An exact microscopic multiphonon approach to nuclear spectroscopy

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An exact microscopic multiphonon approach to nuclear spectroscopy. N. Lo Iudice Universit à di Napoli Federico II Naples (Andreozzi, Lo Iudice, Porrino) Prague (Knapp, Kvasil) Collaboration Tokyo 07. Eigenvalue problem in a multiphonon space. H | Ψ ν > = E ν | Ψ ν > - PowerPoint PPT Presentation

Transcript of An exact microscopic multiphonon approach to nuclear spectroscopy

An exact microscopic multiphonon approach to nuclear

spectroscopyN. Lo Iudice

Università di Napoli Federico II

Naples(Andreozzi, Lo Iudice, Porrino) Prague (Knapp, Kvasil)

Collaboration

Tokyo 07

Eigenvalue problem in a multiphonon space

H | Ψ ν > = Eν | Ψ ν >

| Ψ ν > H = Σn Hn ( n= 0,1.....N )where

Hn ∋ |n, α > ~ | ν1 ν2 …..νn >

| νi > = Σph c ph(νi ) a

†p ah |0>

First goal: Generate the basis states |n, α >

We do it by constructing a set of equations of motion and solving them iteratively

EOM: Construction of the Equations • Crucial ingredient < n; β | [H, a†

p ah] | n-1; α>

• Preliminary step: Derive< n; β | [H, a†

p ah] | n-1; α> = ( Eβ

(n) - Eα

(n-1)) < n; β | a

p ah| n-1; α >

(LHS) (RHS)

The RHS comes from

* property < n; β | a†

pa

h | n’; γ > = δ

n’,n-1 < n; β | a†

pa

h | n-1; γ >

** request

< n; β | H | n; α > = E α

(n) δαβ

Commutator expansion

< n; β | [H, a†p ah] | n-1; α > =

(εp- εh) < n; β | a†p ah| n-1; α >. +

1/2 Σijp’

Vhjpk < n; β | a†p’ ah a

†i aj | n-1; α > + …

Linearization

< n; β | [H, a†p ah] | n-1; α > =

Σp’h’γ [(εp- εh) δγβ

+ 1/2 Σij

Vhjp’k’ < n-1; γ | a†i aj | n-1; α > +

………] < n; β | a†p’ ah’ | n-1; γ >

= Σp’h’γ

Aαγ(n)(ph;p’h’) < n; β | a†

p’ ah’ | n-1; γ >

Î = Σ γ |n-1; γ >< n-1; γ|

Equations of Motion : LHS

LHS=RHS AX = EX

jγ Aαγ(n) ( ij) Xγβ

(n)(j) = (Eβ

(n) - Eα(n1)) Xαβ

(n)(i) where

Xαβ(n) (i )

= < n; β | a†

p ah| n-1; α >

Aαγ(n) ( ij) = [εp–εh] δij

(n-1) δαβ (n-1)

+ [VPHρH + VHPρP + VPPρP + VHHρH ]αiβj

ρH ≡ {< n,γ|a†

hah’|n,α>} ρP ≡ {< n,γ|a†

pap’|n,α>} n =1 (ρP = 0 ρH = δhh’)

A(1) Xα(1) = (Eα

(1) - E0 (0) ) Xα

(1) Tamm-Dancoff

A(1) (ij)= δij[εp–εh] + V(p’hh’p)

Structure of multiphonon states: Overcompletness

The eigenvalue equations yield states of the structure

|n; β> = Σ α ph cα

ph a†

p ah | n-1; α >

• Problem: The multiphonon states are not fully antysymmetrized !!!

a†p ah | n-1; α > ≡

p h p h

The multiphonon states form an overcomplete set

Solution of the redundancy Reminder

Insert |n; β> = Σα ph Cαph a†

p ah | n-1; α >

Xαβ

(n)(ph)

= < n; β | a†

p ah | n-1; α >

X = DC (AD)C = H C = E DC

where

Dij = < n-1; α’| ah’ a†

p’ a†

p ah | n-1; α> overlap or metric matrix

Problem due to redundancy Det D = 0

γ Aαγ(n) Xγβ(n)= (Eβ

(n) - Eα(n-1 ) )Xαβ(n)

A X = E X

Solution of the redundancy *Removal of redundancy: Choleski decomposition(no

diagonalization)

D Ď ** Matrix inversion

Exact eigenvectors

|n; β> = Σ αph Cαph a

p ah | n-1; α > H n (phys)

● Now compute in Hn

i. X(n)

ii. ρ(n) recursive formula

X= D C

ρ(n) = C X (n) + C ρ (n-1) X (n)

HC = (Ď-1AD)C = E C

Iterative generation of phonon basis Starting point |0>

Solve Ĥ(1) C(1)

= E(1)

C(1)

|n=1, α> X(1) ρ(1)

Solve Ĥ(2) C (2) = E (2) C (2) |n=2,α> X(2) ρ(2)

……… X(n-1) ρ(n-1)

Solve Ĥ(n) C (n) = E (n) C (n)

X(n) ρ(n) |n,α

.>

The multiphonon basis is generated !!!

H: Spectral decomposition, diagonalization

H = Σ nα

E α

(n) |n; α><n;α| + (diagonal)

+ Σ nα β

|n; α><n;α| H |n’;β><n’;β| (off-diagonal)

n’ = n ±1, n±2

Off-diagonal terms: Recursive formulas

< n; α | H| n-1; β > = Σphγ

ϑαγ

(n-1)(ph) Xγβ

(n)(ph)

< n; α | H| n-2; β > = Σ V pp’hh’

Xγβ

(n) (ph)

Xγβ

(n-1) (p’h’)

Outcome of diagonalization H |Ψν> = Eν |Ψν>

|Ψν> = Σnα Cα(ν) (n) | n;α> |n;α> = Σγ Cγ

(n) | n-1;γ>

Numerical test: A = 16 ∙ Calculation up to 3-phonons and 3ħω

Hamiltonian

H = H0 + V = Σi hNils

(i)+ Gbare

( VBonnA

⇨ Gbare)

• CM motion (F. Palumbo Nucl. Phys. 99 (1967))

H H + Hg

Hg = g [ P2/(2Am) + (½) mA ω2 R2 ] • Consistent choice of ph space: It must includes all ph

configurations up to 3ħω

Ground state

|Ψ0> = C(0)

0 |0>

+ Σλ C

λ

(0) |λ, 0>

+ Σ λ1λ2 Cλ1λ2

(0) |λ 1 λ

2, 0

>

|λ, 0> |λ 1 λ

2, 0

>

1 = < Ψ0|Ψ

0> = P

0 + P

1 + P

20

10

20

30

40

50

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80

0ph 2ph 4ph 6ph

EMNocoreHJ

0

10

20

30

40

50

60

70

80

0ph 1ph 2ph

P(n) %noCMcorr

16O negative parity spectrum • Up to three phonons

IVGDR

|1->IV ~ |1(p-h) (1ħω)>

ISGDR |1->IS ~ |1(p-h) (3 ħω)> + |2(p-h) (1ħω + 2ħω)> + |3(p-h) (1ħω )>

Toroidal

Octupole modes |3->IS ~ |1(p-h) (3 ħω)> + |2(p-h) (1ħω + 2ħω)> + |3(p-h) (1ħω )>

Low-lying

Effect of CM motion

Effect of the CM motion

Concluding remarks• The multiphonon eigenvalue equations

- have a simple structure - yield exact eigensolutions of a general H

• The 16O test shows that - an exact calculation in the full multiphonon space is feasible at least up to 3 phonons and 3 ħω.

• To go beyond - Truncation of the space needed !!! - Truncation is feasible (the phonon states are correlated).

- A riformulation for an efficient truncation is in progress

THANK YOU

E2 response up to 3 ħω: Running sum

Sn = Σn (En – E0(0) ) Bn(E2,0→2+

n)

Rn = Sn/SEW(E2)

• It is necessary to enlarge the space!!

E2 response up to 3 ħω

S(ω,E2) = Σn Bn(E2,0→2+n) ρΔ(ω-ωn)

ρΔ(x) = (Δ/2π) / [x2 + (Δ/2)2]

M(E2μ) = Σ(p)ep rp2 Y 2μ

Effect of CM on the E2 response

Growing evidence of multiphonon excitations

* Low-energy M. Kneissl. H.H. Pitz, and A. Zilges, Prog. Part. Nucl. Phys. 37, 439 (1996); M. Kneissl. N.

