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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:

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 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

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: