Nuclear Structure

Nuclear StructureErich Ormand

N-Division, Physics and Adv. Technologies DirectorateLawrence Livermore National Laboratory

This work was carried out under the auspices of the U.S. Department of Energy by the University of California, Lawrence Livermore National Laboratory under contract No. W-7405-Eng-48.

This document was prepared as an account of work sponsored by an agency of the United States Government. Neither the United States Government nor the University of California nor any of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise, does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government or the University of California. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States Government or the University of California, and shall not be used for advertising or product endorsement purposes.

15thNational Nuclear Physics15thNational Nuclear PhysicsSummer SchoolSummer School

June 15-27, 2003June 15-27, 2003

Lecture #3

22

Electro-magnetic transitions

• Ok, now what do I do with the states?

• Well, for one excited states decay!

• Electromagnetic decays— Electric multipole (EL)

– J given by |Ji-L| Jf Ji+L

– Parity change (-1)L

— Magnetic multipole (ML)

– J given by |Ji-L| Jf Ji+L

– Parity change (-1)L+1

( ) ( )

0 ;1

r̂1

==

= ∑=

np

A

kkLM

Lkk

ee

YreELMO

€

O MLM( ) = μN gksr s +

2gkl

L +1

r l

⎡

⎣ ⎢

⎤

⎦ ⎥⋅

r ∇rk

LYLMˆ r k( )

k=1

A

∑

gps = 5.6; gn

s = −3.8; gpl =1; gn

l = 0

Free chargesFree charges

Free g-factorsFree g-factors

33

Transition life-times

• Define the “B”-values

• The transition rate is

— Note the important phase-space factor, E2L+1

€

B E(M)L;Ji → J f( ) =ΨJ f

O E(M)L( ) ΨJi

2

2Ji +1

ΨJ fO E(M)L( ) ΨJi =

ψα O E(M)L( ) ψ β

2L +1αβ

∑ ΨJ faα

+ ⊗ ˜ a β[ ]L

ΨJi

€

T E(M)L;Ji → J f( ) =8π L +1( )

L 2L +1( )!![ ]2

Eγ2L +1

h hc( )2L +1 B(E(M)L;Ji → J f )

Radial wave functions go here!Radial wave functions go here!

One-body transition densityOne-body transition density

44

Electromagnetic transitions

• How well does the shell-model work?— Not well at all with free electric charges!— Ok, with free g-factors

• So, where did we do wrong?????!!!!!— Remember we renormalized the interaction

– This accounts for excitations not included in the active valence space

— What about the operators?– We also have to renormalize the transition operators!

– ep~1.3 and en~0.5

– Free g-factors for M1 transitions are not bad (but some renormalization is needed like adding [xY2]1)

• Only with these renormalized (effective) operators, we can get excellent agreement with experiment

55

Estimates for electromagnetic transitions

• Weisskopf estimates

• Assume constant over nuclear volume, zero outside

€

=3

R3 r ≤ R

€

ji O EL( ) j f =e2 rL 2

2 j f +1( ) 2L +1( ) 2 ji +1( )

4π

j f

− 12

L

0

ji

− 12

⎛

⎝ ⎜

⎞

⎠ ⎟1+ −1( )

l f +L +li

2

T ML;Ji = L ± 12 → J f( ) =

2 2 +1( )

L 2L +1( )!![ ]2

3

L + 3

⎛

⎝ ⎜

⎞

⎠ ⎟2

e2

hcR2Lc

Eγ

hc

⎛

⎝ ⎜

⎞

⎠ ⎟

2L +1

=13.3 L +1( )

L 2L +1( )!![ ]2

3

L + 3

⎛

⎝ ⎜

⎞

⎠ ⎟2 1.2Eγ

197.3 MeV

⎛

⎝ ⎜

⎞

⎠ ⎟

2L +1

A2L

3 1021s−1

T EL;Ji = L ± 12 → J f( ) =

1.1 L +1( )

L 2L +1( )!![ ]2

3

L + 3

⎛

⎝ ⎜

⎞

⎠ ⎟2 1.2Eγ

197.3 MeV

⎛

⎝ ⎜

⎞

⎠ ⎟

2L +1

A2L−2

3 1021s−1

66

Estimates for electromagnetic transitions

• Use the Weisskopf estimates to determine

€

T ωL +1( )T ωL( )

= ?

