15thNational Nuclear Physics
Summer School
June 15-27, 2003
Nuclear Structure
Erich Ormand
N-Division, Physics and Adv.
Technologies Directorate
Lawrence Livermore National
Laboratory
Lecture #3
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.
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.
Electro-magnetic transitions
•
•
•
Ok, now what do I do with the states?
Well, for one excited states decay!
Electromagnetic decays
— Electric multipole (EL)
A
OELM    ek rkLYLM r̂k 
k 1
e p  1;
en  0
Free charges
– DJ given by |Ji-L|  Jf  Ji+L
– Parity change (-1)L
— Magnetic multipole (ML)
 s
2gkl  L
OMLM    N gk s 
l  rk YLM rˆk 
L  1 

k1
A
g sp  5.6;
gns  3.8;
g lp  1;
gnl  0
– DJ given by |Ji-L|  Jf  Ji+L
– Parity change (-1)L+1

2
Free g-factors
Transition life-times
•
Define the “B”-values
BE(M)L;J i  J f 
J f OE(M)L Ji
2Ji  1
J f OE(M)L Ji  

•

2
Radial wave functions go here!
 OE(M)L  
2L  1
J f a  a˜   Ji
L
The transition rate is
T E(M)L;Ji  J f 
One-body transition density
8 L  1
L2L  1!!
2
Eg2L 1
 c
— Note the important phase-space factor, Eg2L+1
3
2L 1
B(E(M)L;Ji  J f )
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 [sxY2]1)
•
Only with these renormalized (effective) operators, we can get
excellent agreement with experiment
4

Estimates for electromagnetic transitions
•
•
Weisskopf estimates
Assume  constant over nuclear volume, zero outside

3
R3
rR
j i OEL j f 

2
e r
L 2
2 j
f
 12L  12 j i  1  j f L j i 1 1l f L li
 1
1 
4
2
 2 0  2 
2L 1
2 2


E


3
e 2L
g
T ML;J i  L  12  J f 
R
c




2


L

3
c
c


L2L  1!!
22  1
2L 1
2
13.3L  1  3   1.2E g 

 

2 


L

3
197.3
MeV


L2L  1!!
2L
3
A 10 21 s1
2L 1 2L2
2


1.2E


1.1
L

1
3


g
21 1
T EL;J i  L  12  J f 
 
 A 3 10 s
2 
L2L  1!! L  3  197.3 MeV 
5
Estimates for electromagnetic transitions
•
Use the Weisskopf estimates to determine
T L  1
?
T L 
T ML 
?
T EL
T ML  1
?
T EL
6
How big are nuclei?
•
Electron scattering
ueg  ue JNucEM
— Current-current interaction
•
Charge form factor
FC q 
protons
n
p
i
f i q
neutrons
p
n
i
fi
p(n )
q 
n
i
f in q
i

 r 2drRip(n ) r M 0p(n ) qr
2
0
M
p(n )
0
1
4
qr 

G

GEp(n ) q 2 4mN2 
j 0 qr

2
2

 1 q 4mN
p(n )
E
q
2
4m
2
N
 2G q
p(n )
M
2

7
4m
2
N

2q 2 4mN2 j1 qr
s  
qr


How big are nuclei?
•
Electron scattering
-1
10
-2
10
1/2
R0p (r)
-3
10
-4
10
Neutron
Proton
0
10
Data
HF
HF+SE
-1
10
-5
10
0
5
10
15
20
r (fm)
-2
2
FC (q)
10
11C
-3
10
11B
-4
10
12C
-5
10
n  
-6
10
-7
10
Sn
Sp
12
C a  iA 11  iA 11 a
12
C
i
0
1
2
3
4
Separation energy from intermediate
state
-1
q (fm )
FC q  0  r
Determines the asymptotic
behavior of the radial wave functions
2
8
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!
0+
0+
e-
e-
+
Z
0+
•
e-
0+
e-
Parity-violating electron scattering also provides a test of the
Standard Model
9
The Signal
•
PV elastic electron scattering:
ALR

 GF q 2 FCN q 2  FCs q 2 
ds   ds 

 
 E 2  Rq

E
2


ds   ds 
4

2
F
q

 C   FC q 

protons
E(N )
C
F

 n Q
p
neutrons
E(N ) p
p

f

1
4
Rq  1%
f p(n ) q 
— Charges:
q   nnQnE(N ) f n q

 r drR r j qr
2
p(n )
2
0
QpE  1;
QnE  0
QpN  1 4 sin 2 W  ; QnN  1


10
– Include intrinsic electric and
magnetic form factors
– Neutral current couples more
strongly to the neutron distribution
Nuclear Structure effects
•
With isospin symmetry
n p  n n and Rp r  Rn r hence, f p q  f n q
ALR  A
0
LR
 GF q 2  2
6 2
 
