As we all know, a gauge invariant theory is of the form $$ \mathcal{L} = \bar{\psi} \gamma^\mu \left( i\partial_\mu + A_\mu^a T^a\right) \psi.$$ The multiplet $\psi$ and gauge field $A_\mu = A^a_\mu T^a$ transform as follows under a gauge transformation: $$\psi \rightarrow G \psi, \quad \quad A_\mu \rightarrow G A_\mu G^{-1} - (\partial_\mu G) G^{-1},$$ where $G(x)$ is an element of the gauge group. I know that $T^a$ are supposed to be the generators of the gauge group, i.e. they are a basis for the associated Lie Algebra. However, it seems to me that actually this fact is not essential for $\mathcal{L}$ to be gauge invariant! I mean, any old matrices $T^a$ will do; we just say that under a 'gauge transformation' $A_\mu$ transforms as above.

So I am confused. Suppose I was ignorant and all I want to do is construct a gauge invariant theory. It would seem that I should be able to take any matrix (of the right dimension) for $T^a$, or am I wrong? If I am right then, from a constructionist point of view, what is the prime reason for choosing $T^a$ as we normally do?

  • $\begingroup$ To clarify, are you asking why $A_\mu$ are members of the lie algebra associated to the gauge group? $\endgroup$ – J. Murray Apr 5 '20 at 21:59
  • $\begingroup$ You seem to be suggesting that “any old matrices” are a basis for some Lie algebra. This is not the case. $\endgroup$ – G. Smith Apr 5 '20 at 22:03
  • $\begingroup$ @J.Murray Yes please. Why must it be that $A_\mu$ is a general element of the associated Lie Algebra. $\endgroup$ – Rudyard Apr 5 '20 at 22:06
  • $\begingroup$ @G.Smith No I am asking why does $A_\mu$ have to be from the Lie Algebra at all. My point is it doesn't seem necessary for gauge invariance. $\endgroup$ – Rudyard Apr 5 '20 at 22:06

Suppose that the lagrangian $$ \newcommand{\cL}{{\cal L}} \newcommand{\opsi}{{\overline \psi}} \newcommand{\pl}{\partial} \cL=\opsi\gamma^\mu(i\pl_\mu+A_\mu)\psi \hskip2cm A_\mu := \sum_a A^a_\mu T^a \tag{1} $$ is invariant under gauge transformations$^\dagger$ \begin{gather} \psi\to G\psi \tag{2}\\ (i\pl_\mu+A_\mu)\to G(i\pl_\mu+A_\mu)G^{-1} \tag{3} \end{gather} for all $G$ in some matrix group, where the $T^a$ are matrices of the same size. In order for this to make sense, the right-hand side must end up with the same matrices $T^a$ as the left-hand side, because the fields $A_\mu^a$ are the only things being transformed in the last equation. (The components of the matrices are just fixed coefficients in the Lagrangian, like the coefficient $m$ in a mass term.) This gives the requirements \begin{gather} G\pl_\mu G^{-1} = \text{linear combination of }T^a\text{s} \tag{4}\\ GT^a G^{-1} = \text{linear combination of }T^a\text{s}. \tag{5} \end{gather} By taking $G$ to be infinitesimally close to the identity, equation (4) implies that $G$ is generated by the $T^a$s, and equation (5) implies that the commutator of two $T^a$s must be a linear combination of $T^a$s.

$^\dagger$ The second equation in (3) expresses how $A_\mu$ transforms. The partial derivatives on the right-hand side act on both $G^{-1}$ and whatever stands to the right of $G^{-1}$, just like the partial derivative on the left-hand side acts on whatever stands to the right of the closing parenthesis.


Since a gauge field is spacetime-dependent, a translation in spacetime is necessarily accompanied by some change in gauge. Thus, you can think of the gauge covariant derivative ($D_\mu$) as an infinitesimal spacetime translation ($\partial_\mu$) along with an infinitesimal transformation in gauge space (any other terms appearing in $D_\mu$). Then it should hopefully be more clear that the other terms in $D_\mu$ must somehow relate to the gauge group, and cannot be written in terms of completely arbitrary matrices.

To be more concrete, recall/note that an element ${g}$ of a Lie group $G$ is given by \begin{equation} {g}=\exp(iA_\mu^aT^a)\,, \end{equation} where $A_\mu^a$ are the continuous parameters of $G$, and ${T}^a$ its generators. We can use a Taylor expansion to write this as: \begin{equation} {g}={I}+\sum_{n=1}^\infty \frac{1}{n!}\,(iA_\mu^a{T}^a)^n={I}+iA_\mu^a{T}^a+\mathcal{O}\left((A_\mu^a)^2\right), \end{equation} where $I$ is the identity element. For the infinitesimal transformation one takes the leading term in the expansion ($iA_\mu^a{T}^a$), and this is what you see in your example: \begin{equation} \mathcal{L}={\bar\Psi}i\gamma^\mu {D}_\mu {\Psi}={\bar\Psi}i\gamma^\mu (\partial_\mu+iA_\mu^a {T}^a) {\Psi}\,. \end{equation}


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