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
- 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 $b>0$, $\bar{\phi}\neq 0$.
- 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!
