For a lagrangian: $$ \mathcal{L}=\partial^\mu\phi_i^*\partial_\mu\phi_i-m_i^2|\phi_i|^2+\lambda(\phi_2^3\phi_1+\text{h.c.}). $$ where summation over $i=1,2$ is understood. I am trying to find the two point correlation function: $$ \left<\Omega\middle|T\phi_1(x)\phi_1(y)\middle|\Omega\right>=\frac{\left<0\middle|\phi_{1I}(x)\phi_{1I}(y)\exp\!\big(-i\int dt\,H_I(t)\big)\middle|0\right>}{\left<0\middle|\exp\!\big(-i\int dt\,H_I(t)\big)\middle|0\right>}\approx\frac{\left<0\middle|\phi_{1I}(x)\phi_{1I}(y)(1-i\int dt\,H_I(t))\middle|0\right>}{\left<0\middle|(1-i\int dt\,H_I(t))\middle|0\right>} $$ Up to order $\lambda$. But the denominator has a term: $$ i\lambda\int d^4z(\phi_2(z))^3\phi_1(z) $$ Which, since as far as I understand $[\phi_2^-,\phi_1^+]=0$ (where $\phi_i^\pm$ are the terms in $\phi_i$ containing the annihilation\creation operators respectively), would mean that the term has no possible contraction, and therefor by wick's theorem would give: $$ i\lambda\int d^4z(\phi_2(z))^3\phi_1(z)=0 $$ Which I think make the whole things trivial. Is my asumption regarding commutation relations correct? If so, does this mean that the denominator, up to order $\lambda$, is indeed 1?
Edit: thinking about this, could this be related to the fact that in this particular theory there are no vacuum bubbles?
