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Reading papers and books about QFT, the term current is often mentioned with examples like the quark current in QCD or the electromagnetic current in QED. I was wondering, if there is a precise definition of what a current is in QFT.

  • Is there a mathematical definition of what makes an operator a current?
  • Speaking in terms of Feynman diagrams, what characterises a current?
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    $\begingroup$ Generically, it is the coefficient of the term linear in the gauge field in a gauge-invariant action, so it is the "stuff" a gauge field propagator abuts to in a Feynman diagram. Depending on the symmetry realization, it may be bilinear, or linear, or messy-linear in the quantum fields, and classically it corresponds to a conserved global-invariance Noether current associated to a conserved charge. In the gauge theory, it is the linchpin of gauge-invariance and covariant derivative completion. $\endgroup$ – Cosmas Zachos Dec 29 '20 at 16:10
  • $\begingroup$ @CosmasZachos That's great. If you put this in an answer with some equations for, e.g. the QED case and a non-Abelian case, I should like to mark it as the answer to the question. $\endgroup$ – Thomas Wening Dec 30 '20 at 12:54
  • $\begingroup$ Linked, and also, and further. $\endgroup$ – Cosmas Zachos Dec 30 '20 at 20:41
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Generically, it is the coefficient of the term linear in the gauge field in a gauge-invariant action, so it is the "matter stuff" a gauge field propagator abuts to in a Feynman diagram. Classically, it corresponds to a conserved global-invariance Noether current associated to a conserved charge.

In the gauge theory, it is the linchpin of gauge-invariance and covariant derivative completion. QFT texts focus on describing the global-to-local symmetry conversion elegantly, but leave the obvious stuff implied. I will illustrate the obvious stuff with U(1), QED, and a complex scalar field (since it contains an extra ingredient (seagulls) lacking in the spinor case), and then only cursorily extend to the non-abelian case, differing only in group theoretical aspects, covered amply in the texts. I may be cavalier with some coefficients.

$$ {\cal L}_{matter}= \tfrac{1}{2} \partial_\mu \phi^* \partial^\mu \phi + ... , $$ where the ellipsis ... indicates mass and interaction terms without derivatives, important in the E-L eqns of motion, but irrelevant to the gauge invariance we'll be discussing. This action is invariant under $\phi \mapsto e^{i\theta} \phi$, for constant angle θ, but not for spacetime dependent θ(x). The Noether current of the global symmetry is $$ J^\mu_N= \tfrac{i}{2} (\phi \partial^\mu \phi^* -\phi^* \partial^\mu \phi ), \tag{1} $$ conserved on-shell, i.e., by use of the E-L eqns of motion (which also involve the omitted non-derivative terms).

However, for local θ(x), there are extra uncancelled pieces in the transformation, $$ \delta {\cal L}_{matter}=\partial^\mu \theta ~ J^N_\mu + \partial_\mu \theta \partial^\mu\theta ~ \phi^* \phi. \tag{2} $$

Still, these can be cancelled by first adding a gauge action $$ {\cal L}_{gauge}= -\tfrac{1}{4} F_{\mu\nu} F^{\mu\nu}, $$ where $A_\mu\mapsto A_\mu + \tfrac{1}{e} \partial_\mu \theta$, rendering it locally invariant, and also coupling terms cancelling (2), $$ {\cal L}_{int}= -e A^\mu j^N_\mu + \frac{e^2}{2} A_\mu A^\mu \phi^* \phi\\ =-\frac {i e}{2} A^\mu \left ( \phi (\partial_\mu +ieA_\mu )\phi^* -\phi^* (\partial_\mu -ieA_\mu )\phi \right ) , \tag{3} $$ rendering
$$ {\cal L}_{matter} + {\cal L}_{int} = \tfrac{1}{2} \left((\partial_\mu -ieA_\mu) \phi\right )^* (\partial^\mu -ieA^\mu) \phi + ... , $$ separately gauge invariant. (Off-shell: one never needs to know about the E-L equations.) A fortiori, it is globally U(1) invariant, with the gauge field not transforming here. (In the non-abelian case, it only needs transform like a matter field.)

Now, note $J^N_\mu$ is not gauge invariant—it needs the bilinear ("seagull") term in (3) to cancel its gauge variation. People usually call it the symmetry current, and, as you saw, the easiest way to isolate it is to look for the coefficient of $\partial_\mu \theta$ in a gauge variation of the action.

