First year grad student here, I've taken two terms QFT.(adsbygoogle = window.adsbygoogle || []).push({});

I'm studying some effective field theories, and one of the techniques I've seen used for writting down the effective lagrangian is identifying some fields or components of fields that are "small" and removing them from the lagrangian by plugging in their equations of motion. So for example if you had something like

[tex] \frac{1}{2}\partial_{\mu}f\partial^{\mu}f + \frac{1}{2}\partial_{\mu}g\partial^{\mu}g + fg\phi(x)[/tex]

you would as have equations of motion

[tex] \partial^2f = g\phi(x),\quad\quad \partial^2g = f\phi(x)[/tex]

so if you wanted to eliminate g you would substitute

[tex] \frac{1}{2}\partial_{\mu}f\partial^{\mu}f + \frac{1}{2}\partial_{\mu}(\partial^2f/\phi(x))\partial^{\mu}(\partial^2f/\phi(x)) + f\partial^2f[/tex]

My problem is I'm having a hard time wrapping my head around what exactly I'm doing when I do this. If this were just a regular classical functional minimization problem where I was minimizing something like volume or cost rather than an action, I think I haven't changed anything, I've just hidden information about g. The solutions to f's (new) equations of motion should give me the correct mimization of the original problem, and then I could for example use g's equations of motion to obtain g from f.

However, in field theory I'm not really sure what I've done. A source I'm reading says that I've "removed g at tree-level". I can understand this in the sense that as I guessed above I should still have the right answer classically, so I haven't mess up tree-level. But where exactly along the line did I break the quantum corrections? Does this have something to do with the fact that off-shell particles don't solve their equations of motion, so the assumption that g will always be on shell is wrong?

Any comments about what's going on here would be extrememly helpful, thanks!

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# Effective field theories, eliminating fields using equations of motion

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