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Random-Matrix Approach to RPA Equations X. Barillier- Pertuisel, IPN, Orsay O. Bohigas, LPTMS, Orsay H. A. Weidenmüller, MPI für Kernphysik, Heidelberg kshop on “Random-Matrix Theory and Applications: Fr ber Theory to Mesoscopic Physics”, June 25 – 27, Or

Random-Matrix Approach to RPA Equations X. Barillier-Pertuisel, IPN, Orsay O. Bohigas, LPTMS, Orsay H. A. Weidenmüller, MPI für Kernphysik, Heidelberg

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Random-Matrix Approach to RPA Equations

X. Barillier-Pertuisel, IPN, Orsay

O. Bohigas, LPTMS, Orsay

H. A. Weidenmüller, MPI für Kernphysik, Heidelberg

Workshop on “Random-Matrix Theory and Applications: FromNumber Theory to Mesoscopic Physics”, June 25 – 27, Orsay.

Contents:

1. Motivation 2. RPA Equations 3. Random-Matrix Approach 4. Pastur Equation 5. Solutions. Numerical Results 6. Critical Strength 7. Summary

1. MotivationRandom-phase approximation (RPA) is a standard tool ofmany-body physics. First used in condensed-matter physicsbut later also in other areas and especially in nuclear-structurephysics.

Study RPA equations in most general form by taking matrixelements as random. Yields unitarily invariant random-matrixmodel.

Study spectrum using (generalized) Pastur equation as afunction of coupling between states at positive and at negativeenergies.

When does the spectrum become instable (i.e., when doeigenvalues become complex)? Distribution of eigenvaluesin complex energy plane?

2. RPA Equations N-dimensional space for particle-hole pairs. RPA equations have dimension 2N :

Matrix A is Hermitean. Matrix C is symmetric.Total matrix not Hermitean.Eigenvalues not necessarily real.Non-real eigenvalues: Instability.

If is eigenvalue then also (- *) isan eigenvalue.

Two symmetries:

If is an eigenvalue then also * is aneigenvalue.

Result: Real and purely imaginary eigenvalues come in pairswith opposite signs. Complex eigenvalues with non-vanishingreal and imaginary parts come in quartets arrangedsymmetrically with respect to the real and to the imaginaryenergy axis.

Hermitean matrix A0 causes repulsion amongst positiveeigenvalues, matrix – (A0)* causes repulsion amongst negativeeigenvalues. For C = 0 all eigenvalues real.

Role of C: Eliminate negative-energy subspace and obtainHermitean effective operator - C (E + (A(0)*)-1 C* inpositive-energy subspace.

Causes additional level repulsion amongst positive and amongstnegative eigenvalues and level attraction between positive andnegative eigenvalues (Trace of that operator is negative).

With increasing strength of C, pairs of real eigenvalues withopposite signs coalesce at E = 0 and then move along theimaginary axis in opposite directions.

Instability of the RPA equations.

3. Random-Matrix ApproachDegenerate particle-hole energies at r, A0 = 1N r + A wherematrix A contains particle-hole interaction matrix elements,belongs to GUE with

and is invariant under A → U A (U*)T. Matrix C obeys

C is invariant under C → U C UT. The RPA matrix becomes

Generalized unitary invariance: Ensemble is invariant under

The matrices are not Hermitian. Ensemble does not belongto ten generic random-matrix ensembles listed by Altlandand Zirnbauer. We have also looked at the orthogonal case.

Aim: To study average spectrum for N → 1. Model dependson two dimensionless parameters: = 2 / 2 measures relativestrength of C versus A. For = 0, spectrum consists of twosemicircles. And x = r / (2gives distance of centers ofsemicircles from origin at E = 0. Spectral fluctuations notstudied; are very likely to be same as for GUE in each branch.Two limiting cases: C = 0 and A = 0.Find value of for which RPA equations become instable.

Distribution of eigenvalues in the complex energy plane

N = 20, x = 4

4. Pastur EquationDo this by calculating

Imaginary part of average Green’s function yields level density.Expand in powers of H.

where

Average each term in sum separately. For N large, keep onlynested contributions. Yields Pastur equation

Define for i = 1,2 the spectral density of subspace i in totalspectrum (projection operators Qi onto the two subspaces)

and take trace of Pastur equation. Yields two coupledequations for 1 and 2,

1 = / (E – r - (1 - 2)), 2 = / (E + r - (2 - 1)).

For = 0 these are two separate equations. Imaginary partsof solutions yield two semicircles of radius 2 centered at rand at – r. Two dimensionless variables

= 2 / 2 , x = r / (2 ) .

Study spectrum as function of for fixed x.

5. Solutions. Numerical ResultsTo gain understanding of solutions, consider first = 0.With i = (E – (-)i r) / (2 ), spectrum given by

Im(i) = {1 - 2i}1/2 .

Usual semicircle law. Two branch points at = § 1. Eachi defined on Riemann surface with two sheets.

For 0, solution defined on Riemann surface with four sheets.But which sheet to choose for physically relevant solution? Take very small, use perturbation theory, find that we need pair(1, 2) of solutions for which imaginary parts have oppositesigns and Im(1) > 0. Then total level density (E) given by

(E) = (N / ( )) Im (1 + 2) .

Eliminate 2 and obtain fourth-order equation for 1. Then2 obtained from solution 1 via linear equation.

Pairs of solutions:

● Four pairs of complex solutions: Domain (I)

● Two pairs of real and two pairs of complex solutions with equal signs for Im(1) and Im(2): Domain (II)

● Two pairs of real and two pairs of complex solutions with opposite signs for Im(1) and Im(2): Domain (III)

● Four pairs of real solutions: Domain (IV)

Physically interesting solutions only in domains (I) and (III).

Case (a): x = 1.2 Case (b): x = 2.0 Case (c): x = 4.0

N = 50, 100 realizations

Comparison of resultsfor Pastur equation withthose of matrixdiagonalization.

Evolution of two spectrawith increasing .

Upper panels: x = 1.2

Lower panels: x = 4

Normalization integral of total level density taken over realenergies versus . For = 0, integral is normalized to unity.

X = 1.2

6. Critical StrengthFind smallest value of for which average spectra touch:Imaginary parts of a pair of physically acceptable solutions(1, 2) have non-vanishing values at E = 0.

Determine crit analytically from solutions of fourth-orderequation for 1 at E = 0. Find

crit = x2 – 1.

7. SummaryRandom-Matrix model for RPA equations with “Generalizedunitary invariance”.

Level repulsion between levels of same sign, level attractionbetween levels with opposite sign. Latter causes coalescenceof pairs of eigenvalues with opposite signs at E = 0 andinstability of RPA equations.

Use Pastur equation to derive two coupled equations for 1, 2.Surprisingly simple structure. Criterion for physically relevantsolutions. Get average level density from (1, 2). Twoparameters: x and .

With increasing , semicircles are deformed and move towardeach other. RPA instability for average spectrum: Thedeformed spectra touch. Critical strength crit = x2 – 1.