Graduate Understanding Goldstone's Theorem Proof: Volume Limit and Commutator

[Aleolomorfo]

Hello everybody!
I have a question regarding the first step of the quantistic proof of the Goldstone's theorem. Defining
$$a(t) = \lim_{V \rightarrow +\infty} {\langle \Omega|[Q_v(\vec{x},t),A(\vec{y})]| \Omega \rangle}$$
where $|\Omega\rangle$ is the vacuum state of the Fock space, $Q_v$ is the conserved charge of the relative current $J^\mu$ and $A$ is a local operator. The first step is to prove that actually $a(t)$ does not depend on $t$.
$$\frac{da(t)}{dt} = \lim_{V \rightarrow +\infty} \partial_t {\langle \Omega|[Q_V(\vec{x},t),A(\vec{y})]| \Omega \rangle} = $$
$$= \lim_{V \rightarrow +\infty} \partial_t \int_V d\vec{x} {\langle \Omega|[J^0(\vec{x},t),A(\vec{y})]| \Omega \rangle} = $$
$$= \lim_{V \rightarrow +\infty} \int_V d\vec{x} {\langle \Omega|[\partial_t J^0(\vec{x},t),A(\vec{y})]| \Omega \rangle} = $$
I substitute the conservation of the current $\partial_\mu J^\mu$
$$= - \lim_{V \rightarrow +\infty} \int_V d\vec{x} {\langle \Omega|[\nabla\cdot \vec{J}(\vec{x},t),A(\vec{y})]| \Omega \rangle} = $$
Stoke's theorem
$$= - \lim_{V \rightarrow +\infty} \int_S d\vec{n} {\langle \Omega|[ \vec{J}(\vec{x},t),A(\vec{y})]| \Omega \rangle} = $$
The conclusion is: the commutator is zero since we are doing the limit $V \rightarrow +\infty$, this means that $|\vec{x}-\vec{y}| \rightarrow +\infty$.

I do not understand two things.
The first one is the last sentence, I do not see why sending $|\vec{x}-\vec{y}| \rightarrow +\infty$ means that the commutator is zero.
Secondly, essentialy in this proof we change $J^0$ to $\vec{J}$, why the same argument made about $\vec{J}$ to sent the volume to $+\infty$ was not used to $J^0$?
Thanks in advance for the help!

 

[vanhees71]

Applying Stokes's theorem transforms the volume integral in the formula before the last formula to a surface integral over its boundary. Letting the volume "go to infinity", meaning taking the limit that $V$ becomes the entire space, the integral vanishes under the assumption that the limit of the volume integral exists, i.e., that the integrand is going sufficiently quickly to 0 in the limit when the entire surface goes to infinity.

You cannot use the argument directly to $\dot{J}^0$ but you need to convert it via the continuity equation to $-\vec{\nabla} \cdot \vec{j}$ in order to be able to convert the volume integral to the surface integral over this volume's boundary and then apply the argument with the vanishing of this surface integral when all parts of the surface are pushed to infinity.