Ph 406: Elementary Particle Physics Problem Set 9

Ph 406: Elementary Particle Physics

Problem Set 9K.T. McDonald

[email protected] University

Due Monday, December 1, 2014 (updated September 14, 2016)

1. Double-Beta Decay

a. In 1935, Goeppert-Mayer made a calculation1 of the process A → A′eeνν, so-called 2-neutrino double-beta decay, which was first definitively observed in1987.2 Make a quick order-of-magnitude estimate (not a Fermi-theory calcu-lation) of the lifetime of a nucleus against double-beta decay, starting from thefact that the lifetime of the neutron is 880 s. What is the characteristic timescale/frequency for the collision of two quarks within a nucleon of radius ≈ 1fermi? What fraction of these collisions result in a beta-decay of a neutron?

b. An important comment by Majorana3 was that neutrinos might not be “Dirac”particles as considered in the Notes (Lectures 6, 7, 19, etc.),4 but rather whatare called the (Dirac) neutrino and antineutrino are combined in a single particle,the neutrino, which then is its own antiparticle (and which implies that leptonnumber is not conserved).

It was realized by Furry5 that, if neutrinos behave according to Majorana’s view,there could exist the phenomenon of neutrinoless double-beta decay, A →A′ee. Further, if the nuclear matrix elements were similar for 2-neutrino double-beta decay and neutrinoless double-beta decay, then the rate for the latter wouldbe much larger (and the lifetime much shorter), because the 3-body phase space(Lecture 11 of the Notes) for the final state of the latter reaction is much largerthan the 5-body phase space for the former.

1M. Goeppert-Mayer, Double Beta-Disintegration, Phys. Rev. 48, 512 (1935),http://physics.princeton.edu/~mcdonald/examples/EP/goeppert-mayer_pr_48_512_35.pdf.

2S.R. Elliott et al., Direct Evidence for Two-Neutrino Double-Beta Decay in 82Se, Phys. Rev. Lett. 59,2020 (1987), http://physics.princeton.edu/~mcdonald/examples/EP/elliott_prl_59_2020_87.pdf.

3E. Majorana, Teoria simmetrica dell’elettrone e del positrone, Nuovo Cim. 14, 171 (1937),http://physics.princeton.edu/~mcdonald/examples/EP/majorana_nc_14_171_37.pdf.http://physics.princeton.edu/~mcdonald/examples/EP/majorana_nc_14_171_37_english.pdf.F. Wilczek, Majorana Returns, Nature Phys. 5, 614 (2009),http://physics.princeton.edu/~mcdonald/examples/neutrinos/wilczek_np_5_614_09.pdf.

4Massless Dirac fermions are sometimes called Weyl fermions, and the theory of massless Dirac neutrinois often called the “two component” theory, following H. Weyl, Elektron und Gravitation. I, Z. Phys. 56,330 (1929), http://physics.princeton.edu/~mcdonald/examples/EP/weyl_zp_56_330_29.pdf.Gravitation and the Electron, Proc. Nat. Acad. Sci. 15, 323 (1929),http://physics.princeton.edu/~mcdonald/examples/GR/weyl_pnas_15_323_29.pdf.

5W.H. Furry, On Transition Probabilities in Double Beta-Disintegration, Phys. Rev. 56, 1184 (1939),http://physics.princeton.edu/~mcdonald/examples/EP/furry_pr_56_1184_39.pdf.

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To date, no neutrinoless double-beta decay has been observed, with limits on thelifetime (> 1025 yr) being longer than that for observed 2-neutrino decays.6

Draw a Feynman diagram for neutrinoless double-beta decay, as due to the decayof two d-quarks. Include arrows on the spin-1/2 particle lines indicating thenominal helicities expected for the V -A model of the weak interaction (with 1−γ5

coupling).7 Then, recalling the spin-1/2 propagator (p. 28, Lecture 3 of the Notes),the matrix element includes a factor (between the final-state electron spinors),

(1 − γ5)qμγμ −mν

q2 −m2ν

(1 − γ5). (1)

Show that this reduces to

−2mν

q2 −m2ν

(1 − γ5) ≈−2mν

q2(1 − γ5). (2)

Thus, the matrix element is proportional to the mass mν of the neutrino, and thedecay rate is proportional to the square of the mass, so the rate vanishes if theneutrino were massless.8

In neutrinoless double-beta decay the four momentum qμ of the virtual neutrinois small, but we expect that q2 � m2

ν while also q2 � m2e. This suggests that

the rate for neutrinoless double-beta decay is suppressed relative to that for 2-neutrino double-beta decay by a factor (mν/E)2 where E is very roughly of order1 keV. This suppression may be sufficient to overcompensate for the larger phasespace for neutrinoless double-beta decay, and make its rate smaller than that for2-neutrino double-beta decay.

In searches for neutrinoless double-beta decay (negative to date), a factor of twoimprovement in the limit on the neutrino mass requires a factor of four improve-ment in the limit on the lifetime, etc.

While Majorana neutrinos could lead to neutrinoless double-beta decay, this neednot be the only explanation for the nonconservation of lepton number in the decay.If neutrinoless double-beta decay were detected, additional effort would still berequired to identify its cause.

The virtual neutrino in neutrinoless double-beta decay can be a mixture of thethree mass eigenstates ν i. This implies that the mν written above is actually∑

i U2eimi, where Uαi is the CKMNS mixing matrix for neutrinos (p. 356, Lecture 19

of the Notes). Hence, it may be that neutrinoless double-beta decay is furthersuppressed by the numerical values of the neutrino-mixing matrix.

6See, for example, R. Henning, Current status of neutrinoless double-beta decay searches, Rep. Phys. 1,29 (2016), http://physics.princeton.edu/~mcdonald/examples/neutrinos/henning_rp_1_29_16.pdf.

7R.P. Feynman and M. Gell-Mann, Theory of the Fermi Interaction, Phys. Rev. 109, 193 (1958),http://physics.princeton.edu/~mcdonald/examples/EP/feynman_pr_109_193_58.pdf.

