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?


1 Answer 1


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.

  • $\begingroup$ yeah the numerator does come out as 0, which I suppose makes sense given the Feynman Diagram. My question was regarding the denominator, since it seemed to give something too close to the non-interacting picture. Thanks! $\endgroup$
    – Nitzan R
    Jan 7 at 15:57
  • $\begingroup$ Right - up to linear order in $\lambda$ this is true for both numerator and denominator. The interesting stuff only starts happening at second order. $\endgroup$
    – nox
    Jan 22 at 2:47

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