When one tries to compute the deep inelastic scattering for the process:
where $l$ is a lepton with incoming momentum $k$ and outgoing $k'$, $h$ is an hadron with momentum $P$, $q$ denotes some quark and $\gamma^*$ represents a virtual photon with momentum $p$. After the scattering the results is an outgoing lepton and a buch of hadrons that here are called $X$.
The amplitude for this process can be written, choosing for example the covariant gauge, as:
$$ i\mathcal{M}=\frac{ie^2}{p^2} \bar{u}(k')\gamma_\mu u(k) \langle h|J^\mu(0)|X\rangle $$
here for convinience the polarization coefficients are neglected, and $J^\mu(x):=e\bar\psi(x)\gamma^\mu\psi(x)$ is the electromagntic current.
What gives me truble is the use of $J^\mu(0)$, while to me seem more natural and correct to use $\hat J^\mu(p)$ in momentum space, being $p$ the momentum of the virtual photon. However if I use the latter and try to recover the equation above I get: \begin{align} i\mathcal{M}&=\frac{ie^2}{p^2} \bar{u}(k')\gamma_\mu u(k) \langle h|\hat J^\mu(p)|X\rangle\\ &= \frac{ie^2}{p^2} \bar{u}(k')\gamma_\mu u(k) \int dx~e^{iqx}\langle h|J^\mu(x)|X\rangle. \end{align} I saw the amplitude equation in Quantum Chromodynamics at High Energy - Y.V. Kovchegov, E. Levin but I recall to have seen similar use of $J^\mu(0)$ for interaction of photon with generic initial and final states.
If someone can give some explanation for the use of $J^\mu(0)$ I would be more then happy to hear.
Some naive guess is that we use $J^\mu(0)$ because we are using $\int dp' \hat J^\mu(p')=J^\mu(0)$, i.e. we integrate over all possible momenta for the virtual photon, where then the right momentum $p$ will be selected by the on-shell condition and conservation of mememntum for the external states. For this I need: \begin{align} \langle h|\hat J^\mu(p')|X\rangle=& \delta^{(4)}((P-P_X)-p') \langle h|\hat J^\mu(P-P_X)|X\rangle\\ =& \delta^{(4)}(p-p') \langle h|\hat J^\mu(p)|X\rangle \end{align} but I am not really sure how to justify such result.