T ML( )T EL( )

= ?

T ML +1( )T EL( )

= ?

77

How big are nuclei?

• Electron scattering— Current-current interaction

• Charge form factor

€

u eγμ ueJμ

Nuc−EM

€

FC q( ) = nip f i

p q( )i

protons

∑ + nin f i

n q( )i

neutrons

∑

f ip(n ) q( ) = r2dr Ri

p(n ) r( )[ ]2M0

p(n ) qr( )0

∞

∫

M0p(n ) qr( ) =

1

4π

GEp(n ) q2 4mN

2( )

1+ q2 4mN2

j0 qr( ) ⎧ ⎨ ⎪

⎩ ⎪

+ GEp(n ) q2 4mN

2( ) − 2GM

p(n ) q2 4mN2

( )[ ]2q2 4mN

2 j1 qr( )qr

σ ⋅l ⎫ ⎬ ⎭

88

How big are nuclei?

• Electron scattering

12C

11B

11CSn

Sp

functions waveradial theofbehavior

asymptotic theDetermines

state

teintermedia fromenergy Separation

CC 12111112μμμ aan

i

Ai

Ai∑ ==+=

0 1 2 3 410-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Data HF HF+SE

FC2(q)

q (fm-1)

0 5 10 15 20

10-5

10-4

10-3

10-2

10-1

Neutron Proton

R0p1/2

(r)

r (fm)

€

FC q = 0( ) → r2

99

Is there anyway to probe the neutrons?

• Yes, again with electron scattering— But we must look to the parity violating part— Neutral current - Z-boson!

• Parity-violating electron scattering also provides a test of the Standard Model

0+

0+

e-

e-

0+

0+

e-

e-

Z

+

1010

The Signal

• PV elastic electron scattering:

— Charges:

– Include intrinsic electric and magnetic form factors

– Neutral current couples more strongly to the neutron distribution

€

ALR =dσ + − dσ −

dσ + + dσ −= −

GFq2

4πα 2

⎛

⎝ ⎜

⎞

⎠ ⎟

FCN q2

( )

FCE q2

( )+

FCs( ) q2( )

FCE q2

( )+ R q( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

FCE(N ) = nμ

pQpE(N ) fμ

p q( )μ

protons

∑ + nμnQn

E(N ) fμn q( )

μ

neutrons

∑

fμp(n ) q( ) =

1

4πr2dr Rμ

p(n ) r( )( )2

j0 qr( )∫R q( ) << 1%

€

QpE =1; Qn

E = 0

QpN = 1− 4sin2 θW( ) ; Qn

N = −1

1111

Nuclear Structure effects

• With isospin symmetry

— No nuclear structure effects – low q 1% measurement of sin2qW

– Deviations are a signature for “new” Physics

– Exotic neutral currents

– Strangeness form factor, FC(s) (higher q)

€

nμp = nμ

n and Rμp r( ) = Rμ

n r( ) hence, fμp q( ) = fμ

n q( )

ALR = ALR0 =

GFq2

πα 2

⎛

⎝ ⎜

⎞

⎠ ⎟ sin2 θW = 3.22 ×10−6q2

Deviations from q2 behavior could signal new Physicsor be due to Nuclear Physics

1212

Nuclear Structure effects

• Isospin-symmetry is broken— Coulomb interaction (larger)— strong interaction (smaller): isotensor or charge-dependent interaction

v(pn) - (v(pp) + v(nn))/2

• Mix states with Tmax=2

• Effect on observables:

€

ALR = ALR0 q( ) 1+ Γ q( )( ) Γ q( ) → 0, q → 0

( ) ( ) ( )

( ) 22

22

3

2

2

1

1

2

3 1

2

1

rmrV

R

eZr

R

eZrV

N

C

ω=

−+

−−=

0 5 10 15 20

10-5

10-4

10-3

10-2

10-1

Neutron Proton

R0p1/2

(r)

r (fm)Valence statesnm

p nmn

1p-1h ExcitationsRm

p Rmn

Hartree-Fock

0s

0p

0d-1s

0f-1p

0s

0p

0d-1s

0f-1p

1313

Nuclear Structure effects

• Correction due to breaking of isospin symmetry

— Overall agreement with recent ab initio calculation (Navratil and Barrett)

— G(q) < 1% for q < 0.9 fm-1 (1% measurement for 0.3 < q < 1.1 fm-1, Musolf & Donnelly, NPA546, 509 (1992).

( ) ( )( )qqAA LRLR Γ+= 1 0

(q) < 1% for q < 0.9 fm-1 and q=2.4± 0.1 fm-1

0 1 2 3 40

2

4

6

Shell Model Result Standard Model

ALR

(x10

5)

q (fm-1)

0 1 2 3 410-3

10-2

10-1

100

101

102

|( )| (%)q

(q fm-1)

1414

Neutron Radii

• What good is parity violation?— Assume Standard Model correct to 1% level - infer neutron distribution

• Very little precision data regarding the distribution of neutrons

— Useful for mean-field models - improve extrapolation to the drip line— Hadron scattering - strong in-medium effects

e-

e-

e-

e-

Z

€

ALR ≈GFq2

4πα 2

⎛

⎝ ⎜

⎞

⎠ ⎟

Z − N

Z− 4sin2 θW + q2 1

6

N

Zrn

2 − rp2

( ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥

Experiments planned for 208Pb at TJNAF

1515

The weak interaction in the shell model

• -decay and neutrino absorption— -decay

– Partial half-life

– Fermi (F):

– Gamow-Teller (GT):

– gA=1.2606 0.0075

– GT is very dependent on model space and shell-model interaction– Spin-orbit and quasi SU(4) symmetry

– Meson-exchange currents modify B(GT)– For an effective operator, GT must be renormalized: multiply by ~ 0.75

– Total half life: Branching ratio:

€

f = dWW (W −1)1/ 2(W0 −W )2 Fcorr (W )1

W0

∫ ∝W05

€

t1/ 2 =1

t1/ 2i→ f

f

∑

€

BR i→ f =t1/ 2

t1/ 2i→ f

€

t1 2i→ f 6170 ± 4s

B F( )i→ f+ gA

2B GT( )i→ f( ) f i→ f

€

B(F) =1

2Ji +1Ψ f t± Ψi

2= T T +1( ) − TziTzf[ ]δ if

€

B(GT) =1

2Ji +1gA

2 Ψ f σt± Ψi

2

1616

The weak interaction in the shell modelIsospin-symmetry violation

• Isospin is approximately conserved (~ 1% level)• For transitions, isospin violation enters in two places

— One-body transition density as no longer has good isospin

— One-body matrix element

– Note we have a proton(neutron) converted to a neutron(proton)

– Due to the Coulomb interaction protons and neutrons have different radial wave functions, so we need the overlap:

– Important for high-precision tests of the vector current (~0.4%)

– For GT this effect can be large, and is essentially contained in the global factor of 0.75 obtained empirically

– Mirror transitions are no longer the same!!!

τατα ±± or

if aa +α

12 <=∫ pnRRdrrN αα

1717

Superallowed Fermi -decay

• Test of the Standard Model

• Cabibbo-Kobayashi-Maskawa matrix

€

ft1 2 =K

GV2 MF

2

f = dWpW W0 −W( )1

W0

∫

MF = f τ + i = 2 for T =1

€

MF

2= 2 1−δc( )

δC = δIM + δRO

δIM : ψ i = α iψ i,T + β j , ′ T ψ j, ′ T

j, ′ T

∑

r2drRn r( )∫ Rp r( ) <1

δRO ≈1− r2drRn r( )∫ Rp r( ) <1

1

222 =++

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

ubusud

S

S

S

tbtstd

cbcscd

ubusud

W

W

W

vvv

bsd

vvvvvvvvv

bsd

1818

Superallowed Fermi -decay

0 5 10 15 20 25 300.0

0.2

0.4

0.6 δRO δC

δ (%)

Z

3150

3160

3170 fRt1/2 Ft1/2

ft1/2( )s

δδCC for for 1010C: 0.15(9)% C: 0.15(9)%

Ab initio calculationAb initio calculation

Unitarity condition: 0.9956(8)stat(7)sys

Current SituationCurrent Situation

0.1211 0.0933 0.0654 0.0466 0.0231 (%)

8 6 4 2 0

CδΩΩΩΩΩ hhhhh

1919

Neutrino absorption

• As for -decay, neutrino absorption requires Fermi and Gamow-Teller matrix elements

• We carried out calculations for 23Na and 40Ar— Ormand et al., PLB308, 207 (1993), Ormand et al., PLB345, 343 (1995)— ICARUS and proposed bolometric detectors

• For 40Ar Gamow-Teller is very important— Total is twice as large as Fermi contribution— Counter to original design assumption

• Can we trust the calculation?— -decay of analog

€

ν e +ZAA→Z +1

A A + e−

€

i Eν( ) =GV

2

π 3ch4B F( )0→i

+ gA2B GT( )0→i[ ]Ee

i kei

2020

Checking the calculation

• For 40Ar, look at -decay of 40Ti— -delayed proton emitter

• Calculated half-life: 55 5 ms— Exp: 52.7 1.5 ms and 54 2 ms

0 1 2 3 4 5 6 7 8 9 10

Excitation Energy (MeV)

0

10

20

30

40

Branching Ratio (%)

Gamow-Teller

Fermi

2121

Checking the calculation

• But there are problems with B(GT) strength

W. Liu et al., PRC58, 2677 (1998)W. Liu et al., PRC58, 2677 (1998)

=14.3=14.30.30.31010-43-43 cm cm

M. Bhattacharya et al., PRC58, 3677 (1998)M. Bhattacharya et al., PRC58, 3677 (1998)

=14.0=14.00.30.31010-43-43 cm cm

Theory:Theory: =11.5=11.50.70.71010-43-43 cm cm

There is no substitute for experiment when available

2222

What about heavier nuclei?

• Above A ~ 60 or so the number of configurations just gets to bed too large ~ 1010!

• Here, we need to think of more approximate methods

• The easiest place to start is the mean-field of Hartree-Fock— But, once again we have the problem of the interaction

– Repulsive core causes us no end of grief!!

– We will, at some point use effective interactions like the Skyrme force

2323

Hartree-Fock

• There are many choices for the mean field, and Hartree-Fock is one optimal choice

• We want to find the best single Slater determinant 0 so that

• Thouless’ theorem— Any other Slater determinant not orthogonal to 0 may be written as

— Where i is a state occupied in 0 and m is unoccupied

— Then

€

0 H Φ0

Φ0 Φ0

= minimum δΦ0 H Φ0 + Φ0 H δΦ0 = 0

Φ0 = ai+ 0

i=1

A

∏

€

=exp Cmiam+ ai

mi

∑ ⎡

⎣ ⎢

⎤

⎦ ⎥Φ0

€

δ0 = Cmiam+ ai Φ0

im

∑ and Φ0 δΦ0 = Cmi Φ0 am+ ai Φ0 = 0

im

∑

2424

Hartree-Fock

• Let i,j,k,l denote occupied states and m,n,o,p unoccupied states

• After substituting back we get

• This leads directly to the Hartree-Fock single-particle Hamiltonian h with matrix elements between any two states α and

