mon netrino ν μ ß 0 0 ta fi ta netrino ν fi ß 0 0 Particles have corresponing antiparticles which have the opposite spin an charge. Each of the

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1 LECTURE 13 Elementary Particle Physics (The chart on the classroom wall can be fon at sm large.html.) For Fnamental Forces As far as we know, there are jst for fnamental forces in natre: strong, weak, electromagnetic, an gravitational. When we think of these forces in terms of qantm mechanics, we say that each of these forces is meiate by the exchange of a particle. These meiating particles are bosons (integer spin). Yo are familiar with the electromagnetic force. The meiator is the photon an it has spin 1. The photon is massless an has no electric charge. The gravitational force is meiate by the graviton which has spin 2, is massless, an has no electric charge. The fact that the graviton an the photon are massless means that the electromagnetic an gravitational forces are long range. The strong force hols protons an netrons together in the ncles. Until abot 1972, the strong force between ncleons was thoght to be meiate by pions. This is still a goo pictre at low energies. As we shall see, the proton an the netron are each mae of 3 qarks; the strong force hols these qarks together. The strong force between qarks is meiate by the glon which is massless an has spin 1 (like the photon). Even thogh the glon is massless, the strong force is short range becase the glons attract each other an wol rather clmp p an not sprea over all space. The weak force is responsible for particle ecays sch as the ecay of the netron: n! p + e +μν e (1) The weak force is meiate by spin 1 intermeiate vector bosons: W +, W, an Z 0. The sperscripts enote their electric charges. The W + an W each have a mass of 80,280 MeV/c 2 while the Z 0 has a mass of 91,188 MeV/c 2. The relative strengths of the 4 forces are strong 10 electromagnetic 10 2 weak gravitational Leptons The leptons are fermions that are insensitive to the strong force an have spin 1/2. The most familiar lepton is the electron. There are 6 known leptons: Lepton Symbol Mass (MeV/c 2 )) Charge electron e electron netrino ν e ß 0 0 mon μ 105-1

2 mon netrino ν μ ß 0 0 ta fi ta netrino ν fi ß 0 0 Particles have corresponing antiparticles which have the opposite spin an charge. Each of the leptons we liste has an antilepton. For example there are antinetrinos. Antielectrons are calle positrons. Antiparticles are enote by a bar over their symbol. For example μν e is an electron antinetrino. Helicity: If netrinos are massless, then they move at the spee of light. Sppose we choose the irection of motion as the z axis, an look at S z, the component of spin along the z axis. If S z = +1=2 so that the z component of spin points along the irection of motion, then the helicity is +1 an the netrino is right hane. If S z = 1=2 so that the z component of spin points opposite to the irection of motion, then the helicity is 1 an the netrino is left hane. All netrinos are left hane an all antinetrinos are right hane (if netrinos are massless). Harons an Qarks The lifetime of states which ecay via the strong force is on the orer of secons. Particles are efine as those entities which livemch longer than this, say secons or longer. All particles which are sensitive to the strong force are calle harons. Harons are ivie into 2 classes: baryons an mesons. The baryons are fermions an have half integer spin. The mesons are bosons an have integer spin. Harons are mae of qarks. A qark is a fermion with spin 1/2. A baryon consists of 3 qarks, while a meson consists of a qark an an antiqark. (All the harons an qarks have antimatter conterparts.) There are 6 types or flavors" of qarks: p, own, charm, strange, top an bottom. The qarks have fractional charges. Qark Symbol Mass (MeV/c*c) Charge p 5 2/3 own 10-1/3 charm c /3 strange s 180-1/3 top t 180,000 2/3 bottom b /3 For example, a proton consists of. If we a the charges of the constitent qarks, we get Q = +1. If we a the spins of the qarks, S=1/2 is allowe. A netron consists of. The sm of the charges is Q = 0 an the spin is 1/2. A ß + meson consists of μ ;ithascharge +1 an spin 0. A ß meson consists of μ; ithascharge -1 an spin 0. A ß meson is the antiparticle of a ß + meson. A ß 0 meson consists of (μ μ )= p 2; it has charge an spin 0, an is its own antiparticle. Notice that there are 6qarksan 6 leptons which is a nice symmetry. 2

