The leading mass corrections are one-vertex <tadpole diagrams>. In index notation, one contraction closes a freely summed component loop and is proportional to $N$; the two exchange contractions force the internal component to equal the external one and are not proportional to $N$. Including their multiplicities gives
$$
\Delta\mu^2=4(N+2)g_0
\int_{\Lambda/\zeta}^{\Lambda}\frac{d^dq}{(2\pi)^d}\frac1{q^2+\mu_0^2}.
$$
Thus the $N$ in $N+2$ comes from the closed index loop, while the $2$ comes from the two same-component contractions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 3 / a / ii / Solution

 Solution

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