# Topics in Contemporary Physics Basic concepts 6 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 28, 2015.

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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

L. R. Flores CastilloCUHK January 28, 2015 PART 1 Brief history Basic concepts Colliders & detectors From Collisions to papers The Higgs discovery BSM MVA Techniques The future 2 5σ

L. R. Flores CastilloCUHK January 28, 2015 … 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 3

L. R. Flores CastilloCUHK January 28, 2015 Reminder: interactions 4 QED: QCD: Weak: W/Z:W/Z/ γ: Cabibbo-Kobayashi-Maskawa matrix NO Flavor-Changing-Neutral-Currents SM Particle Content

L. R. Flores CastilloCUHK January 28, 2015 Reminder: Relativistic kinematics 5 Maxwell equations c for all observers Lorentz transformations is Lorentz-invariant covariant contravariant Scalar product: Four-vector time-position: x μ = (ct, x, y, z) proper velocity: η μ =dx μ /dτ = γ(c, v x, v y, v z ) energy-momentum: p μ = mη μ = (E/c, p x, p y, p z ) Energy-momentumUseful: For v=c, E = hv

L. R. Flores CastilloCUHK January 28, 2015 Symmetry, conservation laws, groups 1917: Emmy Noether’s theorem: Every symmetry yields a conservation law Conversely, every conservation law reflects an underlying symmetry 6 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 -1 a = I Associativity: a(bc) = (ab)c if commutative, the group is called Abelian

L. R. Flores CastilloCUHK January 28, 2015 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 In QM, –“Spin” interpretation no longer valid –All 3 components cannot be measured simultaneously; and most we can measure: the magnitude of L ( L 2 = L  L ).Allowed values: j(j+1)ħ 2 one component (usually labeled “z”)Allowed values: -j,…,j in integer steps –Differences: Ket notation: “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 7 Orbital angular momentum ( l )Spin angular momentum ( s ) Allowed valuesintegerinteger or half integer For each particle typeany (integer) valuefixed

L. R. Flores CastilloCUHK January 28, 2015 Angular Momentum … in QM: Orbital (L) Allowed values: –magnitude (L 2 ): l(l+1)ħ 2 where l is a nonnegative integer (0, 1, 2,…) –one component: m l ħ Spin (S) Allowed values: –magnitude (S 2 ): s(s+1)ħ 2 with s half-integer or integer –one component: m s ħ 8 where m l is an integer between –l and l: m l = -l, -l+1, …, -1, 0, +1, …, l-1, l where m l is an integer between –l and l: m s = -s, -s+1, …, -1, 0, +1, …, s-1, s 2l+1 or 2s+1 possibilities for the measured component.

L. R. Flores CastilloCUHK January 28, 2015 Angular momentum Example: l =2 L 2 = 2(2+1)ħ 2 ; L=√6 ħ = 2.45ħ L z can be 2ħ, ħ, 0, -ħ, -2ħ L z cannot be oriented purely in z Similarly, with spin ½: S 2 = (1/2 *3/2 ) ħ 2 ; S = 0.866 ħ S z can be –ħ/2, +ħ/2 9

L. R. Flores CastilloCUHK January 28, 2015 Angular momentum Bosons (integer spin)Fermions (half-integer spin) Spin 0Spin 1Spin 1/2Spin 3/2 -MediatorsQuarks, Leptons- Elementary Pseudoscalar mesons Vector mesonsBaryon octetBaryon decuplet Composite 10 “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 The connection between spin and statistics is a deep result from QFT.

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Angular momentum states represented by ‘kets’: Example: an electron in a hydrogen atom occupying: –orbital state | 3 -1 > : l=3,m l =-1 –spin state |½ ½ > : s=½,m s =½ (although specifying s would be unnecessary ) We may need the total angular momentum L+S How to add the angular momenta J 1 and J 2 ? J = J 1 + J 2 Again, we can only work with one component and the magnitude 11

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta What do we get if we combine states |j 1 m 1 > and |j 2 m 2 > ? The z components are simply added: m = m 1 + m 2 But the magnitudes depend on J 1, J 2 relative orientations –If they are parallel, the magnitudes add –If antiparallel, they subtract In general, in between: j = |j 1 -j 2 |, |j 1 -j 2 |+1, …, (j 1 +j 2 )-1, (j 1 +j 2 ) Examples: –A particle of spin 1 in an orbital state l =4: j=5 (J 2 =30ħ 2 ), j=4 (J 2 =20ħ 2 ), j=3 (J 2 =12ħ 2 ) –a quark and an antiquark bound in a state with zero orbital angular momentum: j=½+½=1, j=½-½=0. 12 “Vector” mesons (ρ, K*, φ, ω) “Pseudoscalar” mesons (π, K’s, η, η’)

L. R. Flores CastilloCUHK January 28, 2015 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) 13

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Adding three angular momenta: first two, then the third Adding three quarks in a state with zero orbital angular momentum: 14 From two quarks (each with spin ½) ½ + ½ = 1 ½ - ½ = 0 Adding the third one (also spin ½) 1 + ½ = 3 / 2 1 – ½ = ½ 0 + ½ = ½ Decuplet Octets

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Besides total angular momentum, sometimes we need the specific states: 15 Clebsch-Gordan coefficients (Particle Physics Booklet, internet, books, etc.)

