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Lecture 5 Symmetries

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Lecture 5 Symmetries Lecture 5 Symmetries • Light hadron masses • Rotations and angular momentum • SU(2 ) isospin • SU(2 ) flavour • Why are there 8 gluons ? • What do we mean by colourless ? FK7003 1 Where do the light hadron masses come from ? Proton (uud ) mass∼ 1 GeV. Quark Q Mass (GeV) π + ud mass ∼ 130 MeV (e) () u- up 2/3 0.003 ∼ u, d mass 3-5 MeV. d- down -1/3 0.005 ⇒ The quarks account for a small fraction of the light hadron masses. [Light hadron ≡ hadron made out of u, d quarks.] Where does the rest come from ? FK7003 2 Light hadron masses and the strong force Meson Baryon Light hadron masses arise from ∼ the stron g field and quark motion. ⇒ Light hadron masses are an observable of the strong force. FK7003 3 Rotations and angular momentum A spin-1 particle is in spin-up state i.e . angular momentum along an 2 ℏ 1 an arbitrarily chosen +z axis is and the state is χup = . 2 0 The coordinate system is rotated π around y-axis by transformatiy on U 1 0 ⇒ UUχup= = = χ down 0 1 spin-up spin-down No observable will change. z Eg the particle still moves in the same direction in a changing B-field regardless of how we choose the z -axis in the lab. 1 ∂B 0 ∂B 0 ∂z 1 ∂z Rotational inv ariance ⇔ angular momentum conservation ( Noether) SU(2) The group of 22(× unitary U* U = UU * = I ) matrices with det erminant 1 . SU (2) matrices ≡ set of all possible rotation s of 2D spinors in space. α 1 0 α ' α Spinor: ψαβ== + , = U ( θ ) β 0 1 β ' β U(θ )= SU (2) rotation matrix θˆ θ U(θ )= cos I − i () θ • σ sin 2 2 θ = direction angle of rotation (right-hand sense) σ=() σ1, σ 2 , σ 3 = Pauli matrices - "generators" of the tr ansformation. Mapping out the states ℏ Eigen values/conserved quantities: S χ= σ χ z 2 3 ℏ 1 0 Eg Sz = ± corresponding to , 2 0 1 Ladder operators: 101 1 00 σ+ =+=() σ12i σ σ− =−=() σ 12 i σ 200 2 10 01 0 1 0 σ 1 σ χ= = = χ + + down00 1 0 up 1 0 σχ−down= σχ + up = 0 Ladder operators "map out" the possible states. 1 0 In this case a doublet , 0 1 Combining states Combine eg e− and e + : 22 = 4 orthogonal states possible. 1 1 Start with the straightforward state: 11 ≡ 0 e− 0 e + 111 011 0 Ladder operator: σ − →1 0 ≡ + 00ee− + 2 100 eee− + −1 e − 1 0110 00 σ − + →1 -1 ≡ 2 1001eeee− + − + 11 ee− + 00 11 σ−= σ + = 0 ⇒ Spin-1 triplet 11ee−+ 00 ee −+ 1 01 1 0 Deduce singlet state: 0 0 ≡ − 2 1 e−0 e + 0 e − 1 e − 1 01 1 0 Checks: + orthogonality σ ± − = 0 1− 0 + 01 − + 2 e e e e In language of group theory : 2⊗ 2 → 31 ⊕ Check orthogonality Eg 1 1 11 ≡= ↑ ↑ and e− e + 0 e− 0 e + 101 1 0 1 0 0 ≡ − =↓↑−↑↓−+ −+ ()ee ee 21e− 0 e + 01 e − e + 2 1 11|00 = ↑↑−+ | () ↓↑−↑↓ −+ −+ 2 ee ee ee 1 1 = ↑↑↓↑−+−+| − ↑↑↑↓ −+−+ | = 0 2eeee 2 eeee = 0-0 = 0 Positronium Write combined states as 111 0110 00 11 ≡ ; 10 ≡ + ; 1-1 ≡ 00ee− + 2 1001 eeee− + − − 11 ee− + Orthopositronium (spin 1 e+ e − state) 1 0 1 1 0 0 0 ≡ − 2 1 e− 0 e + 0 e− 1 e − Parapositronium (spin 0 e+ e − state) H positronium Summary so far… Angular momentum is conserved Multiplets defined by angular momentum values. Invariance to a rotation/SU (2) transformation in physical space. Re-use the template A quantity X is conserved Multiplets defined by values of X. Invariance to a rotation/SU (2) transformation in a new space. SU(2) - isospin ● Proton and neutron have similar masses: mp=0.9383 GeV m n = 0.9396 GeV ● Heisenberg postulated that proton and neutron are two states of the same particle – isospin doublet. Mass differences due to electromagnetic effects. 11 11 p=| > n =−> | (5.34) 22 22 | I I3 > I = isospin quantum number, I3 = quantum number for 3rd component of iso spin. ● Proton and neutron have different projections in internal ”isospin space” but are the same particle in the eyes of the strong force. ● Strong force invariant under rotations in isospin space. ● Isospin conserved for strong interactions (Noether’s theorem). FK7003 12 SU(2) -isospin with quarks Almost exact symmetry of the strong forc e. Assume a two-quark world : u, d . "SU(2) isospin space" mapped out by u , d (fundame ntal representation). 