8This result was perhaps first obtained, by a different argument, on p. 314 of K.M. Case, Reformulationof the Majorana Theory of the Neutrino, Phys. Rev. 107, 307 (1957), some 9 months before the emergenceof the V -A theory, http://physics.princeton.edu/~mcdonald/examples/neutrinos/case_pr_107_307_57.pdf.

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2. Helicity of Neutrinos. (M. Goldhaber et al., Phys. Rev. 109, 1015 (1958),http://physics.princeton.edu/~mcdonald/examples/EP/goldhaber_pr_109_1015_58.pdf;see also p. 1423 ff in Alpha-, Beta-, and Gamma-Ray Spectroscopy, ed. by K. Siegbahn.)

Suppose a spin-0 nucleus A decays via electron capture, A + e− → B� + νe, to aspin-1 excited state B� of nucleus B along with emission of a neutrino (Gamow-Tellertransition). Then, suppose the spin-1 state B� decays via emission of an electric-dipolephoton to the spin-0 ground state of B. We wish to infer the helicity of the neutrinoby observation of the angular distribution of the photon. To do this, the direction ofthe neutrino in the laboratory must be singled out by some feature of the apparatus.This is possible by an ingenious argument of Goldhaber et al.

Take the z axis along the direction of the neutrino in the decay A → B�, such that theneutrino has S(ν)

z = ±12

(with S(ν)z = −1

2in the V − A theory of the weak interaction,

which is to be confirmed). The nucleus B� could have S(B�)z = −1, 0 or 1, leading to Jz

for the final state of −32, −1

2, 1

2or 3

2. (Recall that Lz = 0 for any two-body state where

the z axis is along the direction of momentum in the rest frame.) But, the initial statehas only spin 1

2from the electron, assuming electron capture from an S-wave orbital

(K capture), so if S(ν)z = −1

2only S(B�)

z = 0 or 1 are possible for nucleus B�, while ifS(ν)

z = 12

only S(B�)z = −1 or 0 are possible

Let θ′ be the angle of emission of the photon with respect to the −z axis in the restframe of state B�. (In the lab frame, B� moves along the −z axis.) Use the spin-1 rotation matrix, http://pdg.lbl.gov/2015/reviews/rpp2015-rev-clebsch-gordan-coefs.pdf, to showthat the angular distribution of the E1 photons is sin2 θ′ if S(B�)

z = 0, and (1+cos2 θ′)/2when S(B�)

z = 1. Remember that the photon can only have S(γ)z′ = ±1 along the z′

axis, which is along the photon’s direction in the B� rest frame.

In particular, show that photons emitted at θ′ = 0 or 180◦ can only have S(γ)z = +1 if

S(ν)z = −1

2, and only S(γ)

z = −1 if S(ν)z = 1

2.

Let EK be the energy of the neutrino emitted in the (K-capture) decay of A, and E0

be the excitation energy of B� with respect to ground state B (MB� = MB + E0 �E0, EK). Deduce the energy E ′

γ of the photon in the rest frame of the B�, andtransform this to the lab frame, expressing Eγ as a function of θ′, EK and E0. Whatare the minimum and maximum values for Eγ? You should find that the highest photonenergy occurs for θ′ = 0, for which these photons have Sz = 1 (and negative helicityas these photons are moving along the −z axis) if the neutrino has negative helicity;and that if the neutrino had positive helicity, the highest-energy photons would havepositive helicity also.

So if we can measure the helicity of the highest-energy photons, we determine thehelicity of the neutrino.

The helicity of photons can be determined by passing them through a filter consist-ing of magnetized iron, which attenuates photons of +1 and −1 helicity by differentamounts. The reaction here is just Compton scattering of polarized electrons and pho-tons. Because the electron has spin 1

2, an electron can only absorb a photon whose

spin is opposite, which flips the spin of the intermediate electron prior to the radiation

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(scattering) of the final photon.

We want to determine the helicity of only the highest-energy photons, so a final trickis needed. Suppose the photons from the B� decay impinge upon other ground-stateB nuclei that are at rest in the lab. Calculate the energy of the photons such that anucleus B at rest can be excited to the level B�. The latter states decay back to theground state B by photon emission, in effect scattering only a certain subset of thephotons from the first B� decay into the detector.

For the historical experiment, A = Eu152, and B = Sm152, for which EK = 840 keV,while E0 = 961 keV. Due to recoil effects, you should have found that even the highest-energy photons from the first B� decay have insufficient energy to re-excite B nucleiat rest. However, the lifetime of the spin-1 Sm152 excited state was measured tobe 7 × 10−14 s. Convert this to a width in eV. What fraction of the Breit-Wignermass/energy distribution (recall p. 11, Lecture 1 of the Notes) of the short-lived B� inthe reaction γ+B → B� lies with the energy distribution of the photons from the firstB� → B + γ decay (assuming that distribution to be flat between its min and maxenergies)?

If the lifetime of the B� level had been too long, the overlap would be too small for theexperiment to work. Also, if the lifetime were long, the B� atom might have collidedwith another atom and changed its momentum prior to the photon decay. Then, thecorrelation between the decay-photon helicity and the neutrino helicity would havebeen lost.

So it’s a small miracle that any system exists in nature which permits this measurement!

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3. Spurions

The spurion concept9 is that in weak decays of an s quark the hadronic isospin characterof the final state can be predicted by supposing the s quark absorbs a fictitious particle,the spurion, that is an isospin state |I = 1

2, I3 = −1

2〉 (and that an s quark absorbs an

antispurion that is an isospin state |I = 12, I3 = 1

2〉). Use this model to predict

ΓΞ0→Λπ0

ΓΞ−→Λπ−, and

ΓΩ−→Ξ0π−

ΓΩ−→Ξ−π0

. (3)

Compare with data reported at http://pdg.lbl.gov/2013/tables/contents_tables_baryons.html.

4. The charmed mesonD0 can decay toKπ via the Cabibbo-favored W -exchange diagram(with gluons not shown),

If this were the only possible diagram, predict the ratio of branching ratios:

Γ(D0 → K−π+)

Γ(D0 → K0π0)

. (4)

Draw any other Cabibbo-favored diagrams for these decays.

Assuming that the decay D0 → K+π− proceeds via a diagram similar to the above,predict the ratio of branching ratios:

Γ(D0 → K+π−)

Γ(D0 → K−π−). (5)

Compare with data reported at http://pdg.lbl.gov/2013/tables/rpp2013-tab-mesons-charm.pdf.

