= Solution
Choose the current convention corresponding to $\delta\varphi=i\alpha\varphi$. <Noether's theorem> gives
$$
J^\mu=i\bigl[(\partial^\mu\varphi^*)\varphi-\varphi^*\partial^\mu\varphi\bigr]
=\frac{2r^2}{f}\partial^\mu\pi.
$$
For the requested terms with no $\sigma$ and $f=\sqrt2v$, the <Lagrangian> is $\tfrac12(\partial\pi)^2$ and the current becomes $J^\mu=f\partial^\mu\pi$. The supplied covariant oscillator normalization cancels the invariant measure when the <annihilation operator> contracts with $a^\dagger(p)$, so
$$
\langle0|\pi(x)|\pi(p)\rangle=e^{-ip\cdot x},\qquad
\boxed{\langle0|J^\mu(x)|\pi(p)\rangle=-ifp^\mu e^{-ip\cdot x}.}
$$
The <momentum> is on the massless shell, $p^2=0$. Differentiating once more therefore gives
$$
\boxed{\langle0|\partial_\mu J^\mu(x)|\pi(p)\rangle=-fp^2e^{-ip\cdot x}=0.}
$$
This <Goldstone current matrix element in a phase-only scalar theory> shows that a conserved current can have a nonzero <matrix> element proportional to a null <momentum>. Thus current conservation is fully consistent with the current creating a massless one-particle state from the <vacuum>; the nonzero coefficient $f$ is its Goldstone-current <coupling>. It is the divergence that vanishes, not the current <matrix> element. Reversing the current convention reverses the nonzero <matrix> element but leaves this zero divergence unchanged.
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