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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.

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

1 Answer 1

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Thanks to @Duepietri, I've come to a conclusion. Taking into account that $D_{\mu}$ needs to be gauge invariant, the minimal coupling

$$D_{\mu} = \partial_{\mu} + ieA_{\mu}$$

would give some remaining terms, of which we could compose an interaction:

$$\mathcal{L}_{int} = -e\bar{\psi}\gamma^{\mu}A_{\mu}\psi -ieA_{\mu}\left[\left(\partial^{\mu}\phi^*\right)\phi-\phi^*\partial^{\mu}\phi\right]-e^2A_{\mu}A^{\mu}\phi^*\phi$$

since we can identify $\mathcal{L}_{free}$ with the same expression as in the original post, with $\partial_{\mu}$ instead of $D_{\mu}$.

For those who wish to read a bit more about this, I would recommend the online note by Matthew Schwartz (2012) on Scalar QED or his book, Quantum Field Theory and The Standard Model by Matthew D. Schwartz.

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