Nuclear Structure RPA The Shell Model · INT November 2004 12 Computational Nuclear Structure RPA as generalized harmonic oscillator From the h.o. creation/annihilation operators

INT November 2004 1

ComputationalNuclear Structure

RPAand

The Shell ModelCalvin W. Johnson San Diego State University

Ionel Stetcu Louisiana State University & University of Arizona

I. Stetcu and C. W. Johnson, Phys. Rev. C 66 034301 (2002)C. W. Johnson and I. Stetcu, Phys. Rev. C 66, 064304 (2002)I. Stetcu and C. W. Johnson, Phys. Rev. C. 67, 043315 (2003) I. Stetcu and C. W. Johnson, Phys Rev. C 69, 024311 (2004)Johnson, Vazquez, and Stetcu, in preparation

This work was supported by grants from the Department of Energy

INT November 2004 2

ComputationalNuclear StructureMotivational Slide

ab initio methods

0hω shell model

density functionaltheory

INT November 2004 3


Even MORE Motivation:

“We believe there are good prospects to develop a better global theory [of

nuclear energies] ...treating correlation energies by RPA.”*

* Bertsch & Hagino nucl-th/0006032Phys.Atom.Nucl. 64 (2001) 588/ Yad.Fiz. 64 (2001) 646

INT November 2004 4

ComputationalNuclear StructureSystematic Investigation of RPA

orOutline of my talk!

Review of RPA

Overview of previous “benchmarking” of RPA

Shell model context (in brief) and our code SHERPA

Four elemental tests:

(1) Correlation energies

(2) Scalar observables and “restoration of symmetries”

(3) Transitions in RPA and pnRPA and sum rules

(4) “Collapse” of RPA

Conclusions ... and future work

INT November 2004 5


Review of RPARPA models excited states as small oscillations about the mean-field.

There are many ways to derive the random phase approximation (RPA):

eqns of motion, time-dependent Hartree-Fock, linear response...

INT November 2004 6


The nuclear landscape

Hartree-Fock based upon variational principle:minimize ΨΨ H

Nc

ccc

M3

2

1 0s1/2, m=+ ½Here Ψ is a Slater determinant, a product of single-particle wfns which can be written as a vector :

0s1/2, m=- ½

0p3/2, m=+ 3/2

Find minimum on nuclear energy surface = Hartree-Fock!

INT November 2004 7



There are many ways to derive the random phase approximation (RPA):eqns of motion, time-dependent Hartree-Fock, linear response...

I prefer quantization of the energy surface:

2002

10 ))(()()( xxxVxVxV −′′+≈

ΨHF

We often approximate a potentialby a harmonic oscillator

INT November 2004 8



To parameterize the energysurface, exponentiate the particle-hole operator:

0* ˆˆexp Ψ

∑mi

immi aaZ )(Zr

Ψ=

Thouless’ theoremThen consider

)()(

)(ˆ)()(

ZZ

ZHZZE rr

rrr

ΨΨ

ΨΨ= Hartree-Fock energy

is at the minimum, Z=0

E(Z)This defines an energy surface:

INT November 2004 9



)()(

)(ˆ)()(

ZZ

ZHZZE rr

rrr

ΨΨ

ΨΨ=

( )

+=

****

21

)0(

ZZZZ

E

r

rrr

ABBA

We can expand to quadraticorder about Z=0:

We can treat this as a harmonic oscillatorby replacing numbers Z with (boson) operators b

( )

+=

++

bbbb

E

ˆˆˆˆ

)0(

**21

ABBA

INT November 2004 10



The final step (much math) is to write this in diagonal form:

( )∑ +Ω+− +

λλλλ ββ 2

121 ˆˆ)0( hATrE

∑ +−=mi

mimimi mibYbX ˆˆˆ

,, λλλβusing a Bogoliubov (quasiboson) transformation

Ω=

−− λ

λλ

λ

λ

YX

YX

r

r

hr

r

** ABBAand

The RPAmatrix equation

There is a correction to the Hartree-Fock energy due to “zero-point motion” or, correlations among the nucleons:

( )∑ Ω+−=λ

λ 21

21)0( hATrEERPA



Ami,nj= <m i-1| H | n j-1 >

= <ΨHF| a†i am H a†

n aj| ΨHF>

= matrix elements of H between 1p-1h states

Interpretation of matrix A:

Solving A alone is the Tamm-Dancoff Approximation (TDA)

Bmi,nj= <ΨHF| H | m i-1 n j-1 >

<ΨHF| H a†m ai a†

n aj| ΨHF>

= matrix elements of H between 2p-2h statesand the HF state

Interpretation of matrix B:



RPA as generalized harmonic oscillator

From the h.o. creation/annihilation operators βλ, β†

λ, we can go back to generalized coordinates/momenta Pλ, Qλ

2221

2ˆ

2

ˆλλλ

λ λ

λ QMMPH RPA Ω+=∑

Ω−≈Ψ ∑

λλ

λλ 2

2exp QM

RPAh

The RPA wfn can then be written as a Gaussian....IF Ωλ > 0

There can be zero-frequency modes...in the nuclear landscape, the HF energy does not change in those directions. These arise frombroken symmetries (translation, rotation).

