# Conserved generalization of isospin I_3

1. Mar 14, 2009

### Jonathan Scott

The Gell-Mann - Nishijima relation can be rearranged to show that there is an absolutely conserved quantity which is like a generalization of I3 for anything made of quarks:

$$Q - \frac{B}{2} = I_3 + \frac{(S + C + B^* + T)}{2}$$

Since the LHS terms are absolutely conserved individually, the RHS is as well.

Leptons can also be classified by a similar "weak isospin" quantity which distinguishes the members of a pair of the same flavour, such as electron and electron neutrino:

$$Q + \frac{L}{2} = I_{3W}$$

This suggests that these two results could be combined to produce an overall rule as follows:

$$Q - \frac{B}{2} + \frac{L}{2} = I_{3G}$$

where I3G represents a generalized absolutely conserved equivalent of the third component of isospin, equal to the sum of the first RHS for quarks plus the second RHS for leptons.

Note also that this quantity is half-integer for fermions and integer for bosons.

Is this the same as the third component of the "weak isospin" as usually defined to cover all types of particle? I've had difficulty pinning down the exact definition of that term, but I got the impression that it wasn't necessarily assumed to be absolutely conserved, whereas the quantity defined as above must be absolutely conserved.

(I've long thought that it should be possible to find a quantity like this associated with any fermion wave function which would make it single-valued. That is, if you consider the spinor component of a fermion wave function, and follow it round a complete circuit, it switches sign, making it two-valued. If however the wave function also had some complex scalar phase factor which also changed phase at the same rate but was normally invisible because of some gauge invariance or similar, that factor would also change sign, and the product would effectively give a single-valued fermion wave function.)

2. Mar 14, 2009

### blechman

okay, so far

Your lepton formula is rubbing me the wrong way pretty hard! The formula is most certainly not true for the "weak isospin" that is gauged in the SM (right handed electron: $Q=+1,~L=-1,~I_{3W}=0$)

where are there any bosons in any of this? and why should the fermions have half-integer isospin? what does isospin have to do with spin-statistics?! or are you saying this is just a coincidence?

Last edited: Mar 14, 2009
3. Mar 14, 2009

### blechman

Your lepton formula is correct if you replace L/2 with "-Y", where "Y" is the "weak hypercharge" (note the minus sign).

4. Mar 14, 2009

### Jonathan Scott

I've found inconsistent definitions of the "weak isospin" which is one reason why I've been asking about this. I think the case where I originally found that formula was ignoring the "right-handed" particles and "left-handed" antiparticles (presumably on the grounds that before neutrino oscillations they were assumed not to exist). Does that make it more accurate?

And yes, I'm aware that the weak isospin is defined in terms of the weak hypercharge in the same way as the relationship between ordinary isospin and hypercharge, but as far as I could see, the weak hypercharge is defined in terms of the weak isospin, and I couldn't find a more fundamental definition!

If the relation which I've quoted for leptons is NOT weak isospin (which answers part of my question somewhat more negatively than I anticipated), then assume that what I called I3W is some other quantity defined by that equation, and you still get something which is an absolutely conserved generalization of I3 anyway.

The equation holds for any collection of quarks, antiquarks, leptons and antileptons, not just for individual baryons and leptons.

Yes, I think there is a connection between this quantity and spin. It is certainly odd/even multiples of 1/2 at the same time as the spin, and it is obviously closely related to charge.