we reproduce this argument below.

Summing over intermediate states, and using Lorentz invariance, the vacuum expectation value of the commutator may be expressed as $$\langle 0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle = \int d^4 p
\left( \rho^{a,\mu}_i(p) e^{-i p \cdot (y-x)} -
\tilde{\rho}^{a,\mu}_i(p) e^{ip \cdot
  (y-x)}\right),$$ where the spectral densities $\rho^a_i(p), \tilde{\rho}^a_i(p)$ are defined as $$\begin{aligned}
\rho^{a,\mu}_i(p) &=& \sum_N \langle 0| J^{a,\mu}_0(0)|N\rangle \langle
N| \phi_{i,0}(0)|0\rangle \delta^4(p-p_N), \\
\tilde{\rho}^{a,\mu}_i(p) &=& \sum_N \langle 0|\phi_{i,0}(0)|N\rangle \langle
N| J^{a,\mu}_0(0)|0\rangle \delta^4(p-p_N),\end{aligned}$$ and $p_N$ is the total four-momentum in the state $\mid N \rangle$.

By Lorentz invariance and non-negativity of energy, these densities are of the form $$\begin{aligned}
\rho^{a,\mu}_i(p)&=& p^\mu \rho^a_i (p^2)\theta(p^0), \\
\tilde{\rho}^{a,\mu}_i(p)&=& p^\mu \tilde{\rho}^a_i (p^2)\theta(p^0),\end{aligned}$$

which implies that $$\begin{aligned}
\lefteqn{\langle 0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle =}
\nonumber \\
&& i \frac{\partial}{\partial y_\mu} \int dM^2 \left( \rho^a_i (M^2)
\Delta_+(y-x;M^2) +\tilde{\rho}^a_i (M^2) \Delta_+(x-y;M^2) \right)\end{aligned}$$ where $\Delta_+(x;M^2)$ is the standard two-point Wightman function: $$\Delta_+(x;M^2) = \int d\mu(p) e^{-i p \cdot x}, \quad {\rm where}
\quad d\mu(p) = \frac{d^3 p}{2p_0}, \quad p_0 = \sqrt{\vec{p}^2 + M^2}.$$

Since $\Delta_+(x;M^2)$ and $\Delta_+(-x;M^2)$ are equal for $x$ spacelike, we can write, for such $x-y$, $$\langle 0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle =
i \frac{\partial}{\partial y_\mu} \int dM^2 \left( \rho^a_i (M^2)+
\tilde{\rho}^a_i (M^2) \right) \Delta_+(y-x;M^2).$$

For spacelike separations, the commutator vanishes, so that 