5. There are four spin-0 mesons that contain one bottom quark: B+u = ub, B0

d = db, B0s =

sb, and B+c = cb. These decay via the weak interaction by two graphs with roughly

equal strength, as sketched below (the “spectator” model):

Here we consider only nonleptonic final states. Suppose the four final-state quarks formexactly two mesons (as happens a few percent of the time). List the two dominanttwo-body decays for each of the four bottom mesons.

9Attributed to G. Wentzel, Heavy-Meson Decays and the Selection Rule ΔI ≤ 1/2, Phys. Rev. 101,1215 (1956), http://physics.princeton.edu/~mcdonald/examples/EP/wentzel_pr_101_1215_56.pdf.

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A complication arises for the Bc meson. The charm quark has a slightly shorter lifetimethan the bottom quark. Hence, there are two more prominent two-body decays of theBc involving c→ Wq rather than b →Wq transitions. List these.

According to the measured values of the C-K-M matrix elements

Vub

Vcb≈ Vus

Vud≈ Vcd

Vcs≈ λ = Cabibbo angle.

List (or indicate on diagrams) the prominent two-body nonleptonic decays of the fourbottom mesons that are suppressed by one power of λ in the matrix element (and henceby λ2 ≈ 1/25 in rate).

Note that D+ = cd, D0 = cu, and D0s = cs. If a meson is produced from, say, a dd

state it could be a π0, η, ρ0, or ω0. Here it is sufficient to list only the ρ0 as empiricallyρ produced seems favored...

Compare with data reported at http://pdg.lbl.gov/2013/tables/rpp2013-tab-mesons-bottom.pdf.

6. Both the B0d = db and B0

d = db can decay to common final states, such as J/ψK0S as

you found in Prob. 5. Hence, there are transitions between B0 and B0 and so theseare not the states of definite mass and lifetime, which latter states can be written as

B0± = pB0 ± qB

0, where |p|2 + |q|2 = 1. (6)

Taking into account the weak interactions, one writes the 2 × 2 Hamiltonian (in the

|B0〉-|B0〉 basis) as

H = M − i

2Γ, (7)

where the mass matrix M and the decay matrix Γ are Hermitian. (Since neutral B’sdecay, H itself is not Hermitian.) CPT invariance implies that the diagonal componentsof H are equal, and if CP is conserved M and Γ are real. Allowing for the possibilityof CP violation, the Hamiltonian can be written as

H =[m M12

M∗12 m

]− i

2

[γ Γ12

Γ∗12 γ

]. (8)

Unlike the eigenstates K0L and K0

S of the neutral K system, the eigenstates B0± of the

neutral B system have roughly equal lifetimes, such that we can approximate matrixelement Γ12 as zero.

Show that this implies that q/p is a pure phase (with magnitude 1), so that

B0± ≈ eiφ±

B0 ± B0

√2

. (9)

The time evolution of the eigenstates can then be written as |B±〉 → e−Γt/2eim±t|B±〉.Show that the time evolution of initially pure B0 and B0 states can be written as

|B0〉 → Fc(t)|B0〉 + iq

pFs(t)|B0〉, |B0〉 → i

p

qFs(t)|B0〉 + Fc(t)|B0〉, (10)

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where Fc and Fs are proportional to the cosine and sine of Δmt/2, respectively, withΔm ≡ m+ −m−.

If a B0-B0

pair is produced at an electron-positron collider via the reaction e+e− →B0B

0, it is produced in the entangled10 state (|B0

1〉|B02〉 − |B0

1〉|B02〉)/

√2 with negative

charge conjugation. Later, B1 decays at time t1 and B2 decays at time t2 (where t1can be larger than t2 if we suppose that B1 travels to the “left” and B2 travels to the“right”). Consider the time evolution of the entangled state from (0,0) to (t1, t2) toshow if the two B ′s happened to decay at the same time t1 = t2 then one must be a

B0

and the other a B0 (although if t1 = t2 they have nonzero amplitudes to be both

B0’s or both B0’s).11

Such persistence of the initial quantum correlation for spacelike-separated events puz-zled Einstein, Podolsky and Rosen,12 and remains ever impressive.13

A subtle consequence of this “EPR” quantum correlation is that if the lab frame ofthe e+e− is the same as the center-of-mass frame, as typically the case, CP violationcould not be detected in certain decay modes. To make such decay modes useful forthe study of CP violation, so-called asymmetric B-factories were built in which thecenter of mass is in motion in the lab frame.

10The term entangled was introduced in a quantum context by E. Schrodinger (1935) in sec. 14 of hisfamous “cat” paper, http://physics.princeton.edu/~mcdonald/examples/QM/schroedinger_cat.pdf.Schrodinger lived with two women, one of whom was his legal wife. The concept of quantum-mechanicalentanglement first appeared in J. von Neumann, Mathematische Grundlagen der Quantenmechanik (Springer,1932), Mathematical Foundations of Quantum Mechanics, (Princeton U. Press, 1955). For commentary bythe author, see Prob. 5 of http://physics.princeton.edu/~mcdonald/examples/ph410problems.pdf.

11This effect also occurs for K0-K0

produced in the reaction e+e− → φ → K0K0, as discussed by

H.J. Lipkin, CP Violation and Coherent Decays of Kaon Pairs, Phys. Rev. 176, 1715 (1968),http://physics.princeton.edu/~mcdonald/examples/EP/lipkin_pr_176_1715_68.pdf.

12A. Einstein, B. Podolsky and N. Rosen, Can Quantum Mechanical Description of Physical Reality BeConsidered Complete? Phys. Rev. 47, 777 (1935),http://physics.princeton.edu/~mcdonald/examples/QM/einstein_pr_47_777_35.pdf.

13Some people say things like the decay of one B instantaneously forces the “other” B to be its antiparticle,and the “other” B then subsequently evolves according to the single-B prescription (10). This is not a goodway to think about the nature of quantum correlations in entangled states.