Pietralla, and A. Zilges, J.Phys. G, 32, R217 (2006) : • Two- and three-phonon multiplets Q2 × Q3|0>, Q2×Q2×Q3|0>• Proton-neutron (F-spin) mixed-symmetry states (N. Pietralla et al. PRL 83, 1303 (1999))

[Q2(p) - Q2

(n)] (Q2(p) + Q2

(n)) N|0>,

** High-energy (N. Frascaria, NP A482, 245c(1988); T. Auman, P.F. Bortignon, H. Hemling, Ann. Rev. Nucl.

Part. Sc. 48, 351 (1998))

• Double and (maybe) triple dipole giant resonances D × D |0>

From TDA to RPA

A(n)

B(n)

X(n)

= (E(n)

- E(n-1)

)

B(n)

A(n)

-Y(n)

Aαγ(n) (ph;p’h’)=δhh’δpp’δαγ(n-1)[εp–εh]

+ Σh1 V(p’h1h

’p) ραγ

(n-1) (h1h)

+ Σp1 V(p’hp

1h’) ραγ

(n-1) (pp1)

+ δhh’ [ Σp2p3 V(p’p2pp3)ραγ

(n-1) (p2p3)

+ Σh2h3 V(p’h2ph3)ραγ

(n-1)(h2h3)]

+ δpp’ [ Σh2h3 V(hh2h

’h3)ραγ

(n-1)(h2h3)

+ Σp2p3 V(hp2h

’p3)ραγ

(n-1)(p2p3)]

Bαγ

(n) (ph;p’h’)=

Σh2 V(p p’h2 h

’)ραγ

(n-1)(h2h)

+ Σp2 V(h’hp2p

’)ραγ

(n-1)(pp2)

AX = EXwhere

Aαγ

(n)

(ph;p’h’)=δhh’δpp’δαγ

(n-1)[εp–εh]

+ Σh1 V(p’h1h’p) ραγ

(n-1)(h1h)

+ Σp1 V(p’hp1h’) ραγ

(n-1)(pp1)

+ δhh’1/2 Σp2p3 V(p’p2pp3) ραγ

(n-1)(p2p3)

+ δpp’1/2 Σh2h3 V(hh2h’h3) ραγ

(n-1)(h2h3)

(ραγ

(n)(ij) = <n,α|a†

iaj|n,α>)

Choleski decomposition• Any real non negative definite

symmetric matrix can be written as

D = L LT

( L {lij} lower triangular matrix) Det{D} = (Det{L})2

= l112 l22

2 ...lii2…..

= λ21 …λ2

i… (D | λi > = λi | λi > )

Definition of L: Recursive formulas

l211 = d11

l11 lj1 = dj1 j=2,….,nl2

ii = dii – Σk=1,i-1 l2

ik

lii lji = dji – Σk=1,i-1 lik ljk

(It reminds the Schimdt orthogonalization method)

The decomposition goes on until lnn = 0

lnn = 0 → Det{L} =0→Det{D} =0 ⇨

⇨ | λn > (D | λi > = λi | λi > ) linearly dependent ⇨ to be discarded

• For numerical stability we need maximum overlap

λj ≤ λi j > i

• Sequence order

lii ≤ ljj j > i

Once lnn = 0 → lii = 0 i >n

1) We can stop at the nth step

2) We get Ď (Nn < N

r) with maximum determinant

(overlaps)

Elimination of CM spuriosity F. Palumbo Nucl. Phys. 99 (1967)

• HSM H = HSM + Hg

• Hg = g [ P2/(2Am) + (½) mA ω2 R

2 ] = g/A [ Σi hi + Σi<j vij ]

hi = pi

2/(2m) + (1/2) m(Aω)2ri

2

vij = (1/m) pi· pj + m (Aω)2ri · rj

Hg effective only in Jπ = 1

- ph channel.

H Ψ = (HSM + Hg) Ψ =E Ψ

In the full space

Ψ = ψ

int Φn

CM

For the physical states

Ψn = ψn Φ0

CM

with CM energy

E0

CM = 3/2 g ħω

For the spurious ones

Ψn

(1)

= ψn

Φ1

CM

at the very high CM energy

E1

CM

- E0

CM

= g ħω

They can therefore be tagged and eliminated