Disconnected 2nd order terms in 2-point correlation function in $\phi^3$ theory

Question

I'm trying to figure out a detail on the calculation of correlation functions in the $\phi^3$ theory. So, I know we can calculate a 2-point correlation function as:

$$G(x_1, x_2)=\frac{\langle0|\mathcal{T}\phi_{1I}(x_1)\phi_{2I}(x_2)e^{-i\int dz\phi³}|0\rangle}{\langle0|\mathcal{T}e^{-i\int dz\phi³}|0\rangle}$$

Now, I understand the VEV of a time ordering of operators will make all terms that are not fully contracted disappear, and since the 1st order expansion leaves us with an odd number of operators there are no 1st order terms. My problem, however, appears in developing the 2nd order expansion. One of the contractions I find (i'm using $[\phi.\psi]$ for contractions, wasn't sure how to get standard notation here) is:

$$\langle0|\mathcal{T}\phi_{1I}(x_1)\phi_{2I}(x_2)e^{-i\int dz\phi³}|0\rangle\approx\mathcal{T}\phi_{1I}(x_1)\phi_{2I}(x_2)(1+\int dzdz'\phi^3(z)\phi^3(z'))=[\phi_{1I}.\phi(z)][\phi_{2I}.\phi(z')][\phi(z).\phi(z)][\phi(z').\phi(z')]$$

Which would correspond to a diagram like

Now, obviously I can't get a term like this in the denominator and so I can't argue that it cancels out of the final result. Yet, I'm reading in my class notes that $G(x_1, x_2)$ is given by the sum of all connected feynman diagrams until such and such order. Clearly this term does not correspond to a connected diagram, but I cannot find any mathematical reasoning to get rid of it when doing the Wick contractions. What am I doing wrong?

Answer 1

Since that is a disconnected diagram, the issue is not really with the two-point function, but with the one-point function. The diagram you have drawn obviously just factorizes into two of the problematic one-point functions.

The issue is that $\phi^{3}$ theory is unstable, and that means that zero-momentum particles can disappear. (Note that the one-point diagram is cannot exist unless the momentum on the external leg is zero, because of momentum conservation at the vertex.) To understand what is going on, it makes more sense to talk about a stable theory that has a cubic vertex—like the linear $\sigma$ model with $O(N)$ symmetry. When the symmetry is spontaneously broken, there are both $\phi^{3}$ and $\phi^{4}$ vertices. [More precisely, these are $(\phi_{i}\phi_{i})\phi_{1}$ and $(\phi_{i}\phi_{i})^{2}$ vertices, where $\phi_{1}$ is the massive excitation, pointing in the direction in field space of the symmetry-breaking vev $v$.] The $\phi^{3}$ vertex arises when one of the bosons involved in the four-boson vertex comes from the vev $v$ instead. The symmetry breaking has led to a macroscopic number of particles condensing into a zero-momentum state; that is what produces $v$. So a nonzero amplitude for a one-point diagram corresponds to a zero-momentum propagating particle disappearing into (or emerging from) the Bose-Einstein condensate.

Because of the loop, the amplitude for this disappearance is naively infinite, but it is usual to renormalize the theory so that the one-point function vanishes. Physically, this means choosing $v$ to be the true physical value of the condensate vev, which is not further renormalized. However, this is not generally possible to accomplish in the unstable $\phi^{3}$ theory, because of the ultimate stability problem. You can keep adding to the condensate without running into then energy barrier that the $\phi^{4}$ term eventually imposes in the linear $\sigma$ model. Instead, the theory can exhibit runaway condensation.