Random Matrices Hieu D. Nguyen Rowan University Rowan Math Seminar 12-10-03 Historical Motivation Statistics of Nuclear Energy Levels - Excited states of an atomic nucleus Level Spacings {E1 , E2 , E3 ,...} – Successive energy levels Si Ei 1 Ei – Nearest-neighbor level spacings Wigner’s Surmise Level Sequences of Various Number Sets Basic Concepts in Probability and Statistics Statistics {x1 , x2 , x3 ,..., xN } – Data set of values 1 N xi N i 1 – Mean 1 N ( xi )2 N i 1 2 – Variance Probability x f ( x) 0 b a f ( x)dx 1 – Continuous random variable on [a,b] – Probability density function (p.d.f.) – Total probability equals 1 Examples of P.D.F. f ( x) 1 f ( x) 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0.2 0.4 0.6 b 0.8 P (a x b) f ( x)dx a xf ( x)dx 2 x 2 f ( x)dx 1 -3 -2 -1 1 e 2 1 x2 2 2 3 – Probability of choosing x between a and b – Mean – Variance Wigner’s Surmise Notation {E1 , E2 , E3 ,...} – Successive energy levels Si Ei 1 Ei – Nearest-neighbor level spacings D Si – Mean spacing si Si D – Relative spacings Wigner’s P.D.F. for Relative Spacings pW ( s ) s exp s 2 , 2 4 s S D Are Nuclear Energy Levels Random? Poisson Distribution (Random Levels) Distribution of 1000 random numbers in [0,1] 175 150 125 100 75 50 25 0.001 0.002 0.003 0.004 0.005 0.006 175 150 125 100 75 50 25 0.001 0.002 0.003 0.004 0.005 0.006 How should we model the statistics of nuclear energy levels if they are not random? Distribution of first 1000 prime numbers 250 200 150 100 50 5 10 15 20 25 30 35 Distribution of Zeros of Riemann Zeta Function 1 z n 1 n ( z) (Re z 1) Fun Facts 1. 2. 1 1 1 2 (2) 2 2 2 ... 1 2 3 6 (3) is irrational (Apery’s constant) 3. ( z ) can be analytically continued to all z 1 4. (1 z ) 2(2 ) z cos( z / 2)( z) ( z) (functional equation) 5. Zeros of ( z ) Trivial Zeros: Non-Trivial Zeros (RH): z {2, 4,...} 1 z i n 2 (critical line) Distribution of Zeros and Their Spacings First 105 Zeros First 200 Zeros 30 25 20 15 10 5 100 200 300 400 50 40 30 20 10 1 2 3 4 5 6 7 Asymptotic Behavior of Spacings for Large Zeros Question: Is there a Hermitian matrix H which has the zeros of ( z ) as its eigenvalues? Model of The Nucleus Quantum Mechanics H – Hamiltonian (Hermitian operator) H i Ei i i Ei – Bound state (eigenfunction) – Energy level (eigenvalue) Statistical Approach H – Hermitian matrix H i Ei i (H * H ) (Matrix eigenvalue problem) Basics Concepts in Linear Algebra Matrices a11 a A (a jk ) 21 ... an1 a12 a22 ... an 2 ... a1n ... a2 n ... ... ... ann n x n square matrix Special Matrices Symmetric: A A 1 2 A 2 1 Hermitian: A* A 1 i A i 1 Orthogonal: T AT A I ( A1 AT ) cos A sin sin cos Eigensystems 1 2 A 2 1 Ax x – Eigenvalue x – Eigenvector 1 1 1, x 1 2 3, Similarity Transformations (Conjugation) A A UAU 1 Diagonalization UAU 1 D(1 , 2 ,..., n ) 1 x 1 Gaussian Orthogonal Ensembles (GOE) H (h jk ) – random N x N real symmetric matrix Distribution of eigenvalues of 200 real symmetric matrices of size 5 x 5 Level spacing Eigenvalues 50 40 40 30 30 20 20 10 10 -4 -2 0 2 4 1 2 3 4 Entries of each matrix is chosen randomly and independently from a Gaussian distribution with 0, 1 