Such zero-frequency modes must be treated with care...



Degenerate nuclear landscapes

translational

rotational

We will revisit this issuein a few minutes...



Historical validation of RPA



Benchmarking RPA

Despite its widespread use, RPA has generally been onlytested against toy models

A typical exampe are Lipkin-type models: Parikh & Rowe, Phys Rev 175 (1968) 1293Hagino & Bertsch, PRC C 61 (2000) 024307

acts like parity conservation

The 2-body interaction promotes or demotes 2 particles at a time2e

Your basic Lipkin model has only 2 independent parameters:N, the number of particles and ratio of 2-body to single-particlespitting, V/ε

N particles, each either up or down...simple quasispin Hamiltonian



Benchmarking RPA

0 0.5 1 1.5(N-1)V/e

-4.4

-4.2

-4

g.s.

ene

rgy

exact"HF""HF+RPA"

I’ll return to this pointlater...



Benchmarking RPA

Other tests of RPA...

two-level pairing Hagino & Bertsch, Nucl. Phys. A679 (2000) 163

schematic interactionin small SM space Ullah & Rowe, Phys. Rev. 188 (1969) 1640.

+ handful of others...



The Shell-Model Contextand a flexible RPA code in the shell model

Diagonalization of a Hamiltonian in a shell-model basis yields “exact” (for that space) and nontrivial numerical results

Let’s compare RPA againstthese numerical shell-modelresults



How a shell-model code works

Shell-model codes, such as OXBASH, ANTOINE, or REDSTICK, writes the Schrodinger eqn as a matrix eigenvalue equations

...selects a valence space...One defines a single-particle basis ...

....puts in valence nucleons...

...possibly assuming an inert core.

Let’s askan expert....




The basis is the set of Slater determinantsof all (sort of) possible configurationsin the valence space

The interaction Hamiltonianis specified as

single-particle energiesplus

two-body matrix elements(the “residual interaction”)

These are read in to the program as a list of numbers




The hard part is actually computing efficientlythe many-body matrix elements from the two-body matrix elements

The final result is the low-lying energy spectrumand the corresponding wavefunctions (the coefficients in the Slater determinant basis)



I have an idea! Let’s write an RPA code usingexactly the same shell-model input!



SHEll-model RPA code (Stetcu PhD LSU 2003)

Shell-model input compatible with REDSTICK:list of single-particle orbits (0s1/2, 0p3/2 etc.)list of two-body matrix elements < ab; JT |H|cd;JT >fair to compare output with REDSTICK results

Fully self-consistent Hartree-Fock:no restrictions on Slater determinant → arbitrary deformations within model space(except, wfns purely real)

Standard RPA:solve matrix RPA equationssee rotation of deformed HF state as zero-frequency modes;option to do pnRPA



The First Element:RPA Correlation Energies

(g.s. Binding energies beyond the mean-field)



Results: Correlation energies

−202468

∆E (

MeV

)

0369

2832

3436

20 2224

28

2426

NeMg

Si SAr

−3

0

3

6

∆E (

MeV

)

0

3

6

1921

2327

2123 23

25

29

25 27 27F

Ne NaMg Al

Si P S Cl

Ar29 31 33 2733

35 3537

Lower energies = HF+RPA correlation energy

Upper energies = HF energy

All energies relative to “exact” SM diagonalization g.s.

Poorest results for single-species calculations (oxygen, calcium isotopes)



Results: Correlation energies

space # nuclides rms err (keV)sd (p+n) 41 870sd oxygen 6 1800pf (p+n) 11 480pf calcium 7 730

Comments:

no obvious correlation of error with:magnitude of deformation (results better for deformed than spherical)goodness of HF state (overlap with exact SM g.s.)

not clear that HFB+QRPA will improve the situation:if we set all pairing matrix elements = 0, results get worse.



The Second Element:Scalar observables

and “restoration” of broken symmetries



In the same way as computing the ground state correlation energy, one can derive RPA corrections to the g.s. expectation value of scalar (“Hamiltonian-like”) operators

Example: g.s. values of < J2 >, < Q2 >, < S2 > or < r2 >

This provides a useful test of RPA... for example, a deformed HF state has < J2 >HF ≠ 0.... is < J2 >HF+RPA = 0 ?