€

H =pi

2

mi

∑ +1

2V ri − r j( )

ij

∑ = T +1

2Vij

ij

∑

€

α h β = α T β + jα V jβA

j

occupied

∑

= α T β + α U β

€

m T i + jm V jiA

j

occupied

∑ = 0

2525

Hartree-Fock

• We now have a mechanism for defining a mean-field— It does depend on the occupied states— Also the matrix elements with unoccupied states are zero, so the first

order 1p-1h corrections do not contribute

• We obtain an eigenvaule equation (more on this later)

• Energies of A+1 and A-1 nuclei relative to A

€

m T i + jm V jiA

j

occupied

∑ = m h i = 0

€

h i = ε i i

E = Φ0 H Φ0 = i T i +1

2ij V ij

Aij

∑i

∑ =1

2i T i + ε i[ ]

i

∑

iAAmAA EEEE εε −=−=− −+ 11

2626

Hartree-Fock – Eigenvalue equation

• Two ways to approach the eigenvalue problem— Coordinate space where we solve a Schrödinger-like equation— Expand in terms of a basis, e.g., harmonic-oscillator wave function

• Expansion— Denote basis states by Greek letters, e.g., α

— From the variational principle, we obtain the eigenvalue equation

€

i = Ciα αα

∑

Ciα* C jα

α

∑ = δij Ciα* Ciβ

i

∑ = δαβ

€

α T β + αj V βjA

j

occupied

∑ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥Ciβ

β

∑ = ε iCiα

or

α h β Ciβ

β

∑ = ε iCiα

2727

Hartree-Fock – Solving the eigenvalue equation

• As I have written the eigenvalue equation, it doesn’t look to useful because we need to know what states are occupied

• We use three steps1. Make an initial guess of the occupied states and the expansion

coefficients Ciα

• For example the lowest Harmonic-oscillator states, or a Woods-Saxon and Ciα=δiα

2. With this ansatz, set up the eigenvalue equations and solve them

3. Use the eigenstates |i from step 2 to make the Slater determinant 0, go back to step 2 until the coefficients Ciα are unchanged

The Hartree-Fock equations are solved self-consistently

2828

Hartree-Fock – Coordinate space

• Here, we denote the single-particle wave functions as i(r)

• These equations are solved the same way as the matrix eigenvalue problem before

1. Make a guess for the wave functions i(r) and Slater determinant 0

2. Solve the Hartree-Fock differential equation to obtain new states i(r)

3. With these go back to step 2 and repeat until i(r) are unchanged

€

−h2

2m∇1

2φi r1( ) + φ j* r2( )V r1 − r2( )φ j r2( )d3r2∫

j

occupied

∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟φi r1( ) − φ j

* r2( )V r1 − r2( )φ j r1( )∫j

occupied

∑ φi r2( )d3r2 = ε iφi r1( )

Direct or Hartree term: UDirect or Hartree term: UHHExchange or Fock term: UExchange or Fock term: UFF

Again the Hartree-Fock equations are solved self-consistently

2929

Hartree-Fock

• M. Moshinsky, Am. J. Phys. 36, 52 (1968). Erratum, Am. J. Phys. 36, 763 (1968).

