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Topics in Contemporary Physics Basic concepts 6 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 28, 2015 PART 1 • Brief history • Basic concepts • Colliders & detectors 5σ • From Collisions to papers S ATLA (*) 15 GeV d Selecte s=7 2000 1800 TeV, s=8 ò TeV, ò Ldt = -1 5.9 fb 1600 1400 1200 1000 800 600 10 ATLAS 400 5 150 0 GeV) sample 2 126.5 and 201 fit (m H = 2011 usive Data Bkg incl -1 Sig + nomial 4.8 fb er poly Ldt = 4th ord n diphoto 2400 2200 / Events -1 fb t = 4.8 V: òLd -1 5.8 fb Ldt = TeV: ò Te s=7 s=8 ®4l 100 nary Prelimi 200 250 0 ] [GeV 200 m100 4l - Bkg Even • The Higgs discovery c. Un Syst. 20 H®ZZ Data V ts/5 Ge (*) Data ZZ round s, tt Backg Z+jet round V) Backg 25 Ge l (m H=1 Signa 25 150 140 -100 100 160 V] mg g [Ge 130 0 120 110 • BSM • MVA Techniques • The future L. R. Flores Castillo CUHK January 28, 2015 2 … last time: • Review of relativistic kinematics – Relativistic collision examples – Sticky collision – Explosive collisions – Collider vs fixed target • Symmetries – Symmetry groups – SU(n) • Angular momentum – Orbital and spin angular momenta – Addition of angular momenta L. R. Flores Castillo CUHK January 28, 2015 3 Reminder: interactions QED: QCD: SM Particle Content Cabibbo-Kobayashi-Maskawa matrix Weak: NO Flavor-Changing-Neutral-Currents W/Z: L. R. Flores Castillo W/Z/γ: CUHK January 28, 2015 4 Reminder: Relativistic kinematics Maxwell equations æ v ö t ' = g çt - 2 x ÷ è c ø x ' = g (x - vt) y' = y z' = z Lorentz transformations c for all observers x 0 ' = g (x 0 - b x1 ) Four-vector x m ' = Lnm xn x1 ' = g (x1 - b x 0 ) x2 ' = x2 x 0 = ct, x = x, 1 x3 ' = x3 x 2 = y, x =z 3 bº time-position: proper velocity: energy-momentum: v c E = g mc2 = mc 2 + 12 mv2 + 83 m vc4 +... 4 contravariant I º (x 0 )2 - (x1 )2 - (x 2 )2 - (x3 )2 I = xm x is Lorentz-invariant Energy-momentum a × b º am bm Useful: p = g mv pm pm = (mhm )(mh m ) = m2 c2 E2 pm p = 2 - p2 c m L. R. Flores Castillo Scalar product: m covariant CUHK E = m c +p c 2 xμ = (ct, x, y, z) ημ=dxμ/dτ = γ(c, vx, vy, vz) pμ = mημ = (E/c, px, py, pz) 2 4 2 2 E = g mc 2 For v=c, January 28, 2015 p / E = v / c2 v = pc 2 / E E = hv 5 Symmetry, conservation laws, groups • 1917: Emmy Noether’s theorem: Every symmetry yields a conservation law Conversely, every conservation law reflects an underlying symmetry • A “symmetry” is an operation on a system that leaves it invariant. i.e., it transforms it into a configuration indistinguishable from the original one. The set of all symmetry operations on a given system forms a group: • Closure: If a and b in the set, so is ab • Identity: there is an element I s.t. aI = Ia = a for all elements a. • Inverse: For every element a there is an inverse, a-1, such that aa-1 = a-1a = I • Associativity: a(bc) = (ab)c if commutative, the group is called Abelian L. R. Flores Castillo CUHK January 28, 2015 6 Angular Momentum • Classically, orbital (rmv), spin (Iω) not different in essence. • In QM, – “Spin” interpretation no longer valid – All 3 components cannot be measured simultaneously; and most we can measure: • the magnitude of L ( L2 = L L ). Allowed values: j(j+1)ħ2 • one component (usually labeled “z”) Allowed values: -j,…,j in integer steps – Differences: Orbital angular momentum (l) Allowed values For each particle type • Ket notation: l ml , Spin angular momentum (s) integer