Particle Physics, Fall 2012 Solutions to Final Exam December 11, 2012

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1 Particle Physics, Fall Solutions to Final Exam December, Part I: Short Answer [ points] For each of the following, give a short answer (- sentences, or a formula). [5 points each]. [This one might be har for Sam]: hat is your name?. The cross section ratio ee harons ee is usually about flat, but occasionally unergoes steps up, where the constant value increases to a new, higher constant. hat is causing those steps up? The cross-section is proportional to the sum of all the quark charges less than the appropriate energy. As you increase past the successive quark masses, the ratio increases.. Suppose you ha calculate the amplitue for some quantum mechanical process * involving photons, say X Y, an it takes the form i f, where f is something complicate, an is the polarization of the photon. hat coul you o to check if your answer is gauge invariant? Gauge invariance can be checke by replacing the polarization with its corresponing momentum an see if the result vanishes. In other wors, you check if f q. 4. Isospin is an abbreviation of isotopic spin. hat sort of isotopes can be relate by isospin? For example, can it tell you how the energy levels of 4 He nuclei relate to energy levels of 8 U nuclei? Isospin converts protons to neutrons. Hence it can relate ifferent isotopes with the same mass number A, but it can t relate nuclei with ifferent A. In particular 4 He an 8 U can t be relate. 5. hich of the isospin or SU() flavor is a better approximation to reality? Is either of them a perfect symmetry? Give an argument for why they are imperfect, or not. Isospin is a better symmetry than SU(). e can tell neither is perfect, since if it were,,, an K woul all have the same mass, which they on t. In fact, the an are close in mass (suggesting isospin is goo but not perfect) while the K is rather ifferent.

2 6. e use SU() twice in this class, an calle them SU() F an SU() C. Suppose I ha a re up quark. hat sort or particle might SU() F relate it to? hat sort of particle might SU() C relate it to? SU() F relates quarks of ifferent flavors. So SU() F can relate a re up quark into a re own or re strange quark. SU() C relates quarks of ifferent color, so it can relate a re up quark to a blue or green up quark. 7. The is a strangeness - baryon with charge +. hat is its quark content? Assume it contains only some subset of the three lightest quarks. Since it has strangeness -, it must have one strange quark with charge. Since the total charge is +, the other two quarks must have charge to make the charge work out, so they are up quarks. So uus. 8. Free quarks have never been iscovere. hat happens, qualitatively, if you try to separate the quarks, say, in a meson, like u? If you try to separate them, a flux tube will form between them. As you move them farther apart, the flux tube will have so much energy that it can create a new quark anti-quark pair, an we will just en up with two mesons. 9. A * spin baryon with spin z S woul have its spin an quark content escribe as *, u, s, s s, u, s s, s, u. But this appears to be completely symmetric, yet it involves only fermions. How is this apparent iscrepancy resolve? In aition to the spin an flavor, quarks have color. Though the spin an flavor part is symmetric, the color part, which is not written above, is completely anti-symmetric. So it s okay.. How many colors are there for the electron, electron neutrino, up quark, own quark, gluon, an photon? The electron, electron neutrino, an photon are colorless, so there is just one color (white). There are three colors for the up an own quark. There are eight colors for the gluons.

3 . As you go to higher energies, it is known that the electromagnetic fine structure constant gets stronger. Is this true of the strong coupling as well? No, the strong coupling constant gets weaker at higher energy, leaing to what is calle asymptotic freeom. S. The first theory of weak interactions we came up was the Fermi theory, with a coupling G F which le to Feynman amplitues like i ~ G F u 5 v u 5v. hy i we ultimately abanon this theory? The theory is non-renormalizable, as is evient from the fact that G F is of imension mass to the minus two. As a consequence, the probability increases as you increase your energy, ultimately passing one.. hy are weak interactions weak? In other wors, why are processes like e e so slow? The interaction involves an intermeiate boson, which has a high mass. This leas to a large enominator in the propagator which suppresses the amplitue. 4. hat is the ifference between charge current an neutral current weak interactions? hat particles are responsible for each of them? Charge current interactions always involve particles changing charge by one unit, like or u conversions. Neutral currents o not, they always connect fermions to other fermions of the same type f f. Charge currents are always meiate by -bosons, an neutral current by Z-bosons. 5. hat is the gauge group of the stanar moel? The gauge group is SU SU U, or just SU SU U c L Y.

