Conserved currents in quantum electrodynamics

Question

A general Noether theorem in fields theory says that an infinitesimal symmetry of the action leads to a conserved current $j^\mu$, i.e. $\partial_\mu j^\mu=0$.

Below I would like to consider a minor generalization of the following well known situation in QED. The Lagrangian density in QED is $$\mathcal{L}=\bar\psi (i\gamma^\mu(\partial_\mu+ieA_\mu)-m)\psi)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$ $\mathcal{L}$ is invariant under the global symmetry

\begin{eqnarray} \psi\to (1+ie\epsilon)\psi,\\ \bar\psi\to (1-ie\epsilon)\bar\psi\\ A_\mu\to A_\mu, \end{eqnarray} where $\epsilon$ is an infinitesimal parameter. It is well known that application of the Noether theorem leads to conserved current $$j^\mu=-e\bar\psi\gamma^\mu\psi.$$ The operator $Q=\int d^3x j^0(x)$ is interpreted as electric charge operator and commutes with the Hamiltonian.

However it is equally well known that $\mathcal{L}$ is invariant under more general local transformations. Hence one could try to construct a Noether current for each such transformation. More precisely fix an arbitrary function $f(x)$. Then $\mathcal{L}$ is invariant under \begin{eqnarray} \psi\to (1+ief\epsilon)\psi,\\ \bar\psi\to (1-ief\epsilon)\bar\psi,\\ A_\mu\to A_\mu-(\partial_\mu f)\epsilon, \end{eqnarray} where again $\epsilon$ is an infinitesimal parameter. (Notice that the previous transformation corresponds to $f\equiv 1$.) It is easy to compute the Noether current: $$j^\mu=-fe(\bar\psi\gamma^\mu\psi)+\partial_\nu f(\partial^\mu A^\nu-\partial^\nu A^\mu).$$

Does the latter current has any physical interpretation? Does it play any role in the theory? More generally in any gauge theory (e.g. QCD) one can get similar currents for any gauge transformation. Are they useful?

ADDED. Let me add a comment to make my question more precise. When one has a Lie group preserving the action, the application of the Noether theorem and construction of charge operators using these currents lead presumably to a representation of the Lie algebra of this group (e.g. this is the case with the Poincare group, inner symmetries and more generally with supersymmetries). Hence if the gauge group preserves the action, one would expect that its Lie algebra acts on the Hilbert space of the theory. In the case of QED one would get a representation in the Hilbert space of the theory of the commutative (infinite dimensional) Lie algebra of real valued functions on $\mathbb{R}^{3+1}$. I am wondering if this is indeed the case. What I have seen in the literature is only one charge operator $Q=-e\int d^3x \cdot \bar\psi(x)\gamma^0\psi(x)$ corresponding to the current $-e\bar\psi\gamma^\mu\psi$ which corresponds to the rotation with the constant phase I mentioned first in the post.

Answer 1

Sweet question.

Please allow mo to say that the global infinitesimal transformations that you write a the beginning are nothing more that the infinitesimal forms of the following global $U(1)$ transformation.

Now, the Lagrangian that you show is trivially invariant under such global transformations, but it is also invariant under the local ones: $$ \psi\rightarrow{}e^{ie\Lambda(x)}\psi\,, $$

and something similar for $\bar{\psi}$ plus the additional one:

$$ A_\mu\rightarrow{}A_\mu-\partial_\mu\Lambda $$

where $\lambda$ is now a function (please forget and correct any silly sign and factors of i mistakes).

In fact, the lagrangian that you are interested in is usually gotten as follows, you first look at the Dirac lagrangian

$$ {\cal{L}}_D=\bar{\psi}\left[i\gamma^\mu\partial_\mu-m\right]\psi $$

and notice that it is invariant under the global $U(1)$ transformations. Then wonder what happens if you think on local transformations and realize that the Dirac lagrangian is not longer invariant for the simple reason that the derivative pics an extra term coming for the $U(1)$ factor. Then you jump to an attempt to recover the symmetry by modifying the derivative operator by the minimal thing you might think of, adding a vector field and praying for this add up to restore the symmetry. The symmetry is indeed recovered if you impose that the vector field $A_\mu$ transforms as $A_\mu\rightarrow{}A_\mu-i\partial_\mu\Lambda$ (again, I might be making sign mistakes because is very late in the night...HAHAHA, but I am sure you will correct them).

Finally you think ... "HEY, I have introduced a new field (the vector field) and I would love it t have a physical dynamics as well, for the system to be closed. But the transformation rule I have just found is exactly what we found when studying electrodynamics so I will just add the electromagnetic lagrangian FF to the original pure Dirac field to see what happens".

What you get is the Lagrangian that you began with and by direct comparison with standard electrodynamics you get the interpretation that $\bar{\psi}\gamma^\mu\psi$ is the electromagnetic four current, and so, you are forced to regard $\bar{\psi}\gamma^0\psi$ as the electric charge density.

There' s still more in this. From the mathematical point of view, $A_\mu$ is the connection of a principal $U(1)$ bundle (ask Juan Carlos) and $\psi$ is a section of an associated bundle. The local $U(1)$ functions we have been talking about are the " transition functions" of the bundle and the transformation rules, the " guge transformations".

As for your calculation, it is sweet as I already told you, but what you did was (up to boundary conditions) the proof that the action of electrodynamics coupled wth a Dirac field (it might represent electrons and their antiparticles) is gauge invariant (i.e. invariant under the local transformations)

There is also the question of why on earth did we get interested in local transformations in the first place. Well, I love to imagine a lot of physicists making experiments in different labs, it is clear that if they are observing the same physics they would all try to describe what any of them see as charged particles and their antiparticles should be described by Dirac should be described by a Dirac field, quantum mechanics says that the physics should be invariant under constant phase changes i.e. global $U(1)$ transformations, but it is very, very hard to imagine that all physicists in their respective labs choose exactly the same phase change with respect to one another, it is far more reasonable to think that each of them chooses a phase as pleases (local $U(1)$ transformations)

I hope I might have been of any help.

Best regards

Mario

Answer 2

I think that in QED one usually fixes a gauge before quantizing (e.g. $A^0=0$ or $A^3=0$; the same can be said on the Gupta-Bleuler quantization of a free EM field). This destroys many of the gauge symmetries of the Lagrangian, hence one cannot construct a Noether current for them.