PX435 Neutrino Physics 2018 with Sols PDF

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PX435 Neutrino Physics 2018 with Sols...

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PX4350

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THE UNIVERSITY OF WARWICK FOURTH YEAR EXAMINATION: Summer 2018 NEUTRINO PHYSICS

Answer 2 questions Time Allowed: 1.5 hours Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book. Each question of a section is equally weighted. The nominal mark assigned to each part of a question is indicated by means of a bold figure enclosed by curly brackets, e.g. {2} , immediately following that part. This mark scheme is for guidance only and may be adjusted by the examiner. Calculators may be used for this examination

The following information may be used:

Unless otherwise specified, assume that natural units, h ¯ = c = 1, are used. The mass of the electron is me = 0.511 MeV. 2 The values for the mass splitting and mixing angle in the 23−sector are : ∆m23 = (2.54 ± −3 2 ◦ 0.08) × 10 eV and θ23 = (45 ± 7) .

1 Solar Neutrino Unit (SNU) equals 10−36 interactions per target atom per second. The anticommutation relation for any of the γ matrices with γ 5 : {γ µ , γ 5 } = 0. The square of the γ 5 matrix is the identity matrix, 1. In the two flavour approximation, the probability of observing a neutrino of flavour β of energy E (measured in MeV) at a distance L (measured in metres) from a 100% pure να neutrino source can be written as L P (να → νβ ) = sin2 (2θ) sin2 (1.27∆m2 ) E where θ is the mixing angle and ∆m2 = m12 − m22 (measured in eV2 ) with mi being the mass of the i-th mass eigenstate. PX4350

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NEUTRINO PHYSICS Time allowed: 1.5 hours 1.

a) (i) What are cosmological, or relic, neutrinos and why would it be desirable to detect them? Why has it not so far been possible? {3} Answer: (Bookwork : ) Relic neutrinos are leftovers from the processes occurring very soon after the Big Bang[1]. These neutrinos decoupled from thermal equilibrium with electrons and positrons much earlier than the photons we observe in the CMB. Observation of these relics would therefore tell us much about the very early Universe.[1] Due to red-shift, however, the energy of relic neutrinos are on the order of 10−4 eV, making detection practically impossible both from the tiny cross section point of view and the practical difficulties in detecting such low energy neutrinos[1]. (ii) Neutrinoless double-β decay searches are actively pursued in several laboratories with different techniques. Write the decay scheme for an isotope with proton number Z and mass number A and explain why this decay is forbidden in the Standard Model. Describe a typical experiment based on kinematic techniques, and highlight the advantages and disadvantages of this technique. {5} Answer: (Bookwork : ) This decay mode is forbidden in the Standard Model because if violates lepton number conservation[1]. The decay scheme is (Z, A) → (Z + 2, A) + e− + e− [1]. A kinematic experiment explicitly images the final state electrons using standard particle detection techniques[1]. The advantage of this is that it has good background rejection properties as the electrons must be emitted back-to-back[1]. However, these experiments typically have low source mass and are constrained by low statistics[1]. (iii) How can neutrino interactions be detected using the Cerenkov technique? Discuss the merits and liabilities of using this technique and name an experiment which uses this method. {4} Answer: (Bookwork : ) (Maximum of 4 marks) When a charged particle moves through a material faster than the speed of light in that material it radiates light in the form of a cone of Cerenkov radiation with an opening angle that depends on the particle velocity[1]. Water cerenkov detectors exploit this by surrounding a transparent medium (water) with photosensors to detect this cone of light. From the pattern of light detected, the particle type, direction and interaction vertex and particle energy can be measured[1]. The advantage is that this allows huge low cost targets to be used[1]. Additionally any such detector has 4π acceptance[1]. The disadvantages are that only low multiplicity events can be studied[1], neutral particles or particles below the Cerenkov threshold are not visible[1] and that only low rate beams can be studies[1]. b) (i) Give the action of a parity transformation on an arbitrary wavefunction, φ(x, t). {2}

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PX4350 Answer: (Bookwork : ) If Pˆ is the parity operator, then Pˆφ(x, t) → φ(−x, t)[2] (ii) Consider an arbitrary scalar current, S = φφ, and an arbitrary pseudo-scalar current, P = φγ 5 φ. Show that the scalar current is unchanged and the pseudoscalar current changes sign under application of a parity transformation using the parity operator, γ 0 . {6} Pˆ

Answer: (Unseen Problem : ) Under a parity transformation the spinor φ → γ 0 φ, and ˆ