integer or half integer any (integer) value fixed s ms , j mj • “A particle with spin 1” : – a particle with s=1 – simple label, not the magnitude of its spin angular momentum: – notice that the magnitude is always a bit larger than the maximum z component L. R. Flores Castillo CUHK January 28, 2015 7 Angular Momentum … in QM: • Orbital (L) Allowed values: – magnitude (L2): l(l+1)ħ2 where l is a nonnegative integer (0, 1, 2,…) where ml is an integer between –l and l: – one component: ml ħ ml = -l, -l+1, …, -1, 0, +1, …, l-1, l • Spin (S) Allowed values: – magnitude – one component: msħ (S2): s(s+1)ħ2 æ 1 3 5 ö with s half-integer or integer çè 0, 2 ,1, 2 , 2, 2 ,...÷ø where ml is an integer between –l and l: ms = -s, -s+1, …, -1, 0, +1, …, s-1, s 2l+1 or 2s+1 possibilities for the measured component. L. R. Flores Castillo CUHK January 28, 2015 8 Angular momentum Example: l=2 • L2 = 2(2+1)ħ2; L=√6 ħ = 2.45ħ • Lz can be 2ħ, ħ, 0, -ħ, -2ħ • Lz cannot be oriented purely in z Similarly, with spin ½: • S2 = (1/2 *3/2 ) ħ2; S = 0.866 ħ • Sz can be –ħ/2, +ħ/2 L. R. Flores Castillo CUHK January 28, 2015 9 Angular momentum Bosons (integer spin) Fermions (half-integer spin) Spin 0 Spin 1 Spin 1/2 Spin 3/2 - Mediators Quarks, Leptons - Elementary Pseudoscalar mesons Vector mesons Baryon octet Baryon decuplet Composite • “Fermion” and “Boson” refer to the rules for constructing composite wavefunctions for identical particles – Boson wavefunctions should be symmetric – Fermion ones should be antisymmetric • Consequences: – Pauli exclusion principle – Statistical properties L. R. Flores Castillo CUHK The connection between spin and statistics is a deep result from QFT. January 28, 2015 10 Addition of Angular Momenta • Angular momentum states represented by ‘kets’: l ml , s ms • Example: an electron in a hydrogen atom occupying: – orbital state | 3 -1 > : – spin state |½ ½ > : l=3, s=½, ml=-1 ms=½ (although specifying s would be unnecessary) • We may need the total angular momentum L+S How to add the angular momenta J1 and J2? J = J1 + J2 • Again, we can only work with one component and the magnitude L. R. Flores Castillo CUHK January 28, 2015 11 Addition of Angular Momenta What do we get if we combine states |j1 m1> and |j2 m2> ? • The z components are simply added: m = m1 + m2 • But the magnitudes depend on J1, J2 relative orientations – If they are parallel, the magnitudes add – If antiparallel, they subtract In general, in between: j = |j1-j2|, |j1-j2|+1, …, (j1+j2)-1, (j1+j2) Examples: – A particle of spin 1 in an orbital state l=4: j=5 (J2=30ħ2), j=4 (J2=20ħ2), j=3 (J2=12ħ2) – a quark and an antiquark bound in a state with zero orbital angular momentum: j=½+½=1, j=½-½=0. “Vector” mesons (ρ, K*, φ, ω) L. R. Flores Castillo CUHK “Pseudoscalar” mesons (π, K’s, η, η’) January 28, 2015 12 Addition of Angular Momenta Adding three angular momenta: first two, then the third • For the two quarks, adding orbital angular momentum l>0: l+1, l, l-1. • Orbital quantum number has to be an integer, hence: – All mesons carry integer spin (and are bosons) – All baryons (3 quarks) must have half-integer spin (fermions) L. R. Flores Castillo CUHK January 28, 2015 13 Addition of Angular Momenta Adding three angular momenta: first two, then the third Adding three quarks in a state with zero orbital angular momentum: From two quarks (each with spin ½) Adding the third one (also spin ½) 1 + ½ = 3 /2 ½+½=1 Decuplet 1–½= ½ Octets ½-½=0 L. R. Flores Castillo 0+½= ½ CUHK January 28, 2015 14 Addition of Angular Momenta Besides total angular momentum, sometimes we need the specific states: j1 m1 j2 m2 = j1+ j2 å Cmj jm1 1j2m2 j m , with m = m1 + m2 j= j1 - j2 Clebsch-Gordan coefficients (Particle Physics Booklet, internet, books, etc.) L. R. Flores Castillo CUHK January 28, 2015 15 Addition of Angular Momenta (a square root sign over each number is implied) m=5/2 m=3/2 m=1/2 m= -1/2 m= -3/2 m= -5/2 j1 m1 j2 m2 = j1+ j2 å Cmj jm1 1j2m2 j m , with m = m1 + m2 j= j1 - j2 L. R. Flores Castillo CUHK January 28, 2015 16 Addition of Angular Momenta (a square root sign over each number is implied) m=5/2 m=3/2 Example: m=1/2 e in a H atom in orbital state |2 -1>, m= -1/2 spin state |½ ½>. If we measure J2, what values might we get, and what is the probability of each? m= -3/2 m= -5/2 Possible values: l+s = 2+½ = 5/2, l-s = 2 – ½ = 3/2 z component: m = -1 + ½ = –½ L. R. Flores Castillo CUHK January 28, 2015 17 Addition of Angular Momenta (a square root sign over each number is implied) m=5/2 m=3/2 Example: m=1/2 e in a H atom in orbital state |2 -1>, m= -1/2 spin state |½ ½>. If we measure J2, what values might we get, and what is the probability of each? m= -3/2 m= -5/2 Possible values: l+s = 2+½ = 5/2, l-s = 2 – ½ = 3/2 z component: m = -1 + ½ = –½ L. R. Flores Castillo CUHK January 28, 2015 18 Addition of Angular Momenta (a square root sign over each number is implied) m=5/2 m=3/2 Example: m=1/2 e in a H atom in orbital state |2 -1>, m= -1/2 spin state |½ ½>. If we measure J2, what values might we get, and what is the probability of each? 2 -1 L. R. Flores Castillo CUHK 1 2 1 2 = 2 5 5 2 m= -3/2 m= -5/2 - 12 - 3 3 5 2 January 28, 2015 - 12 19 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. L. R. Flores Castillo CUHK January 28, 2015 20 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 1 2 1 2 1 2 1 2 1 2 1 2 - 12 - 12 - 12 =11 = 12 1 0 + 1 2 0 0 1 2 = 1 2 0 0 1 2 1 2 1 2 1 2 1 2 1 2 10 - - 12 = 1 -1 Equivalently (solving for states with j=0,1): 11 = 1 2 1 2 10 = 1 2 éë 12 1 -1 = 1 2 0 0 = L. R. Flores Castillo 1 2 1 2 1 2 1 2 - 12 éë 12 Symmetric under 12 1 2 1 2 1 2 CUHK - 12 + 1 2 - 12 1 2 1 2 ùû Spin 1 states “triplet” - 12 1 2 - 12 - 1 2 - 12 1 2 1 2 ùû Antisymmetric Spin 0 state “singlet” January 28, 2015 21 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 1 2 1 2 1 2 1 2 1 2 1 2 - 12 - 12 - 12 =11 = 12 1 0 + 1 2 0 0 1 2 = 1 2 0 0 1 2 1 2 1 2 1 2 1 2 1 2 10 - - 12 = 1 -1 Equivalently (solving for states with j=0,1): 11 = 1 2 1 2 10 = 1 2 éë 12 1 -1 = 1 2 0 0 = 1 2 L. R. Flores Castillo 1 2 1 2 1 2 - 12 éë 12 1 2 1 2 1 2 CUHK - 12 + 1 2 - 12 1 2 1 2 ùû 1 2 - 12 1 2 1 2 ùû - 12 1 2 - 12 - N.B.: we could have read these coefficients from the Clebsch-Gordan table (it works both ways) January 28, 2015 22 Spin ½ • Most important case (p, n, e, all quarks, all leptons) • Illustrative for other cases • For s=½, 2 states: ms=½ (“spin up”, ) or ms= –½ (“spin down”, ) • Better notation: Spinors – two-component æ 1 ö æ 0 ö 1 1 1 1 ÷ ÷ column vectors: 2 2 =ç 2 - 2 =ç è 0 ø è 1 ø • “a particle of spin ½ can only exist in one of these states” Wrong! its general state is æ a ö æ 1 ö æ 0 ö çç ÷÷ = a ç ÷+ bç ÷ è 0 ø è 1 ø è b ø Where α and β are complex numbers, and L. R. Flores Castillo CUHK a + b =1 January 28, 2015 2 2 23 