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Solutions

1. a. 2-Neutrino Double-Beta Decay

A neutron has radius r ≈ 1 fermi, so the time scale for collisions of quarks insideit is r/c ≈ 3 × 10−23 s. The neutron lifetime is about 103 s, so the probability ofa weak interaction occurring during each collision is about 3 × 10−26.

The probability of two such weak interactions occuring in a nucleus within a singlecollision time is the square of this, and hence the lifetime for double-beta decay isabout 3× 1025 times the neutron lifetime, i.e., ≈ 3× 1028 s. Recalling that a yearcontains about π× 107 s, we estimate that the lifetime against double-beta decayis 1021 years (which agrees with experiment to within an order of magnitude).

b. Neutrinoless Double-Beta Decay

A Feynman diagram for neutrinoless double-beta decay, A → A′ee from two dquarks of neutrons in the nucleus A, is, in the V -A theory of the weak interaction,

As such, the matrix element includes a factor, related to the exchanged virtualneutrino νi, of ∑

i

U2ei(1 − γ5)

qμγμ −mi

q2 −m2i

(1 − γ5), (11)

recalling the form of the spin-1/2 propagator mentioned on p. 28, Lecture 3 ofthe Notes. Here, we consider 3 neutrino mass eigenstates ν i, which are related tothe neutrino flavor states να via the 3-neutrino CKMNS matrix Uαi. The factors1−γ5 indicate that the virtual neutrino couples to lefthanded (chirality) electrons,e−L .

In eq. (11), the term in the numerator involving the neutrino mass mi is,

(1 − γ5)mi(1 − γ5) = mi(1 − 2γ5 + γ25) = 2mi(1 − γ5), (12)

recalling from p. 116, Lecture 7 of the Notes that γ25 = 1.

The other term in the numerator of eq. (11) includes the product

(1 − γ5)γμ(1 − γ5) = γμ − γ5γμ − γμγ5 + γ5γμγ5 = γμ − γ25γμ = 0, (13)

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recalling from p. 116, Lecture 7 of the Notes that γ5γμ = −γμγ5. Hence, thefactor (11) reduces to

− 2∑

i

U2eimi

q2 −m2i

(1 − γ5) ≈ −2∑

i

U2eimi

q2(1 − γ5) , (14)

with the implication that the rate for neutrinoless double-beta decay varies as thesquare of the effective neutrino mass

∑i U

2eimi.

If we interpret the lower beta-decay as involving emission of the virtual neutrino,then in the V -A theory this is a righthanded (chirality) Dirac antineutrino. On theother hand, in the upper beta-decay, the virtual neutrino is absorbed, and mustbe a lefthanded Dirac neutrino. Since a Dirac antineutrino is distinct from a Diracneutrino, the neutrinoless double-beta decay is not possible for Dirac neutrinos.But it is possible for Majorana neutrinos, for which a lefthanded “neutrino” isa 50-50 combination of a lefthanded Dirac neutrino and a righthanded Diracantineutrino (as discussed in the Digression below).

It remains that the arrowheads on the two ends of the virtual-neutrino line pointin opposite directions in the diagram. If we interpret the arrowheads as indicatingthe needed helicity, then the diagram indicates that the antineutrino is emittedas chirality state vR in the lower beta-decay with nominally positive helicity, butmust be absorbed in the upper beta-decay (as a neutrino in chirality state uL)with nominally negative helicity.

An argument is sometimes made14 that a righthanded chirality antiparticle state,vR, has a small relative amplitude, ≈ m/2E, to have negative helicity, ratherthan the nominal positive helicity (recall eq. (70), Prob. 2, Set 4). This suggeststhat there is an amplitude ≈ mi/2q0 that the arrowhead for the lower neutrinoin the diagram actually points downward, which implies that the amplitude forneutrinoless double-beta decay varies as neutrino mass, so that the rate varies asthe square of the neutrino mass.

However, as discussed in the Digressions below, a Majorana neutrino from a beta-decay is a 50-50 mixture of a lefthanded Dirac neutrino and a righthanded Diracantineutrino, so there is no suppression of the type speculated above. The author’sview is that it’s better to stick with to the discussion based on the form of thespin-1/2 propagator.

Digression: Majorana 4-Spinors.15

We saw in the digression on p. 5, Prob. 2, Set 4 that the charge-conjugationtransformation ψ = iγ2ψ

∗ of a 4-spinor state ψ leads to its antiparticle state ψ.

14See, for example, p. 17 of B. Kayser, Neutrino Mass, Mixing and Flavor Change (Feb. 2002),https://arxiv.org/abs/hep-ph/0211134.This argument may have been inspired by Case, footnote 8.

15This digression follows Prob. 7.51 of Griffiths’ text:http://physics.princeton.edu/~mcdonald/examples/EP/griffiths_particles_08.pdfhttp://physics.princeton.edu/~mcdonald/examples/EP/griffiths_particles_08_solutions.pdf.

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Majorana spinors (when constructed from spinors that obey the Dirac equation)are their own antiparticles, which suggests that they are combinations of Diracparticles u and antiparticles v. So, recalling eqs. (17)-(18) of Set 4, we consider ageneral Majorana spinor of the form,

ψ√E+

= au+ bv = a e−ipx

⎛⎜⎝ χ

p·σE+m

χ

⎞⎟⎠+ b eipx

⎛⎜⎝

p·σE+m

χ

χ

⎞⎟⎠ , (15)

where χ and χ = −iσ2χ∗ (χ = iσ2χ

∗, from eq. (28), Prob. 2, Set 4) are 2-spinorswith unit normalization, E+ = E + m, and |a|2 + |b|2 = 1, so that ψψ = 2m.Then, the requirement that this state be its own antiparticle, ψ = ψ = iγ2ψ

∗,implies

ψ√E+

=ψ√E+

= au+ bv = a∗u+ b∗v

=

⎛⎜⎝ 0 iσ2

−iσ2 0

⎞⎟⎠⎡⎢⎣a∗ eipx

⎛⎜⎝ χ∗

p·σ∗E+m

χ∗

⎞⎟⎠+ b∗ e−ipx

⎛⎜⎝ p·σ∗

E+mχ∗

χ∗

⎞⎟⎠⎤⎥⎦

= a∗ eipx

⎛⎜⎝ p·σ

E+m(−iσ2χ

∗)

