I've read this post and also this one but I couldn't find my answer.

My question is a stupid one. I know that matter fields in Yang mills theory can be transformed in any representation of gauge group, for example in fundamental representation of in adjoint representation.

I know also that in fundamental representation scalar fields $\phi$ transform according to this rule:

$$\phi_i \xrightarrow{} \phi_i + i \theta^a (T^a_{fund})_{ij} \phi_j.$$

But since dimension of adjoint representation is $N^2-1$ and we have $N$ scalar fields in above expression, how we could transform fields in adjoint representation? Could anyone give me the rule for transformation of matter field in adjoint representation as above?

  • $\begingroup$ " I know that matter fields in Yang mills theory can be transformed in any representation of gauge group" where did you get this from? Usually the matter field has a specific representation. Often it is the fundamental representation. $\endgroup$ Commented Feb 7, 2022 at 9:46

1 Answer 1


A single field does not transform in all representations. A particular field transforms in a particular representation and you can have more than one field, each transforming in their own representations.

A fundamental field $\phi_i$ transforms as $$ \phi_i \to \phi_i - i \theta^a (T^a_{fund})_{ij} \phi_j + O(\theta^2). $$ An adjoint field $\phi^a$ transforms as $$ \phi^a \to \phi^a - i \theta^a(T^a_{adj})^{bc} \phi^c + O(\theta^2) = \phi^a + f^{abc} \theta^b \phi^c + O(\theta^2) . $$ etc.

As a side note, it is often said that a gauge field transforms in the adjoint, but its transformation is not the same as above. This is because the gauge field is a connection. It transforms as $$ A_\mu^a \to A_\mu^a + f^{abc} \theta^b \phi^c + \partial_\mu \theta^a + O(\theta^2) $$ Notice that this transformation is basically the same as that of $\phi^a$ but it has an extra term.

  • $\begingroup$ so a field that transforms in adjoint representation must contain $N^2-1$ components? $\endgroup$ Commented Feb 7, 2022 at 10:18
  • $\begingroup$ yes! A field transforming in representation $R$ has $\text{dim}R$ number of components. $\endgroup$
    – Prahar
    Commented Feb 7, 2022 at 10:18

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