# Interaction for QED with charged, scalar particles

Let $$\mathcal{L}$$ be the Lagrangian for usual QED with scalar, charged particles (with photons and electrons as well):

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m \right)\psi - \partial_{\mu}\phi^*\partial^{\mu}\phi-M^2\phi^*\phi$$

I was trying to show that, because of symmetries, that the Lagrangian above could be written in the same way if $$\partial_{\mu} \to D_{\mu}$$. However, I have to find $$D_{\mu}$$. The symmetry I am considering is such that

$$\psi \to \psi e^{i\Lambda(x)}$$ $$\phi \to \phi e^{i\Lambda(x)}$$ $$A_{\mu} \to A_{\mu}-\frac{1}{e}\partial_{\mu}\Lambda(x)$$

My attempt

Since $$\psi$$ and $$\phi$$ have no crossed terms between them, I figured that we should add a $$\mathcal{L}_{int}$$ (interactions) such that it would cancel the term

$$-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$$

that results from the fermionic part. Hence, $$\mathcal{L}_{int} = -e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$$ seems to do the trick. However, the scalar particle bit does not seem to trivially simplify, as there are some remaining terms:

$$ieA_{\mu}\left[\left(\partial^{\mu}\phi^*\right)\phi-\phi^*\partial^{\mu}\phi\right]+e^2A_{\mu}A^{\mu}\phi^*\phi$$

This would give $$D_{\mu} = \partial_{\mu}+ieA_{\mu}$$.

Is my procedure correct? Is there a more intuitive way of solving this?

Edit: Corrected misuse of concepts as pointed in the comments.

• How can the fields not be coupled if both have charges? Because A has spin 1?
– user303670
Jun 7, 2021 at 17:29
• @Duepietri My bad, wrong use of concepts. I should've said that there are no crossed terms between them. I apologise. Jun 7, 2021 at 17:32
• I would just comment for mu!
– user303670
Jun 7, 2021 at 17:47
• @Duepietri Already took care of that :) Jun 7, 2021 at 17:53
• Look how the Dirac Lagrangian is written here: quantummechanics.ucsd.edu/ph130a/130_notes/node508.html
– user303670
Jun 7, 2021 at 19:41

Thanks to @Duepietri, I've come to a conclusion. Taking into account that $$D_{\mu}$$ needs to be gauge invariant, the minimal coupling
$$D_{\mu} = \partial_{\mu} + ieA_{\mu}$$
$$\mathcal{L}_{int} = -e\bar{\psi}\gamma^{\mu}A_{\mu}\psi -ieA_{\mu}\left[\left(\partial^{\mu}\phi^*\right)\phi-\phi^*\partial^{\mu}\phi\right]-e^2A_{\mu}A^{\mu}\phi^*\phi$$
since we can identify $$\mathcal{L}_{free}$$ with the same expression as in the original post, with $$\partial_{\mu}$$ instead of $$D_{\mu}$$.