Is vacuum expectation value equivalent to the sum of tadpole diagrams in the QFT?

Question

I am recently learning Spontaneously Symmetry Breaking in the QFT. For example let us just focus on the potential $$V[\phi]=a\phi^2 + b\phi^4\tag{1}$$ where $a<0,b>0$ and denote the vacuum expectation value $$\bar{\phi}:=\langle\Omega|\hat{\phi}|\Omega\rangle\tag{2}$$ 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{3} \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 $$\langle\Omega|\hat{\phi}|\Omega\rangle = \sum\text{Tadpole Diagrams}.\tag{4}$$ 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.(3) may break down in this case for some reason. However, recall that when we derive the stationary condition for $\bar{\phi}$: $$\frac{\delta \Gamma(\phi)}{\delta\phi}\bigg|_{\phi=\bar{\phi}}=0\tag{5}$$ where $\Gamma[\phi]$ is quantum action, the $\bar{\phi}$ is just defined by $$\bar{\phi}=\frac{\delta W[J]}{\delta J}\bigg|_{J=0}\tag{6}$$ where $W[J]$ is the generating functional for connected diagrams and it is exactly Eq(3)! The Eq(3) must be correct or the theory is not consistent!

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

Answer 1

1. We expand the field $$\phi(x)~=~ \underbrace{\bar{\phi}}_{\text{VEV}}+\underbrace{\sigma(x)}_{\text{quant. fluct.}},\tag{A}$$ a la Ref. 1.

2. The 1-point function $$\bar{\phi}~:=~\langle\phi\rangle_{J=0}~=~\langle\Omega|\phi|\Omega\rangle_{J=0}~\neq~ 0\tag{B}$$ is non-zero, since the vacuum state $|\Omega\rangle$ spontaneously breaks the $\mathbb{Z}_2$-symmetry $$\phi(x)~\to~ -\phi(x).\tag{C}$$ Superficially, the fact that (i) the action $S[\phi]$ is $\mathbb{Z}_2$-invariant, and (ii) the path integral measure ${\cal D}\phi$ (including boundary conditions) formally looks $\mathbb{Z}_2$-invariant would suggest that the 1-point function (B) should vanish, cf. e.g. my Phys.SE post here. However, the main point is that the path integral $$Z~=~\int\!{\cal D}\phi ~\exp\left\{\frac{i}{\hbar}S[\phi]\right\}\tag{D}$$ is now implicitly defined relative to the vacuum state $|\Omega\rangle$, so that the path integral measure ${\cal D}\phi$ is actually not $\mathbb{Z}_2$-invariant.

3. On one hand, it makes sense to consider $$\begin{align}0~\stackrel{(B)}{=}~& \langle\phi(x)\rangle_{J=0}-\bar{\phi}\cr ~\stackrel{(A)}{=}~&\langle\sigma(x)\rangle_{J=0}\cr ~=~&\left.\frac{\delta W_c[J]}{\delta J(x)}\right|_{J=0}\cr ~=~&\left(\fbox{conn.}- -\stackrel{\sigma}{-}- -\right)\cr ~=~&\sum\text{tadpole diagrams in $\sigma$},\end{align}\tag{E}$$ where $J(x)$ is a source for $\sigma(x)$; not $\phi(x)$. E.g. one 1-loop tadpole diagram is $O\!-$, due to a cubic $\sigma^3$-vertex, cf. Ref. 1.

4. On the other hand, OP is considering $$\begin{align}\bar{\phi}~\stackrel{(B)}{=}~& \langle\phi(x)\rangle_{J=0}\cr ~=~&\left.\frac{\delta W_c[J]}{\delta J(x)}\right|_{J=0}\cr ~=~&\left(\fbox{conn.}- -\stackrel{\phi}{-}- -\right)\cr ~=~&\sum\text{tadpole diagrams in $\phi$},\end{align}\tag{F}$$ where $J(x)$ is a source for $\phi(x)$; not $\sigma(x)$. Due to the $\mathbb{Z}_2$-symmetry (C) there is indeed no perturbative tadpole diagrams in $\phi$, as OP correctly observes. The problem is that we are trying to perform a perturbative diagrammatic calculation around a wrong vacuum state.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; Sec. 11.1 p. 348.

Answer 2

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.

Answer 3

To expand on the answer of @Qmechanic: so you just insist on expanding around the "wrong" vaccuum. Why is that so bad?

You cannot get $Z_2$-breaking effect to any order in pertubation theory if you expand around the $Z_2$-symmetric point, as was pointed out. The state will however be unstable against any infinitesimal $Z_2$-breaking perturbation, and you can indeed extract that susceptibility with a wrong sign from the correlation functions. To be concrete, for instance you will get a negative pressure, signalling mechanical instability. So we can at most show that the slightest bit of fluctuation will send the system irreversibly downhill.

Beyond that, finding the true ground state is a non-perturbative affair. One cannot hope to infer any property of the true ground state even if you can calculate and resum all terms to infinite order in perturbation theory.