• Two identical spin-1/2 particles in a spin singlet interact via the Hamiltonian

• Use the coordinates and to show the exact energy and wave function are

• Note that since the spin wave function (S=0) is anti-symmetric, the spatial wave function is symmetric

Hard homework problem:

€

H = 12 p1

2 + r12

( ) + 12 p2

2 + r22

( ) + χ 12

r r 1 −

r r 2( )

2

[ ]

( ) 2rrr 21

rrr−= ( ) 2rrR 21

rrr+=

€

E = 32 h 1+ 2χ +1( )

Ψ r,R( ) =1

πh( )3 4 exp −R2 2h( )

2χ +1

πh

⎛

⎝ ⎜

⎞

⎠ ⎟

3 4

exp −2χ +1

2hr2

⎛

⎝ ⎜

⎞

⎠ ⎟

3030

Hartree-Fock

• The Hartree-Fock solution for the spatial part is the same as the Hartree solution for the S-state. Show the Hartree energy and radial wave function are:

Hard homework problem:

€

E H = 3h χ +1

Ψ r1,r2( ) =χ +1

πh

⎛

⎝ ⎜

⎞

⎠ ⎟

3 2

exp −χ +1

2hr1

2 ⎛

⎝ ⎜

⎞

⎠ ⎟exp −

χ +1

2hr2

2 ⎛

⎝ ⎜

⎞

⎠ ⎟

0 5 10 15 20 25 300

5

10

15

20

E/h

χ

Eexact

EH

3131

Hartree-Fock with the Skyrme interaction

• In general, there are serious problems trying to apply Hartree-Fock with realistic NN-interactions (for one the saturation of nuclear matter is incorrect)

• Use an effective interaction, in particular a force proposed by Skyrme

— P is the spin-exchange operator

• The three-nucleon interaction is actually a density dependent two-body, so replace with a more general form, where α determines the incompressibility of nuclear matter

€

v12 = t0 1+ x0Pσ( )δr r 1 −

r r 2( ) +

1

2t1 1+ x1Pσ( ) δ

r r 1 −

r r 2( )

1

2i

r ∇12 −

r ∇ 22

( ) +1

2i

s ∇12 −

s ∇ 22

( )δr r 1 −

r r 2( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥+

t2 1+ x2Pσ( )1

2i

s ∇1 −

s ∇ 2( ) ⋅δ

r r 1 −

r r 2( )

1

2i

r ∇1 −

r ∇ 2( ) + W0

r σ 1 +

r σ 2( ) ⋅

1

2i

s ∇1 −

s ∇ 2( ) ×δ

r r 1 −

r r 2( )

1

2i

r ∇1 −

r ∇ 2( ) +

t3δr r 1 −

r r 2( )δ

r r 2 −

r r 3( )

€

1

6t3 1+ x3Pσ( )δ

r r 1 −

r r 2( )ρ α r

r 1 +r r 2( ) 2( )

3232

Hartree-Fock with the Skyrme interaction

• One of the first references: D. Vautherin and D.M. Brink, PRC5, 626 (1972)

• Solve a Shrödinger-like equation

— Note the effective mass m*

— Typically, m* < m, although it doesn’t have to, and is determined by the parameters t1 and t2

– The effective mass influences the spacing of the single-particle states

– The bias in the past was for m*/m ~ 0.7 because of earlier calculations with realistic interactions

€

− r

∇ ⋅h2

2mτ z

* r r ( )

r ∇ + Uτ z

r r ( ) − i

r W τ z

r r ( ) ⋅

r ∇ ×

r σ ( )

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥φτ z

i r r ( ) = ε iφτ z

i r r ( ) ττzz labels protons or labels protons or

neutronsneutrons

€

h2

2mτ z

* r r ( )

=h2

2m+

1

4t1 + t2( )ρ

r r ( ) +

1

8t2 − t1( )ρ τ z

r r ( )

3333

Hartree-Fock calculations

• The nice thing about the Skyrme interaction is that it leads to a computationally tractable problem

— Spherical (one-dimension)— Deformed

– Axial symmetry (two-dimensions)

– No symmetries (full three-dimensional)

• There are also many different choices for the Skyrme parameters— They all do some things right, and some things wrong, and to a large

degree it depends on what you want to do with them— Some of the leading (or modern) choices are:

– M*, M. Bartel et al., NPA386, 79 (1982)

– SkP [includes pairing], J. Dobaczewski and H. Flocard, NPA422, 103 (1984)

– SkX, B.A. Brown, W.A. Richter, and R. Lindsay, PLB483, 49 (2000)

– Apologies to those not mentioned!