3 Qantm Nmbers In the early ays of particle physics people fon a whole plethora of particles (things living mch longer than secons). They trie to make sense of these particles by arranging them in sensible ways. They also trie to nerstan why they saw the reactions they i an why they in't see other reactions. So they came p with selection rles that sai that certain qantities were conserve in a reaction or ecay. Here is a list of the conserve qantities: ffl Energy an Momentm. ffl Total Anglar Momentm = sm of spin an orbital anglar momentm. ffl Electric Charge Q. ffl Baryon Nmber B: Baryons have baryon nmber B = +1 an their antiparticles have B = 1. ffl Lepton Nmber L: Leptons have lepton nmber L = +1 an their antiparticles have L = 1. So L =+1forthe electron an for ν e, an L = 1 for the positron an for the electron antinetrino. Until recently it was believe that electron lepton nmber L e was conserve, as was mon lepton nmber L μ an ta lepton nmber L fi. However, recent evience for netrino oscillations inicates that one type of netrino can become another type of netrino. It trns ot that this can only happen if netrinos acqire mass. We will learn more abot this later. The 4 qantm nmbers Q, B, an L are goo" in the sense of being exactly conserve in every known reaction. (By the way, there is no conservation of mesons. For example a ß o can ecay into 2 photons.) So if we have a reaction that has the following form A + B! C + D (2) then the total Q, B, anl before the reaction will eqal the total Q, B, an L after the reaction. We will now list qantities that are conserve in strong interactions bt not conserve in electromagnetic an/or weak interactions. ffl Strangeness S an Hypercharge Y : The strangeness qantm nmber was invente to accont for the nsally long ecay lifetime of certain baryons (e.g., Λ o, the ±'s, an the Ξ's,) an certain mesons (kaons). It trns ot that these particles all have at least one strange qark. If a particle has one strange qark, it has S = 1. If it has 2 strange qarks, then its S = 2, etc. If a particle has a strange antiqark, then it has S = +1, etc. Hypercharge Y is efine as the sm of the baryon nmber an the strangeness: Y = B + S (3) 3

4 This is jst a special case of the fact that strong an electromagnetic interactions conserve qark flavor, bt the weak interaction oes not. For example, the strong interaction is involve in the reaction ß (μ)+p + ()! K + (μs)+± (s) (4) Notice that the total strangeness (S=0) is the same on both sies. In ecays, however, the nonconservation of strangeness is very conspicos, becase for many particles this is the only way they can ecay. The Λ, for instance, is the lightest strange baryon; if it is to ecay into lighter particles, strangeness cannot be conserve. Ths the ecay occrs via the weak interaction. Its ecay paths are Λ(s)! p + ()+ß (μ) 64% of the time, an Λ(s)! n()+ß o ((μ μ )= p 2) 36% of the time. Similarly charmness, bottomness, an topness are conserve in strong an electromagnetic interaction, bt not in weak interactions. Hypercharge was efine before charm, bottom, an top were iscovere. It shol be pgrae, bt it's easier jst to think in terms of the constitent qarks. ffl Isotopic spin I an I 3 : In nclear physics isotopic spin I an its thir component I 3 were introce so that the netron an the proton, with nearly eqal masses, col be treate as charge states of the same particle, the ncleon. Ths the netron an the proton were part of an isospin oblet with I =1=2. The proton has I 3 =1=2 an the netron has I 3 = 1=2. In particle physics also the observe masses of the harons form clsters at particlar vales; there are many families (n, p; Ξ, Ξ o ; ±, ± o, ± + ) within which the masses are nearly the same. Apparently, the strong interactions are nearly inepenent ofcharge. It is convenient to regar these families as isospin mltiplets. We se the algebra of orinary spin. So a mltiplet with isotopic spin I has a mltiplicity 2I +1. The thir component I 3 can be se to etermine the charge of the particle sing the formla Q = I 3 + Y 2 (5) This is tre as long as the particle oes not have charm, bottom, or top qarks. The p qark has I 3 = 1=2, the own qark has I 3 = 1=2. They comprise an I = 1=2 oblet. The other qark flavors have I = 0. Electromagnetic interactions o not conserve I, e.g., the proton an the netron both have I = 1=2 bt they behave ifferently electromagnetically (one has charge an the other oesn't). ffl C, P, an T: Charge conjgation C changes a particle to its antiparticle. In particlar, it reverses the sign of the electric charge an magnetic moment of the particle. Any particle that is its own antiparticle has a C parity associate with its wavefnction. The C parity of a state is o if the wavefnction changes sign ner charge conjgation, an is even otherwise. The C parity of a photon is 1 becase the electric fiel proce by a charge (say an electron) changes sign if the particle 4

