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I am recently reading E.Witten's review for $1/N$ expansion of QCD. In there, considering the main contribution of quark bilinears like $\bar{q}q$, then He mentions that in free field theory there is only one-loop diagram which is

one-loop diagram for quark bilinears

Sorry for much lack of information if there is, but I am confused here because from my understanding, the all connected diagram such as this contributes to a vacuum, not correlation functions, and $\bar{q}q$ seems for me to be like a sum of diagrams with two quark sources like

loop correction diagram

So, I regard $\bar{q}q$ like a propagator. is it?

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  • $\begingroup$ Link to abstract page? $\endgroup$
    – Qmechanic ♦
    Commented Jun 13 at 0:41

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There is a quark propagator which you can write as $\left < \bar{q}(x) q(y) \right >$ but that's not what Witten is computing. He's considering $\bar{q} q$ as a composite operator at position $x$ and then inserting another one at position $y$. So this two point function is $\left < \bar{q}q(x) \bar{q}q(y) \right >$ with four quarks.

The free theory contribution to this is found by just using Wick's theorem where you connect the top pair with one propagator and the bottom pair with another. This produces the diagram you drew. It looks like a closed loop but it is not a vacuum diagram because $x$ and $y$ are fixed positions. Even if you integrated over them, this would not be a vacuum diagram because $\bar{q} q$ is not an interaction vertex of the theory.

As we know, this free contribution is not a good approximation at low energy because the gauge coupling is not an expansion parameter in QCD. This is why Witten uses $1/N$ as a small parameter instead. This requires more diagrams to be considered but only the planar ones. Inserting the interaction vertex in all possible ways consistent with planarity is what leads to nice properties when you cut open the graphs and see what mesons are allowed to be exchanged.

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