0
$\begingroup$

I have a Lagrangian which contains the term $$\operatorname{tr} \left(D_\mu\phi^{\tilde{a}} D^\mu \phi^{\tilde{a}} \right),$$ where $\phi^{\tilde{a}}=\phi^{\tilde{a}a}T^a$ with the indices $a$ and $\tilde{a}$ denoting the gauge and flavor group respectively. Furthermore, $D_\mu$ is the gauge covariant derivative which can be written as $$ \partial_\mu+ig A_\mu^aT^a $$ I have a hard time writing the above term explicitly, i.e., I am not sure how to assign color labels (without the tilde) to the fields. Intuitively I expand it as $$ \begin{aligned} &\operatorname{Tr}\left( \left[\partial_\mu \phi^{\tilde{a}}+ig A_\mu \phi^{\tilde{a}} \right] \left[ \partial^\mu \phi^{\tilde{a}}+ig A^\mu \phi^{\tilde{a}} \right] \right)\\ & \operatorname{Tr}\left(\partial_\mu \phi^{\tilde{a}a}\partial^\mu\phi^{\tilde{a}b}T^aT^b+2ig\partial_\mu \phi^{\tilde{a}a}A^{\mu b}\phi^{\tilde{a}c}T^aT^bT^c-g^2A_\mu^{a}\phi^{\tilde{a}b}A^{\mu c}\phi^{\tilde{a}d}T^aT^bT^cT^d\right)\\ &=\partial_\mu \phi^{\tilde{a}a}\partial^\mu\phi^{\tilde{a}a}+2ig\partial_\mu \phi^{\tilde{a}a}A^{\mu b}\phi^{\tilde{a}c}\operatorname{Tr}\left(T^aT^bT^c\right)-g^2A_\mu^{a}\phi^{\tilde{a}b}A^{\mu c}\phi^{\tilde{a}d}\operatorname{Tr}\left(T^aT^bT^cT^d\right) \end{aligned} $$ Please correct me if I am wrong.

$\endgroup$

1 Answer 1

0
$\begingroup$

You first need to know under which representation your field transforms, the flavor indices play no role here. Assuming $\phi$ is a scalar that transforms in the adjoint representation of the color group, sort of implied by the expression the OP writes. First we need to understand how the gauge covariant derivative acts on the field. Recall $T^a$ are infinitesimal generators of the Lie algebra, which means the only product that formally exists between them is the Lie bracket $$[T^a, T^b] = \sum_c {\rm i} f^{abc}T^c$$ where $f^{abc}$ are the structure constants. Then we have for example for a given flavor $f$, (which does not need to be manipulated) \begin{align} D_\mu \phi_f(x) &= \sum_a D_\mu^a \phi_f^a(x) T^a \\ &= \sum_a \partial_\mu \phi_f^a(x) T^a + \sum_{a,b} {\rm i}\,g\, \phi^a_f(x) A_\mu^b(x) [T^b,T^a]\\ &= \sum_a \partial_\mu \phi_f^a(x) T^a - \sum_{a,b,c}\,g f^{bac}\phi^a_f(x) A_\mu^b(x) T^c \end{align}

With that you should be able to work out the rest. The issue is, you need to identify the representation of your field. If $\phi$ transformed in the fundamental rep. then it would only have an extra index but no factors of $T^a$.

$\endgroup$
7
  • $\begingroup$ Thank you, I should have mentioned that it indeed concerns the adjoint representation. However, if the only product is the lie-bracket, how come that I frequently encounter terms like $Tr(T^aT^bT^c)$? And for example $Tr{T^aT^b}=\delta^{ab}$? Might be a silly question and I have to get a better understanding of Lie-groups. $\endgroup$
    – user214621
    Sep 1, 2021 at 15:18
  • $\begingroup$ Because in physics we are sloppy and those expressions are actually in a representation $tr[\rho(T^a)\rho(T^b)]$ with the representation $\rho$ and then the multiplication of matrices is the usual one, but only after the group element has been represented $\endgroup$
    – ohneVal
    Sep 1, 2021 at 15:48
  • $\begingroup$ So whenever you write down geometrical and properly defined objects, you only have the bracket. I am afraid within physics, there can be intermediate steps, where it is not obvious that what you have is a bracket, because in practice you use $[A,B]=AB-BA$. $\endgroup$
    – ohneVal
    Sep 1, 2021 at 15:50
  • $\begingroup$ I get it, thank you very much! $\endgroup$
    – user214621
    Sep 1, 2021 at 15:54
  • $\begingroup$ I am still struggling with a short question. For example the non-abelian field strength contains a term $-i g\left[A_{\mu}, A_{v}\right]$. This would be written as (using the trace of the commutator of the generators) $-ig A^a_\muA^b_\nu Tr([T^a,T^b])=gf^{abc}T^c A^a_\mu A^b_\nu,$ but how is this equal to $g f^{a b c} A_{\mu}^{b} A_{v}^{c}$? $\endgroup$
    – user214621
    Sep 1, 2021 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy