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  Computation of the QCD vector two point function

+ 3 like - 0 dislike

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, particularly I am interested on where do the traces come from.

asked May 9, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

1 Answer

+ 5 like - 0 dislike

This is a tree level result, just write $\langle\Omega|T\{j_{\mu}(x)j^{\mu}(0)\}|\Omega\rangle$ explicitly in terms of quark field operators, and note that quark operators carry both spinor indices and color indices. Then do Wick contractions as usual. Do not contract two quark fields in the same current since by symmetry  $\langle\Omega|T\{j_{\mu}(x)\}|\Omega\rangle$ must be renormalized/normally ordered to  0. The trace will naturally appear as a result of contracting indices, in fact the trace will be over both color and spinor indices. The color part of the trace is quite trivial, it's just a trace over an $N_c\times N_c$ identity matrix, hence a factor $N_c$ appears as an overall coefficient and you are left with only the spinor trace. You may have seen these in similar QED calculations, it's exactly the same story.

answered May 9, 2015 by Jia Yiyang (2,640 points) [ revision history ]
edited May 9, 2015 by Arnold Neumaier

@Jia Yiyang, can you please elaborate more one the fact that we need only take contraction of fields at different spacetime points?

@silvrfuck, when you contract within the same current, you are essentially calculating $\langle\Omega|j_{\mu}(0)|\Omega\rangle$. Due to Lorentz invariance of the vacuum, we have

 $\langle\Omega|U^{-1}(\Lambda)j_{\mu}(0)U(\Lambda)|\Omega\rangle=\langle\Omega|j_{\mu}(0)|\Omega\rangle$. On the other hand,  $U^{-1}(\Lambda)j_{\mu}(0)U(\Lambda)=\Lambda^{\nu}_{\ \mu}j_\nu(0)$, hence we conclude $\langle\Omega|j_{\mu}(0)|\Omega\rangle= \Lambda^{\nu}_{\ \mu}\langle\Omega|j_\nu(0)|\Omega\rangle$, and this can only be true if $\langle\Omega|j_{\mu}(0)|\Omega\rangle=0$. You should take this as a kind of renormalization condition.  

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