a close up of the spinning nucleus s. frauendorf department of physics university of notre dame, usa...
TRANSCRIPT
A close up of the spinning nucleus
S. Frauendorf
Department of Physics
University of Notre Dame, USA
IKH, Forschungszentrum Rossendorf
Dresden, Germany
How is the nucleus rotating?
What is rotating?
The nuclear surface
Nucleons are not on fixed positions.
Collective model accounts for the appearance of rotational bands E I(I+1), Alaga rules for e.m. transitions andmany more phenomena.
Bohr and Mottelson
2
HI+small arrays
HI+large arrays
Decay+detector
Collectiverotation
Interplaybetweencollective and sp.degrees offreedom
Nucleonicorbitals –gyroscopes
Spinning clockwork of gyroscopes
Nucleonicorbitals –gyroscopes
3
Aspects of the close up
• How does orientation come about?
• How is angular momentum generated?
• Examples: magnetic rotation, band termination and recurrence
• Weak symmetry breaking at high spin
• Examples: reflection asymmetry, chirality
4
How does orientation come about?
Orientation of the gyroscopes
Deformed density / potential
Deformed potential aligns thepartially filled orbitals
Partially filled orbitals are highly tropic
Nuclus is oriented – rotational band
Well deformed Hf174 -90 0 90 180 2700.0
0.2
0.4
0.6
0.8
1.0
over
lap
5
How is angular momentum generated?
Moving massesor currents in a liquidare not too useful concepts HCl
rigid
irrotational
Myth: Without pairing the nucleus rotates like a rigid body. 6
Angular momentum is generated by alignment of the spin of the orbitals with the rotational axisGradual – rotational bandAbrupt – band crossing, no bands
Microscopic cranking Calculations do well inreproducing the momentsof inertia.With and without pairing.
Moments of inertia for I>20 (no pairing) differ strongly from rigid body value
M. A. Deleplanque et al. Phys. Rev. C69, 044309 (2004)
7
Magnetic Rotation
-90 0 90 180 2700.0
0.2
0.4
0.6
0.8
1.0
over
lap
Weakly deformed Pb199
8
TAC
Long transverse magnetic dipole vectors, strong B(M1)
The shears effect
9
2
2/132/92/13
ihit qqqQ
Better data needed for studying interplay between shape ofpotential and orientation of orbitals.
10
Terminating bandsA. Afanasjev et al. Phys. Rep. 322, 1 (99)
Orientation of the gyroscopes
Deformed density / potential
11
Instability after termination
After termination, several alignments, substantial rearrangement of orbitals
Coexistence of sd, hd, with wd
new shape, bandsinstability
M. RileyE. S..Paulet al.@Gammasphere
Calculations:I. Ragnarsson
termination
12
Symmetries at high spin
Combination of Shape (time even)With Angular momentum (time odd)
Determine the parity-spin-multiplicity sequence of the bands
13
Th223
Parity doubling
Best case of reflection asymmetry. Must be better studied!
4 6 8 10 12 14 16
-0.08
-0.06
-0.04
-0.02
0.00
0.02
0.04
0.06
0.08
223Th
E-0
.007
4I2 [M
eV]
I
erpp(+,+) ermp(-,+) erpm(+,-) ermm(-,-)
<60keV
Tilted reflection asymmetric nucleus
14
Good simplex
Several examples in mass 230 region
Substantial staggering
I
i
z
e
)(parity
simplex ||
1)(
S
PRS
15
Weak reflection symmetry breaking
Driven by rotation
0 5 10 15 20 25 30 35
0.0
0.2
0.4
0.6
0.8
240Pu
222Th226Th
S=
(E--E
+)[
MeV
]
I
StaggeringParameter S
Changes sign!
16
Condensation of non-rotating vs. rotatingoctupole phonons
0.00 0.05 0.10 0.15
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
j=3 phonon condensation
n=3
n=2
n=0
n=1
E' n-
E' 0
+
-
+
-
+
-
j=0 phonon j=3 phonon
3/vibph rot
Angular momentum rotational frequency
crotph
17
exp
n=0
n=1n=2
n=3
0 5 10 15 200.0
0.5
1.0
1.5
2.0
2.5
3.0
E
I
E0 E1 E2 E3
0 5 10 15 200.0
0.5
1.0
1.5
2.0
2.5
3.0
E
I
Ea Eb Ec Ed
n=0
n=1n=2
n=3
harmonic (non-interacting) phonons
an harmonic (interacting) phonons
0-2 1-3
Data: J.F.Smith et al.PRL 75, 1050(95)Plot :R. Jolos, Brentano PRC 60, 064317 (99)
Ra220
c
18
02468
1012141618202224262830
0.0 0.1 0.2 0.3
222Th3.0
2.5
1.7
_
+
226Th
240Pu
[MeV]
I
0 5 10 15 20 25 30 35
0.0
0.2
0.4
0.6
0.8
240Pu
222Th226Th
S=
(E--E
+)[
MeV
]I
Rotating octupole does not completely lock to the rotating quadrupole.
