Neutrinoless Double Beta Decay
Student: Alina HriscuSupervisors: Olaf Scholten
Gerco Onderwater
30 November 2005
Summary
-Beta decay(-review-) -Two neutrinos double beta decay (2ν2β) -Zero neutrinos double beta decay(0ν2β) -Majorana particles -Calculus of half-life time
-elementary particle problem -nuclear structure point of view
-Experiments on neutrinoless double beta decay -Conclusions -References
Beta decay-Decay of a neutron in a nucleus, into a proton
-Theory of β decay -parity is violated
-neutrinos exist in nature only as “left-handed ” particles (antineutrinos “right-handed”)and only these can interact
eeZAZA ~)1,(),(
Fermi theory -analogy with electromagnetic interaction
-Dirac-Pauli representation with γ matrixes
-Matrix elements are calculated
0
0
nosantineutri handed-rightfor )1(
neutrinos handed-leftfor )1(
5
5
5
I
I
u
u
Other possible decays
n p
e-W-
β- decay
p n
e+W+
+ decay
(A,Z+1) (A,Z)
W+
e-
EC
p n
e+
W+
e~
neutrino conversion
-In nature-only in supernovae-
e~
(A,Z)
(A,Z+1)
Q
β decay(A,Z)
(A,Z+2)
Q
sequential β decay
(A,Z+1)
(2ν)ββ decay
2νββ decay-Simultaneous transmutation of two neutrons in two protons inside a nucleus
thanks to β-decay
ex: 54Xe136 -> 56Ba136 + 2e-
-It is possible whenever beta decay is forbidden by energy conservation or by angular momentum mismatch
-Conserves the lepton number-Allowed in SM
2νββ decay
-Very rare process -It is possible only for heavy nuclei (nuclei which can
ββ decay have complicated nuclear structure) -It has been observed experimentally -Half-life time of order of y241810
The ββ emitting isotopes
Isotope Half-time T1/ 2,2ν (y) exp.
48Ca ~ 4.0 1019
76Ge ~ 1.4 1021
82Se ~ 0.9 1020
96Zr ~ 2.1 1019
100Mo ~ 8.0 1018
116Cd ~ 3.3 1019
128Te ~ 2.5 1024
136Xe not observed yet
150Nd ~ 7.0 1018
0νββ decay If the two neutrinos are missing…
Violates the lepton conservation rule (left side has lepton nb=0,right side equal to 2)=>beyond SM
Possible only if neutrinos are Majorana particles neutrinos have non-zero mass
Existence of right-handed currents in weak interaction Has not been (yet) observed experimentally
Feynman diagram for 0νββ decay
~
ν
e-
e-
p
p
n
n
W+
W+
ν->e-
eν~ (in SM)
Calculating the half-life time Two aspects:
Elementary particle Nuclear structure
Half-life time (for Majorana neutrinos)
So,if we know the half-life time and the matrix elements=> we can obtain the NEUTRINO MASS
Elem. particle properties: neutrino masses and mixing Nuclear strucure calculations: the matrix elements
mass neutrino effective
elementsmatrix nuclear -M
charge)nuclear on the depends(which integral space-phase theis g where
22)(
12/1
m
moMZgoT
Majorana Particles Majorana particles –the particle is the same as the antiparticle, opposite as
the Dirac particles Ex: Dirac particles: electron,proton Majorana particles: photon,*all particles with spin ½ known (fermions) are Dirac particles
We define left- and right-handed components of Dirac 4-spinor by:
Construct left-handed neutrinos - charge conjugate field, which for neutrinos is also neutral
00 ,
2
1 5,
RL
Rc
L
c
i
i
C
)()(
0
0 where,
0
0 where,iC
operator charge the-C *,
c
22
222
Neutrino-fields-can be linear combination of ,since
We define independent Majorana neutrino fields that are their own charge conjugate (antiparticles)
-Since the helicity flips,Majorana particles have nonzero mass
handed-right iscLhanded-left is and c
R
c,
Neutrinos masses and mixing-the See-saw mechansim
If neutrinos have masses,flavour is mixed, and a leptonic mixing matrix will appear
-Since we can write the neutrino fields as linear combination of
Mass term in Lagrangian can couple these two kinds of fields
themselves and to eachother:
c,
2
)(,
2
)( cRR
R
cLL
L
two thecouples that mass Dirac is and masses Majorana are , where
)(
DRL
LRRLDRRRLLLm
MMM
MMML
If ML and MR are zero, the Majorana ν-left pair with ν-right to form N Dirac neutrinos
The Seesaw mechanism: assuming a hierarchy in the values of elements of with μ negligible or zero,one (set)of particle(s)become
heavy,while another becomes light In the simple seesaw model, there are as many right- as left-handed
neutrinos such that we have three light and three heavy neutrinos, the lightest of which has mass
LD MMMM RM :~
N*N now are ~
,~
,~
matrices thewhere
~~
~~M~
,~
)(L
:matrix Seesaw the-
matrix ain arranged becan masses theseneutrinos of flavours NFor
M
DRL
RD
DL
R
LRL
MMM
MM
MMM
RD MM /2
Neutrino mixing Mixes of light neutrinos: For three active neutrinos,the mixing matrix can be written:
The effective neutrino mass:
mass definite with where, m, statesU mmll
particles Majoranaonly affect phases other two theandmatrix, CKM in the
phase the toanalogus phase Dirac is , angles mixing the,cosc,sins where
1,,
ijij
2
2
2
132313231223121323122312
132313231223121323122312
13131213121
ijijij
ii
ii
ii
i
eediag
ccescsscesccss
csesssccessccs
escscc
U
ej
j
)2(2
33
)(2
22
2
112
Uamplitudewith
absorbed and emitted,m mass of neutrino Majorana exchanged and outgoing
112
ie
i
eeejj eUmeUmUmUmm
Calculating the nuclear matrix elements
If the 0νββ decay will be observed,it is important to have accurate values of nuclear matrix elements in order to obtain quantitative results
The hadronic part contributing to the half time must be evaluated between initial and final states in the intermediate nucleus summed over
Many body techniques which lead to such results are: QRPA (neutron-proton Quasiparticle Random Phase Approximation)
-treats a large fraction of nucleons as active and allows these a large single-particle space to move in-suitable for collective motion
SHELL MODEL-Treats a small fraction of the nucleons in a limited single-particle space, but allows nucleons to corrrelate in arbritary ways
These methods have been applied to 2νββ decay(which was observed)->result: RPA model gave the most precise n.m.e –same order of magnitude-it is expected to give better results for 0νββ, too –one order of magnitude difference
Why is that so difficult?
