support seems to grow like
pN ...
Rescale the matrices M 7! N1/2M :
Joint prob. measure:
1
#Nexp
⇢�N Tr
1
2
M2
��dM,
dM =
Y
j<k
dM (R)jk dM (I)
jk
NY
j=1
dMjj
Still unitarily invarant...
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There are many di↵erent matrix models.
1
ˆZN
exp {�NTr [V (M)]} dM
1
ˆZN
exp
⇢�NTr
1
2
A2+B2
+ C2 � it (ABC +ACB)
��dA dB dC
Example 3: A coupled “Matrix Model”
Example 2: N ⇥N Symmetric matrices
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Example 4: Normal Matrix Model:
A weaker symmetry requirement: [M,M
⇤] = 0.
Given Q = Q(x, y) = Q(z, z), one defines
1
ZNexp
�N
t
Tr (Q (M,M
⇤))
�dM
Induced measure on eigenvalues:
1
ˆ
ZN
e
�Nt
PNj=1 Q(zj ,zj)
Y
1j<kN
|zj � zk|2NY
j=1
dxjdyj
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Induced measure on eigenvalues:
1
ˆZN
exp {�NTr [V (M)]} dMExample 1:
1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
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Induced measure on eigenvalues:
1
ˆZN
exp {�NTr [V (M)]} dMExample 1:
1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
1
ZNexp
�N
tTr (Q (M,M⇤
))
�dM
1
ZN
e
�Nt
PNj=1 Q(zj ,zj)
Y
1j<kN
|zj � zk|2NY
j=1
dxjdyj
Example 4:
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A Physical Interpretation
The asymptotic behavior for N ! 1:
low temperature and many particle limit
1
ˆ
ZN
exp
8<
:�N
t
NX
j=1
Q(zj , zj) +
X
j 6=k
log |zj � zk|
9=
;
NY
j=1
dxjdyj
Eigenvalues are a system of particles
Logarithmic interaction potential,log |zj � zk|In the presence of an external field with potential Q(z, z)
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N=200
1000 experiments
(10000)
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Theoretical Physics, String Theory, Combinatorics:
Finance/Genetics/Statistics: Detecting correlations in data sets
Applications of random matrix theory
Form large matrices from huge data sets, compare to our benchmark.
Cellular networks: theoretical predictions of capacity
Information Theory: based on random strings of data
Feynman diagrammatic expansions + explicit asymptoticsyield predictions that can even be PROVEN!
Random growth models / Stochastic interface models
Probabilistic models of interfacial growth discrete matrix models
Hele-Shaw / Laplacian Growth / IntegrabilityNormal Matrices: Support set obeys Laplacian growth (deforming parameters)
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What does one want to calculate
as N ! 1?
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Mean density of eigenvalues
What is its average behavior?
Consider the random variable 1N # { �j � B}.
E�
1N
# { �j � B}⇥
�(N)1 is called the mean density of eigenvalues
Question: behavior when N �⇥???
=
Z
B⇢(N)1
⇢dx
dxdy
�
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Movie of �(N)1 for N = 1 through N = 50.
1
#
N
e
x
p
⇢ �NT
r
1
2
M2
�� dM,
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Movie of �(N)1 for N = 1 through N = 50.
1
#
N
e
x
p
⇢ �NT
r
1
2
M2
�� dM,
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Movie of �(N)1 for N = 1 through N = 50.
1
#
Ne
x
p
{�NT
r
[
MM
⇤ ]}dM
,
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Movie of �(N)1 for N = 1 through N = 50.
1
#Nexp
⇢�N Tr
MM⇤
+
T 2
2
⇣M2
+ (M⇤)
2⌘��
dM,
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1
#Nexp
n
�N Tr
h
M2(M⇤
)
2io
dM,
Movie of ⇢(N)1 for N = 1 through N = 60.
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Question: behavior when N �⇥???
Occupation Probabilities
Prob {B has k eigenvalues}
F (B, z) =1X
k=0
Prob {B has k eigenvalues} zk
F (B, z) is called the occupation probability generating function.
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Example 4: Normal Matrix Model:
n⇥ n matrices with [M,M
⇤] = 0.
Given Q = Q(x, y) = Q(z, z), one defines
1
ZNexp
�N
t
Tr (Q (M,M
⇤))
�dM
Induced measure on eigenvalues:
1
ˆ
ZN
e
�Nt
PNj=1 Q(zj ,zj)
Y
1j<kN
|zj � zk|2NY
j=1
dxjdyj
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Orthogonal Polynomials
Their asymptotic behavior is wide open.
