Deep Inelastic Scattering - electromagnetic current

Question

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.

Answer 1

The matrix element $\langle X | J^{\mu}(0) | h \rangle$ encodes the hard scattering part of the DIS process, with $J^{\mu}(0)|h \rangle$ accounting for the injection of an electromagnetic current at spacetime location $0$ in the hadron $h$. The lorentz invariant inclusive cross section $\sigma$, is such that$$\sigma \propto \sum_X \delta(P_X - P - p)| \langle X | J^{\mu}(0) | h \rangle|^2 = \sum_X \delta(P_X - P - p) \langle h | J^{\mu}(0) | X \rangle \langle X | J^{\mu}(0) | h \rangle$$ Now, from the integral representation of the delta function, $$\delta(P_X - P-p) = \int \frac{\text{d}^4 x}{(2\pi)^4} e^{-\mathrm{i}x(P_X - P - p)}$$ and that under spacetime translations, $$\langle h | J^{\mu}(x) | X \rangle = \langle h | J^{\mu}(0) | X \rangle e^{\mathrm{i}x(P-P_X )}$$ we arrive at $$\sigma \propto \text{Disc}\int \text{d}^4 x \,e^{\mathrm{i}px}\, \langle h | J^{\mu}(x) J^{\nu}(0) | h \rangle,$$ which is a fourier transform of a non local forward matrix element involved in the computation of the forward scattering process $\gamma^* q \rightarrow \gamma^* q$ with an inclusive sum over all possible final states, as the $\text{Disc}$ entails.