Remember: we expand these operators only to 2nd order



Results: g.s. expectation values

<J2>Nucleus HF RPA SM

20Ne 16.06 -0.45 024Mg 20.13 -2.52 0

23Na 19.42 11.87 3.7525Mg 23.87 14.51 8.75

22Na 25.57 14.57 1246V 39.56 20.00 0

20O 18.46 12.41 022O 0.00 7.99 0

Conclusion: RPA is often an improvement over HF, but not a great improvement

Notice unphysical value!

this is because of truncatingapproximation at 2nd order(not fully treating Pauli principle)

and (rarely) RPA can makeexpectation value worse



Hey! I’m sure I read somewherethat RPA “restores broken

symmetries”! What’s going on?Shouldn’t I get J2 exact?

A: RPA does “restore broken symmetries”but in a restricted fashionSpecifically, it restores the symmetry to the energy landscape

but not in the RPA wavefunction!



Remember the energy landscape

The Hartree-Fock energyis independent of orientation...

since TDA and RPA are expansionin all particle-hole excitations, this independence ought to be reflected as Goldstone modes(zero-energy excitations)

TDA does not yield Goldstone modes,

but RPA does...this is the restoration of the broken symmetry



What about the RPA wavefunction?

The RPA wavefunction is nota symmetry-projected wavefunction...

RPA is a perturbative expansion about the mean-field state...and so RPA only restores the symmetry in the vicinityof the mean-field state, not globally. (Marshalek & Weneser, Ann. Phys. (N.Y.) 53, 569 (1969))

Remember that RPA wavefunction is essentially a Gaussian(harmonic oscillator) so cannot model a global wfn for, say, linear momentum (plane wave) or angular momentum (Wigner D-function)

I have some ideas!

To truly restore a broken symmetry,one needs a global or topological approach....



The Third Element:Transitions

and “sum rules”


ComputationalNuclear StructureRPA transition strengths

0.0

0.1

0.2

0.3

0.4

0.0

0.1

0.2

0.3

Tra

nsiti

on S

tren

gth

/ MeV

0 10 20 30Ex. Energy (MeV)

0.0

0.1

0.1

0.2

0.2

0.2 Isovector E2

GT

SF

“exact” shell modelresults

Notice the “bump” foreach transition: a resonance

RPA results



0.0

0.1

0.2

0.3

0.4

0.0

0.1

0.2

0.3

Tra

nsiti

on S

tren

gth

/ MeV

0 10 20 30Ex. Energy (MeV)

0.0

0.1

0.1

0.2

0.2

0.2 Isovector E2

GT

SF

“exact” shell modelresults

RPA results

0 5 10 15 20Ex. Energy (MeV)

0.00

0.05

0.10

Tra

nsiti

on S

tren

gth

/ MeV

Isoscalar E2

RPA results“missing” strength



0 5 10 15 20Ex. Energy (MeV)

0.00

0.05

0.10

Tra

nsiti

on S

tren

gth

/ MeV

Isoscalar E2

Hey! “Missing strength”? What is this? Isn’t there some sort of energy-weighted

sum rule that should be going here?

“missing” strength


ComputationalNuclear StructureSum Rules and Rule-Breakers

Hey! “Missing strength”? What is this? Isn’t there some sort of energy-weighted

sum rule that should be going here?

Let F be a transition operator; then the energy-weighted sum rule states that

[ ][ ] ∑ Ω=λ

λ λ2

21 ˆ0ˆ,ˆˆ

RPAFHFFHFHF h

This theorem is proven in many text-books...but is wrong!



The “proof” assumes no Goldstone(zero-energy) modes

if one rederives it using those Goldstone modes one gets a correction

[ ][ ] ∑ Ω=λ

λ λ2

21 ˆ0ˆ,ˆˆ

RPAFHFFHFHF h

[ ]∑=Ω

+0,

2ˆ,2

1ν

νν

PFM

Correction term(Stetcu, 2003)



[ ][ ] ∑ Ω=λ

λ λ2

21 ˆ0ˆ,ˆˆ

RPAFHFFHFHF h


if one rederives it using those Goldstone modes one gets a correction

[ ]∑=Ω

+0,

2ˆ,2

1ν

νν

PFM

The missing strength can be interpreted as transitions within a rotational band (that is,

within the intrinsic state) while RPA modelstransitions within a vibrational band

This is bolstered by the fact that we see missing strength (in even-even nuclides)

for E2 transitions but not for, say, spin-flip (∆J=1) transitions

(because rotational band only allows ∆J ≥2 )

Correction term(Stetcu, 2003)




The missing strength can be interpreted as transitions within a rotational band (that is,

within the intrinsic state) while RPA modelstransitions within a vibrational band

In addition, if we choose a spherical state (no Goldstone modes) rather than a deformed state, we regain the missing strength



We also looked at Gamow-Teller transitions in pnRPA

here there have been previous detailed comparisons withthe shell model, but using spherical pnQRPA

the central question: which is more important,

pairing or deformation?