4 6. Naively, you woul think there is a an up quark fiel, with both left- an righthane parts, an a own quark fiel, with both left- an right-hane quarks. But if it weren t for symmetry breaking, these associations woul be meaningless. hat are the three actual fiels that account for the up quark an own quark? The left hane up an own quarks are together in a weak oublet q L, while the right quarks are separate fiels, u R an R. 7. Because the Higgs fiel is a complex oublet, you woul think that there woul be four egrees of freeom, an hence four istinct particles. But only a single egree of freeom (the Higgs boson) has been iscovere, an this is all that is expecte. hat happene to those other egrees of freeom? The other three egrees of freeom are eaten by the an Z boson, which allows then to become massive, an makes them gain a thir polarization or egree of freeom. 8. It is known experimentally that there is CP violation. here in the stanar moel can we fin a source of CP violation? It exists only in the CKM matrix, that escribes how up-type quarks an own-type quarks are connecte by the -boson. 9. The most common ecay of the Higgs is H bb. Yet this ecay was not the (primary) one that was use at the LHC to etect the Higgs. hy is this ecay ifficult to istinguish? Quarks are prouce in copious quantities via strong interactions, an therefore there is a massive backgroun for this process. This makes it very ifficult to pick out this signal from the noise.. hen we calculate cross-sections like ee ee, we sum over the final spins, average over the initial electron spin, an sum over the initial neutrino spin. hy is this the right thing to o? Neutrinos all are left-hane in the stanar moel. Fortunately, the couplings are zero for the wrong helicity, so summing over both spins only as zero to the total, which is the correct thing to o.

5 Part II: Calculation [ points] Each problem has its corresponing point value marke. Solve the equations on separate paper.. [] Draw the Feynman iagrams relevant for electron-electron scattering, e pe p e pe p4. rite the Feynman amplitue, incluing the correct relative sign. Assume you are at low enough energy that e p e p only QED contributions are important. p p e p e p 4 There are two iagrams, sketche at right. Since they iffer only by switching external fermion lines, there will be a e p e p relative minus sign between them. The electron has charge -, so p p4 the coupling is ie, an the resulting amplitue is e p e p 4 ig i ie u u u u u u u u 4 4 p p p p4 u uu4u u4 uuu ie p p p p 4. ig. [] In SU() F notation, the operators on. particle is B. Fin the effects of all six Ti j e recall that Ti j turns i into j if it acts on a lower inex, an it turns j into i an as a minus sign if it acts on an upper inex. Hence T B B, T B, T B B, T B, T B B B, T B.. [5] Draw all relevant tree-level Feynman iagrams for a pair of gluons to collie an make a quark/anti-quark pair, gq gq q p q p. Correctly label any intermeiate momenta that arise. You on t have to calculate the amplitue. gq g q q p p q q p gq g q q p p q q p q gq q q p g q q p above. The three iagrams, together with the appropriate intermeiate momenta, are rawn

6 4. [] The three pions,, I form an isospin triplet, I =, for which we have, I. (a) ork out I an I for all three particles. The isospin operators always increase or ecrease the charge by exactly one unit. The factor is clearly going to be, so we have I I I,,, I I I,,. (b) The is an isospin singlet, so Ia. If isospin is a perfect symmetry, argue that I. If isospin is a perfect symmetry, then I commutes with the Hamiltonian. Hence I I. (c) Show from part (b) that there is a simple relationship between an. e first note that I. Taking the Hermitian conjugate of this equation, we see that e therefore have I, I,.

7 5. [5] Consier the hypothetical process for prouction of top quarks by colliing neutrinos with positrons, e p p t p f p e 4 where f represents some sort of final state fermion (a) hich possible fermions coul it be? There shoul be multiple correct answers. hich fermion is most likely to be prouce? By conservation of charge, the initial charge was + =, an the quark has charge, which means that the missing charge must be. This implies it must be the anti-particle of one of the own-type quarks, so it is, s, or b. The relevant Feynman iagram is sketche at right There will be a CKM matrix element involve, which will favor the bottom quark over any others in this process. t p (b) Draw the relevant Feynman iagram. (c) rite the Feynman amplitue for this process. For the moment, let q be the momentum of any intermeiate particle that you nee for the process. The Feynman iagram is above. The amplitue can be foun from the rules on the hanout, so we have i g q q M ta q M ie i V v u u v sin e p e p q A p 4 () rite q in two ways in terms of the p i s. From straightforwar conservation of momentum, we have q p p p p4. (e) Your propagator may contain terms that look something like qq M X. Argue that if we treat the positron an neutrino as massless, this term actually oesn t contribute. The relevant term contains a factor of v 5 u q, which we can rewrite as v q u v p p u v p u v p u

