Schwartz's and Zee's proof of Goldstone theorem In Refs. 1 & 2 the Goldstone theorem is proven with a rather short proof which I paraphrase as follows.

Proof: Let $Q$ be a generator of the symmetry. Then $[H, Q] = 0$ and we want to consider the case in which $Q | 0 \rangle \neq 0$. As a consequence of the null commutator the state $Q | 0 \rangle$ has 0 energy. We know that $Q = \int d^{D} x ~J^{0} ( \vec{x}, t )$. Then we consider the state $| s \rangle = \int d^{D} x ~e^{- i \vec{k} \vec{x}} J^{0} ( \vec{x}, t )| 0 \rangle$ which has spatial momentum $\vec{k}$. In the zero momentum limit this state goes to $Q |0 \rangle$ which we know has 0 energy. We thus conclude that $| s \rangle$ describe a massless scalar particle with momentum $\vec{k}$. $\Box$

The problem with this proof is that the operator $Q$ is not well-defined because of the Fabri-Picasso theorem. So $Q |0 \rangle$ is not even a state of the Hilbert space. Is it possible to fix this proof so that it becomes rigorous maybe through the use of some regularization of the charge?
I must say I'm not asking for alternative rigorous derivation of the theorem such as the original one or something that exploits the effective action. I'm asking to provide a rigorous proof along the line of the Zee one.
References:


*

*M.D. Schwartz, QFT & the standard model, 2014, section 28.2, p.563-64. 

*A. Zee, QFT in a nutshell, 2010, p. 228.  
 A: In this answer we give a proof$^1$ of Goldstone's theorem at the physics level of rigor following Ref. 1:

*

*We are given a self-adjoint spacetime-translation-covariant 4-current
$$\hat{J}^{\mu}(x)~=~e^{i(\hat{H}t-\hat{\bf P}\cdot {\bf x})}
\hat{J}^{\mu}(0)e^{i(\hat{\bf P}\cdot {\bf x}-\hat{H}t)} \tag{1}$$
that satisfies the continuity equation
$$d_{\mu}\hat{J}^{\mu}(x)~=~0. \tag{2}$$
It is furthermore assumed that the vacuum state $|\Omega \rangle$ is spacetime-translation-invariant.


*In order to avoid the fallacy of the Fabri–Picasso theorem, let us introduce a bounded spatial integration region $V \subseteq  \mathbb{R}^3$. Define a volume-regularized charge operator
$$\hat{Q}_V(t)~:=~\int_V\! d^3{\bf x}~\hat{J}^0(x), \qquad V~\subseteq ~\mathbb{R}^3. \tag{3}$$


*The assumption of spontaneous symmetry breaking (SSB)
is implemented via the existence of a self-adjoint observable $\hat{A}$
such that
$$\begin{align}
{\rm Im}a_V(t)~\stackrel{(7)}{=}~&\frac{1}{2i}\langle \Omega | [\hat{Q}_V(t),\hat{A}]|\Omega \rangle\cr
\quad \longrightarrow& \quad a~\neq~0\quad\text{for}\quad V~\to ~\mathbb{R}^3. \end{align}\tag{4}$$


*We may assume w.l.o.g. that
$$ \langle \Omega |\hat{A}|\Omega \rangle~=~0 \tag{5}$$
in eq. (4) by performing the redefinition
$$ \hat{A}~ \longrightarrow~\hat{A}^{\prime}~:=~\hat{A}-|\Omega \rangle \langle \Omega |\hat{A}|\Omega \rangle \langle \Omega | \tag{6}$$
(and afterwards remove prime from the notation).


*On the rhs. of eq. (4) we have defined
$$\begin{align}
a_V(t)~:=~&\langle \Omega | \hat{Q}_V(t)\hat{A}|\Omega \rangle\tag{7} \cr
~\stackrel{(3)}{=}~&\int_V\! d^3{\bf x}~\langle \Omega | \hat{J}^0(x) \hat{A} |\Omega \rangle\tag{8} \cr
~=~&\int_V\! d^3{\bf x}~\sum_n\langle \Omega | \hat{J}^0(x)|n \rangle\langle n |\hat{A}|\Omega \rangle \tag{9} \cr
\stackrel{(1)+(5)}{=}&\int_V\! d^3{\bf x}~\sum_{n\neq\Omega} e^{i( {\bf P}_n\cdot {\bf x}-E_nt)}c_n, \cr
&\quad c_n~:=~\langle\Omega | \hat{J}^0(0)|n \rangle\langle n |\hat{A}|\Omega \rangle, \tag{10}\cr\cr
~ \longrightarrow& \sum_n  (2\pi)^3 \delta^3({\bf P}_n) e^{-iE_n t}c_n \tag{11}\cr
~\stackrel{(13)}{=}~&\sum_E e^{-iE t} f(E)\tag{12}\cr
&\quad\text{for}\quad V~\to ~\mathbb{R}^3,
\end{align}$$
where
$$ f(E)~:=~\sum_{n\neq\Omega}^{E_n=E}  (2\pi)^3 \delta^3({\bf P}_n)  c_n,\tag{13}$$
and where $|n \rangle$ are a complete set of states  with definite 4-momentum $(E_n,{\bf P}_n)$.