5 is replace by its antiparticle (a positron). The C parity of the spinless ß o is even becase the ß o ecays into 2 photons, the C parity of 2 photons is ( 1) 2 = +1, an C parity is conserve in electromagnetic interactions. As we have seen, the parity operation P reverses the irection of any spatial vector ~r, that is, P~r = ~r. Invariance of an interaction ner P implies that the interaction is symmetric ner mirror reflection. The intrinsic parity of a particle is o or even epening on whether the wavefnction of the particle is o or even ner P. The parity of a fermion is opposite to that of its antiparticle while the partiy of a boson is the same as its antiparticle. Qarks have positive intrinsic parity, so antiqarks have negative parity. The parity of a composite system in its gron state is the proct of the parities of its constitents. For excite states, there is an extra factor of ( 1)` where ` is the orbital anglar momentm. Ths, in general, the mesons carry a parity of ( 1)`+1. So in the gron state, a pion has P = 1, 2 pions have P = +1 an 3 pions have P = 1. The intrinsic parity of a photon is 1. T stans for time reversal. Experimentally, both strong an electromagnetic interactions are fon to be invariant ner the operations C, P, an T carrie ot separately. The weak interaction is not invariant ner P. (T. D. Lee an C. N. Yang receive the 1957 Nobel Prize for preicting this.) Most reactions are fon to be invariant to the combine operations of CP, except for the very special bt important case of kaon ecay an, more recently, B o meson ecay. (Val Fitch an James Cronin won the 1980 Nobel Prize for iscovering CP violation, observe in K o ecay.) The CPT theorem states that all interactions are invariant ner the combine operation CPT. Its proof is base on fnamental assmptions se in fiel theory. As a conseqence of the CPT theorem, CP violation implies T violation. CP Violation Let s iscss CP violation in more etail. We will iscss the kaon system bt similar consierations apply to the B o system. CP violation was iscovere by stying the netral kaon system. Kaons are typically proce by the strong interactions, in eigenstates of strangeness. K o has strangeness +1 an its antiparticle K o has S = 1. The weak interaction allows these particles to interconvert: K o *) K o (6) As a reslt the particles we normally observe in the laboratory are not K o an K o, bt rather some linear combination of the two. In particlar, we can form eigenstates of CP as follows. The K o 's are pseoscalars which means they are o ner parity: P jk o i = jk o i; P jk o i = jk o i (7) 5

6 Uner charge conjgation, So if we apply CP, we obtain C jk o i =j K o i; CjK o i = jk o i (8) CPjK o i = jk o i; CPjK o i = jk o i (9) So we can constrct normalize eigenstates of CP: Ths jk 1 i = 1 p 2 jk o i jk o i jk 2 i = 1 p 2 jk o i + jk o i (10) CPjK 1 i = jk 1 i an CPjK 2 i = jk 2 i (11) Netral kaons ecay via the weak interaction. Assming CP is conserve in the weak interactions, K 1 can only ecay into a state with CP=+1, whereas K 2 mst ecay into a state with CP= 1. Typically netral kaons ecay into two or three pions. Bt CP conservation reqires that K 1 ecay into two pions (never 3); K 2 ecays into 3 pions (never 2): K 1! 2ß; K 2! 3ß (12) This is becase both 2 an 3 pion states have C =+1bt 2ß has P =+1an 3ß has P = 1. The 2ß ecay is mch faster becase greater energy is release (corresponing to greater phase space for the final states). In fact K 1 has a lifetime of 0: sec while K 2 has a lifetime of sec. So if we start with a beam of K o 's jk o i = 1 p 2 (jk 1 i + jk 2 i) (13) the K 1 component will qickly ecay away, an own the line we shall have a beam of pre K 2 's. Near the sorce where the kaons are mae, we expect to see a lot of 2ß events, bt farther along we expect only 3ß ecays. If at this point we observe a 2ß ecay, we shall know that CP has been violate. Sch an experiment was reporte by James Cronin an Val Fitch in At the en of a beam 57 feet long, they observe 45 two pion events in a total of 22,700 ecays. That's a tiny fraction (1 part in 500), bt it provie nmistakable evience of CP violation. Eviently the long live netral kaon K L is not a perfect eigenstate of CP after all, bt contains a small amixtre of K 1 : 1 jk L i = (jk 2 i + ffljk 1 i) (14) q1+jfflj 2 The coefficient ffl is a measre of natre's epartre from perfect CP invariance; experimentally its magnite is abot 2: Frther measrements have confirme the 6