Z
CPk,N (z)Pl,N (z)e�NQ(z,z)
dA(x, y) = hk,N�kl
⇢(N)1 (z, z) = e�NQ(z,z)
N�1X
`=0
h�1`,N |P`,N (z)|2
KN (z, w) = e�1N (Q(z)+Q(w)) 1
N
N�1X
k=0
h�1k Pk,N (z)Pk,N (w)
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1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
Random Hermitian Matrices
Orthgonal Polynomials
Z
Rp
j
(x)p
k
(k)e
�NV (x)dx = �
jk
reproducing kernel,
K
N
(x, y) = e
�N2 (V (x)+V (y))
N�1X
`=0
p
`
(x)p
`
(y).
We want to establish (first) the following formula:
1
ˆ
Z
N
e
�N
PNj=1 V (�j)
Y
1j<kN
|�j
� �
k
|2 = det
0
B@K
N
(�1,�1) · · · K
N
(�1,�N
)
.
.
.
.
.
.
.
.
.
K
N
(�
N
,�1) · · · K
N
(�
N
,�
N
)
1
CA .
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1
ZN
e
�Nt
PNj=1 Q(zj ,zj)
Y
1j<kN
|zj � zk|2NY
j=1
dxjdyj
Random Normal Matrices
KN (z, w) = e�1N (Q(z)+Q(w)) 1
N
N�1X
k=0
h�1k Pk,N (z)Pk,N (w)
1
ZN
e�NPN
j=1 Q(zj ,zj)Y
1j<kN
|zj � zk|2
= det
0
B@KN (z1, z1) · · · KN (z1, zN )
.... . .
...KN (zN , z1) · · · KN (zN , zN )
1
CA .
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Y
1j<kN
(�k � �j) = det
0
BBB@
1 · · · 1�1 · · · �N...
. . ....
�N�11 · · · �N�1
N
1
CCCA.
Thursday, May 15, 14
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Y
1j<kN
(�k � �j) = det
0
BBB@
1 · · · 1⇡1 (�1) · · · ⇡1 (�N )
.... . .
...⇡N�1 (�1) · · · ⇡N�1 (�N )
1
CCCA.
Y
1j<kN
(�k � �j) = det
0
BBB@
1 · · · 1�1 · · · �N...
. . ....
�N�11 · · · �N�1
N
1
CCCA.
Thursday, May 15, 14
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Y
1j<kN
(�k � �j) = det
0
BBB@
1 · · · 1⇡1 (�1) · · · ⇡1 (�N )
.... . .
...⇡N�1 (�1) · · · ⇡N�1 (�N )
1
CCCA.
Y
1j<kN
(�k � �j) =1
QN�1j=0 j
det
0
BBB@
0 · · · 0
1⇡1 (�1) · · · 1⇡1 (�N )...
. . ....
N�1⇡N�1 (�1) · · · N�1⇡N�1 (�N )
1
CCCA.
Y
1j<kN
(�k � �j) = det
0
BBB@
1 · · · 1�1 · · · �N...
. . ....
�N�11 · · · �N�1
N
1
CCCA.
Thursday, May 15, 14
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Y
1j<kN
(�k � �j) =1
QN�1j=0 j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA.
Thursday, May 15, 14
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Y
1j<kN
(�k � �j) =1
QN�1j=0 j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA.
Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
Thursday, May 15, 14
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
Thursday, May 15, 14
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
Y
1j<kN
|zk � zj |2 =1
QN�1j=0 |j |2
det
0
BBB@
p0 (z1) · · · p0 (zN )p1 (z1) · · · p1 (zN )
.... . .
...pN�1 (z1) · · · pN�1 (zN )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
Thursday, May 15, 14
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
P(�1, . . . ,�N ) =1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
Thursday, May 15, 14
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
P(�1, . . . ,�N ) =1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
P(�1, . . . ,�N ) =1
ZNQN�1
j=0 2j
det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
Thursday, May 15, 14
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
P(�1, . . . ,�N ) =1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
P(�1, . . . ,�N ) =1
ZNQN�1
j=0 2j
det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
=1
ZNQN�1
j=0 2j
⇥ det
0
BBB@
e�N2 (V (�1)+V (�1)))
PN�1`=0 p`(�1)p`(�1) · · · e�
N2 (V (�1)+V (�N ))
PN�1`=0 p`(�1)p`(�N )
e�N2 (V (�2)+V (�1))
PN�1`=0 p`(�2)p`(�1) · · · e�
N2 (V (�2)+V (�N ))
PN�1`=0 p`(�2)p`(�N )
.... . .
...
e�N2 (V (�N )+V (�1))
PN�1`=0 p`(�N )p`(�1) · · · e�
N2 (V ((�N )+V (�N ))
PN�1`=0 p`(�N )p`(�N )
1
CCCA,
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Y
1j<kN
(�k � �j)2 =
1QN�1
j=0 2j
det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )
.... . .
...pN�1 (�1) · · · pN�1 (�N )
1
CCCA
P(�1, . . . ,�N ) =1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
P(�1, . . . ,�N ) =1
ZNQN�1
j=0 2j
det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
T
·
· det
0
BBB@
e�N2 V (�1)p0 (�1) · · · e�
N2 V (�N )p0 (�N )
e�N2 V (�1)p1 (�1) · · · e�
N2 V (�N )p1 (�N )
.... . .
...
e�N2 V (�1)pN�1 (�1) · · · e�
N2 V (�N )pN�1 (�N )
1
CCCA
K
N
(x,
y
) =e
�N2(V
(x)+
V
(y))
N
�1X
`
=0
p
`
(x)p `
(y).
=1
ZNQN�1
j=0 2j
⇥ det
0
BBB@
e�N2 (V (�1)+V (�1)))
PN�1`=0 p`(�1)p`(�1) · · · e�
N2 (V (�1)+V (�N ))
PN�1`=0 p`(�1)p`(�N )
e�N2 (V (�2)+V (�1))
PN�1`=0 p`(�2)p`(�1) · · · e�
N2 (V (�2)+V (�N ))
PN�1`=0 p`(�2)p`(�N )
.... . .
...
e�N2 (V (�N )+V (�1))
PN�1`=0 p`(�N )p`(�1) · · · e�
N2 (V ((�N )+V (�N ))
PN�1`=0 p`(�N )p`(�N )
1
CCCA,
Thursday, May 15, 14
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1
ZN
e�NPN
j=1 V (�j)Y
1j<kN
|�j � �k|2NY
j=1
d�j
Random Hermitian Matrices
Orthgonal Polynomials
Z
Rp
j
(x)p
k
(k)e
�NV (x)dx = �
jk
reproducing kernel,
K
N
(x, y) = e
�N2 (V (x)+V (y))
N�1X
`=0
p
`
(x)p
`
(y).
We want to establish (first) the following formula:
1
ˆ
Z
N
e
�N
PNj=1 V (�j)
Y
1j<kN
|�j
� �
k
|2 = det
0
B@K
N
(�1,�1) · · · K
N
(�1,�N
)
.
.
.
.
.
.
.
.
.
K
N
(�
N
,�1) · · · K
N
(�
N
,�
N
)
1
CA .
Thursday, May 15, 14
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1
ZN
e
�Nt
PNj=1 Q(zj ,zj)
Y
1j<kN
|zj � zk|2NY
j=1
dxjdyj
Random Normal Matrices
1
ZN
e�NPN
j=1 Q(zj ,zj)Y
1j<kN
|zj � zk|2
= det
0
B@KN (z1, z1) · · · KN (z1, zN )
.... . .
...KN (zN , z1) · · · KN (zN , zN )
1
CA .
KN (z, w) = e�1N (Q(z,z)+Q(w,w)) 1
N
N�1X
k=0
h�1k Pk,N (z)Pk,N (w)
Thursday, May 15, 14
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K
N
(x, y) = e
�N2 (V (x)+V (y))
N�1X
`=0
p
`
(x)p`
(y).
Verify thatZ
RK
N
(x, y)KN
(y, z)dy = K
N
(x, z),
andZ
RK
N
(x, x)dx = N
Thursday, May 15, 14
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KN (z, w) = e�1N (Q(z,z)+Q(w,w)) 1
N
N�1X
k=0
h�1k Pk,N (z)Pk,N (w)
Verify thatZ
RKN (z, w)KN (w, s)d2w = KN (z, w),
andZ
RKN (z, z)d2z = N
Thursday, May 15, 14
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Small collection of examples for which calculations are explicit
• Q(x, y) = Q(x
2+ y
2): Show: Pn = cn,Nz
n, and if Q(x, y) = x
2+ y
2, then
KN (z, w) =
1
⇡
e
�N(
|z|2+|w|2)
/2N�1X
j=0
(Nzw)
j
j!
and (still assuming Q = x
2+ y
2) then
⇢N =
1
⇡
e
�N |z|2N�1X
j=0
�N |z|2
�j
j!
Thursday, May 15, 14