P φ = φ† γ 0 → (Pˆ φ)† γ 0

= (γ 0 φ)† γ 0 = φ† γ 0,† γ 0 = φ† γ 0 γ 0 = φγ 0 [2] Hence, the scalar current transforms as PˆjS = Pˆ (φφ) = (φγ 0 )(γ 0 φ) = φφ = jS [2] since (γ 0 )2 = 1. In the case of the pseudoscalar current PˆjP = Pˆ(φγ 5 φ) = (φγ 0 )γ 5 (γ 0 φ) = −φγ 5 γ 0 (γ 0 φ) = −φγ 5 φ = −jP [2] as γ 5 anticommutes with all other gamma matrices. (iii) Outline an argument as to why the current combination S ± P could be used to introduce parity violation into a model of the weak charged current interaction. Why, instead, have we chosen the V ± A combination of vector and axial-vector currents? {5} Answer: (Bookwork : ) To include parity violation in the standard model one needs to combine a current which change sign under parity, and one which does not. The interference of these terms then will be the parity violating part of the interaction[1]. In the case of the weak charged current, the matrix element could be a combination of two currents j1 and j2 which have a scalar-pseudoscalar form : j ∝ (S ± P ) Under a parity transformation the probability of the interaction goes M 2 ∝ (S ±P ) ∗ (S ±P ). Under a parity transformation, the SS and P P combinations PX4350

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are unchanged, but the SP combination changes sign. This implements parity violation in the standard model[2]. We have chosen to use the combination V ±A because it fits the data better[1]. The angular distribution of outgoing particles in the weak interaction is better described by the vector currents than the scalar. Any answer pointing out that gauge bosons of the theory are required to be vector particles to ensure that electroweak unification works will also attract full marks.

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2.

a) (i) Discuss the merits and liabilities of using scintillators for neutrino detection and name an experiment which uses this method. {4} Answer: (Bookwork : ) A scintillating material emits light when a charged particle passes through it. This light can be detected using photomultiplier tubes or other photosensors. The advantage of using scintillators is that there is no threshold to scintillation; a particle with a very low energy can still cause the material to scintillate[1] - this is good for solar neutrino detectors which have to detect low energy neutrinos. A disadvantage is that the scintillation light is isotropic[1]; scintillator detectors have no intrinsic sensitivity to the direction of the neutrino[1]. A furthur disadvantage is that many scintillators are toxic. An example of an experiment which used this method of detection is KamLAND[1]. (ii) Conventional neutrino beams have a wide energy spread, which can be useful for some experiments but is a problem in oscillation experiments where the oscillation details depend on energy. Outline one method of reducing the energy spread of a beam, and comment on its benefits and liabilities. {4} Answer: (Bookwork : ) (Maximum of 4 marks) There are two ways to answer this: • Move the detectors off-axis to create an OA beam[2]. This method utilises the kinematics of two body pion decay. The neutrino energy spectrum becomes pseudo-monochromatic as one moves away from the beam axis.[1] The advantage is that one can tune the beam the detector sees to a specific energy, and one still has the on-axis beam to perform other tasks[1]. The disadvantage is that the flux drops quite a lot, and that one becomes very sensitive to mismeasurements in the angle.[1] • The other option is : narrow-band beam[2]. In this configuration a magnetic chicane is used to select a narrow range of the secondary meson momenta before the mesons decay to give the neutrino[1]. The advantage is that one understands the beam and absolute flux well[1]. The disadvantage is that the neutrino flux is much lower than for the on-axis or even the off-axis experiments[1]. (iii) What was the atmospheric neutrino anomaly? How was this problem eventually resolved? {4} Answer: (Bookwork : ) Muon and electron neutrinos are produced in cosmic ray air showers. In the 1980’s several experiments measured significantly less muon neutrinos than were expected from air shower models[1]. This deficit was larger for up-coming neutrinos that for down-going neutrinos, and changed depending on zenith angle[2]. The puzzle was resolved using neutrino flavour oscillations[1]. b) You have been asked to design an experiment to measure the mixing angle θ13 by

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making a measurement of the electron neutrino survival probability, 2 P (νe → νe ) = 1 − sin2 2θ13 sin2 (1.27∆m23

