Massless QED modified Lagrangian

Question

Consider a massless theory of QED, with Lagrangian $$\mathcal{L}_{QED}= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\Psi}i\gamma^{\mu}\partial_{\mu}\Psi+ e\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi$$

Is there any reason to expect/allow for an additional interaction term of the form $$\frac{e}{4E}\bar{\Psi}S^{\mu\nu}F_{\mu\nu}\Psi$$ turning, thus, my Lagrangian to $$\mathcal{L}_{QED}\rightarrow\mathcal{L}_{QED}+ \frac{e}{4E}\bar{\Psi} S^{\mu\nu}F_{\mu\nu}\Psi$$

Also, if something like that is not prohibited, what physical phenomenon could it correspond to?

Note: $\Psi(x)$ is the massless fermion field and $A_{\mu}(x)$ denotes the photon field, whereas $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is the usual electromagnetic field strength. $S^{\mu\nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$ is the spin operator and $E$ is some energy scale so that the additional Lagrangian term has consistent units of mass.

Any help will be appreciated.

Answer 1

Note that the operator you wrote down has mass-dimension $5$, so including it makes your theory non-renormalizable. This is not inherently a problem, this situation is of course very natural when considering Effective Field Theories. In your notation, $E$ plays the role of the heavier scale that you have integrated out (usually the mass of some particle). Note that this procedure will not only produce the operator you wrote, but will produce every single higher dimensional operator consistent with the symmetries of the original system.

In the context of QCD, the operator $\overline{\psi} \sigma_{\mu \nu} F_{\mu \nu} \psi$ is known as the chromomagnetic operator. I don't know any nice reviews, but here's a reference (it also appears in HQET, for example here). I don't know what it's called in QED (I would like to know). I think we can write down a contrived example, if we introduce a heavy charged massive scalar:

$$\mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \overline{\psi} (i D_\mu \gamma^\mu ) \psi + \frac{1}{2}\phi^\dagger (D_\mu D^\mu - M^2) \phi + g \overline{\psi} \psi \phi$$

if you integrate out the massive $\phi$ particle, you should obtain the dimension 5-operator in the resulting EFT. $\color{red}{\text{Note that you will also generate a mass term for the fermion $\overline{\psi} \psi$.}}$ To put it another way, the massless-QED lagrangian as a $U(1)$-axial symmetry that is broken by terms like $\overline{\psi} \psi \phi$, and $\overline{\psi} \sigma_{\mu \nu} F_{\mu \nu} \psi$.

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As fewfew4 pointed out in comments, it's sometimes called the Pauli term, e.g. here is a paper explicitly working out its phenomenological consequences in massive QED.

Answer 2

Such a term is not allowed, if we insist that the $U(1)$ axial symmetry should be respected as in the massless QED.

Specifically, the massless QED Lagrangian $$ \mathcal{L}_{QED}= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\Psi}i\gamma^{\mu}\partial_{\mu}\Psi+ e\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi $$ enjoys the $U(1)$ axial symmetry $$ \Psi_L \rightarrow e^{\theta i}\Psi_L \\ \Psi_R \rightarrow e^{-\theta i}\Psi_R $$ because terms like $$ \bar{\Psi}i\gamma^{\mu}\partial_{\mu}\Psi $$ and $$ e\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi $$ only couple spinors with the SAME chirality.

However, a term such as $$ \overline{\Psi} \sigma_{\mu \nu} F_{\mu \nu} \Psi= \overline{\Psi}_L \sigma_{\mu \nu} F_{\mu \nu} \Psi_R + \overline{\Psi}_R \sigma_{\mu \nu} F_{\mu \nu} \Psi_L $$ couples spinors with the opposite chirality (right-handed to left-handed), hence the $U(1)$ axial symmetry is broken by such a term, just like a mass term $$ m\overline{\Psi} \Psi= m\overline{\Psi}_L \Psi_R + m\overline{\Psi}_R \Psi_L $$

Therefore, if we insist that the $U(1)$ axial symmetry should be respected as in the original massless QED, then such a term is not allowed.

On the other hand, if we lift the $U(1)$ axial symmetry restriction, then such a term is allowed, see @QCD_IS_GOOD's nice answer for more details.