7 violation of CP. In particlar althogh 34% of all K L 's ecay by the 3ß moe, some 39% go to (a) ß + + e + ν e (b) ß + e + + ν e (15) Both reactions have CP = 1. Notice that CP takes (a) into (b), so if CP were conserve, an K L were a pre eigenstate, (a) an (b) wol be eqally probable. Bt experiments show thatk L ecays more often into a positron than into an electron, by a fractional amont 3: Here for the first time is a process that makes an absolte istinction between matter an antimatter, an provies an nambigos, convention free efinition of positive charge: it is the charge carrie by the lepton preferentially proce in the ecay of the long live netral K meson. CP violation may be responsible for the matter antimatter asymmetry observe in the niverse. As we mentione above, there is no evience of CPT violation which is base on very general assmptions like Lorentz invariance, qantm mechanics an the iea that interactions are carrie by fiels. If CPT is conserve an CP is violate, then time reversal invariance (T) mst be violate. Haron Spectroscopy Back in the ays when people jst knew abot 3 qark flavors (,, an s), Gell Mann pointe ot that there was a systematic way to pt the 3 qarks together to get the known isospin mltiplets. This has been bbe The Eightfol Way." Recall that when we iscsse aing anglar momenta, we sai that aing 2 spin 1/2 objects gives a singlet (S =0)an a triplet (S = 1). So when we combine a qark an an antiqark to make a meson, we canhave a singlet (pseoscalar meson) or a triplet (vector meson). (A pseoscalar changes sign ner a parity operation.) In grop theory notation we can write 2 2 = 1+3 where yo can regar the nmbers as enoting the nmber of components. (Actally, it's the imension of the matrices in the representation of the grop. Spin 1/2 objects are represente by the 2 2 Pali matrices which is a representation of the grop calle SU(2)." ^S x, ^Sy, ^Sz an the ientity matrix are the elements of the grop SU(2). In general, spin is escribe by SU(2); this is tre for any vale of the spin, not jst spin 1/2.) If one regars the 3 qark flavors as components of amltiplet (they are escribe by the grop SU(3)), it trns ot that grop theory tells s that a qark an an antiqark can combine into an octet an a singlet: 3 μ3=8+1. That means that there is an octet an a singlet of pseoscalar mesons, an an octet an a singlet of vector mesons. We can represent this pictorally by ptting the 3 qark flavors at the corners of a triangle: 7

8 Qarks S=0 S=-1 s Q=2/3 Q=-1/3 where S is the strangeness qantm nmber. Similarly the antiqarks are at the corners of an psie own triangle. S=1 s Antiqarks S=0 Q=-2/3 Q=1/3 Now we can sperpose the triangles to form an octet pls a singlet: s s S=+1 S=0 ss s s s Q=+1 S=-1 Q=-1 Q=0 These give the spinless mesons: 8