L ) E

where θ is the mixing angle and ∆mij2 = mi2 − mj2 with mi being the mass of the 2 . i-th mass eigenstate, at a baseline consistent with the atmospheric mass scale ∆m23 In this equation L is measured in units of metres, E is measured in units of MeV and 2 is measured in units of eV2 . ∆m23 (i) A pure source of νe is hard to find, but a nuclear reactor is a copious generator of νe . What fundamental symmetry or combination of symmetries ensures that P (νe → νe ) = P (νe → νe )? Justify your answer. {2} Answer: (Bookwork : ) CPT[1]. Consider P (νe → νe ). T swaps the time direction : T (P (νe → νe )) = P (νe → νe ). P swaps the chirality of the anti-neutrinos turning them into left-handed antineutrinos, and C turns the antineutrinos into neutrinos: CP T (P (νe → νe )) = P (νe → νe )[1]. Note: CP also works here and would receive one mark. (ii) Why might it be preferable to use a reactor experiment to measure θ13 over an accelerator-based long baseline design. {3} Answer: (Bookwork : ) In principle a reactor experiment is a disappearance measurement that is sensitive only to θ13 [1]. It is not sensitive to δCP or the mass-hierarchy (although the latter depends on the baseline). Hence one has a pure measurement of θ13 [1]. In addition, the reactor experiments know their neutrino flux better than the accelerator experiments.[1] (iii) The peak of the energy spectrum of antineutrinos produced in nuclear reactors occurs at 2.0 MeV. At what distance from the reactor source should the far detector be placed? {3} Answer: (Unseen Problem : ) (similar to problems in problem sets) We would like to put our detector at a position to maximise the disappearance probability[1]. This is maximised when 2 1.27∆m23

π L = [1] E 2

Experimentally we would like the deficit to appear at the peak energy of the neutrino spectrum as this maximises the sensitivity. To fulfill the probability constraint, then, the far detector should be at L=

E π 2 1.27∆m223

This evaluates to L = 990m from the source[1]. (iv) The anti-neutrinos generated by reactors are usually detected using the inverse β-decay process ν e + p → e+ + n. The energy of the final-state positron is of the order of 2 MeV. Could one use the water Cerenkov technique to detect PX4350

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this positron, given that the refractive index of water is approximately 1.3? In general, detectors for this type of experiment use liquid scintillator rather than water. Why might this be the case? {5} Answer: (Unseen Problem : ) The Lorentz factor for a 2 MeV electron is γ = E/m = 2/0.511 = 3.91[1] giving a β of 0.966[1]. The electron will produce Cerenkov radiation if the β > 1/n = 1/1.3 = 0.77[1]. So in principle water cerenkov could be used. However, at these energies a Cerenkov ring is difficult to detect.[1] Scintillator, on the other hand, allows the detection of the final-state neutron as well. This significantly cuts down on the background[1].

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a) The mass of a Dirac-type fermion appears in the Standard Model through a mass term mψψ where ψ is the fermion field and m is the fermion mass. (i) Assuming that the neutrino were a Dirac particle with non-zero mass, show that this mass term implies the existence of a chirally right-handed neutrino. {3} Answer: (Bookwork : ) The fields can be broken up into their chiral components mψψ = m(ψL + ψR )(ψL + ψR ) = mψL ψL + mψL ψR + mψR ψL + mψR ψR [1] Now, we note that 1 − γ5 1 − γ5 ψ) ψ 2 2 1 − γ5† 1 − γ5 γ0 ψ ψ† 2 2 1 − γ5 1 + γ5 ψ† γ0 ψ 2 2 1 ψ† (1 − γ 25 )γ0 ψ 4 0[1]

ψL ψL = ( = = = =

Similarly for the term ψR ψR . Hence mψψ = mψL ψR + mψR ψL [1] and terms involve a chirally right-handed neutrino. (ii) How does the introduction of the Majorana neutrino help with this problem? {1} Answer: (Bookwork : ) Chirally right-handed neutrinos do not exist in the Standard Model. However, chirally right-handed antineutrinos do. If neutrinos were identical to their anti-particles (i.e. Majorana particles) then a chirally righthanded neutrino field could be accommodated naturally into the theory.[1] b) Explain how the effects of CP violation (CPV) in the lepton sector could be measured. How many neutrino families are needed for CPV to manifest in neutrino oscillations? Which parameter(s) of the PMNS matrix should be different from zero for its observation? Could an experiment observe a CPV effect if all neutrinos had identical mass? Justify all your answers. {4} Answer: (Bookwork : ) The existence of CP violation implies that particles behave differently from antiparticles. In the case of neutrinos, we need to compare the oscillation probabilities of neutrinos and antineutrinos[1]. CP violation can only take place if there are at least 3 neutrino families, and if all mixing angles are larger than zero[2] (if reference is only given to θ13 , only one mark will be awarded). If all neutrinos had identical mass the oscillations would not proceed, so CP violation would not be visible, even if the CP violation parameter were large[1]. PX4350