+
-
+
-
+
-
rotph rot
19
0.00 0.05 0.10 0.15
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
j=3 phonon condensation
n=3
n=2
n=0
n=1
E' n-
E' 0
X. Wang, R.V.F. Janssens, I. Wiedenhoever et al. to be published.
Preliminary
20
Consequence of chirality: Two identical rotational bands.
Chirality
21
band 2 band 1134Pr
h11/2 h11/2
Come as close as 20keV
StrongTransitions2 -> 1
K. Starosta et al.K. Starosta et al.Results of the Results of the GammasphereGammasphereGS2K009 GS2K009 experiment.experiment.
22
Soft chiral vibrations
Shape
Microscopic RPA calculations (D. Almehed’s talk)
Decreasing energy (about 2 units of alignment)Strong transitions 2->1, weak 1->2Tiny interaction between 0 and 1 phonon states (<20 keV)Systematic appearance of sister bands
Difficult to explain otherwise.
UnharmonicitesMust be even,because symmetryis spontaneouslybroken
32
Triaxial Rotorwith microscopicmoments of inertia
Rigid shape
IBFFM
Soft shape
A. Tonev et al. PRL 96, 052501 (2006) 2/ 10 tt QQC. Petrache et al.PRL 24
0 0.25 0.5 0.75 1 1.25 1.50
0.25
0.5
0.75
1
1.25
1.5
02 Q
2Q
Transition Quadrupolemoment
other for deforms ,30o
larger
smaller
25
Summary
• Close up refined our concept of how nuclei are rotating: assembly of gyroscopes
• Rich and unexpected response as compared to non-nuclear systems
• Rotation driven crossover between different discrete symmetries resolved
• Chirality of rotating nuclei appears as a soft an harmonic vibration
26
Congratulations!
27
Loss and onset of orientation
Geometrical picture vs. TAC
Chiral vibratorFrozen alignment
2/1 2
1
12
12
)()(
12
2/1
2
1
1
222
233
211
IIA
j
Ij
JAjJAjJAH
ii
W
J
J JJJ
J
Harmonic approximation
02468
101214161820222426
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
chiral rotor
chiral vibrator
jp,j
n frozen
=30o
[MeV]
J
om1 om2
Full triaxial rotor + particle + hole (frozen)
[8] K. Starosta et al., Physical Review Letters 86, 971 (2001)
8 10 12 14 16 18 20-200
-100
0
100
200
300
400
500
600
134Pr
V<25keV
E2-
E1[
keV
]
I
E2E1TPR(J-TAC) E2E1exp
0
2
4
6
8
10
12
14
16
18
20
22
24
0 100 200 300 400 500 600
TPR (J-TAC)
134Pr
[keV]
J
om1e om2e om1o om2o
02468
1012141618202224
0 100 200 300 400 500 600
134Pr
[kEV]
J
om1e om2e om1o om2o
134Pr - a chiral vibrator,which does not make it.
Experiment
Calculation:Triaxial rotor with Cranking MoI +particle+hole
Frozen alignment Coupling to particles
pj
hj
J
pj
hj
J
hj
pj
J
Additional alignment
Tiny interaction between states!
8 10 12 14 16 18 20-200
-100
0
100
200
300
400
500
600
134Pr
V<25keV
E2-
E1[
keV
]
I
E2E1TPR(J-TAC) E2E1exp keVV 18||Ru 112
)17(25||Pr 134 keVkeVV
keVV 1||Rh 104
But strong cross talk!!??
4 irreducible representations of group2 belong to even I and 2 to odd I. For each I, one is 0-phonon and one is 1-phonon.
)(iR)(sR )(lR1
hD2
The 1-phonon goes below the 0-phonon!!!
hj
pj R
8 10 12 14 16 18 20
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
TPR(J-TAC)
B(M
1,I-
>I-
1)[
N
2 ]
I
BM1o11 BM1o12 BM1021 BM1o22
8 10 12 14 16 18 200.0
0.1
0.2
0.3
0.4
0.5
0.6
TPR(J-TAC)
B(E
2,I-
>I-
2)[e
b2 ]
I
BE2so11 BE2so12 BE2so21 BE2so22
vib rotvib rot
Strong interband
Strong decay 2->1 weak decay 1->2 .
Cross over of the two bands (Intermediate MoI maximal)
Almost no interaction between bands 1 and 2 (manifestation of D_2)
Evidence for chiral vibration
Problem: different inband B(E2)
Coupling to deformation degrees of freedom seems important
Two close bands, same dynamic MoI, 1-2 units difference in alignment
Do not cross
Conclusions
• So far no static chirality – look at TSD
• Evidence for dynamic chirality
• Chiral vibrators exotic: One phonon crosses zero phonon
• Coupling to deformation degrees
)()(
frame fixedbody in
rervrv xLB
Deformed harmonic oscillatorN=Z=4 (equilibrium shape)
)(/)()(
fieldvelocity
rrjrv mL
Moment of inertiahas the rigid body value
generated by thep-orbitals
rotational alignmentBackbends
K-isomers
M. A. Deleplanque et al. Phys. Rev. C69, 044309 (2004)
Moments of inertia for I>20
Combination of many orbitals-> classical periodic orbits
Velocity field in body fixed frame of unpaired N=94 nuclides