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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.

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    $\begingroup$ Your added term does not appear to be Lorentz invariant. Is that a problem? Also what is $E$? $\endgroup$
    – mike stone
    Sep 19, 2023 at 14:28
  • $\begingroup$ Hi @mikestone. I edited about $E$. Why do you support that it does not appear to be Lorentz invariant? Oh yes, sorry, I see. Is it because of the $\gamma^0$ floating around in the additional term?? $\endgroup$
    – schris38
    Sep 19, 2023 at 14:32
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    $\begingroup$ You have an explicit $\gamma^0$ so its not Lorentz invariant. Apart from that, there is no reason to not include it. This term (without the $\gamma^0$) is typically present in SQED theories so its not that crazy to include it. $\endgroup$
    – Prahar
    Sep 19, 2023 at 14:35
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    $\begingroup$ yes, that's correct. $\endgroup$
    – Prahar
    Sep 19, 2023 at 14:40
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    $\begingroup$ They are not supposed to incorporated. Representations of the Lorentz group appear in the Lagrangian, but representations of the little group appear in the S-matrix. This is because one-particle states are labelled by little group reps whereas field are labelled by Lorentz group reps. $\endgroup$
    – Prahar
    Sep 19, 2023 at 18:55

2 Answers 2

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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 $\overline{\psi} \sigma_{\mu \nu} F_{\mu \nu} \psi$.


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.

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    $\begingroup$ On-shell, the $\overline{\psi} \sigma_{\mu \nu}F^{\mu \nu}\psi$ term is equivalent to the magnetic part of the Gordon decomposition. Thus, we can expect it to induce an additionary correction to the electron's magnetic moment. So I think calling it a "magnetic term" is not a bad thing. $\endgroup$ Sep 19, 2023 at 15:50
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    $\begingroup$ I've seen this interaction referred to in the literature as a "Pauli term". $\endgroup$
    – fewfew4
    Sep 20, 2023 at 2:53
  • $\begingroup$ @fewfew4 thanks! $\endgroup$ Sep 20, 2023 at 13:50
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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.

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    $\begingroup$ Axial symmetry is broken in the quantum theory anyway so it makes no sense to insist on preserving it. $\endgroup$
    – Prahar
    Sep 20, 2023 at 20:30
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    $\begingroup$ @Prahar, about your comment let me give you an example: the Peccei–Quinn symmetry is also an anomalous symmetry broken in the quantum theory. However, the axion modelers take pains to construct Lagrangians to preserve the PQ symmetry, which is then broken via the spontaneous symmetry breaking process. You don't think their work is in vain, right? A side note, the quantum anomaly renders the axions pseudo-NG bosons, instead of regular NG bosons. $\endgroup$
    – MadMax
    Sep 20, 2023 at 21:11

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