Spin ½ æ a ö æ 1 ö æ 0 ö çç ÷÷ = a ç ÷+ bç ÷ è 0 ø è 1 ø è b ø • The measurement of Sz can only return +½ħ and -½ħ • with probabilities |α|2 and |β|2, respectively • If we now measure Sx or Sy on a particle in this state, – what possible results may we get? +½ħ and -½ħ – what is the probability of each result? L. R. Flores Castillo CUHK January 28, 2015 24 Spin ½ In general, in QM: • Construct the matrix Â representing the observable A • The allowed values of A are the eigenvalues of Â – Eigenvalues: ei, eigenvectors: i • Write the state of the system as a linear combination of the eigenvectors of Â Y = åci i • The probability that a measurement of A would yield the (assuming Y is normalized) value ei is |ci|2 L. R. Flores Castillo CUHK January 28, 2015 25 • Construct matrix Â representing observable A • Allowed values of A are the eigenvalues of Â • Write state as linear combination of these eigenvectors • The probability to measure ei is |ci|2 Spin ½ • Eigenvalues of Sx are ±ħ/2, corresponding to normalized eigenvectors: æ 1 2 ç c± = ç ±1 2 è ö ÷ ÷ ø • any spinor can be written as a linear combination of these eigenvectors: æ a ö æ 12 ö æ 12 ö ÷ + bç ÷ çç ÷÷ = a ç 1 1 è b ø ç 2 ÷ ç - 2 ÷ è by choosing L. R. Flores Castillo a= CUHK 1 2 ø è (a + b ); b = ø 1 2 (a - b ) January 28, 2015 26 Spin ½ • In terms of the Pauli spin matrices: æ 0 1 ö æ 0 -i ö æ 1 0 ö sx =ç ÷, s y = ç ÷, s z = ç ÷ è 1 0 ø è i 0 ø è 0 -1 ø the spin operators can be written as • Effect of rotations on spinors: where U (q ) = e-i(q×s )/2 æ a' ö æ a ö ç ÷ ç ÷ ç b ' ÷ = U (q ) ç b ÷ è ø è ø θ is a vector pointing along the axis of rotation, and its magnitude is the angle of rotation. e A =1+ A+ 12 A2 + 3!1 A3 +... L. R. Flores Castillo CUHK January 28, 2015 27 Spin ½ • These matrices U(θ) are Unitary, of determinant 1. i.e., they constitute the group SU(2) • Spin-½ particles transform under rotations according to the two-dimensional representation of the group SU(2) • Particles of spin 1, described by vectors, transform under the three-dimensional representation of SU(2) • Particles of spin 3/2: described by 4-component objects, transform under the 4d representation of SU(2) • Why SU(2)? the group is very similar (homomorphic) to the SO(3) (the group of rotations in three dimensions). • Particles of different spin belong to different representations of the rotation group. L. R. Flores Castillo CUHK January 28, 2015 28 Flavor Symmetries • Shortly after the discovery of the neutron (1932) • Heisenberg observed that the neutron is very similar to the proton – mp = 938.28 MeV/c2; mn = 939.57 MeV/c2. • Two “states” of the same particle? (the nucleon) • Maybe the mass difference was related to the charge? (it would be the other way around: p would be heavier) • The nuclear forces on them are likely identical • Implementation: æ a ö ÷ N = çç ÷ b è ø L. R. Flores Castillo CUHK æ 1 ö æ 0 p = çç ÷÷ , n = çç è 0 ø è 1 January 28, 2015 ö ÷÷ ø 29 Flavor symmetries • Drawing an analogy with spin, Heisenberg introduced the isospin I (for ‘isotopic’ spin; better term: ‘isobaric’ spin) • I is not a vector in ordinary space (no corresp. to x, y, z); rather, in abstract ‘isospin space’. • Components I1, I2, I3. • Borrowing the entire machinery of angular momentum: – Nucleon carries isospin ½ – The third component, I3, has eigenvalue +1/2, -1/2. p= L. R. Flores Castillo 1 2 1 2 , n= CUHK 1 2 - 12 