9 o + K ( s) K ( s ) S=+1 π- ( ) η η π () S=0 o π K- ( s ) K o ( s ) Q=+1 S=-1 Q=-1 Q=0 By sperposing 3 qark triangles, one can get the baryon ecplet an baryon octet. (see pages in The New Physics). In grop theory langage we wol write 3 3 3= The baryon ecplet has 10 baryons, each with spin 3/2. The baryon octet has 8 baryons, each with spin 1/2. This makes sense since we can get total spin 1/2 or 3/2ifwe pt together 3 qarks which each have spin 1/2. It trns ot that there is only one octet of baryons, not two octets, becase there is only one way to write a completely antisymmetric wavefnction for the octet. ψ = ψ(color)ψ(flavor)ψ(spin)ψ(space) (16) Color In the ecplet there are baryons sch as ++ which is mae of (), which consists of (), an Ω which consists of (sss). This appears to violate the Pali exclsion principle becase we appear to be ptting 3 ientical qarks in the same state. To get ot of this problem, the qarks were given an aitional qantm nmber calle color". Each qark flavor comes in 3 colors (re, ble, an green, say). Antiqarks have anticolors. To make a baryon, we simply take one qark of each color, then the three 's in ++ are no longer ientical (one is re, one is green, an one is ble). Since the exclsion principle only applies to ientical particles, the problem evaporates. The allowe qark combinations mst follow the rle that all natrally occrring particles are colorless. By colorless" I mean that either the total amont of each color is zero or all three colors are present in eqal amonts. (The latter case mimics the fact that white light is mae of many colors.) A meson is colorless becase it has a qark of one color (say, ble) an an antiqark with the anticolor (antible). This rle means that we won't fin a particle mae of 2 qarks or 4 qarks. It also tells s that we won't fin inivial qarks in natre. The only colorless combinations yo can make are q μq (mesons), qqq (baryons), an μq μq μq (antibaryons). Yo col have 6 qarks, bt we wol interpret that as a bon state of 2 baryons. 9

10 Actally yo can't separate qarks becase the strong interaction between 2 qarks increases as the istance grows. So it takes too mch energy to separate qarks. With all that energy, qark-antiqark pairs are create. glons qark q q qark q q q q qark qark antiqark qark The glons each carry a color an an anticolor. Since 3 μ3 = 8 + 1, there is an octet of glons: (r μ b + bμr)= p 2, (rμg + gμr) p 2, (rμr b μ b)= p 2, etc. The color singlet ( 1") is given by (rμr + b μ b + gμg)= p 3. The color singlet is not a glon bt is represente by the colorless mesons which meiate the strong interaction at low energies. For example pions meiate the strong interaction between ncleons. Weak Interactions Both qarks an leptons are affecte by weak interactions. There are two kins of weak interactions: charge (meiate by W ± ) an netral (meiate by the Z 0 ). Let's look at leptons first. Leptons: The fnamental charge vertex looks like this: time ν l l- W- A negative lepton(e, μ,orfi )isconverte into the corresponing netrino with the emission of a W (or absorption of a W + ): `! ν` + W. This implies that `+! μν` + W + is also allowe. We can combine the primitive vertices together to make more complicate reactions. For example, μ + ν e! e + ν μ is represente by the iagram: time ν µ e µ W - ν e 10

11 Sch a netrino mon scattering event wol be har to set p in the laboratory, bt with a slight twist essentially the same iagram escribes the ecay of the mon, μ! e + ν μ +μν e, which happens all the time: time ν ν e µ e µ W - Note that a particle traveling backwars in time is an antiparticle. So in the iagram the netrino ν e traveling backwars in time is the antinetrino μν e. The fnamental netral vertex is: time l l Z 0 In this case ` can be any lepton, incling netrinos. The Z 0 meiates netral weak processes sch as netrino electron scattering (ν μ + e! ν μ + e ): time ν µ e ν µ Z 0 e Netrino scattering is extremely ifficlt to observe experimentally. Presmably scattering between 2 electrons can occr with the exchange of a Z 0, bt this is maske by the mch stronger electromagnetic interaction involving the exchange of a photon. Experiments at DESY (in Hambrg) stie the reaction e + e +! μ + μ + at very high energy an fon nmistakable evience of a contribtion from the Z 0. So netral weak processes o occr. Note that the leptonic weak vertices connect members of the same generation: e converts to a ν e with emission of a W, or μ! μ with the emission of a Z 0, bt e never goes to μ nor oes μ go to ν e. In this way electron nmber, mon nmber, an ta nmber conservation are enforce at the leptonic weak vertices. Qarks: For qarks the fnamental charge vertex is: 11

12 time q +2/ 3 q -1/ 3 W - A qark with charge 1=3 (i.e.,, s, or b) converts into the corresponing qark with charge +2=3(, c, ort, respectively), with the emission of a W. The otgoing qark carries the same color as the ingoing one. Bt the flavor changes becase flavor is not conserve in weak interactions. The far en of the W line can cople to leptons (a semileptonic" process), or to other qarks (a prely haronic process). The most important semileptonic process is + ν e! + e: time e W - ν e Becase of qark confinement, this process wol never occr in natre as it stans. However, if we combine the qark with a an a qark, we obtain the beta ecay of the netron (n! p + + e +μν e ): proton ν e e W - netron The weak interaction is not confine to a qark generation. This allows strangeness changing weak interactions, sch as the ecay of the lamba (Λ! p + + ß ), which involves the conversion of a strange qark into an p qark. 12

