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I studied free particle field like Dirac field and Klein Gordon field. My question is about interaction. How can I put a potential term in the Lagrangian density?

$\mathcal{L} =\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{m^2}{2}\phi^2$

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    $\begingroup$ Super-short answer: Interactions (usually) involve higher powers of the field, so you'd have terms like $u \phi^4$ $\endgroup$ – Lagerbaer Oct 19 '13 at 21:55
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Naively, the answer, as was already mentioned, is adding terms of higher powers in the field to the Lagrangian. This means that the equations of motion, in the case of a simple scalar field given by the Klein-Gordon equation, acquires additional terms. So quite generally, one can write

$$\mathcal{L}=\frac12\partial_\mu\phi\partial^\mu\phi-\frac{m^2}{2}\phi^2+\sum_{n\gt2}\frac{g_n}{n!}\phi^n,$$ which results in the equation of motion being

$$(\partial_\mu\partial^\mu+m^2)\phi=\sum_{n\gt2}\frac{g_n}{(n-1)!}\phi^{n-1}.$$

A natural question arises: just because I can add all these terms, is doing so justified? Are there any restrictions on the amount and form of terms I can add? It turns out there are. Some are easy to see, some are more subtle and complicated. I will mention just a few key points, for details simply refer to standard literature on quantum field theory.

  • Stability - As the Lagrangian basically gives you the energy of a system, its precise form can show pathologies from the beginning. Take for example a term of third order: $\phi^3$. If we allow the field to take on negative values, the energy can become arbitrarily negative. This is something one does not want, therefore we drop terms with this property.
  • Dimensional analysis - From the fact that the spacetime integral of the Lagrangian (Density) defines the action, one the prefactors of the interaction terms have to be fixed in such a way that that the overall dimension of the action, $[\hbar]$, is preserved. One can link the dimensionality of these new terms to some "fundamental energy scale" (Planck energy) in such a way that it is possible to argue that at low energies, terms of powers beyond a certain value are negligible. This this context, one speaks of "relevant", "irrelevant" and "marginal" operators. See David Tong's lectures for an explanation.
  • Renormalization - In order to have the theory produce meaningful results which can be both adjusted to and compared to experiment, one has to modify the original Lagrangian, i.e. allow both the parameters of the theory (in our case $m$ and the coupling constants $g_n$) and the fields to change as a function of the energy scale. This leads to the concept of the renormalization group, which allows us to determine the way in which these parameters change. The most famous example is the strong interaction, where we observe that the coupling constant is large at low energies and small at high energies. This phenomenon, referred to as asymptotic freedom, is captured mathematically by the renormalization group equations of QCD, but it can also be found in simple scalar field theory: ignoring the aforementioned instability, a theory which is third order in the fields also shows asymptotic freedom.

I hope this gives you a vague idea of what might be involved. I can only recommend books like Peskin&Schroeder and Srednicki, which you should definitely study if you want to develop a thorough understanding of the subject.

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    $\begingroup$ The most general (local) terms are of course products of $\partial^n \phi^m$, contracted in any suitable way. That is, if you think of the Lagrangian as an effective description and don't impose renormalizability (the old viewpoint). $\endgroup$ – Vibert Oct 31 '13 at 15:22
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You can either have higher order terms such as the one Lagerbaer mentioned. That term will give self-interactions. If there is a second field/particle type involved, you could include a term that involves polynomials of the terms multiplied by a coupling constant.

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  • $\begingroup$ Is there a typical method to determine coupling costants? $\endgroup$ – m.mybo Oct 20 '13 at 8:02

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