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c) Describe one approach being followed to determine the absolute neutrino mass. {3} Answer: (Bookwork : ) A way of measuring the neutrino mass is to study the electron energy spectrum from β-decay [1]. One method currently being tried (by Katrin) is to use tritium as the source and a large spectrometer to measure the integral electron energy spectrum[1]. A massive neutrino should suppress the end point of this spectrum below that expected from a massless neutrino[1]. One could also mention neutrinoless double beta decay here. d) The Homestake experiment pioneered the field of solar neutrino detection and measured the solar neutrino flux for about 30 years. (i) Describe the experiment and identify three drawbacks to the technique of radiochemical solar neutrino detection. {4} Answer: (Bookwork : ) The Homestake experiment was the first of the radiochemical solar neutrino experiments. Basically a big tank filled with 615 tonnes of cleaning fluid (perchloroethylene, C2 Cl4 ). The idea was that an electron neutrino from the sun would react with the 37 Cl in the liquid to form radioactive 37 Ar via the reaction[1] νe + 37 Cl → 37 Ar + e− The 37 Ar atoms were then chemically extracted from the tank and counted to estimate the number of solar neutrinos that had interacted.[1] There are many drawbacks to this kind of detection technology. The experiments are usually extrememly low rate[1] (Homestake saw one Argon atom a day) and only have the capability of measuring the integrated neutrino flux above the energy threshold of reaction[1]. Any information on the primary neutrino direction, energy or arrival time is unavailable[1] (ii) The observed rate of interactions was 2.56 SNU. Suppose the cross-section for the absorption of an electron neutrino by 37 Cl is σ = 10−42 cm2 . Compute the measured electron neutrino flux. {4} Answer: (Unseen Problem : ) The number of interactions observed in an experiment is given by the equation[1] Nevent = Φν σν NT T where Φν is the neutrino flux in unit of neutrinos per square centimetre per second, σν is the cross section of a neutrino interacting with one target, NT are the number of targets and T is the total amount of observation time in seconds. Given one target atom and one second of observation time, the event rate is then Nevent = Φν σν events per target atom per second[1]. Converting to SNU : Nevent = Φν σν /10−36 in units of 10−36 interactions per target atom per second. The absorption cross section is 10−42 cm2 , so Nevent = 1 × 10−6 Φν SNU[1]. If the event rate were 2.56 SNU, then the neutrino flux must be 2.56/(1 × 10−6 = 2.56 × 106 neutrinos per square centimetre per second.[1] PX4350

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(iii) In the absence of flavour oscillations, solar models predict a solar electron neutrino flux of 10.5 SNU. Comparing this prediction to the observed flux 2.56 SNU, compute the oscillation probability of an electron neutrino to a neutrino of a different flavour. Assuming oscillation in vacuum between two neutrino species and a mixing angle of sin2 (2θ) = 0.86, what value of ∆m2 does Homestake suggest? Assume that the distance from the sun to the earth is 150 × 106 km and that the average energy of the neutrinos that Homestake can see is 1 MeV. {5} Answer: (Unseen Problem : ) The predicted rate of solar neutrino capture is 8.5 SNU. The observed rate is 2.56 SNU. If one attributes this to two flavour oscillations, the oscillation survival probability is 2.56/10.5 = 0.24[1]. The two flavour mixing probability is just L(km) P (να → νβ ) = sin2 (2θ)sin2 (1.27∆m2 (eV )2 ). E(GeV )

(1)

A sun-earth distance of 150 × 106 km and an average neutrino energy of 1 MeV suggests an L/E of 150 × 109 (km/GeV)[1]. If the mixing angle were 0.86, a survival probablity of 0.3 would suggest a[2] ∆m2 of p asin( 0.24/0.86) E/L = 3 × 10−12 eV2 (2) ∆m2 = 1.27 2 (iv) The current best fit to solar neutrino data is ∆m12 = 7 × 10−5 eV2 . Compare this value to the one you just computed, and suggest reasons for any discrepancy. {1} Answer: (Bookwork : ) The current best fit is ∆m2 = 7 × 10−5 eV2 . The discrepancy between the measured value and the naive predicted value based on vacuum oscillation can be explained using matter effects. The electron neutrinos, born in the core of the sun, travel outwards towards vacuum. Somewhere between vacuum and the core, the pass through a region with just the right density to cause resonant oscillation within the solar matter[1]

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