# How to determine Gauge Boson charge?

Given the field strength:

$$F^a_{\mu\nu} = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g \; f^{abc} A^b_\mu A^c_\nu$$

We'd get interactions between three and four bosons, former one will look like:

$$g \; f^{abc} (\partial_\mu A^a_\nu - \partial_\nu A^a_\mu) A^{b\mu} A^{c\nu}$$

And the latter one will have $g^2$ and four fields.

For fermions, on other hand the charge enters through this term:

$$i g \; A^a_\mu T^a \psi$$

In this case we say that fermions have $g$ as a charge.

How does one determine charges of gauge bosons in Gauge Theories? Can charge of bosons be related to the charge of fermions?

Apparently it has something to do with global gauge transformation of the fields. Fermions transform as:

$$\psi \rightarrow (1 + i \alpha^a T^a) \psi$$

Gauge bosons:

$$A_\mu^a T^a \rightarrow A_\mu^a T^a + i [\alpha^a T^a, A^b_\mu T^b]$$

In Electroweak theory we are interested following transformation for fermions:

$$exp\left(i \alpha \left( \tau^3 + Y \; \mathbb{1} \right) \right) \psi$$

The infinitesimal transformation of $Z^\pm$ (given $\tau^\pm = \frac{1}{\sqrt{2}} \left( T^1 \pm i T^2 \right)$ and $[\tau^3, \tau^\pm ] = \pm \tau^\pm$) is:

$$Z^\pm \tau^\pm \rightarrow Z^\pm \tau^\pm \pm i \alpha \tau^\pm Z^\pm$$

Thus all in all gauge field should transform as:

$$e^{\pm i \alpha} Z^\pm$$

Is this why we say that $Z^\pm$ boson has charge $\pm 1$? Does it generalize for Yang-Mills in some way?

• See e.g. Schwartz' QFT and the Standard Model, the paragraphs below eq. 29.7 (page 585). – AccidentalFourierTransform Sep 21 '17 at 18:03
• It is funny that you suggested this paragraph since it actually brought me here. Schwartz jumps straight to commutators, and I can't quite understand why... – Darkseid Sep 21 '17 at 18:07
• Funny indeed :-) well, I hope someone clears it up for you. – AccidentalFourierTransform Sep 21 '17 at 18:13

But I think a better perspective is the following. When you say that a field has charge $n$ under the group $U(1)$, you mean that a gauge transformation $e^{i \theta}$ acts on the field by multiplication by $e^{i n \theta}$. In other words, a charge $n$ field transforms in the representation of $U(1)$ labeled by $n \in \mathbb{Z}$. The charge just tells you in which representation of the gauge group the field transforms.
Now the non-abelian generalization is clear : you just have to say in which representation your field transforms. In your example, I think you use $SU(N)$ as gauge group. Then the quarks transform in the fundamental representation, of dimension $N$ (this is why there are three quarks for $SU(3)$, one for each color), the anti-quarks transform in the antifundamental representation, also of dimension $N$. Finally, the gluons transform in the adjoint representation, of dimension $N^2-1$ (for $N=3$, you have 8 gluons).
• It might be worth noting that my original question arose out of the study of Electro Weak interaction, where we select particular gauge boson basis in terms of $Z^a_\mu$ and $B_\mu$. In general does your comment say that it is just a matter of gauge transformation? We transform fermions in fundamental representation and see how bosons must transform to compensate for this? – Darkseid Sep 22 '17 at 3:25
• For your updated question : in electroweak theory, the gauge group is $SU(2) \times U(1)$. If you want to give charges to particles, you have to select a maximal commutative subgroup, isomorphic to $U(1)^2$. One of these $U(1)$s is the electromagnetic one, and indeed observing how the fields transform under it gives you their electromagnetic charge. This is how you see that $W^\pm$ has charge $\pm 1$, and $Z^0$ has charge $0$. – Antoine Sep 24 '17 at 9:34