500 matrices of size 5 x 5 100 150 80 125 100 60 75 40 50 20 25 -4 -2 0 2 4 6 1 2 3 4 5 1000 matrices of size 5 x 5 140 150 120 125 100 100 80 75 60 50 40 20 25 -6 -4 -2 0 2 4 6 1 2 3 4 5 10 x 10 matrices 350 300 200 250 150 200 150 100 100 50 50 -5 -2.5 0 2.5 5 1 7.5 2 3 4 20 x 20 matrices 300 400 250 300 200 150 200 100 100 50 -10 -5 0 5 10 1 2 3 4 Why Gaussian Distribution? Uniform P.D.F. Gaussian P.D.F. 0, 1 0 200 150 150 100 100 50 50 -2 -1 0 1 2 -5 -2.5 0 2.5 5 7.5 350 250 300 200 250 200 150 150 100 100 50 50 0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 Statistical Model for GOE H (h jk ) – random N x N real symmetric matrix Assumptions 1. Probability of choosing H is invariant under orthogonal transformations 2. Entries of H are statistically independent Joint Probability Density Function (j.p.d.f.) for H f jk (h jk ) – p.d.f. for choosing h jk N p( H ) f jk (h jk ) j k – j.p.d.f. for choosing H Lemma (Weyl, 1946) All invariant functions of an (N x N) matrix H under nonsingular similarity transformations H H AHA1 can be expressed in terms of the traces of the first N powers of H. Corollary Assumption 1 implies that P(H) can be expressed in terms of tr(H), tr(H2), …, tr(HN). Observation N tr( H ) i i 1 N tr( H ) i2 2 i 1 ... N tr( H ) iN N i 1 (Sum of eigenvalues of H) Statistical Independence Assume H UHU 1 , cos sin U 0 ... 0 sin cos 0 ... 0 0 0 1 ... 0 ... ... ... ... ... 0 0 0 ... 1 Then H U 1 HU U T U HU U T H U T U UH HU T AH HAT (*) 0 1 1 0 U T A U ... ... 0 0 0 0 ... 0 ... ... ... ... 0 0 ... 0 Now, P(H) being invariant under U means that its derivative should vanish: p ( H ) f kj (hkj ) j k p 0 1 f kj hkj 0 f kj hkj We now apply (*) to the equation immediately above to ‘separate variables’, i.e. divide it into groups of expressions which depend on mutually exclusive sets of variables: {h11 , h12 , h22 },{h1k , h2k } 1 f11 1 f 22 1 f12 (2 h ) (h11 h22 ) 12 f12 h12 f11 h11 f 22 h22 1 f1k 1 f 2 k h2 k h1k 0 f1k h1k f 2 k h2 k k 3 N It follows that say 1 f1k 1 f 2 k h2 k h1k Ck f1k h1k f 2 k h2 k Ck 1 f1k 1 f 2 k h1k f1k h1k h2 k f 2 k h2 k h1k h2 k (constant) It can be proven that Ck = 0. This allows us to separate variables once again: 1 f1k 2a h1k f1k h1k (constant) 1 f 2 k 2a h2 k f 2 k h2 k Solving these differential equations yields our desired result: f1k (h1k ) exp(ah12k ) f 2 k (h2 k ) exp(ah22k ) (Gaussian) Theorem Assumption 2 implies that P(H) can be expressed in terms of tr(H) and tr(H2), i.e. p( H ) exp(a tr( H 2 ) b tr( H ) c) J.P.D.F. for the Eigenvalues of H D U T HU , U TU I , 1 0 ... 0 0 ... 0 2 D ... ... ... ... 0 0 0 N Change of variables for j.p.d.f. F: N O( N ) Sym( N ) ( , U ) Jac( F ) H UDU T , (1 ,..., N ) ( H11 , H12 ,..., H NN ) (1 ,..., N ,U11 ,U12 ,...U NN ) P ( H ) p ( H )dH , Sym( N ) p ( )Jac( F ) dUd F 1 ( ) 1 F () N F 1 ( ) p ( ) Jac( F ) dU d O( N ) p N ( ) d N Joint P.D.F. for the Eigenvalues pN ( ) p( ) Jac( F )dU O ( N ) Lemma Jac( F ) g (U ) k j j k Corollary pN (1 ,..., N ) exp( a i2 b i c) k j j k Standard Form j 1 b xj 2a 2a 1 pN ( x1 ,..., xN ) CN exp xi2 xk x j 2 j k Density of Eigenvalues Level Density We define the