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.


1 Answer 1


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 $N$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.