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?

Update (using answer below):

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?

  • $\begingroup$ See e.g. Schwartz' QFT and the Standard Model, the paragraphs below eq. 29.7 (page 585). $\endgroup$ – AccidentalFourierTransform Sep 21 '17 at 18:03
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    $\begingroup$ 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... $\endgroup$ – Darkseid Sep 21 '17 at 18:07
  • $\begingroup$ Funny indeed :-) well, I hope someone clears it up for you. $\endgroup$ – AccidentalFourierTransform Sep 21 '17 at 18:13

The concept of charge is usually defined in the case of an Abelian (=commutative) gauge group. If the group is non-commutative, one can find a maximal commutative subgroup, and define the charge with respect to this subgroup.

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).

How then to see that on the Lagrangian? Well, just by looking for the right representation! The fundamental acts by multiplication on the left by a matrix of the gauge algebra, while the adjoint is given by the commutator.

  • $\begingroup$ Is it related to any conserved charge? $\endgroup$ – Darkseid Sep 21 '17 at 23:02
  • $\begingroup$ 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? $\endgroup$ – Darkseid Sep 22 '17 at 3:25
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    $\begingroup$ 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 $0$. $\endgroup$ – Antoine Sep 24 '17 at 9:34

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