What’s the most general traceless HERMITIAN 2 2 matrices? c a ib a+ib c a ib a ib c cc and check out: = a +b +c i i The Pauli matrices completely span the space! Are they orthogonal (independent)? You can’t make one out of any combination of the others!
Sure, a more general and OBVIOUSLY linearly independent set is but not simple in the minimalistic sense…should only NEED 3. And although this set of matrices is all different… …not linearly independent!!
SU(3) FOR What’s the form of the most general traceless HERMITIAN 3×3 matrix? a3a3 a 1 ia 2 a 1 ia 2 Diagonal terms have to be real! a 4 ia 5 a 4 ia 5 a 6 ia 7 a 6 ia 7 Transposed positions must be conjugates! Must be traceless! = a 1 +a 2 +a 3 +a 4 +a 5 +a 6 +a 7 +a i 0 i i i i 0 i 0
a1a1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 The most general traceless Hermitian 3×3 matrix can be composed as shown using the following basis of 8 independent matrices, j, j = 1,8 a· = i=1,8 a i i =
Should be able to construct the SU(3) generators from interwoven subgroups of SU(2) Compare with 1 & 2, 4 & 5, 6 & 7 which leaves for 3 & 8 but which are not linearly independent
The usual convention selects combining these two
SU(3) “ rotations ” occur in an 8-dim “ space ” e i · /2 ħ 8-dim vector 8 3 x 3 generators Only TWO of which can be simultaneously diagonalized In the adopted convention these are 3, 8
This means TWO new sets of EIGENVALUES with which we can specify states (2 new identifiable QUANTUM NUMBERS describing eigenstates in this new 8-dim space) We’ll see these get plotted as 8 3 with “raising” and “lowering” operators m,nm-1,n m-½,n-1
c s du p n + + + + 0 0 + + 0 0
1930sstudies of the binding energies of nuclei direct:precision mass measurements of whole vs parts indirect:energy imparted to daughters in decay cross sections: “nuclear alchemy”: bombarding nuclei with p, n, to manufacture heavier nuclei comparing measurements from accelerated beams incident on nuclei: specifically p vs n (beams or targets) Protons and neutrons seem IDENTICAL w.r.t NUCLEAR FORCES (the nuclear force seems unable to distinguish them) Folding in the calculable effect of the Coulomb force is able to explain all differences
11.13width 0.26 MeV3/2 9.73/ /2 6.56width 1.0 MeV5/ / /2 03/2 Excited states of Li 7 Excited states of Be width 0.29 MeV3/ width 1.2 MeV5/ / /2 03/2 The charge symmetry of nuclear forces is illustrated by the existence of mirror nuclei like Li 7 and Be 7 C 14 and O 14
σ TOT (mb) pp, pn π + p, π + n K + p, K + n
PROTONSNEUTRONS experience the same strong force both are SPIN- ½ particles initial measurements even looked like they had the same mass: m p = MeV m n = MeV 1.3 MeV (0.139% diff) is just barely enough to allow -decay! Suggesting a intimate connection between the quantum mechanical states of p, n charge magnetic moment The basic differences: In the absence of E & M fields become nearly indistinguishable
1932(Heisenberg) suggested p, n: different states of the same fundamental particle the NUCLEON, N pn with some QM probability of observing this nucleon as p or n. basis set in some “nucleon charge space” Then we can assume ALL OPERATORS on the NUCLEON state have a 2×2 matrix representation in this basis:
Nucleon states: To argue that STRONG interactions are INDEPENDENT OF CHARGE means any term in the L representing STONG INTERACTIONS for example any strong potential term (indeed any operator dealing with characteristics of strong interactions) must be independent of charge!! unlike the Lagrangian term for e + e coupling: e
It must ALSO mean that ALL matrix elements (which contribute to strong cross sections or decay rates) must be completely insensitive to how much p or n a nucleon state is. Such matrix elements must be invariant under ANY UNITARY TRANSFORMATION of this p, n basis set whatever that PHYSICALLY might mean, we mathematically must require it. M ′ = U M U † ≡ M U M = M U [ M, U ] = 0 All matrices which characterize the strong force must commute with ALL unitary transformations on this 2-dim “nucleon-charge space.”
That’s U(2) !!! Involves exactly the same transformations as those on SPIN-½ spinor space! In analogy we introduce ISOTOPIC or ISOBARIC SU(2) !!! It was the special part det(U)=1 that has guaranteed lengths and distances were preserved under SO(n) rotations and total probability in SU(n) mixing of angular momentum states. “SPIN” ISOSPIN or I-spin and recognize the NUCLEON states as a I = ½ representation of SU(2) isospin isospin “UP” isospin “DOWN”
isospin “UP” isospin “DOWN” Not the same as the | j m = | ½ ½ spin of either particle To help mark the distinction we write the generators of ISOSPIN rotations as Only a notational difference. These are still the Pauli matrices i but act on a different SPACE!
Because the PAULI MATRICES obey the SAME commutation rules as the MOMENTUM OPERATORS, you learned they have eigenvalues: S 2 =s(s+1) S 3 = -s, -s+1, …, s-1, s 2s+1 spin multiplets So in direct analogy our 2-dim irreducible representation of ISOSPACE has eigenvalues: I 2 =I ( I + 1 ) I 3 = -I, -I+1, …, I-1, I i.e. I = ½, | I | = 3/2 I 3 = ½, +½ (a doublet)
For angular momentum, additional multiplets (higher-dimensional reps) exist! ℓ = 0 ℓ = 1 ℓ = 2 ℓ = 3.. singlet triplet 2 ℓ +1 = 5-dim 2 ℓ +1 = 7-dim And of course even for spin: ℓ = 1/2 ℓ = 3/2 ℓ = 5/2.. doublet 4-dim 6-dim What about ISOSPIN? Does it come in more than I= ½ values?
Nature 163, 82 (1949) Starting in 1947 Mountain top cosmic ray experiments started noting tracks like these C.F.Powell, P.H. Fowler, D.H.Perkins Nature 159, 694 (1947)
The pions appearance followed in many high energy collisions, captured in bubble chambers, and determined to be a spin-0 particle (in fact a pseudo-scalar) come in 3 charge “flavors” all three with nearly identical mass! m = MeV m o = MeV 3 ~“degenerate” states w.r.t. the strong potential Sounds like possibly I = ? 3 + = | 1 1 0 = | 1 0 = | 1 1 Is I 3 just charge?!
ISOMULTIPLETS spin isospin mass charge Particle J I MeV/c 2 states Q I 3 N ½ p n 0 1 3/ + +1 0 0 1 + +1 0 0 1 0 0 1/2 +1/2 1/ 1 3/2 +3/2 +1/2 1/2 3/2
Just like angular momentum, we can combine ISOSPIN states (with the very same Clebsch-Gordon coefficients) when building nuclei, for example | N | N can combine to give: | 1 1 | 1 0 | 1 -1 | 0 0 = pp = nn pn ? np ? In this I=1 triplet state pp and nn are clearly symmetric combinations! = (pn + np) 1212 In fact the raising/lowering operators confirm this selection = (pn np) 1212 But only the DEUTERON exists! No 2 He !
| 1 1 | 1 0 | 1 -1 | 0 0 = pp = nn pn ? np ? = (pn + np) 1212 = (pn np) 1212 A deuteron is spin 1 ( symmetric) ~ ( 1) ℓ where ℓ = 0 (excited states seem to disassociate) So must be anti-symmetric! What is the ISOSPIN of a DEUTERON? 0