−iσ2χ∗

⎞⎟⎠ + b∗ e−ipx

⎛⎜⎝ iσ2χ

∗

p·σE+m

(iσ2χ∗)

⎞⎟⎠ (16)

recalling that σ2σ∗ = −σσ2. Hence, b = a∗ (= 1/

√2), and v = u (u = v), such

that Majorana 4-spinors have only 2 independent components (and always containboth a Dirac-particle and -antiparticle spinor).16 A consequence is that the topand bottom 2-spinors, χt and χb, of a Majorana 4-spinor state ψ are related by

ψ√E+

=

⎛⎜⎝ χt

χb

⎞⎟⎠ , χb = −iσ2χ

∗t , χt = iσ2χ

∗b . (17)

First, we consider the spin-up/down helicity 2-spinors χ± (eq. (46) of Set 4),

χ+ =

⎛⎜⎝ cos θ

2e−iφ/2

sin θ2eiφ/2

⎞⎟⎠ , χ− =

⎛⎜⎝ − sin θ

2e−iφ/2

cos θ2eiφ/2

⎞⎟⎠ , (18)

for which χ± = −iσ2χ∗± = ±χ∓ (as found in eqs. (29)-(30) of Set 4). Taking

a = b = 1/√

2, the Majorana helicity 4-spinors ψ± are their own antiparticles,

ψ± = ψ±, with

ψ±√E+/2

= e−ipx

⎛⎜⎝ χ±

± pE+m

χ±

⎞⎟⎠ + eipx

⎛⎜⎝ − p

E+mχ∓

±χ∓

⎞⎟⎠ . (19)

16The Majorana two-component theory applies to particle with mass (but without electric charge), unlikethe Weyl two-component theory of massless fermions (pp. 14-15 of Set 4), and footnote 4 above.

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Majorana chirality 4-spinors.

Following Case, footnote 8, we define these states, for a Majorana state ψ = ψ,as

ψR,L =1 ± γ5

2ψ. (20)

However, these are not quite their own antiparticles. Rather (recalling that γμγ5 =−γ5γμ),

ψR,L = iγ2ψ∗R,L = iγ2

1 ± γ5

2ψ∗ =

1 ∓ γ5

2(iγ2ψ

∗) =1 ∓ γ5

2ψ = ψL,R. (21)

Applying the chirality projection operators (1 ± γ5)/2 to the Dirac equation fora Majorana state ψ yields the coupled equations,

1 ± γ5

2i∂μγμψ = i∂μγμ

1 ∓ γ5

2ψ =

1 ± γ5

2mψ, i∂μγμψL,R = mψR,L, (22)

while their sum ψ = ψR + ψL does satisfy the Dirac equation i∂μγμψ = mψ. Itremains that Majorana’s requirement is that states which obey the Dirac equationare their own antiparticles.

Also (recalling eq. (15)), the righthanded Majorana spinor ψR consists of a right-handed Dirac-particle spinor uR together with a lefthanded Dirac-antiparticlespinor vL. That is, if we write a Majorana spinor (15) as ψ ∝ u + v, where theDirac spinors u and v are related by v = u as seen above (and so have equalamplitudes), then the corresponding Majorana chirality spinors would be

ψR,L =1 ± γ5

2ψ =

uR,L + vL,R√2

=

√E+

2√

2

⎛⎜⎝(1 ± σ·p

E+m

)(e−ipx χ± eipx χ)(

σ·pE+m

± 1)

(e−ipx χ± eipx χ)

⎞⎟⎠ , (23)

recalling the states uR,L and vR,L displayed in eq. (77) of Set 4. Then, uR and vL

have equal amplitudes in ψR, and likewise for uL and vR in ψL.17 According toeq. (86) of Set 4, uR,L = vR,L (and vL,R = uL,R), such that indeed

ψR,L ∝ uR,L + vL,R = vR,L + uL,R = ψL,R, (25)

as in eq. (21).

In a beta-decay n→ p e− ν that produces a Majorana neutrino, the V -A interac-tion, with its 1−γ5 coupling, produces a lefthanded Majorana neutrinoψL that hasequal amplitudes to be a lefthanded Dirac-particle spinor uL and a righthandedDirac-antiparticle spinor vR as in eq. (23).18 Similarly, in the reaction ν n→ p e−,

17Recalling eqs. (82) and (85) of Set 4, the Majorana chirality spinors obey

ψRψR = ψLψL = 0, ψRψL = ψLψR = m. (24)18Since a righthanded Majorana neutrino, ψR, has no weak, electromagnetic or strong coupling, it is a

“sterile” neutrino, as is also the case for a righthanded Dirac neutrino.

11

the Majorana neutrino needs to be lefthanded. Hence, if the virtual neutrino ina “neutrinoless” double-beta decay is a Majorana state, the wavefunction of theneutrino created in the “first” beta-decay “matches” the wavefunction of the neu-trino absorbed in the second beta-decay, and there is no suppression by a factorm/E as sometimes argued (p. 9, above).

In an analysis via 4-spinors, the factor of m in the matrix element for neutrinolessdouble-beta decay comes from the spin-1/2 propagator of the virtual neutrino, andnot from a supposed mismatch between the wavefunctions of the virtual neutrinoas created and absorbed.

Digression. Another convention for Majorana chirality 4-spinors.

In a different convention, Majorana chirality states are their own antiparticles.These could be obtained from eq. (20) by symmetrizing with respect to theirantiparticles states,

ψ(2)R,L =

ψR,L + ψR,L√2

. (26)

Such states will contain all four of 4-spinors uR, uL, vR and vL, and so are notwell aligned with the V −A theory of the weak interaction, in which only spinorsuL and vR participate.