Theorists are making real efforts to reduce the uncertainty in calculated n.m.e
Matrix elements: <Z+2| Oβ Oβ|Z>= <Z+2| Oβ |Z+1> <Z+1| Oβ|Z>
ββdecay sequential β decay Graphical representation
|Z>0
|Z+1>0 sequential β decay
|Z+2>0
n
nn
E
ZZ1
11
Values of the calculated n.m.e for 2νββ decay and experimental ones
Predicted n.m.e. and half-times vs. experimental ones for decay
; WS=Wood-Saxon basis for calculated n.m.e(still QRPA); calculated for decay in ground and excited states of daughter nucleus
KrSe 8282
…and calculated ones for 0νββ
In one of the most recent QRPA model: Rodin(2003)
Compared with the SM results;except of Mo similar results
Predicted life-time for some calculated n.m.e (2ν)
The outliers predict wrong life-time; the n.m.e of Rodin and SM are quite close
Experimental 0νββ?
If an experiment observes 0νββ it will have profound physics implications
=>extraordinary evidence is required Difficulties:
Very slow process(one of the most slowest in nature)=>requires a lot of material(500 kg to 1 tone)
Extremely high energy resolution is required Only very pure material is used (contaminations may give background
signals) The material is difficult to obtain - experimentalists have to enrich
nuclei;also,very expensive The experiment must take place in underground-mines or like others,under
a mountain Even in the best conditions, false peaks may appear(cosmic rays,walls)
-Since enormous blocks of material are used,how can they determine the energy of only one decay
A lot of experiments are running to date,and others are being prepared
One of the most advanced : Heidelberg-Moscow
Energetic resolution 0νββ decay being a 2 body decay, experimentally,only the energy of the two
outgoing electrons needs to be measured Ideal case-infinite resolution Real case-finite resolution
N
E
2νββ0νββ
EN
0νββ2νββ
Importance of energy resolution
Heidelberg-Moscow experiment
German-Russian experiment In Gran Sasso Underground Laboratory in Italy Operating with 76Ge(the sample and the detector)
-experiment is possible due to simultaneous use of large source strengths with high resolution detectors
In the underground lab the flux of cosmic muon is reduced by 6 orders of magnitude
They claim to have “seen” 0νββ decay
H-M experimental setup
What do they have?Peak expected
here
Their results
eVm
yT
39.0
105.1 2502/1
with 50% uncertainty in n.m.e
Their best values
Conclusions If 0νββ decay will be observed, it will reveal the
identity of neutrino,so a fundamental issue will be answered If a nonzero rate is seen=> ν=Majorana particle If no signal is seen=> ν=Dirac particle
Experimental proposal are promising To obtain quantitative results (neutrino masses
and hierarchy) from the experiment, good theoretical results are required uncertainties in calculus of n.m.e must be reduced
References Neutrinoless double beta decay from a modern perspective, J.D.
Vergados (Phys. Rep 361(2002) 1-56) Weak interaction and nuclear-structure aspects of nuclear double beta
decay-Jouni Suhonen,Osvaldo Civitarese (Phys. Rep 300(1998)123-214) Double beta decay –Topical review, S.R. Elliot,J. Engel(Jour. Of Phys G.
30(2004)183-215) Renormalized proton-neutron QRPA and double beta decay of 82Se to
excited states in 82Kr, J. Suhonen, J. Toivanen, A.S. Barabash, I.A. Vanushin, V.I. Umatov, R. Gurriar´an, F. Hubert, Ph. Hubert(Z. Phys. A 358, 297–301 (1997))
Evidence for neutrinoless double beta decay, Klapdor-Kleingrothaus, Modern Physics Letters A [Particles and Fields; Gravitation; Cosmology and Nuclear Physics], Vol. 16, No. 37 (2001) 2409-2420
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