I was reading through Mariano Quirós's lecture notes titled "Finite Temperature Field Theory and Phase Transitions". In Sec. 1.2, the author is calculating the one-loop effective potential at $T=0$. On Page 8, he is doing the calculations when gauge bosons are in the loop, and works with a Lagrangian, $$\mathcal{L}=-\frac{1}{4}\mathrm{Tr}\,F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}\mathrm{Tr}\,(D_\mu\phi_a)^\dagger (D^\mu\phi^a)+\cdots\tag{36}$$ where (as mentioned before in the notes), the $\phi_a$ are $N_s$ complex scalar fields. He then says:

The only vertex which contributes to one-loop is $$\mathcal{L}=\frac{1}{2}(M_{gb}^2)_{\alpha\beta}A_\mu^\alpha A^{\mu\beta}+\cdots\tag{39}$$ where $$(M_{gb}^2)_{\alpha\beta}(\phi_c)=g_\alpha g_\beta\mathrm{Tr}\,\left[(T^i_{\alpha\ell}\phi_i)^\dagger T^\ell_{\beta j}\phi^j\right]\tag{40}$$

where $\phi_c$ is the functional derivative of the connected generating functional wrt the source (ref. eqs. (6) and (15)). He then mentions:

... (ii) $T_\alpha$ are the generators of the Lie algebra of the gauge group in the representation of the $\phi$-fields and the trace in (40) is over indices of that representation.

My confusion: I do not quite follow the term "... in the representation of the $\phi$-fields..." Does the presence of the term $T^\ell_{\beta j}\phi^j$ imply that the $T_\alpha$ are expressed as $N_s\times N_s$ matrices? I am also unsure about whether $i$ and $\ell$ refer to the components of such a matrix.

Please let me know if I need to provide more specific information about my confusion.


1 Answer 1


A gauge theory is specified by some (reductive) Lie group $G$, and some (typically reducible) finite-dimensional representation $R$. The latter describes the matter fields.

For example, QCD corresponds to $G=SU(N)$ (with $N=3$) and $R=\square^{\oplus 6}$, where $\square$ denotes the fundamental ($N$-dimensional) representation, and the six copies corresponds to the six flavours. One could consider alternative models, where $R$ is a different representation. Typical examples are $R=\square\!\square$ (the symmetric representation) or $R=\bar\square\!\square$ (the adjoint representation). Incidentally, for three colours, $\square$ and $\begin{matrix}\square\\[-5ex] \square\end{matrix}$ are isomorphic representations, so one need not consider the anti-symmetric (but for large $N$ expansions, it is sometimes useful to replace the fundamental for the anti-symmetric). Also, gluons transform in the adjoint.

With this in mind, we see that matter fields live in a vector space $V$ that realises $R$. For example, $V(\square)\cong \mathbb C^N$ and $V(\square\!\square)\cong\mathbb C^{\frac12N(N+1)}$. That is to say, a fundamental field can be thought of as an $N$-dimensional column vector, and a symmetric field a $\frac12N(N+1)$-dimensional column vector (although it is more common to think of it as a symmetric $N\times N$ matrix, for obvious reasons).

The trace over $R$ means a trace over this vector space. If $M\colon V\to V$ is a matrix in this space, then $\operatorname{tr}_R(M)\sim\sum_IM^IM^I$, where $I=1,2,\dots,\dim(V)$. The normalisation of the trace depends on conventions, but one that is used very often is $$ \operatorname{tr}_R(M):=\frac{1}{x_R}\operatorname{tr}_V(M) $$ where $x_R$ is the Dynkin index of $R$, defined such that $$ \operatorname{tr}_R(t_R^at_R^b)\equiv\delta^{ab} $$ where $t_R^a$ are the generators of $G$ in the representation $R$, and $a$ is an adjoint index. For example, $x_\square=1$ and $x_{\square\!\square}=N+2$. Other conventions may contain extra factors of $1/2$, or drop $x_R$ entirely.

For example, if $M$ is the mass matrix of fundamental quarks, one has $$ \frac12 \phi^\dagger_i M^{ij}\phi^j\equiv\frac12\operatorname{tr}_\square(\phi^\dagger M\phi) $$ where on the l.h.s. the sum is over all colours, $i=1,2,\dots,N$. In the symmetric representation, one can think of $I$ as a pair of colour indices, $I=(ij)$, symmetrised. So $$ \frac12\phi^\dagger_{(ij)}M^{(ij)(kl)}\phi^{(kl)}\equiv \frac12\phi^\dagger_IM^{IJ}\phi^J\equiv\frac12\operatorname{tr}_{\square\!\square}(\phi^\dagger M\phi) $$

In the end, it is just a matter of notation. If you feel more comfortable using explicit indices, go ahead. But what you have to keep in mind is that for each $a$, $t_R^a$ is a $\dim(R)\times\dim(R)$ matrix. For fundamental fields, this is $N\times N$, but not for other representations. The matrix $t_R^a$ has indices in $V$, so $(t^a_R)_{IJ}$ is a complex number, for each $a=1,2,\dots,\dim(\mathfrak g)$, and each $I,J=1,2,\dots,\dim(R)$.

  • $\begingroup$ Your detailed answer resolves my doubt! Thanks so much. $\endgroup$ Commented Sep 2, 2019 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.