*On one hand,
$$\begin{align}
d_t a_V(t) ~\stackrel{(8)}{=}~&\int_V\! d^3{\bf x}~\langle \Omega | d_0\hat{J}^0(x) \hat{A} |\Omega \rangle \cr
~\stackrel{(2)}{=}~&-\int_V\! d^3{\bf x}~\langle \Omega | {\bf \nabla} \cdot \hat{\bf J}(x) \hat{A} |\Omega \rangle \cr
~=~&-\int_{\partial V}\! d^2{\bf x}~\langle \Omega | {\bf n} \cdot \hat{\bf J}(x) \hat{A} |\Omega \rangle,\tag{14}\end{align}$$
so that
$$\begin{align}  d_t {\rm Im}a_V(t)&\cr
~\stackrel{(14)}{=}~&-\frac{1}{2i}\int_{\partial V}\! d^2{\bf x}~\langle \Omega | [{\bf n} \cdot \hat{\bf J}(x) ,\hat{A}] |\Omega \rangle \cr 
\quad \longrightarrow& \quad 0 \quad\text{for}\quad V~\to ~\mathbb{R}^3,\tag{15}\end{align}$$
because we assume that the observable $\hat{A}$ has a compact spatial support, and commutes with spatially separated (=causally disconnected) operators.
On the other hand,
$$\begin{align} d_t a_V(t)~~\stackrel{(12)}{\longrightarrow}~~& -i \sum_E Ee^{-iE t} f(E)\cr 
&\quad\text{for}\quad V~\to ~\mathbb{R}^3,\end{align}\tag{16} $$
so that
$$ \begin{align}d_t {\rm Im}a_V(t)&\cr
~~\stackrel{(16)}{\longrightarrow}~~&  -\sum_E E\left\{\cos(Et) {\rm Re} f(E) +\sin(Et) {\rm Im} f(E)\right\}\cr &\quad\text{for}\quad V~\to ~\mathbb{R}^3.\end{align}\tag{17} $$
By comparing eqs. (15) & (17) we conclude that
$$f(E)~~\stackrel{(15)+(17)}{\propto}~~ \delta_{E,0}.\tag{18}$$


*Finally
$$\begin{align}0~\neq~ a~ \stackrel{(4)}{\longleftarrow} ~ {\rm Im} a_V(t)
~ \stackrel{(12)}{\longrightarrow}&~
{\rm Im}\sum_E  e^{-iE t} f(E)\cr
~\stackrel{(18)}{=}~&{\rm Im}f(E\!=\!0)\cr 
&\text{for}\quad V~\to ~\mathbb{R}^3.\end{align} \tag{19}  $$
In order to have SSB, we must have $f(E\!=\!0)\neq 0$, i.e. there exists a massless mode $|n \rangle\neq|\Omega \rangle$ with $(E_n,{\bf P}_n)=(0,{\bf 0})$ that couples $c_n\neq 0$ between the current $\hat{J}^0$ and the observable $\hat{A}$. $\Box$
See also this related Phys.SE post.
References:

*

*C. Itzykson & J.B. Zuber, QFT, 1985, Section 11-2-2, p. 520.


*S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 19.2.
--
$^1$ Cartoon version of the proof of Goldstone's theorem (ignoring the Fabri–Picasso theorem):

*

*$\quad |{\bf 0}\rangle ~:=~\hat{Q}|\Omega\rangle~\neq ~0.$ $\quad\hat{H}|\Omega\rangle~=~ 0.$ $\quad [\hat{H},\hat{Q}]~=~ 0.$


*$\quad \hat{H}|{\bf 0}\rangle~=~\hat{H}\hat{Q}|\Omega\rangle ~=~\hat{Q}\hat{H}|\Omega\rangle~=~ 0.$


*$\quad \hat{Q}~:=~\int \! d^3{\bf x}~\hat{J}^0(x).$ $\quad |{\bf k}\rangle ~:=~\int \! d^3{\bf x} ~e^{-i{\bf k}\cdot{\bf x}}\hat{J}^0(x)|\Omega\rangle.$ $\quad |{\bf 0}\rangle~=~|{\bf k}\!=\!{\bf 0}\rangle.$


*$\quad \hat{H}|{\bf k}\rangle ~=~ \sqrt{{\bf k}^2+m^2}|{\bf k}\rangle.$ $\quad \Rightarrow \quad m~=~0.$ $\Box$
