# Two-point correlation function of two complex scalar fields

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?

The denominator will just cancel out the contribution from vacuum bubbles which factorises in the contributions from the numerator. If the denominator is 1 up to order $$\lambda$$ that doesn't mean it doesn't receive corrections at order $$\lambda^{2}$$ (which would involve 6 $$\phi_{2}$$ fields and 2 $$\phi_{1}$$ fields).
Question - do you get anything in the numerator at $$\mathcal{O}(\lambda)$$? It looks like you'll have 3 $$\phi_{1}$$ and 3$$\phi_{2}$$ fields which would give zero?
If you draw the Feynman diagrams corresponding to your process then all of this should become obvious because of the special form of your $$\phi_{2}^{3}$$-$$\phi_{1}$$ vertex.
• Right - up to linear order in $\lambda$ this is true for both numerator and denominator. The interesting stuff only starts happening at second order.