 sin W  3.22 10 q
 2 
— 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)
Deviations from q2 behavior could signal new Physics
or be due to Nuclear Physics
11
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 DTmax=2
0f-1p
1p-1h Excitations
Rmp  Rmn
Hartree-Fock
0d-1s
0p
0s
1 Z  1 e 2 2 3 Z  1 e 2
VC r   
r 
2
R3
2
R
1
VN r   m 2 r 2
2
-1
10
-2
10
1/2
R0p (r)
-3
10
-4
10
Neutron
Proton
-5
10
0f-1p
0
Valence states
nmp  nmn
0d-1s
0p
0s
•
Effect on observables:
0
ALR  ALR
q 1 q
12
5
10
r (fm)
q  0, q  0
15
20
Nuclear Structure effects
•
Correction due to breaking of isospin symmetry
0
q  1  q 
ALR  ALR
6
2
10
1
10
4
|(q)| (%)
5
ALR (x10 )
Shell Model Result
Standard Model
2
0
10
-1
10
-2
10
0
-3
10
0
1
2
3
4
0
1
2
3
4
-1
— Overall agreement
q (fmand
) Barrett)
q (fm ) with recent ab initio calculation (Navratil
— G(q) < 1% for q < 0.9 fm-1 (1% measurement for 0.3 < q < 1.1 fm-1, Musolf &
Donnelly, NPA546, 509 (1992).
-1
(q) < 1% for q < 0.9 fm-1 and q=2.4± 0.1 fm-1
13
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
eee-
ALR
Z
 GF q 2 Z  N
2
2 1 N
2
2 
 

4
sin


q
r

r

W
n
p 

6Z
4  2  Z

e-
Experiments planned for 208Pb at TJNAF

14

The weak interaction in the shell model
•
-decay and neutrino absorption
— -decay
– Partial half-life
t1i2 f

6170  4s

BFi f  gA2 BGT i f f i f
W0
f   dWW (W 1)1/ 2 (W 0 W )2 Fcorr (W ) W 05
1
2
1
f t  i  T T  1  TziTzf if
2Ji 1
2
1
gA2 f st  i
– Gamow-Teller (GT): B(GT)
2Ji  1
– Fermi (F): B(F) 
– 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:
t1/ 2  
f
1
Branching ratio:
i f
1/ 2
t
15
BR i f 
t1/ 2
t1/i2 f
The weak interaction in the shell model
Isospin-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
 f a a i
— One-body matrix element
    or  s  
– 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:
N   r 2 drRn Rp  1
– 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!!!
16

Superallowed Fermi -decay
•
Test of the Standard Model
ft1 2 
M F  21 c 
2
K
2
GV2 M F
C  IM  RO
f   dWpW W 0  W 
j,T 
 r drR rR r  1
  1  r drR rR r  1
2
1
n
M F  f   i  2 for T  1
•
 i   i i,T    j,T  j,T 
IM :
W0
2
RO
Cabibbo-Kobayashi-Maskawa matrix
 dW   vud
  
 sW    vcd
b  v
 W   td
vub   d S 
 
vcb   sS 
vtb   bS 
vus
vcs
vts
vud  vus  vub  1
2
p
2
2
17
n
p
Superallowed Fermi -decay
Current Situation
Ab initio calculation
C for 10C: 0.15(9)%
fRt1/2
Ft1/2
ft1/2(s)
3170
3160
3150
 (%)
0.6
RO
C
0.4
0.2
0
2
4
6
8
 C (%) 0.0231 0.0466 0.0654 0.0933 0.1211
0.0
0
5
10
15
20
25
30
Z
Unitarity condition: 0.9956(8)stat(7)sys
18
Neutrino absorption
 e  ZA AZ 1A A  e
GV2
s i E   3
 c
•

•
4
BF
2
i i

g
B
GT
E
ke



A
e
0i
0i
As for -decay, neutrino absorption requires Fermi and GamowTeller 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
19
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
40
Branching Ratio (%)
Gamow-Teller
Fermi
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
Excitation Energy (MeV)
20
Checking the calculation
•
But there are problems with B(GT) strength
Theory: s=11.50.710-43 cm
W. Liu et al., PRC58, 2677 (1998)
M. Bhattacharya et al., PRC58, 3677 (1998)
s=14.00.310-43 cm
s=14.30.310-43 cm
There is no substitute for experiment when available
21
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
22
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 F0 so that
F0 H F0
 minimum
F0 F0
F0 H F0  F0 H F0  0
A
F0   ai 0
•
i1
Thouless’ theorem
— Any other Slater determinant F not orthogonal to F0 may be written as