Other’s Previous Work: pn-QRPA

A number of papers compared spherical pn-QRPA against “exact” shell –model calculations of Gamow-Teller strengths

Laurtizen, Nucl Phys A489 (1988) 237. Zhao & Brown, PRC 47 (1993) 2641

Running sum of GT strength



Other’s Previous Work: pn-QRPA

Most likely explanation: pn-QRPA fails to sufficiently smear the Fermi surface insufficient fragmentation of GT strength

QRPA ≈ 2p-2hin spherical SM

Auerbach, Bertsch, Brown & Zhao, Nucl Phys A556 (1993) 190



Our Calculations: Deformed pn-RPA

We redid this work, eschewing pairing correlations in favorof unrestricted deformations

exact shell modelour pn-RPApn-QRPA (Lauritzen)




Clearly deformed pn-RPA is superior in total strength as well as general agreement

8 10 12 14 16 18 20E (MeV)

0

1

2

3

4

Σ B

(GT

)

-5 0 5 10 15 20 25E (MeV)

0

2

4

6

8

β+

β−

26Mg

exact shell modelour pn-RPA

pn-QRPA (Lauritzen)

Σ B

(GT

)




Not only deformation, but triaxiality improves the result



Our Calculations: Deformed pn-RPA2 8

10 15 20E (MeV)

0

1

Σ B

(GT

)

−10 0 10 20 30E (MeV)

0

2

4

6

Σ B

(GT

)β+β−

24Na



The Fourth Element:“Collapse” of RPA

at “phase transitions”



“Collapse” at “phase transitions”

Example from the Lipkin model....

0 0.5 1 1.5(N-1)V/e

-4.4

-4.2

-4

g.s.

ene

rgy

exact"HF""HF+RPA"

“spherical” “deformed”

Egs

“collapse” of RPA




0 0.5 1 1.5(N-1)V/e

-4.4

-4.2

-4

g.s.

ene

rgy

exact"HF""HF+RPA"

“spherical” “deformed”

Egs

At the transition point several things happen:

-- At least one RPA frequency ⇒ 0-- the corresponding h-p amplitudes Ymi ⇒ ∞-- the correlation energy ⇒ - ∞

while other observables go to ±∞

“collapse” of RPA




Do we see this with SHERPA?

We induced a shape transition in 28Si (which normally has an oblate HF state) by lowering the 0d5/2 single-particle energyuntil it became spherical

spherical deformed

No collapse of RPA! What’s going on?




No collapse of RPA! What’s going on?

I know what’s happening! I wrote about it in D.Thouless, Nucl. Phys. 22, 78 (1961)




There are first-order and second-order transitions!

First order: coexistence of stable solutions, no collapse

Second order: no coexistence collapse!!




There are first-order and second-order transitions!

Even-parity transitions (such as quadrupole) should be first order!

First order: coexistence of stable solutions, no collapse

Second order: no coexistence = collapse!!

while odd-parity transitions should be

second order!




Quadrupole shape transitions are first order

Lipkin model is 2nd order and is more analogous to mixing of parity across major shells

0 1 2 3 4 5 6shell separation (MeV)

-22

-21

-20

-19

-18

-17

-16

-15

-14

-13

-12

-11

-10

g.s.

ene

rgy

(MeV

)

exact (SM)HFHF+RPA

"16O"in p1/2-d5/2 space

0 1 2 3 4 5 6shell separation (MeV)

0

5

10

15

20

25

30

35

40

45

50

55

60

< Q

Q >

HFHF+RPA

16Oin p1/2-d5/2 space

Example: 0p1/2-0d5/2 model space displays true “collapse”




Quadrupole shape transitions are first order

28Si in sd shell



CONCLUSIONS

We have realized RPA in a non-trivial shell model framework

Tests of RPA show it to be a modest approximation to the full many-body diagonalization

One should speak carefully of topics such as “symmetry restoration” and collapse in RPA

It may be interested to continue this work onto “extended” RPAand other approaches such as generator-coordinate(which may yield “fast, cheap” solutions with ab initio Hamiltonians)

Nuclear Structure RPA The Shell Model · INT November 2004 12 Computational Nuclear Structure RPA as generalized harmonic oscillator From the h.o. creation/annihilation operators

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