# Can we derive the formula $Q=I_3+\frac{1}{2}(B+S)$ instead of accepting it as an empirical relation?

The electric charge of a quark or lepton, $$Q$$, is related to the third component of the weak Isospin $$T_3$$ and weak hypercharge $$Y$$ according to the formula $$Q=T_3+\frac{Y_W}{2}.\tag{1}$$ This, in a sense, is derived in the Standard electroweak theory, by identifying the unbroken diagonal generator after the spontaneous breakdown of $$SU(2)_L\times U(1)_{Y_W}\to U_Q(1)$$ takes place.

On the other hand, the electric charge $$Q$$ of a quark or a hadron also obeys a similar relation $$Q=I_3+\frac{1}{2}(B+S)\tag{2}$$ where $$I_3$$ denotes the third component of the strong Isospin, $$B$$ and $$S$$ denote the baryon number and the strangeness quantum number, respectively.

Relation $$(1)$$, is derived in the Standard Model in the sense explained above. My question is: "is the relation $$(2)$$, to be derived instead of accepting it as an empirical relation?"

• Jun 10 at 16:46

If you take pure QCD with massless $$u$$, $$d$$ and $$s$$ quarks, hadrons form multiplets of $$U(3)$$. This means that they can be characterized by three quantum numbers that are the eigenvalues of the three generators of the Cartan sub-algebra of $$U(3)$$.

If you take the usual Gell-Mann generators, these would be the identity matrix, $$\lambda_3$$ and $$\lambda_8$$, but any linear combination of these will work. In particular you can use

• $$B = \mathrm{diag}(1/3, 1/3, 1/3)$$
• $$I_3 = \mathrm{diag}(1/2, -1/2, 0)$$
• $$S = \mathrm{diag}(0, 0, -1)$$

This is a perfectly valid basis for the Cartan sub-algebra and it is the one most commonly used for hadrons.

The $$U(3)$$ symmetry is however explicitly broken by the electric charge. In particular, a $$U(1)$$ subgroup of $$U(3)$$ is gauged by the photon. This $$U(1)$$ is the one generated by $$Q=\mathrm{diag}(2/3,-1/3,-1/3)$$

Since this $$U(1)$$ is a subgroup of $$U(3)$$, its generator must be linearly dependent on $$B$$, $$I_3$$ and $$S$$ and in particular you find

$$Q=I_3 + \frac{1}{2}(B+S)$$

The bottom line is that this equation is just telling you that the $$U(3)$$ states are described by three quantum numbers. Among these $$Q$$ is special, because it comes from gauging a specific subgroup of $$U(3)$$, while $$B$$, $$S$$ and $$I_3$$ are an arbitrary parametrization of the Cartan sub-algebra. Since this sub-algebra is $$3$$-dimensional, these quantities must be linearly related.