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Typically when you do the counting for large N gauge theory, you rescale fields so that the Lagrangian takes the form \begin{equation} \mathcal{L}=N[-\frac{1}{2g^2}TrF^2+\bar{\psi}_i\gamma^\mu D_\mu \psi_i] \end{equation} where I have chosen the original coupling of the theory to be $\frac{g}{\sqrt{N}}$. From this it is easy to see which vacuum diagrams contribute in the Large-N limit.

However, when you go on to consider connected correlators, people always add a source term $N\sum J_iO^i $ to the Lagrangian. The factor of N out front then determines the N-dependence of the correlators \begin{equation} \langle O_1...O_r \rangle=\frac{1}{iN}\frac{\partial}{\partial J^1}...\frac{1}{iN}\frac{\partial}{\partial J^r}W[J] \end{equation} The N-counting would be different if my source terms were instead just $\sum J_iO^i $.

So my question is, why are we forced to include the factor of N in the source terms? Is it because the original action has been written in terms of rescaled fields and is also proportional to N? If I instead worked with the action in terms of un-rescaled fields, would I not include the factor of N in the source term? Thanks.

Dan
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1 Answers1

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The operator $O^i$ in the source term will in general also contain fields that are rescaled, and the scaling behaviour is supposed to match the rest of the Lagrangian.

If you did not have a factor of $N$ in the source term, you would not need to divide by $N$ when taking functional derivatives. What matters is the result: functional derivatives of the generating functional should produce correlation functions of the operators without any multiplications by $N$.

Frederic Brünner
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