1 0 0 1 u≡ ; d ≡ ; u ≡ ; d ≡− 0 1 1 0 (-ve sign is a technical and (for us) unimportant detail.) I + d u I + u d Use algebra of SU(2) − spin to predict combined qq states. Meson isospin multiplets Triplet and singlet formed from q, q combination: 2231⊗ → ⊕ 1 1 π + =−ud =11 = , 0 0 I 3 0 1 1 1 0 01 π =uu −== dd 1 0 + − 0 + 2 2 01 1 0 π π π 0 0 π − =du =1 -1 = 1 1 η ? π+, π0 , π − all have similar masses ≈ 140 MeV and clearly belong togeth er. 1 1 1 0 01 0 0 =uu += dd − 2 2 01 1 0 A neutral particle with a different mass , η 0 (540 MeV) is a good candidate. Testing isospin ● Best way to understand anything is to look at physical situations ● Two approaches to testing the hypothesis that isospin is a symmetry of the strong force. − Isospin invariance ● If a strong reaction/decay takes place then the reaction/decay of the isospin- rotated particles must also happen at the same rate. ● If a particle within an isospin multiplet is found in nature then the other multiplet members must also exist since they correspond to different projections in isospin space and the strong force is invariant to a rotation in isospin space. − Isospin conservation ● Isospin quantum numbers must ”add up” when considering reactions/decays. ● Both approaches are complementary (Noether’s theorem) FK7003 15 Isospin multiplets If the electromagnetic field could be turned off the masses of the particles within the isospin multiplets would be the same according to isospin symmetry. p (9383 MeV) 1 1 2 2 1 1 n (9396 MeV) − 2 2 π+ (140 MeV) 1 1 π0 (135 MeV) 1 0 π− (140 MeV) 1− 1 I3 Multiplicity of states: N=2I+1 (5.43) FK7003 16 FK7003 17 Rotation in isospin space Slightly more interesting than rotations in real space. A rotation by 180o around the 'y ' -axis in isosp in space converts a proton ↔neutron and π+ ↔ π − . Can measure: (a) ppd+→+π+ and (b) nnd +→+ π − (a) and (b) are the 'same' reaction as seen by the strong force if isospin is a good symmetry. Strong reaction rates are measured to be the same. FK7003 18 Two quarks three quarks Consider u, d , s quarks. Postulate two more "s paces" : I + UV-space and -space d u u↔ d Transformation in I -space. d↔ s Transformation in U -space U + V + u↔ s Transformation in V -space. SU(3) contains 3× SU (2) subspaces. s Only two independent symmetries since UIV++ + ≡ + . ⇒ Expect two quantum numbers. This is SU (3) flavour symmetry. What are we assuming with SU(3) flavour invariance ? In the eyes of the strong force, the up, down and strange quarks are different projections in isospin, U and V space of the same particle. Alternatively : α Quark wave function ψ≡ β γ ⇒ Prob to find a quark in up,down, strange stat es: α2, β 2 , γ 2 Strong force invariant to Uψ transformation. U is 3×3 unitary ( U* U = UU * = I ) matrix with determinant 1. SU (3) transformation. FK7003 20 Scalar meson multiplets in SU(3) Mass Goals: Meson (MeV) (1) Assign spin-0 meson states to +− SU (3) multiplets. π 139.57 π0 134.96 (2) Determine quark content of K+- 493.67 each state. K0 , K0 497.72 2 3 quarks ⇒ 3= 9 qq' combinations. η0 548.8 η0’ 957.6 (Simple) strategy: Apply ladder operators and orthogonality to deduc e multiplet structure. Charged scalar meson states in SU(3) ds us + + I I − V V+ + UU − du ud UU+ + V V− − I −I − su sd Start with, eg, π − ≡ du and apply series of I+−, U +− , V +− st eps. "Map out" states , eg VK+π − = 0 +−+0 − 0 ⇒ ud()π, du() π , usK() , dsK() , usK() , dsK() ⇒ belong to a multiplet. Six states identified - three (neutral) states remaining. Neutral scalar meson states in SU(3) I − V + I + Neutral states V + U + Neutral states at the centre: +−1 +− 1 + 0 1 Iπ =− uuddVK ; =−[] ssuuUK ; =− ssdd 2 2 2 Isopin, U-spin and V -spin individually pre dict a neutral triplet member and a neutral singlet. But there are only 32 - 6 co mbinations left ! Neutral states 0 1 ± π =uu − dd part of isospin triplet with other memb ers π . 2 ⇒ π 0 belongs to the multiplet. 0 Make orthogonal state to π with dd− ss ,[] uu − ss : 8 1 1 η =()dd −+−= ss uu ss dd +− uu2 ss 6 6 Check with la dder operator if η 8 is in the multiplet. Eg I +η8= π + ⇒ η 8 in multiplet 1 1 One more orthogonal state needed ⇒ η =(dd + uu + ss ) 3 U±η1= I ± η 1 = 0 ⇒ η 1 is a singlet. Obs ! There is a freedom in assigning wave functions we picked ours according to convention. Scalar meson multiplets in SU(3) +−+0 − 0 Octet: ud(π), du( π ) , usK( ) , dsK( ) , usK( ) ,, dsK( ) 10 1 0 uudd−()π, dduu + − 2 ss () η 2 6 1 0 Singlet: dd+ uu + ss η ′ 6 ( ) Identify η8≡ η 0 and η 1 ≡ η 0 ′ ⇒ SU(3) flavour: octet and a singlet : 3⊗ 3 → 81 ⊕ Symmetries SU (2) isospin symmetry is a good symmetry because u, d quarks have ∼ the same tiny mass.
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