Understanding Raising and Lowering Operators in SU(3) Representation

1. Announcements • Read 8E-8F, 7.10, 7.12 (me = 0), 7.13 • Friday: 8.2, 8.4, 8.5 • Monday: 9A – 9C Hypercharge: Y = 2(Q – I3) 10/31

2. Raising and Lowering Operators • It is useful to define raising and lowering operators • These are Hermitian conjugates of each other • They raise or lower the charge by one

3. How to think about Raising/Lowering • For the nucleons, there are two particles: • I+increases I3 • I- decreases I3 • The matrix elements tell you what the corresponding factors are I+ I-

4. Other Particles • Anything that interacts strongly must also respect isospin • The commutation relations must still be the same • All possible matrices that satisfy this were worked out in chapter 2:

5. Announcements • Today: 8E-8F, 8.2, 8.4, 8.5 • Monday: 9A – 9C • Wednesday: 8.9, 8.11 11/2

6. SU(3) • The group SU(3) is the set of 3x3 unitary matrices with determinant 1 • Like SU(2), the generators Ta have commutation relations: • The constants fabc are called structure constants • We won’t really be using them much

7. Representations of SU(3) • Just as with SU(2) (isospin or spin), there are a lot of choices of matrices that satisfy the same commutation relations • For SU(2), we normally label them by their “spin”, but we could also label them by their dimension • For isospin, we never get worse than 4x4 matrices • For SU(3), the different choices (representations) are typically denoted by the size of the matrices. • Unfortunately, there are sometimes more than one choice with the same number of dimensions • Some representations of SU(3): • For SU(3)F, we will only be using a very few of these • But I don’t want to write out 8x8 or 10x10 matrices (8 of them!) • Fortunately, a way has been worked out to work use only the conventional 3 representation and figure everything else out

8. Working with the 3 Representation • The 3 or “fundamental” would relate three particles • Because it’s a bigger group, there are two commuting matrices among the eight generators • It’s easy to see how these act on the fundamental: • Similar to SU(2), we combine the remaining six into “raising” and “lowering” operators: • These are not Hermitian: • They turn i into j and give zero on everything else:

9. The 3-bar representation • Consider the anti-particles of these q’s • These will have opposite T3 and T8 charges: • Clearly, this is different from the 3 • But we can work it out from the T’s of the 3 • One change is the minus sign • Another change is that the raising and lowering operators work differently: • Tij turns j into i and throws in a minus sign

10. General Representation • The most general representation has a combination of up and down indices • There can be as many indices as you want • There are some other constraints I won’t explain… • To figure out how T acts on this, you get one term for each index • It affects only one at a time • Tijon a down index turns i to j (or gives zero) • Tij on an up index turns j to i and gives minus • Let’s do a simple problem:

11. The Representations Used • Mesons • Come in octets or singlets • Baryons • Come in octets, decuplets, or singlets • Can be proven: • All the particles in an octet or a decuplet should have the same mass if SU(3) is a good symmetry • The problem: This is false

12. Announcements • Today: 9A – 9C • Wednesday: 8.9, 8.11 • Friday: 9.1, 9.2, 9.4 Problem 8.11(c): Predict mass of eta using naïve formula Redo using correct formula Correct formula should work better, but still doesn’t work that well 11/5

13. Mass Terms • Consider matrix elements for mass of the 8 representation for the baryons • In a manner similar to Lorentz invariance, you can show that the only terms that respect SU(3) symmetry are ij and ijk and ijkand combinations • However, because of the way the B’s are put into combinations, the W-term never contributes • If this expression were valid, it would make all the masses equal to X. • Not experimentally correct

14. Mass Breaking in the 8 • Gell-Mann suggested that there must be another term contributing to the mass • It must commute with isospin (T1, T2 and T3) • It must be proportional to T8 • The mass is therefore

15. Mass For the 10 • A similar argument could be applied to the 10 • There are many ways you can apply the indices, but because B*ijk is completely symmetric on its indices, they are all equivalent • Once again, this would imply that all ten masses are equal (false) • Once again, we need to include a mass-breaking term • Our goal: understand this formula

16. Using the Gell-Mann – Okubo formula • If we knew X and Y, we should be able to predict all ten masses • But many of them are already related by isospin • Let’s try to get a formula for the mass of the ’s • Which one? It doesn’t matter, so pick the easiest • Now let’s go for the *’s: • Because all the terms in m are diagonal, only the matching terms contribute

17. Using Gell-Mann – Okubo (2) • We calculate the remaining two formulas • We now work to eliminate the variables X and Y • The first formula acted as a check… • The second allowed them to predict the mass of the undiscovered - • It was discovered to be at 1672 MeV, 0.5% off

18. Questions from the Reading Quiz “Also, it seems that some particles can be made of the same quarks, like the sigma + and lambda 0, why do we consider those different particles? I feel like the quark composition is essentially what "makes" the particle, like the number of protons in an atom "makes" the element.” • Recall that a particle’s state is defined by its type, momentum, and spin • As fermions, we must also make the state completely anti-symmetric • The momentum part you can think of like the space wave-function • For the lowest energy states, this part is automatically symmetric • Like the 1s state of hydrogen • But we have to make sure the spin is right, and adds up properly for the baryon

19. Questions from the Reading Quiz • A simpler example might be helpful: