Self-energy, 1PI, and tadpoles I'm having a hard time reconciling the following discrepancy:
Recall that in passing to the effective action via a Legendre transformation, we interpret the effective action $\Gamma[\phi_c]$ to be the generating functional of 1-particle irreducible Green's functions $\Gamma^{[n]}$.  In particular, the 2-point function is the reciprocal of the connected Green's function,
$$\tilde \Gamma^{[2]}(p)=i\big(\tilde G^{[2]}(p)\big)^{-1}=p^2-m^2-\Sigma(p)$$
which is the dressed propagator.
But, the problem is this: in the spontaneously broken $\phi^4$ theory, the scalar meson (quantum fluctuations around the vacuum expectation value) receives self energy corrections from three diagrams:
$-i\Sigma(p^2)=$   +  + 
Note that the last diagram (the tadpole) is not 1PI, but must be included (see e.g. Peskin & Schroeder p. 361).  In the MS-bar renormalization scheme, the tadpole doesn't vanish.
If the tadpole graph is included in $\Sigma$, and hence in $\tilde{G}$ and $\tilde\Gamma$, then $\tilde\Gamma$ cannot be 1PI.  If the tadpole is not included, then $\tilde G$ is not the inverse of the dressed propagator (that's strange, too).  What's going on?
 A: I'm going to give an explanation at the one loop level (which is the order of the diagrams given in the question).
At one loop, the effective action is given by
$$ \Gamma[\phi]=S[\phi]+\frac{1}{2l}{\rm Tr}\log S^{(2)}[\phi],$$
where $S[\phi]$ is the classical (microscopic) action, $l$ is an ad hoc parameter introduced to count the loop order ($l$ is set to $1$ in the end), $S^{(n)}$ is the $n$th functional derivative with respect to $\phi$ and the trace is over momenta (and frequency if needed) as well as other indices (for the O(N) model, for example).
The physical value of the field $\bar\phi$ is defined such that $$\Gamma^{(1)}[\bar\phi]=0.$$
At the meanfield level ($O(l^0)$), $\bar\phi_0$ is the minimum of the classical action $S$, i.e.
$$ S^{(1)}[\bar \phi_0]=0.$$
At one-loop, $\bar\phi=\bar\phi_0+\frac{1}{l}\bar\phi_1$ is such that
$$S^{(1)}[\bar \phi]+\frac{1}{2l}{\rm Tr}\, S^{(3)}[\bar\phi].G_{c}[\bar\phi] =0,\;\;\;\;\;\;(1)$$
where $G_c[\phi]$ is the classical propagator, defined by $S^{(2)}[\phi].G_c[\phi]=1$. The dot corresponds to the matrix product (internal indices, momenta, etc.). The second term in $(1)$ corresponds to the tadpole diagram at one loop. Still to one-loop accuracy, $(1)$ is equivalent to
 $$ S^{(1)}[\bar \phi_0]+\frac{1}{l}\left(\bar\phi_1.\bar S^{(2)}+\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}\right)=0,\;\;\;\;\;\;(2) $$
where $\bar S^{(2)}\equiv S^{(2)}[\bar\phi_0] $, etc. We thus find 
$$\bar \phi_1=-\frac{1}{2}\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c. \;\;\;\;\;\;(3)$$
Let's now compute the inverse propagator $\Gamma^{(2)}$. At a meanfield level, we have the meanfield propagator defined above $G_c[\bar\phi_0]=\bar G_c$ which is the inverse of $S^{(2)}[\bar\phi_0]=\bar S^{(2)}$. This is what is usually called the bare propagator $G_0$ in field theory, and is generalized here to broken symmetry phases. 
What is the inverse propagator at one-loop ? It is given by 
$$\Gamma^{(2)}[\bar\phi]=S^{(2)}[\bar\phi]+\frac{1}{2l}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2l}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}, \;\;\;\;\;\;(4)$$
where we have already used the fact that the field can be set to $\bar\phi_0$ in the last two terms at one-loop accuracy. These two terms correspond to the first two diagrams in the OP's question. However, we are not done yet, and to be accurate at one-loop, we need to expand $S^{(2)}[\bar\phi]$ to order $1/l$, which gives
$$\Gamma^{(2)}[\bar\phi]=\bar S^{(2)}+\frac{1}{l}\left(\bar S^{(3)}.\bar\phi_1+\frac{1}{2}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}\right). \;\;\;\;\;\;$$
Using equation $(3)$, we find
$$\bar S^{(3)}.\bar\phi_1= -\frac{1}{2}\bar S^{(3)}.\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c,$$
which corresponds to the third diagram of the OP. This is how these non-1PI diagrams get generated in the ordered phase, and they correspond to the renormalization of the order parameter (due to the fluctuations) in the classical propagator.
A: (I came across this question while looking through the "unanswered" category. Not sure why a 9-month-old question appeared in the list, but there still seems to be a room for my contribution.)
The quantum effective action $\Gamma[\phi]$ generates 1PI diagrams, and a tree-level analysis on it amounts to a full analysis on the original action $S[\phi]$.
The sum of all 2-point 1PI diagrams is what contributes (in addition to $p^{2}-m^{2}$, which is also present in the original action) to the quadratic part of $\Gamma[\phi]$. If there were no term linear in $\phi$ in $\Gamma[\phi]$, this would be the self energy itself; however, if $\Gamma[\phi]$ contained a linear term, one has to do an additional tree-level calculation to obtain the self energy from $\Gamma[\phi]$.
As for Feynman diagrams in the original post, the first two contribute to $\Gamma[\phi]$, and the third one appears when calculating the self energy from $\Gamma[\phi]$.
A: *

