# Is Vacuum Expectation Value equivalent to the sum of Tadpole Diagram in the QFT?

I am recently learning Spontaneously Symmetry Breaking in the QFT. For example let us just focus on the potential $$V[\Phi]=a\Phi^2 + b\Phi^4$$ where $$a<0,b>0$$ and denote the vacuum expectation value $$\bar{\phi}:=\langle\Omega|\hat{\Phi}|\Omega\rangle$$ where $$\Omega$$ is the ground state. It is known that

1. To tree level approximation, the quantum action equals to the classical action and $$\bar{\phi}$$ satisfy the condition $$V'(\bar{\phi})=0$$. For $$a<0$$ and $$b>0$$, $$\bar{\phi}\neq 0$$.
2. In the path integral formulation, $$$$\langle\Omega|\hat{\Phi}|\Omega\rangle = \frac{\int D\phi \,\,\phi \exp(iS)}{\int D\phi \exp(iS)} \tag{1}$$$$ which is non-perturbative. For most of the time we can only evaluate this functional integral perturbatively, i.e., through Feynman diagram expansion. It is clear that for the $$\langle\Omega|\hat{\Phi}|\Omega\rangle = \sum\text{Tadpole Diagrams}$$. However, for the $$V(\phi)$$ here, there is obviously no vertex contribute to Tadpole Diagrams and hence $$\bar{\phi}=0$$, which contradicts the first item.(Actually, from the symmetry it can also easily be seen that the functional integral must be 0)

At the first sight I think that the Eq.(1) may break down in this case for some reason. However, recall that when we derive the stationary condition for $$\bar{\phi}$$: $$\frac{\delta \Gamma(\phi)}{\delta\phi}\bigg|_{\phi=\bar{\phi}}=0$$ where $$\Gamma[\phi]$$ is quantum action, the $$\bar{\phi}$$ is just defined by $$\bar{\phi}=\frac{\delta W[J]}{\delta J}\bigg|_{J=0}$$ where $$W[J]$$ is the generating functional for connected diagrams and it is exactly Eq(1)! The Eq(1) must be correct or the theory is not consistent!

• It is often (I am not sure whether this is always the case) possible to readjust the counterterms in a renormalisation scheme to make the tadpole diagrams vanish. Perhaps someone more knowledgeable can provide a thorough explanation. Commented Jun 30, 2023 at 12:04
• I'm not sure that the VeV is just the sum of the tadpoles. Also why are you assuming no vertices in tadpoles? Commented Jun 30, 2023 at 12:04
• I find that there is a same question in link but I think the answer does not solve the question Commented Jul 1, 2023 at 5:33

• Thanks for your comments. Sorry that I do not get your point. For the case let us just focus on the toy model $S(Φ)$=derivative term $+ aΦ^2+bΦ^4$. We can treat quadratic term as a two-point interaction so there are only two vertices, one contains two legs and the other contains four. It is clear that we can not draw a tadpole diagram which contains only one external leg. Am I wrong ? Commented Jul 1, 2023 at 5:18