For completeness, we return to consideration of a Dirac particle 4-spinor u andits antiparticle v = u, for which the Dirac chirality 4-spinors are,

uR,L =1 ± γ5

2u, vL,R =

1 ± γ5

2v = uL,R. (27)

Then for the general 4-spinor φ = au+ bv, where |a|2 + |b|2 = 1, we have

φR,L = auR,L + bvL,R, φR,L = a∗vR,L + b∗uL,R. (28)

We now define the self-conjugate states19

ψ(2)R,L =

φR,L + φR,L√2

=auR,L + bvL,R + a∗vR,L + b∗uL,R√

2(Convention 2). (29)

The Dirac-type equations of motion for the ψ(2)R,L are

i∂μγμψ(2)R,L = i∂μγμ

auR,L + bvL,R + a∗vR,L + b∗uL,R√2

= mauL,R + bvR,L + a∗vL,R + b∗uR,L√

2= mψ

(2)L,R, (30)

19Convention 2 follows, for example, T.P. Cheng and L.-F. Li, Neutrino masses, mixings, and oscillationsin SU(2) × U(1) models of electroweak interactions, Phys. Rev. D 22, 2860 (1980),http://physics.princeton.edu/~mcdonald/examples/neutrinos/cheng_prd_22_2860_80.pdf, andL.F. Li and F. Wilczek, Physical processes involving Majorana neutrinos, Phys. Rev. D 25, 143 (1982),http://physics.princeton.edu/~mcdonald/examples/neutrinos/li_prd_25_143_82.pdf.

12

recalling eq. (75) of Set 4. That is, both the chirality states (20) and (29) obeythe same coupled Dirac-type equations, (22) and (30). Since these equationscan be deduced from the Lagrangian for noninteracting Majorana particles, thatLagrangian alone does not distinguish between the two conventions.

Note that if a = b = 1/√

2, then ψ(2)R = ψ

(2)L = (uR + uL + vR + vL)/2.

In the V −A theory of the weak interaction, only 4-spinors uL and vR participate,so it seems that the Majorana states ψ

(2)R,L are not well “matched” to this theory,

and we prefer to use the convention (20) for Majorana chirality states.

Digression: Majorana 2-Spinors.

As found in eq. (16), there are only two independent Majorana states, whichimplies that they can be represented by 2-spinors rather than 4-spinors.20,21 Forexample, we could represent Majorana states by the 2-spinor φ =

√E χt, using

the top 2-spinor χt of a Majorana 4-spinor, as in eq. (17) and assuming thatχ†

tχt = 1.

For what it’s worth, the Majorana chirality 2-spinors φR,L, obtained from the top2-spinors of eq. (23), obey coupled Dirac-type equations, i∂μσμφR,L = mφL,R,where σμ = (I,σ). Note that the lefthanded Majorana 2-spinor φL is, to a firstapproximation, an equal mixture of a negative-helicity particle and a positive-helicity antiparticle.

2. Helicity of Neutrinos

In the E1 decay B� → Bγ of a spin-1 excited state B� of a spin-0 nucleus B, thefinal state can only have Sz′ = ±1 along the z′-axis of the photon’s momentum. Thedistribution of angles θ′ of the photon with respect to some z-axis in the rest frame ofthe spin-1 B� nucleus can be determined from the relevant components of the spin-1rotation matrix. In particular, if the B� has Sz = 0, the amplitude to decay to a photonwith Sz′ = ±1 is d1

±1,0 = ∓ sin θ′/√

2, and the angular distribution is (d11,0)

2+(d1−1,0)

2 =sin2 θ′. Similarly, if the B� has Sz = 1, the amplitude to decay to a photon withSz′ = ±1 is d1

±1,1 = (1 ± cos θ′)/2, and the angular distribution is (d11,1)

2 + (d1−1,1)

2 =(1 + cos2 θ′)/2. Of course, if the B� has Sz = −1, the amplitude to decay to a photonwith Sz′ = ±1 is d1

±1,−1 = (1 ± cos θ′)/2, and the angular distribution is(d1

1,−1)2 + (d1

−1,−1)2 = (1 + cos2 θ′)/2 also.

20To discuss Majorana states in, say, beta-decay such as n→ pe−νe, where the n, p and e− are describedby Dirac 4-spinors, it seems best to describe the antineutrino νe, possibly a Majorana state, as a 4-spinoras well.

21Massless neutrinos can be described as two-component Weyl fermions, which also can be representedby 2-spinors (although this is not very useful in calculations where the other spinors are 4-spinors).

13

The spin-1 nucleus B� comes from electron capture, A + e− → B∗ + νe, of a spin-0nucleus A, so Sz of the initial state is that of the electron. If the photon is emittedby the B� opposite to the direction of the νe (or in the same direction as the νe), andthe spins of the e and νe were aligned, then S(B�)

z = 0 = S(γ)z , as shown in the figure

below, which is impossible for a real photon. Rather, the direction of the photon andthe neutrino can be parallel only if the spins of the e and νe are antiparallel, in whichcase the spins of the photon and neutrino are antiparallel.

Turning to the kinematics of the reactions, we start in the rest frame of the B� nucleus,which decays to B + γ. In this frame the photon has energy E ′

γ = P ′γ = P ′

B , and thetotal energy of the B nucleus is

E ′B =

√M2

B + P 2B ≈ MB

P 2B

2MB= MB +

E ′2γ

2MB, (31)

and the photon energy E ′γ is related to the excitation energy E0 = MB� −MB by

E0 = E ′γ + E ′

B −MB ≈ E ′γ +

E20

2MBE ′

γ ≈ E0 − E20

2MB. (32)

The energy-momentum 4-vector of the photon in the rest frame of the B� is

E ′γ(1, sin θ

′, 0,− cos θ′) ≈ E0(1, sin θ′, 0,− cos θ′) (33)

in the convention that angle θ′ is measured with respect to the −z axis (and that theB� moves along the −z axis). To transform this energy to the lab frame, we note thatPB� = Pν = Eν = EK , and the energy of the B� is

EB� =√M2

B� + P 2B� ≈ MB�

(1 +

P 2B�

2M2B�

)= MB�

(1 +

E2K

2M2B�

)≈MB� , (34)

so the velocity of the nucleus B� in the lab frame (where it moves in the −z direction)is related by

βB� =−PB�

EB�

= − Pν

EB�

= − Eν

EB�

= − EK

EB�

≈ − EK

MB�

, |βB�| � 1. (35)

Hence, the Lorentz transformation from the lab frame to the rest frame of the B� has

“boost” γ = 1/√

1 − β2B� ≈ 1, and the energy of the photon in the lab frame is

Eγ ≈ E ′γ(1 − βB� cos θ′) ≈ E0 − E2

0

2MB�

+EKE0

MB�

cos θ′. (36)

14

The range of possible photon energies in the lab frame is therefore

E0 − E20

2MB− EKE0

MB�

< Eγ < E0 − E20

2MB+EKE0

MB�

, (37)

as indicated in Fig. 5, p. 3 (where E1 = E ′γ and both MB and MB� are written as M).

