by translations in space–time: $${{\cal U}}(a)J(x){{\cal U}}^{-1}(a)=J(x+a)\quad{\mbox{\rm and}}\quad{{\cal U}}(a)\vert\Gamma\rangle=
e^{ip_{\Gamma}\cdot a}\vert\Gamma\rangle,$$ and $p_{\Gamma}$ denotes the sum of the energy–momenta of all the particles which define the state $\vert\Gamma\rangle$. The choice $a\equiv -x$ factors out the $x$–dependence of the matrix element into an exponential: $$\langle 0\vert J(x)\vert\Gamma\rangle=e^{-ip_{\Gamma}\cdot x}\,
\langle 0\vert J(0)\vert\Gamma\rangle.$$ All the particles in the state $\vert\Gamma\rangle$ are on–shell. This constrains the total energy–momentum $p_{\Gamma}$ to be a time–like vector: $p_{\Gamma}^2=t$ with $t\ge 0$. With these constrains on $p_{\Gamma}$ we can insert the identity $$\int_{0}^{\infty}dt\int d^{4}p\,\theta(p)\,\delta (p^2 -t)\,\delta^{(4)}
(p-p_{\Gamma})=1$$ inside the sum $\sum_{\Gamma}$ over the complete set of states. Interchanging the order of sum over $\Gamma$ and integration over $t$ and $p$, there appears naturally the definition of the spectral function associated with the $J$–operator $$\sum_{\Gamma}\langle 0\vert J(0)\vert\Gamma\rangle\langle\Gamma\vert
J(0)^{\dagger}\vert 0\rangle\,(2\pi)^4 \delta^{(4)}(p-p_{\Gamma})
\equiv 2\pi\rho(p^2).$$ The spectral function $\rho(p^2)$ is a scalar function of the Lorentz invariant $p^2$ and the masses of the particles in the states $\vert\Gamma\rangle$ only. By construction it is a real function and non–negative $$\rho(p^2)^* =\rho(p^2)\ge 0.$$ We can now rewrite the two–point function in eq. (\[eq:tpf\]) as follows $$\begin{split}
\Pi&(q^2)=\int d^4 x\,e^{iq\cdot x}\int_{0}^{\infty}dt\,\rho(t)\times\\
&\int
\frac{d^4 p}{(2\pi)^3}
\left[i\theta(x)\,e^{-ip\cdot
x}\theta(p)\delta(p^2-t)+i\theta(-x)\,e^{ip\cdot x}\theta(p)
\delta(p^2-t)\right].
\end{split}$$ Here, you may recognize the familiar 