13 proton π- W - s Λ The qark generations are skewe" for the prposes of the weak interactions. Instea of ψ! ψ! c ψ! t s b the weak force coples pairs ψ 0! ψ c s 0! ψ t b 0 where 0, s 0,anb 0 are linear combinations of the physical qarks, s, an b: 0 0 s 0 b 0 1 C A = 0 V V s V b V c V cs V cb V t V ts V tb 1 0 C B This 3 3 matrix is calle the Kobayashi Maskawa matrix. It is close to being the nit matrix; the iagonal elements are close to nity an the off iagonal elements are small. This matrix allows mixing between qark generations. V measres the copling of to, V s the copling of to s, an so on. Netrino Oscillations Until recently it was believe that netrinos are massless. However there is now evience that netrinos may inee carry a small amont of mass. Among these experiments are those being carrie ot by Hank Sobel (UCI) an his collaborators at Sper Kamiokane in Japan. How o yo measre the mass of a netrino? Yo can't stop it an weigh it. It trns ot that if netrinos have mass, they can transform into another netrino flavor after a while. So as the netrino goes along in space, it changes its flavor, e.g., from ν μ to ν fi. These flavor changes are calle netrino oscillations an it is these oscillations that experiments are trying to observe. To seewhyhaving mass correspons to netrino oscillations, let s assme that netrinos have finite mass (m ν 6= 0). Frther let s sppose that the netrino eigenstates of the weak interaction Hamiltonian H W are not the same as the netrino mass eigenstates. Then 0 ν e ν μ ν fi 13 1 C A! s b 1 C A

14 are the eigenstates of H W an 0 are the mass eigenstates. In other wors their energy is given by ν 1 ν 2 ν 3 1 C A E = qp 2 c 2 + m 2 c 4 ß pc + m2 c 3 2p As a reslt, Schroeinger's eqation (17) iμh ψ t = Hψ (18) implies that these mass eigenstates will have a time epenence given by exp( iet=μh). These two representations are relate by a nitary transformation: 0 ν e ν μ ν fi 1 C A = U 0 ν 1 ν 2 ν 3 1 C A As a simple example, consier mixing two netrinos. ψ! ψ νμ cos sin = sin cos ν fi!ψ ν2 ν 3! (We choose ν μ an ν fi becase that's what Sper Kamiokane is looking at.) We can take H 0 = UHU y an plg it into Schroeinger's eqation iμh ψ = H 0 ψ to see how ν t μ an ν fi will evolve in time. We fin jν μ (t)i = jν 2 (t =0)i cos e ie 2t=μh + jν 3 (t =0)i sin e ie 3t=μh jν fi (t)i = jν 2 (t =0)i sin e ie 2t=μh + jν 3 (t =0)i cos e ie 3t=μh (19) (20) The probability P (ν μ ) of etecting ν μ at time t is given by P (ν μ )=jhν μ (t =0)jν μ (t)ij 2 (21) where at jν μ (t =0)i is pre ν μ. Plgging in eq. (19), we fin that P (ν μ )=1 1 2 sin2 2 " 1 cos (E 2 E 3 ) t μh # (22) Using E i ß pc + m2 i c3,we fin that 2p E 2 E 3 = (m2 2 m 2 3) c 3 2p (23) 14

15 Here we are assming that the netrino momentm p 2 ß p 3 ß p fl m ν c. Then P (ν μ )=1 1 2 sin2 2 " 1 cos ψ (m 2 2 m 2 3) c 3 2pμh! t # (24) So the probability of etecting ν μ oscillates in time if netrinos have mass. Sper K has fon that < m 2 < ev 2 at the 90% confience level. There are other implications for massive netrinos. Recall that all massless netrinos are lefthane an all massless antinetrinos are righthane. If netrinos have mass, it also means that there are some righthane netrinos an some lefthane antinetrinos, thogh none have been fon so far. 15