For the reaction γ +B → B� to proceed in the lab frame, energy-momentum conser-vation implies that

pγ + pB = pB�, p2γ + 2pγ · pB + p2

B = p2B� ,

2EγMB +M2B = M2

B� = (MB + E0)2, Eγ = E0 +

E20

2MB. (38)

To have Eγ < Eγ,max from eq. (37), we need

E0 − E20

2MB

+EKE0

MB�

> E0 +E2

0

2MB

, EK > E0MB�

MB

≈ E0. (39)

However, for A = Eu152 and B = Sm152 we have EK = 840 keV, while E0 = 961 keV,so the reaction could not proceed if the excited state B� of Sm152 were long lived, andhad a well-defined mass/energy.

But, since the lifetime of the spin-1 Sm152 excited state is τ = 7 × 10−14 s, its decaywidth is

Γ =hc

cτ=

200 MeV-fermi

3 × 1023 fermi/s · 7 × 10−14 s= 10−2 eV, (40)

Hence, there is some probability that the mass energy of the short-lived excited nucleusis small enough that the reaction can proceed.

From p. 11 of the Notes,

P (E) ∝ 1

(E −ER)2 + Γ2/4, (41)

is the (Breit-Wigner) probability that the short-lived state with nominal (resonant)mass/energy ER appears to have mass/energy E. In the present example, the cen-tral/resonant mass/energy ER of the B� in the reaction γ +B → B� is, from eq. (38),

ER = E0 +E2

0

2MB≈ E0 +

(961 keV)2

2 · 152Mp= E0 + 3.2 eV, (42)

while the highest-energy photon from the decay in flight B� → B+γ has, from eq. (37),

Eγ,max = E0 − E20

2MB+EKE0

MB�

≈ E0 − 3.2 eV +(840 kev)(961 keV)

152 · 940 MeV= E0 − 3.2 eV + 5.6 eV = E0 + 2.4 eV = ER − 0.8 eV. (43)

That is, the maximum photon energy Eγ,max available from the decay in flight, B� →B + γ, is 0.8 eV below the central energy ER for the reaction B� → B + γ. We note

15

that the minimum photon energy Eγ,min available from the decay in flight, B� → B+γ,is

Eγ,min = E0 − E20

2MB− EKE0

MB�

≈ E0 − 3.2 eV − 5.6 eV = E0 − 8.8 eV

= ER − 12.0 eV. (44)

Under the (good) assumption that the distribution of photons from the decay in flight,B� → B+ γ, is uniform in energy, the fraction f of the Breit-Wigner distribution (41)of the energies of the B� in the subsequent reaction γ +B → B� that is overlapped bythe decay photons is

f =∫ Eγ,max

Eγ,min

dE

(E −ER)2 + Γ2/4

/∫ ∞

−∞dE

(E − ER)2 + Γ2/4

=∫ −0.8 eV

−12 eV

dE

E2 + Γ2/4

/∫ ∞

−∞dE

E2 + Γ2/4

=2

Γ

(tan−1 2 · −0.8 eV

Γ− tan−1 2 · −12.0 eV

Γ

)/2π

Γ

=1

π

(tan−1 Γ

1.6 eV− tan−1 Γ

24 eV

)≈ Γ

1.6π eV≈ 10−2

1.6π≈ 0.002. (45)

3. The Ξ0 is an isospin state |I = 12, I3 = 1

2〉 that contains an s quark, so in the

spurion model it combines with the spurion state |I = 12, I3 = −1

2〉, leading to

(|1, 0〉 + |0, 0〉)/√2, while the Ξ− is an isospin state |12,−1

2〉, so it combines with the

spurion leading to |1,−1〉. The final state Λπ0 is the isospin state |1, 0〉, while Λπ− isthe isospin state |1,−1〉. Hence, the spurion model predicts

ΓΞ0→Λπ0

ΓΞ−→Λπ−=

1

2. (46)

The Ξ baryons decay almost exclusively to Λπ, so

ΓΞ0→Λπ0

ΓΞ−→Λπ−≈ τ−Ξτ 0

Ξ

=1.64 × 10−10

2.90 × 10−10= 0.57. (47)

The Ω− is an isospin state |0, 0〉 that contains two s quarks, so in the spurion model itcombines with the spurion state |1

2,−1

2〉, leading to |1

2,−1

2〉. The final state Ξ0π− is the

isospin combination |12, 1

2〉|1,−1〉 = (|3

2,−1

2〉 −√

2|12,−1

2〉)/√3, while the state Ξ−π0 is

the isospin combination |12,−1

2〉|1, 0〉 = (

√2|3

2,−1

2〉 + |1

2,−1

2〉)/√3. Hence, the spurion

model predicts

ΓΩ−→Ξ0π−

ΓΩ−→Ξ−π0

= 2 . (48)

The data are that

ΓΩ−→Ξ0π−

ΓΩ−→Ξ−π0

=23.6

8.6= 2.74. (49)

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4. The Cabibbo-favored W -exchange diagrams for the decays D0 → K−π+ and D0 →K

0π0 are

The decay amplitudes are identical except that the π0 is not uu but (uu− dd)/√

2, so

A(D0 → K0π0) = A(D0 → K−π+)/

√2, and if only these diagrams are relevant,

Γ(D0 → K−π+)

Γ(D0 → K0π0)

= 2. (50)

The present data for D0 branching fractions are

B(D0 → K−π+) = 3.88 ± 0.05%, (51)

B(D0 → K0Sπ

0) = 1.19 ± 0.04%, (52)

B(D0 → K0Lπ

0) = 1.0 ± 0.07%, (53)

Taking K0 ≈ (K0

S − K0L)/

√2, we add the two branching fractions to Kπ0, and have

that

Γ(D0 → K−π+)

Γ(D0 → K0π0)

= 1.77 ± 0.13. (54)

The decay D0 → K+π− is doubly Cabibbo suppressed, with graph:

Hence, we predict

Γ(D0 → K+π−)

Γ(D0 → K−π+)=

sin4 θC

cos4 θC= tan4 13.02◦ = 0.0029. (55)

The branching fraction B(D0 → K+π−) is 0.0137 ± 0.0006%, so the data are that

Γ(D0 → K+π−)

Γ(D0 → K−π+)= 0.0035 ± 0.00015. (56)

17

5. The Cabibbo-favored weak decay of the b quark is to c, so q2 = c.

The mass of the b quark is high enough that the W+ emitted in a weak decay canmaterialize as either ud or cs with Cabibbo-favored coupling cos θC. That is, eitherq3 = d and q4 = u or q3 = s and q4 = c.

