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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, \begin{equation} \langle\Omega|\hat{\Phi}|\Omega\rangle = \frac{\int D\phi \,\,\phi \exp(iS)}{\int D\phi \exp(iS)} \tag{1} \end{equation} 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!

What's wrong with my argument? Look for your valuable comments!

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  • $\begingroup$ 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. $\endgroup$
    – Martin C.
    Commented Jun 30, 2023 at 12:04
  • $\begingroup$ I'm not sure that the VeV is just the sum of the tadpoles. Also why are you assuming no vertices in tadpoles? $\endgroup$
    – LolloBoldo
    Commented Jun 30, 2023 at 12:04
  • $\begingroup$ I find that there is a same question in link but I think the answer does not solve the question $\endgroup$
    – Kong Yeu
    Commented Jul 1, 2023 at 5:33

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I think that in (2) the statement:

there is obviously no vertex contribute to Tadpole Diagrams

Is a false statement. The vertices are dictated by the action appearing into the action. So the tadpoles could contain the interaction vertices cointained into your action. This should be the wrong assumption in your statement.

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  • $\begingroup$ Check for example the tadpoles contained in the SM, where you have an external Higgs coupled to fermion/massive gauge bosons loops. Such tadpoles are emerging from vertex interactions in the SM action $\endgroup$
    – LolloBoldo
    Commented Jun 30, 2023 at 12:15
  • $\begingroup$ 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 ? $\endgroup$
    – Kong Yeu
    Commented Jul 1, 2023 at 5:18

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