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta 16 m=5/2 m=3/2 m=1/2 m= -1/2 m= -3/2 m= -5/2 (a square root sign over each number is implied)

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta 17 m=5/2 m=3/2 m=1/2 m= -1/2 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 Example: e in a H atom in orbital state |2 -1>, spin state |½ ½>. If we measure J 2, what values might we get, and what is the probability of each? (a square root sign over each number is implied)

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta 18 m=5/2 m=3/2 m=1/2 m= -1/2 m= -3/2 m= -5/2 Example: e in a H atom in orbital state |2 -1>, spin state |½ ½>. If we measure J 2, what values might we get, and what is the probability of each? Possible values: l+s = 2+½ = 5 / 2, l-s = 2 – ½ = 3 / 2 z component: m = -1 + ½ = –½ (a square root sign over each number is implied)

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta 19 m=5/2 m=3/2 m=1/2 m= -1/2 m= -3/2 m= -5/2 Example: e in a H atom in orbital state |2 -1>, spin state |½ ½>. If we measure J 2, what values might we get, and what is the probability of each? (a square root sign over each number is implied)

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 20

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 21 Equivalently (solving for states with j=0,1): Spin 1 states Spin 0 state “triplet” “singlet” Symmetric under 1  2 Antisymmetric

L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 22 Equivalently (solving for states with j=0,1): N.B.: we could have read these coefficients from the Clebsch-Gordan table (it works both ways)

L. R. Flores CastilloCUHK January 28, 2015 Spin ½ Most important case (p, n, e, all quarks, all leptons) Illustrative for other cases For s=½, 2 states: m s =½ (“spin up”,  ) or m s = –½ (“spin down”,  ) Better notation: Spinors –two-component column vectors: “a particle of spin ½ can only exist in one of these states” Wrong! its general state is Where α and β are complex numbers, and 23

L. R. Flores CastilloCUHK January 28, 2015 Spin ½ The measurement of S z can only return +½ħ and -½ħ with probabilities |α| 2 and |β| 2, respectively If we now measure S x or S y on a particle in this state, –what possible results may we get? +½ħ and -½ħ –what is the probability of each result? 24

L. R. Flores CastilloCUHK January 28, 2015 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: e i, eigenvectors: Write the state of the system as a linear combination of the eigenvectors of Â The probability that a measurement of A would yield the value e i is |c i | 2 25

L. R. Flores CastilloCUHK January 28, 2015 Spin ½ Eigenvalues of S x are ±ħ/2, corresponding to normalized eigenvectors: any spinor can be written as a linear combination of these eigenvectors: by choosing 26 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 e i is |c i | 2

L. R. Flores CastilloCUHK January 28, 2015 Spin ½ In terms of the Pauli spin matrices: the spin operators can be written as Effect of rotations on spinors: where θ is a vector pointing along the axis of rotation, and its magnitude is the angle of rotation. 27

L. R. Flores CastilloCUHK January 28, 2015 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. 28

L. R. Flores CastilloCUHK January 28, 2015 Flavor Symmetries Shortly after the discovery of the neutron (1932) Heisenberg observed that the neutron is very similar to the proton –m p = 938.28 MeV/c 2 ; m n = 939.57 MeV/c 2. 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: 29

L. R. Flores CastilloCUHK January 28, 2015 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 I 1, I 2, I 3. Borrowing the entire machinery of angular momentum: –Nucleon carries isospin ½ –The third component, I 3, has eigenvalue +1/2, -1/2. 30

L. R. Flores CastilloCUHK January 28, 2015 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. 31

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries Horizontal rows: very similar masses but different charges We assign an isospin I to each multiplet And a particular I 3 to each member of the multiplet: Pion: I=1 : 32

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries For the Λ, I=0: 33

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries For the Δ’s, I = 3 / 2 : 34 Isospin of a multiplet: multiplicity = 2l+1 I 3 =I for the maximum Q

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries Before 1974 (i.e., when only hadrons composed of u, d and s were known), relation between Q and I3: Q = I 3 + 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: (all other flavors carry isospin 0) 35

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries It has dynamical implications; for example: Two nucleons (hence I=1) can combine into –a symmetric isotriplet: – an antisymmetric isosinglet –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. 36

L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries Nucleon-nucleon scattering: From the one in the middle, only the I=1 combination contributes. As a result, the scattering amplitudes are in the ratio: and the cross sections (~ square of s.a.’s): 37

L. R. Flores CastilloCUHK January 28, 2015 38

L. R. Flores CastilloCUHK January 28, 2015 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. 39

Backup 40

L. R. Flores CastilloCUHK January 28, 2015 Eigenvectors and eigenvalues A nonzero vector χ is called an eigenvector of a matrix M if M χ = λ χ for some number λ (the eigenvalue). (notice that any multiple of χ is also an eigenvector of M, with the same eigenvalue) 41

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