Supersymmetry for non Gaussian probability densities

Transcription

1 Sonderforschungsbereich Transregio 12 Lecture Notes 1 Supersymmetry for non Gaussian probability densities Thomas Guhr 1 and Stefanie Kramer 2 1 Fachbereich Physik, Universität Duisburg Essen 2 Fachbereich Mathematik, Ruhr Universität Bochum lecture given at the Langeoog workshop, February 2008

2 Sonderforschungsbereich Transregio 12 Lecture Notes 2 These lecture notes are based on the papers T. Guhr, Norm-Dependent Random Matrix Ensembles in External Field and Supersymmetry, J. Phys. A39 (2006) , math-ph/ T. Guhr, Arbitrary Unitarily Invariant Random Matrix Ensembles and Supersymmetry, J. Phys. A39 (2006) , math-ph/ Again, a comment is in order: to keep the formulae transparent, all unimportant constants are set to one. Hence, if you need the results to be discussed here including those constants, check the papers listed above!

3 Sonderforschungsbereich Transregio 12 Lecture Notes 3 I Introduction We are interested in a model for spectral correlations. Consider discrete spectra, that is in most physics applications spectra of a bound quantum system with Hamilton operator H. The density of states, levels, eigenenergies x n of H is ρ(e) = n 1 δ(e x n ) = trδ(e H) = Im tr E ± iε H, (1) where the limit ε 0 is understood. In the statistical model, we replace H by a N N random matrix. The correlation functions are then k R k (E 1,...,E k ) = d[h]p(h) ρ(e p ). (2) Here we assume a probability density function P(H) for H, we always assume that P(H) is invariant. A convenient choice for P(H) is the Gaussian P(H) exp ( trh 2). (3) For almost all applications in quantum chaos, disordered systems or so, this choice is acceptable, although hardly any physical system shows such a Gaussian form. This is acceptable because of local universality. One is always interested in the correlations transformed onto the local scale of the mean level spacing D which goes like D 1/ N, that is, it becomes ever smaller as we take the limit N. Sometimes one calls that the proper microscope. On this scale, almost all P(H) yield the same, universal correlation functions. Thus, why is it interesting to study P(H) different from Gaussian? Because there are systems such as finance or high energy physics, where the local scale is not the interesting one. The experts will now say: well, but there are powerful methods to calculate R k (E 1 E k ) for non Gaussian P(H), these methods are stated in the Mehta Mahoux theorem. However, to apply it, the probability density function has to separate according to which then implies P(H) = P(x) = N n=1 P(x n ) (4) R k (E 1,...,E k ) = det [K n (E p, E q )] p,q=1,...,k. (5)

4 Sonderforschungsbereich Transregio 12 Lecture Notes 4 The method presented here does not need this separability property, rather it is based on Supersymmetry. Originally, the Supersymmetry method was developed for Gaussian P(H). The question has been asked repeatedly whether Supersymmetry is a priori restricted to Gaussian P(H). For a long time, there was no clear answer. Now there is one: Supersymmetry is not at all restricted to Gaussian P(H). Some history: Guhr (1991): Hubbard Stratonovich transformation identified as two Fourier transforms first in ordinary, then in superspace Lehmann, Saher, Sokolov, Sommers (1995): δ function in matrix superspace used to carry out Hubbard Stratonovich transformation Hackenbroich, Weidenmüller (1995): universality proof for arbitrary probability densities involving supersymmetry and twofold asymptotics, no finite N results Efetov, Schwiete, Takahashi (2004): superbosonization Guhr (2006): algebraic duality, explicit construction for unitary case with Mario Kieburg now also orthogonal and symplectic case Littelmann, Sommers, Zirnbauer (2007): rigorous, threefold way As for the last two references: while the philosophy is the same, the approach is different. II Posing the Problem The correlation functions are obtained from a generating function Z k (E, J) as derivatives with respect to some source variables J p, p = 1,...,k which are put J p = 0 after derivation. Remarkably, we have for the Gaussian case Z k (E, J) = = d[h] exp ( trh 2) k det (H E p J p ) det (H E p iε + J p ) d[σ] exp ( str σ 2) sdet N ( σ E J ) (6) where σ is a 2k 2k supermatrix, while H is N N ordinary. We gain a drastic reduction of the number of integrals and N becomes an explicit parameter (Efetov 1983, Verbaarschot and Zirnbauer 1985). What is a supermatrix? Symbolically, it has the structure [ commuting anticommuting σ = anticommuting commuting ] (7)

