Why should Ward identities only be used with the effective action (as opposed to the generating functional for connected diagrams)? My question is about the derivation of Ward identities. I will sketch it here in the case of an O(N) symmetric model and point out what it bothering me when I am done. I am being very sloppy with the notation. Please ask in the comments if you don't understand it. I assume that you know about the generating functional of connected correlation functions $W[j] = \ln[Z[j]]$ and its Legendre transform $\Gamma[\phi]$.
Consider an action that depends on an $N$ component field, $\phi_a$, and is invariant under rotations in this space,
$$ \phi_a \rightarrow \phi_a = U(\theta_\alpha)_{ab} \phi_b' = \left[\text{e}^{i \theta_\alpha J^\alpha}\right]_{ab} \phi_b' \, .$$
$J^\alpha$ are the generators of the rotations and $\theta_\alpha$ the corresponding "angles".
The ward identities are derived by making a change of variables in the generating functional
$$ Z[j^a] = \int D\phi \, \text{e}^{S\left[ \phi_a \right]+\int j^a \phi_a} = \int D \phi' \, \text{e}^{i S[U(\theta)\phi']+\int j^a U(\theta)^{ab} \phi_b'} \, .$$
After renaming $\phi'\rightarrow \phi$ and exploiting the symmetry of the action $S[U(\theta)\phi']=S[\phi']$ we write
$$ Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a U(\theta)^{ab} \phi_b} \, .$$
Finally we expand to linear order in $\theta$,
$$ Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a \phi_a} \left(1 + i \theta _\alpha \int j^a \left[J^{\alpha}\right]_{ab}\phi^b\right) \, .$$
We conclude that
$$\int j^a \left[J^{\alpha}\right]_{ab} \langle \phi^b \rangle = 0 \, . \tag{1}$$
$\langle \phi^b \rangle$ is the expectation value of $\phi^b$ with the source $j$,
$$\langle \phi^b \rangle = \frac{\delta Z}{\delta j^b} \, .$$
Next this is expressed in terms of the effective action, $\Gamma[\phi] = -\log[Z[j]] + \int j^a \phi_a \, ,$
$$\int \frac{\delta \Gamma}{\delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, . \tag{2}$$
Taking one field derivative of the last equation and evaluation at physical solutions, $j=\delta\Gamma/\delta \phi = 0$ leads to
$$\int \frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, .$$
When the fourier transform of this equation is taken we find Goldstones' theorem
$$\frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a}(p=0) \left[J^{\alpha}\right]_{ab} \phi^b = 0\, . $$
I.e. for each generator of the symmetry there is one mode with zero mass.
My question is the following: As I understand it, Eq. (1) is valid for any choice of the source $j$. However if I use it to constrain correlation functions, I get (after one derivative with respect to $j^c$ and evaluating at $j=0$,
$$\int \left[J^{\alpha}\right]_{cb} \langle \phi^b \rangle = 0 \, .\tag{3}$$
I find that the correlation functions are symmetric, $\langle \phi^b \rangle = 0$. This simply tells me that there is no symmetry breaking. What went wrong here? Why can I use the ward identity in terms of $\Gamma$, Eq. (2) and not the one which is written in terms of $Z$, Eq. (1)?
 A: I) OP is asking about the case where the infinitesimal symmetry transformations
$$ \delta\phi^a~=~t^a{}_b \phi^a \tag{A}$$ 
are linear in the fields in the path integral$^1$
$$Z[J] ~=~\exp\left[\frac{i}{\hbar}W_c[J]\right]~=~\int \! {\cal D}\phi~\exp\left[\frac{i}{\hbar}\left( S[\phi]+J_a\phi^a \right)\right]. \tag{B}$$
Recall the Legendre transformation between the generating functional $W_c[J]$ for connected diagrams and effective action
$$\Gamma[\phi_{\rm cl}]~=~W_c[J]-J_a \phi^a_{\rm cl}.\tag{C}$$
The classical fields becomes equal to the quantum average
$$ \phi^a_{\rm cl}~=~\langle \phi^a \rangle_J. \tag{D}$$
II) The Ward identity reads
$$\forall J:~~ J_a \langle \delta \phi^a \rangle_J ~=~ 0, \tag{E}$$ 
cf. Ref. 1. Combining eqs. (A), (C), (D) and (E), we get
$$\forall \phi_{\rm cl}:~~ \frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}} t^a{}_b ~\phi^b_{\rm cl}~=~ 0. \tag{F}$$ 
Differentiation of eq. (F) wrt. $\phi_{\rm cl}$ yields
$$ \forall \phi_{\rm cl}:~~\Delta^{-1}_{ba}~t^a{}_c ~\phi^c_{\rm cl}~=~ \frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}} t^a{}_b ,\qquad 
\Delta^{-1}_{ab}~=~-\frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}\delta\phi^b_{\rm cl}},\tag{G}$$  
from which we (via standard arguments given in Ref. 1) can deduce for $J=0$ that non-zero VEV leads to zeromodes for the matrix $\Delta^{-1}$, so that the inverse matrix (traditionally denoted $\Delta$) does not exist.
III) On the other hand, combining eqs. (A) and (E), we get 
$$\forall J:~~ J_a ~t^a{}_b \langle \phi^b \rangle_J ~=~ 0, \tag{H}$$
which is OP's eq. (1). Differentiation of eq. (H) wrt. $J$ yields
$$ \forall J:~~ t^a{}_b ~\langle \phi^b \rangle_J 
+J_c ~t^c{}_b ~\Delta^{ba}~=~ 0,\qquad  \Delta^{ab}~=~\frac{\delta^2 W_c[J]}{\delta J_a \delta J_b}\tag{I}.$$
The problem with eq. (I) is that $\Delta$ might become singular for $J=0$, so that one can not drop the second term in eq. (I) to derive OP's eq. (3).
References:


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*S. Weinberg, Quantum Theory of Fields, Vol. 2; Sections 16.1, 16.4, and 19.2. 


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$^1$ We use DeWitt condensed notation to not clutter the notation.