F  exp Cmiam ai F0
mi


— Where i is a state occupied in F0 and m is unoccupied
— Then

F0  Cmiam ai F0
and F0 F0  Cmi F0 am ai F0  0
im
im
23
Hartree-Fock


pi2 1
1
H    V ri  r j  T  Vij
2 ij
2 ij
i m
•
•
Let i,j,k,l denote occupied states and m,n,o,p unoccupied states
After substituting back we get

occupied
mTi 

jm V ji
A
0
j
•
This leads directly to the Hartree-Fock single-particle Hamiltonian h
with matrix elements between any two states  and 

h  T 
occupied

j V j
j
 T  U

24
A
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
occupied

mTi 
jm V ji
A
 m hi 0
j
•
We obtain an eigenvaule equation (more on this later)

h i  i i
E  F0 H F0   i T i 
i
•

1
ij V ij

2 ij
A

1
i T i  i 


2 i
Energies of A+1 and A-1 nuclei relative to A
E A1  E A   m
E A1  E A   i
25
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., 
i   Ci 

C C   
*
i
j
 C  C   
*
i
ij

i
i
— From the variational principle, we obtain the eigenvalue equation

occupied

 T    j V j
 
j
or
  h  C C
i
i
i

26

Ci  iCi
A


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 steps
1. Make an initial guess of the occupied states and the expansion
coefficients Ci
•
For example the lowest Harmonic-oscillator states, or a Woods-Saxon and
Ci=i
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 F0,
go back to step 2 until the coefficients Ci are unchanged
The Hartree-Fock equations are solved self-consistently
27
Hartree-Fock – Coordinate space
•
Here, we denote the single-particle wave functions as fi(r)
occupied
occupied

*
3

 f r1   
f j r2 V r1  r2 f j r2 d r2 
f i r1    f *j r2 V r1  r2 f j r1 f i r2 d 3 r2  if i r1




2m
 j

j
2
2
1 i
Exchange or Fock term: UF
Direct or Hartree term: UH
•
These equations are solved the same way as the matrix eigenvalue
problem before
1. Make a guess for the wave functions fi(r) and Slater determinant F0
2. Solve the Hartree-Fock differential equation to obtain new states fi(r)
3. With these go back to step 2 and repeat until fi(r) are unchanged
Again the Hartree-Fock equations are solved self-consistently
28
Hartree-Fock
Hard homework problem:
•
•
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
H  12 p12  r12  12 p22  r22  
•



1
2
r1  r2  
2




Use the coordinates r  r1  r2  2 and R  r1  r2  2 to show the exact
energy and wave function are

E  32 1 2  1


 2  1 
 2   1 
1
2
r,R 
exp  R 2 
r 2 
 exp
34
2
 
 



34
•
Note that since the spin wave function (S=0) is anti-symmetric, the
spatial wave function is symmetric

29
Hartree-Fock
Hard homework problem:
•
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:
EH  3
 1
   1 3 2
   1 2     1 2 
r1,r2   
r1 exp
r2 
 exp

2
2



 

20
E/h
15

10
Eexact
EH
5
0
0
5
10
15

20
25
30
30
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


1
1
1 2
v12  t 0 1 x 0 Ps  r1  r2   t11 x1Ps  r1  r2  12  22 
1  22  r1  r2 


2
2i
2i
1
1
1
1
t 2 1 x 2 Ps  1  2   r1  r2  1  2  W 0 s1  s2  
1  2   r1  r2  1  2 
2i
2i
2i
2i
t 3 r1  r2  r2  r3 





 




— Ps is the spin-exchange operator
•
The three-nucleon interaction is actually a density dependent twobody, so replace with a more general form, where  determines the
incompressibility of nuclear matter
1
t3 1 x 3Ps r1  r2  r1  r2  2
6
31

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
2


i
i



U
r

iW
r


s
f
r


f






r 



i

*
z
z
z
z
2m z r 






z labels protons
or neutrons
— Note the effective mass m*
2

2m*z r 

2
2m

1
1
t1  t 2 r   t 2  t1 z r 
4
8
— 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
32
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
33
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:
pi2
H 
 VNN ri  rj 
i 2m
i j
 pi2

H    U ri  VNN ri  rj  U ri 
2m
 i j
i 
i
Residual interaction
Perturbation Theory

34
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
35
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)
The tail of the radial density
depends on the separation
energy
S. Mizutori et al. PRC61, 044326 (2000)
36
Drip-line studies
J. Dobaczewski et al., PRC53, 2809
(1996)
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
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!
J. Dobaczewski et al., PRC53, 2809 (1996)
37
Beyond mean field
•
Hartee-Fock is a good starting approximation
— There are no particle-hole corrections to the HF ground state
mT i 
occupied

jm V ji
A
 mhi 0
j
— The first correction is
ij V mn A mn V ij
1

4 ijmn  i   j   m   n
•
A
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-onehole states – Giant resonances
— Tamm-Dancoff approximation (TDA)
— Random-Phase approximation (RPA)
38
You should look these up!
A Shell Model Description of Light Nuclei, I.S. Towner
The Nuclear Many-Body Problem, Ring & Schuck
Nuclear structure in the future
Monte Carlo
Shell Model
Standard shell model
With newer methods and
powerful computers, the future
of nuclear structure theory is
bright!
Ab initio methods
39