January 28, 2015 30 Flavor symmetries • Can this possibly work? Physical content: Heisenberg’s proposition that strong interactions are invariant under rotations in isospin space. if so, by Noether’s theorem, isospin is conserved in strong interactions Specifically • Strong interactions invariant under an internal SU(2) symmetry group • Nucleons belong to the two-dimensional representation (hence isospin ½). • Originally a bold suggestion, but plenty of evidence. L. R. Flores Castillo CUHK January 28, 2015 31 Flavor symmetries • Horizontal rows: very similar masses but different charges • We assign an isospin I to each multiplet • And a particular I3 to each member0 of the multiplet: + p = 1 1 , p = 1 0 , p - = 1 -1 • Pion: I=1: L. R. Flores Castillo CUHK January 28, 2015 32 Flavor symmetries • For the Λ, I=0: L. R. Flores Castillo CUHK L= 0 0 January 28, 2015 33 Flavor symmetries Isospin of a multiplet: multiplicity = 2l+1 I3=I for the maximum Q • For the Δ’s, I = 3/2: D++ = L. R. Flores Castillo 3 2 3 2 , D+ = CUHK 3 2 1 2 , D0 = 3 2 - 12 , January 28, 2015 D- = 3 2 - 23 34 Flavor symmetries • Before 1974 (i.e., when only hadrons composed of u, d and s were known), relation between Q and I3: Q = I3 + ½ (A+S) Gell-Mann–Nishijima formula (A: baryon number; S: strangeness) • Originally just an empirical observation; it now follows from isospin assignment for u and d: u= 1 2 1 2 , d= 1 2 - 12 (all other flavors carry isospin 0) L. R. Flores Castillo CUHK January 28, 2015 35 Flavor symmetries It has dynamical implications; for example: • Two nucleons (hence I=1) can combine into – a symmetric isotriplet: – an antisymmetric isosinglet 1 1 = pp 10 = 1 2 ( pn + np) 10 = 1 2 ( pn - np) 1 -1 = nn – Experimentally, p & n form a single bound state (the deuteron) – There is no bound state of two protons or two neutrons – Therefore, the deuteron must be the isosinglet (otherwise all three states would need to occur). – There should be a strong attraction in the I=0 channel, and not in the I=1 channel. L. R. Flores Castillo CUHK January 28, 2015 36 Flavor symmetries • Nucleon-nucleon scattering: 1 1 1 2 (1 0 +0 0 ) 1 -1 p+ p ® d +p + p+n ® d +p 0 n+n ® d +p - 1 1 1 0 1 -1 • From the one in the middle, only the I=1 combination contributes. As a result, the scattering amplitudes are in the ratio: 1 M a : M b : M c =1: • and the cross sections (~ square of s.a.’s): L. R. Flores Castillo CUHK :1 2 s a : s b : s c = 2 :1: 2 January 28, 2015 37 L. R. Flores Castillo CUHK January 28, 2015 38 Flavor symmetries • In the 50’s, as more particles were found, it was tempting to extend this idea, but it became increasingly hard to argue that they were different states of the same particle • The Λ, Ξ’s and Σ’s could be regarded as a supermultiplet, as if they belonged in the same representation of an enlarged symmetry group • SU(2) of isospin would then be a subgroup, but • what was the larger group? • The Eightfold way was Gell-Mann’s solution: The symmetry group is SU(3), the octets are 8D representations of SU(3) the decuplet a 10D representation, etc. L. R. Flores Castillo CUHK January 28, 2015 39 Backup 40 Eigenvectors and eigenvalues A nonzero vector χ is called an eigenvector of a matrix M if Mχ=λχ for some number λ (the eigenvalue). (notice that any multiple of χ is also an eigenvector of M, with the same eigenvalue) L. R. Flores Castillo CUHK January 28, 2015 41