Propagator for traceless gauge field

[ShayanJ]

I'm reading the $\frac 1 N$ chapter of Sidney Coleman's "Aspects of Symmetry". Equation 3.5, gives the gluon propagator as below:
$A^a_{\mu b}(x)A^c_{\nu d}(y)=\left( \delta^a_d \delta^c_b-\frac 1 N \delta^a_b \delta^c_d \right) D_{\mu \nu}(x-y)$.
Then he explains that the term proportional to $\frac 1 N$ is there because gluon field is traceless. But his notation is not standard and in the standard notation ($A^a_{\mu b}\to A_\mu=A^a_\mu T^a$), the gluon propagator is $-i\delta_{ab} D_{\mu\nu}$.
I want to know how can I get from the propagator in one notation to another because the latter propagator has no term in it that seem to come from the gauge fields being traceless.
Is it really because of the notation? Or is it something deeper?
Thanks

 

[samalkhaiat]

I'm reading the $\frac 1 N$ chapter of Sidney Coleman's "Aspects of Symmetry". Equation 3.5, gives the gluon propagator as below:
$A^a_{\mu b}(x)A^c_{\nu d}(y)=\left( \delta^a_d \delta^c_b-\frac 1 N \delta^a_b \delta^c_d \right) D_{\mu \nu}(x-y)$.
Then he explains that the term proportional to $\frac 1 N$ is there because gluon field is traceless. But his notation is not standard and in the standard notation ($A^a_{\mu b}\to A_\mu=A^a_\mu T^a$), the gluon propagator is $-i\delta_{ab} D_{\mu\nu}$.
I want to know how can I get from the propagator in one notation to another because the latter propagator has no term in it that seem to come from the gauge fields being traceless.
Is it really because of the notation? Or is it something deeper?
Thanks

Pay attention to the indices: The fundamental representation indices $a , b = 1, 2 , \cdots , N$, and the adjoint representation indices $A , B = 1 , 2 , \cdots , N^{2}-1$. The gauge field (Hermitian, traceless) matrix, in any representation, is given by $\mathbb{A}_{\mu}(x) = A^{B}_{\mu}(x) T^{B}$ . The matrix elements of $\mathbb{A}_{\mu}$ in the fundamental representation are, therefore, given by $$\left(\mathbb{A}_{\mu}(x) \right)^{a}{}_{b} = A^{B}_{\mu}(x) (T^{B})^{a}{}_{b} .$$ So $$\langle \left(\mathbb{A}_{\mu}(x) \right)^{a}{}_{b} \ \left(\mathbb{A}_{\nu}(y) \right)^{c}{}_{d} \rangle = \langle A^{B}_{\mu}(x) \ A^{C}_{\nu}(y) \rangle \ (T^{B})^{a}{}_{b} (T^{C})^{c}{}_{d} .$$ Using $\langle A^{B}_{\mu}(x) \ A^{C}_{\nu}(y) \rangle \sim \delta^{BC} \mathcal{D}_{\mu\nu}(x - y)$ , you get $$\langle \left(\mathbb{A}_{\mu}(x) \right)^{a}{}_{b} \ \left(\mathbb{A}_{\nu}(y) \right)^{c}{}_{d} \rangle \sim (T^{B})^{a}{}_{b} \ (T^{B})^{c}{}_{d} \ \mathcal{D}_{\mu\nu}(x - y) .$$
The result follows from the $SU(N)$-identity $$(T^{B})^{a}{}_{b} \ (T^{B})^{c}{}_{d} = \delta^{a}_{d}\delta^{c}_{b} - \frac{1}{N} \delta^{a}_{b}\delta^{c}_{d} ,$$ which I have derived some where on PF, with the normalization $\mbox{Tr}(T^{A}T^{B}) = \delta^{AB}$. You can always find the thing you need in one of my posts on PF .