When using the QCD-improved parton model, we often calculate matrix elements for the scattering of a virtual photon off a quark. $$\gamma^\ast q\rightarrow q$$ In general, we consider the deep inelastic scattering of an electron off a quark and then separate the purely electromagnetic vertex from the hadronic vertex. $$\left(e^-\rightarrow e^-\gamma^\ast\rightarrow\right) \gamma^\ast q\rightarrow q$$ But why can we just neglect the information contained in the electromagnetic vertex and only calculate a transition amplitude from the hadronic vertex? Aren't transition amplitudes calculated from matrix elements only? I do understand why one would separate the two parts of the matrix element, since the electromagnetic vertex is always the same and the entire new information about DIS is contained in the hadronic part.


1 Answer 1


I am loosely following M.D. Schwartz: Quantum Field Theory and the Standard Model, pp.677.

When calculating the spin-averaged transition amplitude for the process $e^-(k)p^+(P)\rightarrow e^-(k^\prime)X(p_X)$ in DIS, one finds that $$\langle|\mathcal{M}_X|^2\rangle=\frac{e^4}{q^4}L_{\mu\nu}K^{\mu\nu},$$ where $q=k-k^\prime$ is the transferred momentum, $L_{\mu\nu}=\frac{1}{2}\mathrm{tr}\left[\gamma_\mu\displaystyle{\not}k\gamma_\nu\displaystyle{\not}k^\prime\right]$ is the leptonic tensor coming from the electron-photon vertex and $K_{\mu\nu}$ is a tensor that comes from the proton-photon-$X$ vertex function. Now, if we calculate the transition amplitude for the process $\gamma^\ast(q)p^+(P)\rightarrow X(p_X)$ with an external off-shell photon $\gamma^\ast$, we find that $$\langle|\mathcal{M}(\gamma^\ast p^+\rightarrow X)|^2\rangle=\sum_\lambda\epsilon^\mu_\lambda(q)\epsilon^{\nu\ast}_\lambda(q)K_{\mu\nu}$$ with the same tensor $K_{\mu\nu}$ because it describes the same vertex function.

The hadronic tensor of the inclusive cross section is defined in terms of $K^{\mu\nu}$ as $$W^{\mu\nu}=\frac{1}{4\pi}\sum_X\int d\Pi_X(2\pi)^4\delta^{(4)}(P+q-X)K^{\mu\nu}.$$

Instead of considering $K^{\mu\nu}$ we can work with the transition amplitude for the process $\gamma p^+\rightarrow X$ since $$\epsilon^\mu_\lambda\epsilon^{\nu\ast}_\lambda W_{\mu\nu} =\frac{1}{4\pi}\sum_X\int d\Pi_X(2\pi)^4\delta^{(4)}(P+q-X)\langle|\mathcal{M}(\gamma^\ast p^+\rightarrow X)|^2\rangle.\quad (\ast)$$

The information contained in the leptonic tensor is not neglected, since it is multiplied with the hadronic tensor in the inclusive cross section, i.e. $d\sigma\sim L_{\mu\nu}W^{\mu\nu}$.

In the QCD-improved parton model, the hadronic tensor is then written in terms of non-perturbative parton density functions $f_i$ and perturbatively calculable partonic tensors $\hat{W}^{\mu\nu}_i$: $$W^{\mu\nu}(x,Q)=\sum_i\int\limits^1_0dz\int\limits^1_0d\xi f_i(\xi)\hat{W}^{\mu\nu}_i(z,Q)\delta(x-z\xi),$$ where $x=Q^2/(2Pq)$ is the Bjorken x, $z=Q^2/(2p_iq)$ is the partonic version of $x$ and $p_i=\xi P$ with $0<\xi<1$ is the parton's momentum, which is taken to be collinear to the proton's momentum, and $i$ specifies the parton at hand.

The partonic tensor $\hat{W}^{\mu\nu}_i(z,Q)$ can now be calculated perturbatively according to $$\epsilon^\mu_\lambda\epsilon^{\nu\ast}_\lambda \hat{W}_{\mu\nu} =\frac{1}{4\pi}\sum_{X_i}\int d\Pi_{X_i}(2\pi)^4\delta^{(4)}(\hat{p}_i+q-\hat{X}_i)\langle|\mathcal{M}(\gamma^\ast p_i\rightarrow X_i)|^2\rangle,$$ where the sum now goes over all possible processes $\gamma^\ast p_i\rightarrow X_i$, e.g. processes such as $\gamma^\ast q_f\rightarrow q_f$ or $\gamma^\ast q_f\rightarrow q_fg$ and so on.

Written out explicitly for the gluon case, e.g. $$\epsilon^\mu_\lambda\epsilon^{\nu\ast}_\lambda \hat{W}^{(gluon)}_{\mu\nu} =\frac{1}{4\pi}\sum_f\int[dp^\prime_f][dp_f]\langle|\mathcal{M}(\gamma^\ast g\rightarrow q_f\bar{q}_f)|^2\rangle\delta^{(4)}(r+q-p_f-p^\prime_f)+\mathcal{O}(\alpha^3_s),$$ where e.g. $[dp_f]=d^4p_f/(2\pi)^4\delta(p^2)_{|p^0\geq0}$ is a Lorentz-invariant phase space element and each of the summands now corresponds to one specific hard parton scattering process, such as $\gamma^\ast q_f\rightarrow q_fg$. The index $f$ denotes the quark flavour.

One sees that this has essentially the structure of an exclusive cross section $d\sigma(\gamma^\ast g\rightarrow q_f\bar{q}_f)$ with a 'flux factor' of $1/(4\pi)$.

Thus, the summary is the following.

  • The inclusive cross section for DIS is factorised like $d\sigma\sim L_{\mu\nu}W^{\mu\nu}$.
  • The hadronic tensor $W^{\mu\nu}$ is then written as a convolution integral in terms of the non-perturbative PDFs $f_i$ and the corresponding perturbatively calculable partonic tensor $\hat{W}^{\mu\nu}_i$.
  • This partonic tensor can then be calculated in the form of a partonic cross section with flux factor $1/(4\pi)$ using partonic matrix elements and equation $(\ast)$.

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