*The full 2-pt function is equal to the full connected 2-pt function plus tadpole contributions:  $$\begin{align}\frac{\hbar}{i}G^{k \ell}~=~&  \langle \phi^k\phi^{\ell} \rangle_{J=0} \cr
~=~&\langle \phi^k\phi^{\ell} \rangle^c_{J=0} +\langle \phi^k \rangle_{J=0} \langle \phi^{\ell} \rangle_{J=0}\cr
~=~&\frac{\hbar}{i} W_{c,2}^{k \ell}   +W_{c,1}^k W_{c,1}^{\ell}.
\end{align}
, \tag{1}$$
cf. e.g. my Phys.SE answer here.


*The self-energy
$$ \Sigma~=~G_0^{-1}-G_c^{-1}\tag{2}$$
in general consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line, cf. e.g. my Phys.SE answer here.
Note in particular that the self-energy (2) may contain non-1PI diagrams with tadpoles.


*The quadratic term in the effective action $\Gamma[\phi_{\rm cl}]$ reads
$$ \begin{align}
(\Gamma_2)_{k\ell}~=~&  ~-~(W^{-1}_{c,2})_{k\ell}\cr 
~-~& W_{c,3}^{pqr}(W^{-1}_{c,2})_{pk}(W^{-1}_{c,2})_{q\ell}(W^{-1}_{c,2})_{rm}W^m_{c,1} \cr 
~+~&{\cal O}((W_{c,1} )^2),\end{align}\tag{3}$$
cf. e.g. my Phys.SE answer here.
Conversely, the full/dressed connected propagator reads
$$ \begin{align}
G_c^{k\ell}~=~&W_{c,2}^{k\ell}\cr
~=~& ~-~(\Gamma^{-1}_2)^{k\ell}\cr 
~-~& \Gamma_{3,pqr}(\Gamma^{-1}_2)^{pk}(\Gamma^{-1}_2)^{q\ell}(\Gamma^{-1}_2)^{rm}\Gamma_{1,m}\cr 
~+~&{\cal O}((\Gamma_1)^2).\end{align}\tag{4}$$
$W_{c,2}^{k\ell}$ and $-(\Gamma_2)_{k\ell}$ becomes inverses of each other when there are no tadpoles $\Leftrightarrow W_{c,1}^k=0\Leftrightarrow \Gamma_{1,k}=0$.
