# 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