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At several points throughout Weinberg QFT Volume I, Weinberg claims that the sum of all diagrams which have in states $\alpha$ and out states $\beta$ and one off-shell photon at position $x$ is given by \begin{equation} \langle \beta|\hat{J}^{\mu}(x)|\alpha\rangle \tag{1} \end{equation} where $J^{\mu}(x)=\frac{\delta L_{QED}}{\delta A^{\mu}(x)}$. For instance he says this at Eq's(10.4.19),(10.5.1) and (13.6.1). However, I am under the impression that the amplitude for single photon emission is rather: \begin{equation} \langle \beta|\hat{A}^{\mu}(x)|\alpha\rangle \tag{2} \end{equation} Below Eq.(10.5.1) he says Eq(1) is the case in theories such as QED where the interaction is linear in $A^{\mu}(x)$. I do not understand this comment as Eq(2) is the amplitude for single photon emission?

My second question: In QCD, does only Eq(2) work for describing single gluon emission, supposing that we insert the gluon operator in place of the photon field operator?

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  • $\begingroup$ Eq. (1) takes already into account (for theories like QED which are linear in the $A_\mu$) the LSZ reduction where the external photon legs is removed (although he needs yet to multiply it by the polarization tensor). Your Eq. (2) instead isn't a scattering amplitude yet, since you need to multiply it by the Fourier tr. of $(p^2-i\epsilon)^{-1}$ (and then the polarization). As for the second question: yes, your Eq. (2) works for single gluon emission (modulo the LSZ reduction mentioned above) $\endgroup$
    – TwoBs
    Dec 28, 2017 at 10:53
  • $\begingroup$ @ TwoBs , LSZ would still require Eq(2) as opposed to Eq(1) to put the photon onshell? It seems that Eq.(1) when Fourier transformed would give rise to two fermions, the current after all being $\bar{\psi}\gamma^{\mu}\psi$ and not one photon? $\endgroup$
    – Luke
    Dec 28, 2017 at 11:18
  • $\begingroup$ Off-shell photon in a scattering amplitude indeed, I understood that's what you asked about.what Weinberg does and I was referring to is a generalization of LSZ as the momentum is not necessarily on shell. Inserting a current is therefore the same (for linear field couple to it). $\endgroup$
    – TwoBs
    Dec 28, 2017 at 11:49
  • $\begingroup$ @ TwoBs I am still confused. Why is (1) not the sum of all diagrams with two fermions (not necessarily onshell) accompanying the $\alpha\rightarrow\beta$ process as opposed to one one photon accompanying the process in the context of QED? $\endgroup$
    – Luke
    Dec 28, 2017 at 15:42
  • $\begingroup$ Because the two fermions meet at the same point forming another local operator which can generate one-particle photon hat couples linearly to it (you can perhaps think of $\ j_\mu$ as mixing with $A_\mu$) . But don't get me wrong: the photon could just be a non-propagating dof and the matrix elements would thus be interpreted as those of the current operator as well. Both pictures are legitimate. The thing that I find instead confusing is the case and the interpretation of non-linear coupling, such as for charged scalars in QED of in non-abelian gauge theories. $\endgroup$
    – TwoBs
    Dec 28, 2017 at 16:34

1 Answer 1

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The equality relies on the Schwinger-Dyson equations. The sum of all Feynman diagrams in which the state $\alpha$ goes to $\beta$ accompanied by the emission/absorption of a not necessarily on-shell photon of momentum $q^{\mu}$ is \begin{equation} \int d^dx \,e^{iq\cdot x}\,\langle\beta|\hat{A}^{\mu}(x)|\alpha\rangle \end{equation} where this amplitude includes the photon propagator. We can get rid of this propagator factor by multiplying through by $\Box_x^{\mu\nu}=\Box g_{\mu\nu}-(1-\frac{1}{\xi})\partial_{\mu}\partial_{\nu}$. So the amplitude without the propagator is \begin{equation} \int d^dx \,e^{iq\cdot x}\,\,\Box_x^{\mu\nu}\langle\beta|\hat{A}_{\nu}(x)|\alpha\rangle=\int d^dx \,e^{iq\cdot x}\,\,\langle\beta|\hat{J}^{\mu}(x)|\alpha\rangle \end{equation} Where the above equality is a Schwinger-Dyson equation for QED. In the case of multiple photon emissions in which there are multiple photon field operators, contact terms emerge at this last equality, however, inspecting these terms, one finds that they correspond to the disconnected components of the S-matrix and hence can be neglected.

In the case of QCD, in which a single gluon is emitted, one cannot make such a correspondence as the Schwinger-Dyson equations give \begin{equation} \int d^dx \,e^{iq\cdot x}\,\,\Box_x^{\mu\nu}\langle\beta|\hat{A}_{\nu}^a(x)|\alpha\rangle=\int d^dx \,e^{iq\cdot x}\,\,\langle\beta|\hat{J}^{\mu}_a(x)+gf^{abc}\partial_{\nu}\big(\hat{A}^{\nu}_b(x)\hat{A}^{\mu}_c(x)\big)|\alpha\rangle \end{equation} where $J^{\mu}_a=-f_{abc}F_c^{\nu\mu}A_{b\mu}-i\frac{\delta \mathcal{L}_{Matter}}{\delta D_{\nu}\psi}t_a\psi$ is the associated conserved current. For further details see Schwartz's QFT textbook Section 14.8

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  • $\begingroup$ The failure of the relation is not only for qcd but also for qed with charged scalars since the right-hand side of the equation of motion for A contains $ J_\mu$ but also $|\Phi|^2A_\mu$. $\endgroup$
    – TwoBs
    Dec 29, 2017 at 18:11

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