— There is also a finite-range potential based on Gaussians due to D. Gogny, D1S, J. Dechargé and D. Gogny, PRC21, 1568 (1980).

• Take a look at J. Dobaczewski et al., PRC53, 2809 (1996) for a nice study near the neutron drip-line and the effects of unbound states

3434

Nuclear structure

• Remember what our goal is:— To obtain a quantitative description of all nuclei within a microscopic

frame work— Namely, to solve the many-body Hamiltonian:

€

H =r p i

2

2mi

∑ + VNN

r r i −

r r j( )

i< j

∑

€

H =r p i

2

2m+ U ri( )

⎛

⎝ ⎜

⎞

⎠ ⎟

i

∑ + VNN

r r i −

r r j( )

i< j

∑ − U ri( )i

∑

Residual interactionResidual interaction

Perturbation TheoryPerturbation Theory

3535

Nuclear structure

• Hartree-Fock is the optimal choice for the mean-field potential U(r)!— The Skyrme interaction is an “effective” interaction that permits a wide

range of studies, e.g., masses, halo-nuclei, etc.— Traditionally the Skyrme parameters are fitted to binding energies of

doubly magic nuclei, rms charge-radii, the incompressibility, and a few spin-orbit splittings

• One goal would be to calculate masses for all nuclei— By fixing the Skyrme force to known nuclei, maybe we can get 500 keV

accuracy that CAN be extrapolated into the unknown region– This will require some input about neutron densities – parity-violating

electron scattering can determine <r2>p-<r2>n.

— This could have an important impact

3636

Hartree-Fock calculations

• Permits a study of a wide-range of nuclei, in particular, those far from stability and with exotic properties, halo nuclei

H. Sagawa, PRC65, 064314 (2002)H. Sagawa, PRC65, 064314 (2002) The tail of the radial density The tail of the radial density depends on the separation depends on the separation energyenergyS. Mizutori et al. PRC61, 044326 (2000)S. Mizutori et al. PRC61, 044326 (2000)

Drip-line studiesDrip-line studiesJ. Dobaczewski et al., PRC53, 2809 (1996)

3737

What can Hartree-Fock calculations tell us about shell structure?

• Shell structure— Because of the self-consistency, the shell structure can change from

nucleus to nucleus

J. Dobaczewski et al., PRC53, 2809 (1996)

As we add neutrons, traditional shell closures are changed, and may even disappear!

This is THE challenge in trying to predict the structure of nuclei at the drip lines!

3838

Beyond mean field

• Hartee-Fock is a good starting approximation— There are no particle-hole corrections to the HF ground state

— The first correction is

• However, this doesn’t make a lot of sense for Skyrme potentials— They are fit to closed-shell nuclei, so they effectively have all these

higher-order corrections in them!

• We can try to estimate the excitation spectrum of one-particle-one-hole states – Giant resonances

— Tamm-Dancoff approximation (TDA)— Random-Phase approximation (RPA)

0==+ ∑ ihmjiVjmiTmoccupied

jA

∑ −−+ijmn nmji

AAijVmnmnVij

εεεε4

1

You should look these up!You should look these up!A Shell Model Description of Light Nuclei, I.S. TownerA Shell Model Description of Light Nuclei, I.S. TownerThe Nuclear Many-Body Problem, Ring & SchuckThe Nuclear Many-Body Problem, Ring & Schuck

3939

Nuclear structure in the future

Ab initio methods

Standard shell modelMean-fie

ld, Hartre

e-Fock

Monte Carlo Shell Model

With newer methods and powerful computers, the future of nuclear structure theory is bright!