2
$\begingroup$

I am following some notes on the computation of the vector two point function in QCD and I would like somebody to make some intermediate steps more explicit. Let's consider

$$\Pi_{\mu\nu}=i\mu^{2\epsilon}\int{}d^dx\,{}e^ {iqx}\langle\Omega|T\{j_{\mu}(x)j_{\nu}(0)\}|\Omega\rangle=(q_{\mu}q_{\nu}-\eta_{\mu\nu}q^2)\Pi,$$

where $\mu$ is a mass scale, $\epsilon$ is the regulator defined in $d=4-2\epsilon$ where $d$ is the dimensionality of space-time and $j_{\mu}(x)=\bar{q}(x)\gamma_{\mu}q(x)$.

The quantity I want to compute is $\Pi$. To do that we first multiply the equation above with $\eta^{\mu\nu}$ on both sides to obtain

$$\Pi=\frac{-i\mu^{2\epsilon}}{(d-1)q^2}\int{}d^dx\,{}e^ {iqx}\langle\Omega|T\{j_{\mu}(x)j^{\mu}(0)\}|\Omega\rangle=\ldots$$

My notes claim that this leads to

$$\ldots=\frac{-iN_c\mu^{2\epsilon}}{(d-1)q^2}\int{}d^dx\,{}e^ {iqx}Tr[S(x)\gamma_{\mu}S(-x)\gamma^{\mu}],$$

where $N_c$ is the number of colors and $S(x)$ is the free quark propagator.

I want somebody to make the steps between the last two equations explicit.

$\endgroup$
2
  • 1
    $\begingroup$ $\uparrow$ Which notes? $\endgroup$
    – Qmechanic
    Commented May 8, 2015 at 21:55
  • $\begingroup$ @Qmechanic sorry but they are not available online $\endgroup$
    – Yossarian
    Commented May 9, 2015 at 12:28

1 Answer 1

0
$\begingroup$

Let's begin with

$$\langle\Omega|T\{j_{\mu}(x)j_{\nu}(0)\}|\Omega\rangle=\langle\Omega|T\{[\bar{q}(x)\gamma_{\mu}q(x)][\bar{q}(0)\gamma^{\mu}q(0)]\}|\Omega\rangle=$$

$$=\langle\Omega|T\{\bar{q}(x)_a^i(\gamma^{ij})_{\mu}q(x)_a^j\bar{q}(0)_b^k(\gamma^{kl})^{\mu}q(0)_b^l\}|\Omega\rangle=\ldots$$

where I have made the $i,j,k,l$ spinor indices and the $a,b$ color indices explicit. Reordering this we can write $$=\langle\Omega|T\{(\gamma^{ij})_{\mu}q(x)_a^j\bar{q}(0)_b^k(\gamma^{kl})^{\mu}q(0)_b^l\bar{q}(x)_a^i\}|\Omega\rangle=\ldots$$

we will work only to first order in perturbation theory. Then, we only save the identity from the interaction Hamiltonian's expansion, and contracting the adjacent quark fields to obtain connected diagrams we get

$$\ldots=(\gamma^{ij})_{\mu}S(x)_{ab}^{jk}(\gamma^{kl})^{\mu}S(-x)_{ba}^{li}=N_CTr[\gamma_{\mu}S(x)\gamma^{\mu}S(-x)]$$

where $S$ is the free fermion propagator and we have taken the trace over color indices obtaining the $N_C$. Plugging this in our original expression and changing the order of the matrices inside the traces taking advantage of the cyclic nature of traces we get

$$\frac{-iN_c\mu^{2\epsilon}}{(d-1)q^2}\int{}d^dx\,{}e^ {iqx}Tr[S(x)\gamma_{\mu}S(-x)\gamma^{\mu}]$$

$\endgroup$

Your Answer

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