5 Sonderforschungsbereich Transregio 12 Lecture Notes 5 where commuting variables are also referred to as bosons and anticommuting ones as fermions or Graßmann variables. For two fermions ζ, η we have ζη = ηζ ζ 2 = 0. (8) What is a function of a fermion? It is defined by its Taylor series (just like the exponential of a matrix). Thus, with a commuting variable/number c we have exp(cζ) = 1 + cζ no further terms! (9) We formally define an integration, which of course has nothing to do with a Riemann integral. We implicitly define it by this example it is called Berezin integral dζ exp(cζ) = dζ(1 + cζ) = dζ + c dζζ (linearity) = 0 + c. (10) Why do we want to call this an integral? In superspace, we have bosons and fermions together, when changing variables we transform and mix them into each other. It turns out that the rules of changing variables in the Riemann integral extend beautifully formally into superspace with the above given definition of the Berezin integral. We notice that the integral above gave c. For a commuting variable ζ, we would have found 1/c. What is str and sdet? These are the proper generalizations of trace and determinant to superspace. A trace of a product of matrices should be invariant under cyclic permutation str is that. The determinant of a product of matrices should be the product of the determinants sdet is that. Thus, we ask: is there an identity of the form Z k (E, J) = = d[h]p(h) k det (H E p J p ) det (H E p + J p ) d[σ]q(σ)sdet N (σ E J) (11) for any (reasonable, well behaved) P(H) and what is then Q(σ)? III Special Case: Norm dependent Ensembles Consider a probability density function P(H) = P (T) (u), u = trh 2 (12)

6 Sonderforschungsbereich Transregio 12 Lecture Notes 6 which depends on H only via trh 2 = u. Then, some Fourier transformations yield and also Q(σ) = Q (T) (w) = 0 P (T) (u + w)u N2 /2 1 du w = str σ 2 (13) P (T) (u) = N2 /2 u N2 /2 Q(T) (u). (14) This is the N 2 /2 1-fold (fractional) integration and N 2 /2-derivative. Consider as an example the bound trace ensemble P (T) (u) = a 0 Θ(a 1 u) Q (T) (w) = (a 1 w) N2 /2 a N2 /2 1 Θ(a 1 w). (15) IV General Case: Extended Hubbard Stratonovich Transformation For the sake of clarity, we simplify the presentation by putting k = 1, (16) but everything generalizes without problems to arbitrary k. We write the determinants as Gaussian integrals Z 1 (E, J) = d[h]p(h) d[z] exp ( iz (H E + J)z ) d[ζ] exp ( iζ (H E J)ζ ), where z and ζ are N component vectors of complex bosons and fermions. Now collecting things, we have Z 1 (E, J) = d[z] exp ( iz ( E + J)z ) d[ζ] exp ( iζ ( E J)ζ ) Φ(K) (18) with the characteristic function (Fourier transform) (17) Φ(K) = d[h]p(h) exp(itrhk), K = zz ζζ. (19) We observe an algebraic duality. Define A = [z ζ], (20)

7 Sonderforschungsbereich Transregio 12 Lecture Notes 7 we have the N N ordinary matrix AA = K (21) and the 2 2 supermatrix A A = B = [ z z z ζ ζ z ζ ζ ] (22) We notice that K contains the dyadic, and B the scalar products! This will lead to the drastic reduction in the number of degrees of freedom. There is now a crucial identity trk m = strb m, m = 1, 2, 3,... (23) which connects the invariants in ordinary and superspace! Hence, if P(H) is invariant, Φ(K) is invariant as well because of the invariance of the measure d[h]. Thus, Φ(K) depends on the trk m only, and viewed as function of the invariants, we have Φ(trK, trk 2, trk 3, ) = Φ(strB, strb 2, strb 3, ). (24) In this sense, we write Φ(K) = Φ(B) = = d[ρ]φ(ρ)δ (4) (ρ B) d[ρ]φ(ρ) d[σ] exp ( i str σ(ρ B)) (25) What is the δ-function of an anticommuting variable? The variable itself, because for a function f(ζ) = f 0 + f 1 ζ one easily checks dζ(ζ η)f(ζ) = dζ(ζ η)(f 0 + f 1 ζ) = dζf 0 ζ + η dζf(ζ) = f 0 + f 1 η = f(η). (26) More tricky are δ functions containing an ordinary commuting variable y and a nilpotent commuting variable ζ ζ assembled from anticommuting ones ζ n, n = 1,..., N. We formally expand δ ( y ζ ζ ) = ν=0 1 ν! δ[ν] (y) ( ζ ζ ) νmax ν 1 = ν! δ[ν] (y) ( ζ ζ ) ν ν=0 (27) because there is a ν max such that ( ζ ζ ) ν = 0 for ν > νmax.

8 Sonderforschungsbereich Transregio 12 Lecture Notes 8 We can now do the z, ζ integral using str σb = [ z ζ ] (σ 1 N ) [ z ζ ] (28) and find Z 1 (E, J) = = d[ρ]φ(ρ) d[σ] exp( i strσρ) sdet N (σ E J) d[ρ] exp ( i str(e + J)ρ) Φ(ρ) I(ρ) (29) with the invariant Ingham Siegel integral I(ρ) = d[σ] exp( i strρσ) sdet N σ = sdet +N ρ r1 >0 = Θ(r 1 )(ir 1 ) N N 1 δ(r 2 ) r N 1 2. (30) Thus, with the convolution theorem we arrive at the desired result Q(σ) = d[ρ] exp(i str σρ) Φ(ρ). (31) The characteristic function is the Fourier transform of both probability density functions in ordinary and superspace, d[h]p(h) exp(itrhk) = Φ(K) = Φ(B) = d[σ]q(σ) exp(i str σb). (32) All that generalizes to arbitrary k. Then, ρ and σ are 2k 2k supermatrices. V Reduction to Eigenvalue Integrals The representation in Fourier superspace has great advantages: as Φ(ρ) and I(ρ) are invariant, we have for all P(H)! a matrix plane wave and can reduce the problem of angular integration to a matrix Bessel function in superspace. We diagonalize ρ = v 1 rv (33) where r = diag(r 11,, r k1, ir 12,, ir k2 ) is the diagonal matrix of the eigenvalues which are bosons, but depend also on the fermions in ρ. The diagonalizing matrix v U(k/k) is in the unitary supergroup, which leaves a generalized length in superspace invariant. The volume element reads [ ] d[ρ] = Bk 2 (r)d[r]dµ(v) with B 1 k(r) = det (34) r p1 ir q2 p,q=1 k

9 Sonderforschungsbereich Transregio 12 Lecture Notes 9 where B k (r) is the superspace generalization of the Vandermonde determinant. We can now do the angular integral and get R k (E 1,...,E k ) = d[r]b k (r) exp( i str Er)Φ(r)I(r). (35) Once we have the characteristic function Φ(K), the problem is thus reduced to 2k ordinary integrals, of which k are trivial because of δ functions. If Φ(r) separates, that is, factorizes in a product, we immediately find the celebrated k k determinant structure of the correlation functions which is thus seen as a built in feature of supersymmetry. VI Relation to Orthogonal Polynomials We define a reduced probability density P (red) (h) = d[r] exp( i str hr)φ(r) (36) as the flat (!!) Fourier backtransform of Φ(r) as function of r. Then we have R k (E 1,...,E k ) = d[h] P (red) (h)r (fund) k (E h), (37) that is a convolution of P (red) (h) with the fundamental correlations R (fund) k (s) = det [ C (fund) (s p1, is q2 ) ] p,q=1 k (38) with a fundamental kernel C (fund) (s p1, is q2 ) = N 1 n=0 (is q2 ) n (s p1) n+1 = 1 (s p1) N (s p1) N (is q2 ) N s p1 is q2 (39) which contains the skeletons of all Christoffel Darboux relations. VII A Simple Example Consider a probability density without factorization of the form. P(H) = ( trh M 1) M2 exp( trh 2 ), M 1, M 2 = 0, 1, 2,... (40)

10 Sonderforschungsbereich Transregio 12 Lecture Notes 10 The correlation functions are linear combinations of determinants R k (E 1,...,E k ) = det [ C mω(p) m ω(k+q) (E p, E q ) ] p,q=1 k {m} ω (41) with coefficients a {m} and C m1 m 2 (E p, E q ) = exp ( E 2 p ) N 1 n=0 1 n! η nm 1 (E p ) ϑ nm2 (E q ) (42) where η nm1 (E q ) and ϑ nm2 (E p ) are linear combinations of Hermite polynomials.

11 Sonderforschungsbereich Transregio 12 Lecture Notes 11 Appendix In our approach, we do not need hyperbolic symmetry. In the non linear σ model, one must use it (Schäfer and Wegner 1980, Efetov 1982, 1983). As one aims at an asymptotic 1/N evaluation, one must start from R 2 (E 1, E 2 ) = d[h]p(h) tr 1 N E 1 + iε H tr 1 N E 2 iε H, (A.1) where the iε are on different sides of the real axis. The ensuing non linear σ model is an integral over Goldstone modes which are parametrized by a coset manifold with non compact degrees of freedom. We are not doing an 1/N expansion, I calculate for finite N. Thus, hyperbolic (non compact) degrees of freedom are not needed, because: Consider R k (E 1,...,E k ) = R k (E 1,...,E k ) = k 1 N d[h]p(h) tr E p + ilp ε H k d[h]p(h) tr δ (E p H). (A.2) Then, the Fourier transforms are r k (t 1,...,t k ) = k d[h]p(h) tr e ihtp (A.3) and, quite remarkably, k r k (t 1,...,t k ) = Θ (L p t p ) r k (t 1 t k ). (A.4) Hence, we can calculate R k (E 1,...,E k ) for the simplest choice Lp = +1, where all iε are on the same side. Then we Fourier transform and identify r k (t 1,...,t k ) in r k (t 1,...,t k ) for Lp = +1. We can now multiply r k (t 1,...,t k ) with arbitrary (!) choices of Θ(L p t p ) with L p = ±1 as we like and by Fourier transform obtain R k (E 1,...,E k ) for any L p = ±1. Since the choice L p = +1 allows one to stay in a fully compact setting, we circumvented hyperbolic symmetry.