probability density of finding a level (regardless of labeling) around x, the positions of the remaining levels being unobserved, to be N ( x) R1 ( x) N ... pN ( x, x2 ,..., xN )dx2 ...dxn Asymptotic Behavior for Large N (Wigner, 1950’s) 300 1 2 2 N x , N ( x) 0 250 x 2N x 2N 200 150 100 50 -10 -5 0 5 20 x 20 matrices 10 Two-Point Correlation We define the probability density of finding a level (regardless of labeling) around each of the points x1 and x2, the positions of the remaining levels being unobserved, to be N! R2 ( x1 , x2 ) ... pN ( x1 ,..., xN ) dx3 ...dxn ( N 2)! We define the probability density for finding two consecutive levels inside an interval ( , ) to be AN ( ; x1 , x2 ) N! ... pN ( x1 ,..., xN )dx3 ...dxN 2!( N 2)! out x j , j 1, 2 x j , j 3,..., N Level Spacings Limiting Behavior (Normalized) We define the probability density that in an infinite series of eigenvalues (with mean spacing unity) an interval of length 2t contains exactly two levels at positions around the points y1 and y2 to be B(t ; y1 , y2 ) lim 2 AN ( ; x1 , x2 ), N t /, y j xj / mean spacing P.D.F. of Level Spacings We define the probability density of finding a level spacing s = 2t between two successive levels y1 = -t and y2 = t to be p( s) 2 B(t; t , t ) Multiple Integration of pN ( x1 ,..., xN ) 1 pN ( x1 ,..., xN ) CN exp xi2 xk x j 2 j k Key Idea Write pN ( x1 ,..., xN ) as a determinant: 1 exp xi2 xk x j c det i ( x j ) 2 j k x2 j ( x) exp H j ( x) (Oscillator wave functions) j 2 2 j! 1 d j x2 H j ( x ) j [e ] dx (Hermite polynomials) Harmonic Oscillator (Electron in a Box) d2 1 2 2 m x n ( x) Enn ( x), 2 2 m dx 1 En (n ) 2 NOTE: Energy levels are quantized (discrete) Formula for Level Spacings? d2 p ( s ) 2 (1 2 ) ds 2 - Eigenvalues of a matrix whose entries are integrals of functions involving the oscillator wave functions The derivation of this formula very complicated! Wigner’s Surmise pW ( s ) s exp s 2 , 2 4 s S D Random Matrices and Solitons Korteweg-de Vries (KdV) equation ut 6uu x u xxx 0 Soliton Solutions 2 u ( x, t ) 2 2 log det( I A( x, t )) x N cm cn ( km kn ) x 4( km3 kn3 )t A( x, t ) e k k n m m,n 1 Cauchy Matrices N 1 A k k n m , n 1 m ( kn 0) - Cauchy matrices are symmetric and positive definite 1 22 1 A 3 2 1 52 1 23 1 33 1 53 1 25 1 35 1 55 Eigenvalues of A: 0.502136, 0.0142635, 0.000267133 0.688884, 4.25005, 8.22776 Logarithms of Eigenvalues: Level Spacings of Eigenvalues of Cauchy Matrices Assumption The values kn are chosen randomly and independently on the interval [0,1] using a uniform distribution 1000 matrices of size 4 x 4 1200 350 1000 300 250 800 200 600 150 400 100 200 50 2 4 6 8 10 Distribution of spacings -15 -10 -5 0 Log distribution 5 Level Spacings First-Order Log Spacings 1000 matrices of size 4 x 4 140 10,000 matrices of size 4 x 4 700 120 600 100 500 80 400 60 300 40 200 20 100 2 4 6 2 4 6 8 Second-Order Log Spacings 70 300 60 250 50 200 40 150 30 100 20 50 10 -6 -4 -2 0 2 -6 -4 -2 0 2 Open Problem Mathematically describe the distributions of these first- and higher-order log spacings References 1. Random Matrices, M. L. Mehta, Academic Press, 1991.