8 6. [] Bottom quarks can be prouce via the process e p e p bp bp. 4 Assume we are working at sufficiently high energies that we nee to consier all processes, not just QED. (a) Show that there are three tree iagrams with ifferent intermeiate particles that contribute to this process in the stanar moel. b p e p e p The three iagrams are sketche above. The possible intermeiate particles are the photon, the Z, an the Higgs boson. (b) Argue that one of them is essentially irrelevant. The last of these three involves Higgs couplings, which are proportional to the mass of the particles. The electron is very light, so we can ignore this iagram, since the Higgs coupling will be so small. (c) rite the Feynman amplitue for the other two Using the rules on the sheet, we have 4 i ie ie v u u v ig q ie i g 4 q q M Z v 5 u u 5v4 cos q MZ 4sin sin. 4sin where the momentum q p p. It isn t that har to show that the qq terms vanish because the electrons are nearly massless, but otherwise there is little we can o to simplify it. i ie q e p b p 4 Z b p e p q e p b p 4 H b p e p v 8 u u v 4sin 4 sin s 6sin cos s MZ v 5 u u 5 v 4. q b p 4

9 7. [5] Calculate the ecay rate H cc. For efiniteness, use m 6 GeV, mc.9 GeV, an v 46 GeV. Get both a formula an a number (in MeV) for the process. You may treat the charm quark as massless compare to the Higgs (but on t get zero for the final answer). Don t forget about colors. H Hq c p c p The relevant iagram is sketche at right. The amplitue is i i v uu e square this an sum over spins to yiel mc mc mc mc i uvvu Tr uuvv Tr pp 4 p p. v v v v e then calculate the necessary ot prouct using the fact that Substituting this in, we have H m c m q p p p p. i m m v Note that we are summing over final spins; there is no averaging to be one here. e therefore procee to the final ecay rate in the usual way, namely D p 4 p mm c H mc p i. M mh 6 m H mh v 4v Because the two final particles are massless, their equal momenta will match their energy, an their total energies will be m H, so p m.hence the final ecay rate is H mm c 8v H c This is to any one color; we want the rate to all colors, which throws in a factor of. e therefore have (incluing all colors), H. mm.9 GeV 6 GeV c H 8 v 8 46 GeV GeV.4 MeV.

10 8. [] The B meson is a bottom anti-own combination, b. It is known experimentally that it can spontaneously change into its b Z b b anti-particle, b. Z Z (a) Explain why the treelevel iagrams sketche b b at right o not contribute. b The Z-coupling is iagonal; it cannot connect quarks of ifferent types, which rules out the first two iagrams. The thir iagram oesn t even make sense, since it has two fermions both with an arrow into a vertex, an one with two arrows out of the vertex. (b) Draw at least one one-loop iagram that oes contribute to this process There are two iagrams, as sketche at right. Technically, each of them represents nine iagrams, since there are three possible intermeiate up-type quarks for each intermeiate line. b uct,, uct,, b b uct,, uct,, b

11 9. [5] Suppose that the stanar moel is right, except that (i) the charges are actually slightly off, an (ii) there are right-hane neutrinos, so that the left-hane fermion content of the theory is given by,, 6,,,,,,,,,, (a) Fin the charge of the left-hane up an own quarks using the,, 6, an the charge of the neutrino an electron using,,. The values of T are, so the charges are Q Y T 6 which works out to charges of Q u, Q. Similarly, for the electron an neutrino, we fin Q Y T, so Q, Q e. (b) Fin the charge of the proton, neutron an hyrogen atom. A proton contains two up an a own quark, a neutron contains two owns an an up quark, an a hyrogen atom is a proton plus an electron. Q Q Q, p u Qn Qu Q, Q Q Q. H p e It is interesting that hyrogen comes out neutral, though this will not be true of any other isotope, since other isotopes will contain neutrons. (c) Check if the anomalies Y an YT vanish. e simply calculate this, keeping track of the number of particles. For anything in a oublet has T 4, whereas for SU() singlets, we have YT T, we note that T. So we have 6 Y 6 6 6, A straightforwar but teious computation will emonstrate that Y as well.