Announcements Read 8E-8F, 7.10, 7.12 (me = 0), 7.13

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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

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

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-

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:

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

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

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

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:

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

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j on 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:

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

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

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 ijk and 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

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

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

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

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

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

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