For the B+u with q1 = u, the left diagram on p. 4 leads to q1q2 = uc = D

0and

q3q4 = ud = ρ+ (π+) or q3q4 = cs = D+. The right diagram leads to either q1q3 =

ud = ρ+ (π+) and q4q2 = uc = D0

or q1q3 = us = K+ (K�+) and q4q2 = cc = J/ψ (ηc).

The spectator model predicts that the leading 2-body hadronic decay modes of the B+u

are D0ρ+(π+), D

0D+

s and K+(K�+)J/ψ(ηc). We anticipate that the decays to two Dsand to K J/ψ are suppressed somewhat due to reduced final-state phase space.

At http://pdg.lbl.gov/2013/tables/rpp2013-tab-mesons-charm.pdf we learn

B(B+u → D

0ρ+) = 1.34 ± 0.18 × 10−2,

B(B+u → D

0π+) = 4.83 ± 0.15 × 10−3,

B(B+u → D

0D+

s ) = 1.00 ± 0.17 × 10−2,

B(B+u → K�+J/ψ) = 1.44 ± 0.08 × 10−3,

B(B+u → K+J/ψ) = 1.03 ± 0.03 × 10−3,

B(B+u → K�+ηc) = 1.0 ± 0.5 × 10−3,

B(B+u → K+ηc) = 0.96 ± 0.11 × 10−3. (57)

Thus, it seems slightly favored to produce vector mesons over scalar mesons.

The leading spectator diagrams for hadronic 2-body decays of the B0d = db meson are

We infer that the most prominent decays are to D−ρ+, D0ρ0, D−D+

s and K�0J/ψ.

Of these, the first and third have very similar branches, while D0ρ0 is somewhat sup-

pressed, being roughly equal to K�0J/ψ which latter has reduced phase space.

The leading spectator diagrams for hadronic 2-body decays of the B0s = sb meson are

18

We infer that the most prominent decays are to D−s ρ

+, D−s D

+s , φJ/ψ and K

�0D

0.

These branches are all similar, and decreasing in the order listed.

The leading spectator diagrams for hadronic 2-body decays of the B0c = cb meson are

We infer that the most prominent decays are to D+D0, ρ+Jψ and D+

s J/ψ. Of these,only the first is reasonably large.

Cabibbo-suppressed decays have diagrams in which b → u or the W+ → us or cd.The leading spectator decays for these are

and

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6. The 2 × 2 Hamiltonian (in the |B0〉-|B0〉 basis) is

H = M − i

2Γ =

[m M12

M∗12 m

]− i

2

[γ Γ12

Γ∗12 γ

]. (58)

Solving |H − λI| = λ2 − 2H11λ +H211 −H12H21 = 0 for the eigenvalues λ, we find

λ = H11 ±√H12H21 = m± Re

√H12H21 − i

2

(γ ∓ 2 Im

√H12H21

). (59)

The weak eigenstates of the B0-B0

system have essentially identical lifetimes, whichimplies that Γ12 = Γ21 = 0, and hence H12H21 = M12M

�12 = |M12|2 is real. CP is

violated if H12 = M12 has an imaginary part.

The masses of the weak eigenstates are

m± = m± Δm

2= m± |M12|1/2 , (60)

and their common decay rate is just Γ = γ.

We define the eigenstates via the complex coefficients p and q, where |p|2 + |q|2 = 1,according to

|B0±〉 = p|B0〉 ± q|B0〉, |B0〉 =

|B0+〉 + |B0

−〉2p

, |B0〉 =|B0

+〉 − |B0−〉

2q. (61)

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The time evolution of the entangled initial B0-B0

pair produced at an e+e− colliderhas negative charge conjugation, and its time evolution is given by

|B01〉|B0

2〉 →(Fc(t1)|B0

1〉 + iq

pFs(t1)|B0

1〉)(

ip

qFs(t2)|B0

2〉 + Fc(t2)|B0

2〉), (68)

|B0

1〉|B02〉 →

(ip

qFs(t1)|B0

1〉 + Fc(t1)|B0

1〉)(

Fc(t2)|B02〉 + i

q

pFs(t2)|B0

2〉), (69)

|B01〉|B0

2〉 − |B01〉|B0

2〉√2

→ ip

q[Fc(t1)Fs(t2) − Fs(t1)Fc(t2)]

|B01〉|B0

2〉√2

+[Fc(t1)Fc(t2) + Fs(t1)Fs(t2)]|B0

1〉|B02〉√

2

−[Fs(t1)Fs(t2) + Fc(t1)Fc(t2)]|B0

1〉|B02〉√

2

+iq

p[Fs(t1)Fc(t2) − Fc(t1)Fs(t2)]

|B01〉|B0

2〉√2

= eim(t1+t2) e−(t1+t2)/2{− ipq

sinΔm(t1 − t2)

2

|B01〉|B0

2〉√2

+ cosΔm(t1 − t2)

2

|B01〉|B0

2〉 − |B01〉|B0

2〉√2

+iq

psin

Δm(t1 − t2)

2

|B01〉|B0

2〉√2

}. (70)

Note that within eq. (70) the amplitudes vanish for the combinations of two like par-

ticles, |B01〉|B0

2〉 and |B0

1〉|B0

2〉, at times t1 = t2 if the B0-B0

is produced in a CP -oddstate. That is, if both B’s decay at the same time, one must be a B0 and the other a

B0.

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Ph 406: Elementary Particle Physics Problem Set 9

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