Diagrammatically, it is what the gauge boson propagator ends at, coupling to φ and φ*. The seagull term has two photons emanating from it, like the wings of a seagull.

As you see from the second line of (3), you may combine everything into a gauge-invariant current involving the gauge field as well ($J_\mu^N-eA_\mu \phi\phi^*$, the Noether current of the full gauge action). You may check that, now, with the full gauge action, it is this gauge-invariant current that is conserved on-shell, while the original $J^N_\mu$ isn't, anymore, by itself. The lagrangian has changed.

Further, note the current need not be bilinear in the matter fields; for fancy interactions like $${\cal L}_{matter}= \tfrac{f(\phi^* \phi)}{2} \partial_\mu \phi^* \partial^\mu \phi $$ you would have, instead $$ J^\mu_N= \tfrac{if(\phi^* \phi)}{2} (\phi \partial^\mu \phi^* -\phi^* \partial^\mu \phi ), $$ for all kinds of baroque functions f(|φ|2). For SSB systems, you'd have the current be proportional to the gradient of the Goldstone boson and less significant higher-order terms in the fields.

  • The extension to the non-Abelian case is straightforward, and basically a test of your appreciation of the group theoretical notation, involving group generator matrices $T_a$, group transformations such as $U=e^{i\theta ^aT_a}$, traces in their spaces, and group covariant derivatives, $$ J^a_\mu = i (\partial_\mu \Phi^\dagger T^a \Phi- \Phi^\dagger T^a \partial_\mu \Phi ) , $$ $${\mathbb D}_\mu =1\!\! 1 \partial_\mu -ig {\mathbb A}_\mu \equiv 1\!\! 1 \partial_\mu -ig T_aA^a_\mu , \\ {\mathbb D}_\mu \mapsto U {\mathbb D}_\mu U^{-1} , ~~\leadsto \\ {\mathbb A}_\mu \mapsto U {\mathbb A}_\mu U^{-1} -\frac{i}{g} (\partial_\mu U )U^{-1} $$ the gauge orbit, as amply summarized in texts. The Noether currents will now need to be extended to gauge covariant, instead of the above gauge invariant ones.
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Let me give you the most general definition which holds for completely generic QFTs (including those which do not have a Lagrangian formulation). The specific case of QED or QCD is described in the answer by @CosmasZachos.

A current operator is any spin-1 operator which satisfies the following Ward identity $$ \nabla^\mu \langle J_\mu(x) O_1(x_1) \cdots O_n(x_n) \rangle = 0 \quad \text{if} \quad x\neq x_i. $$ where $\langle\cdots\rangle$ represents the time-ordered correlation function and $O_i(x_i)$ are local operators (i.e. they commute if they are space-like separated) in the QFT.

All current operators of the type above can be associated to a symmetry by constructing the charge operator $$ Q[\Sigma] = \int_\Sigma n^\mu J_\mu. $$ where $\Sigma$ is any closed $(d-1)$ surface in our spacetime (i.e. $\partial \Sigma = 0$). The Ward identity for the current implies that the charge $Q[\Sigma]$ does not change under deformations of $\Sigma$ (as long as none of the points $x_i$ are crossed under the deformation).

Note that the current operator is not unique. Given any conserved $J_\mu$, we can define a new current $$ J'_\mu = J_\mu + \nabla^\nu K_{\mu\nu}\quad \text{where} \quad K_{\mu\nu} = - K_{\nu\mu}. $$ However, this new current does not correspond to a new symmetry as the charge is still the same $$ Q'[\Sigma] = \int_\Sigma n^\mu J'_\mu = Q[\Sigma] + \int_{\partial\Sigma} dS^{\mu\nu} K_{\mu\nu} = Q[\Sigma]. $$ Noether's theorem is the inverse statement - that any global continuous symmetry implies the existence of a current operator $J_\mu$.

If the QFT can be described by an action $S[\phi]$ then the current can be defined as $$ J_\mu(x) = \frac{\delta S[\phi]}{ \delta \nabla^\mu \phi(x) } \delta_s \phi $$ where $\delta_s \phi$ is the symmetry transformation of the field $\phi$. The currents in QED and QCD can all be derived by the formula above.

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    $\begingroup$ Minor comment to the answer (v2): It seems more natural to write the current as a vector $J^{\mu}$ rather than a co-vector $J_{\mu}$. $\endgroup$ – Qmechanic Dec 31 '20 at 17:08

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