= Solution
Use the normalized Gaussian integral with action $\operatorname{tr}(M^2)/2$. Its <Wick contraction> is
$$
\langle M^i{}_j M^k{}_l\rangle_0=\delta^i{}_l\delta^k{}_j.
$$
There is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized <Hermitian matrix model> integral to first order gives
$$
\langle M^i{}_j M^k{}_l\rangle
=\langle M^i{}_j M^k{}_l\rangle_0-\frac g4\left[
\langle M^i{}_j M^k{}_l\operatorname{tr}(M^4)\rangle_0
-\langle M^i{}_j M^k{}_l\rangle_0\langle\operatorname{tr}(M^4)\rangle_0\right]+O(g^2).
$$
The subtracted term removes <Vacuum Feynman diagrams> disconnected from the external pair. Since the one-point function vanishes by $M\mapsto-M$, the resulting two-point function is a <connected correlation function>.
Write the vertex as $M^a{}_bM^b{}_cM^c{}_dM^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a <tadpole diagram>. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N\delta^i{}_l\delta^k{}_j$ in each case. The other four attach them at opposite positions and give $\delta^i{}_j\delta^k{}_l$, with no free index loop. Therefore
$$
\boxed{\langle M^i{}_j M^k{}_l\rangle_c
=(1-2gN)\delta^i{}_l\delta^k{}_j-g\delta^i{}_j\delta^k{}_l+O(g^2).}
$$
\b[Both index structures are required at finite $N$]. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal <perturbative quantum field theory> expansion; the real integral is convergent for $g\ge0$.
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