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I have a Lagrangian that it have a term $\frac{1}{h}Tr(M^2)$ where $M$ is a 3*3 matrix and scalar field $h$ is one dimension. Is it correct to have such field with negative power in a Lagrangian?

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  • $\begingroup$ Presumably of interest: see the last paragraph in QMechanic's answer here. $\endgroup$ – AccidentalFourierTransform Jan 4 '18 at 10:54
  • $\begingroup$ define "correct" please. $\endgroup$ – Rho Phi Jan 4 '18 at 16:43
  • $\begingroup$ What's the rest of the Lagrangian? $\endgroup$ – Qmechanic Jan 4 '18 at 17:23
  • $\begingroup$ @RhoPhi +1 .... $\endgroup$ – Kite.Y Jan 4 '18 at 18:59
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Yes, it is ok (if the Action is real-valued and has Basic invariances like Lorentz invariance).

Note that you can do the Substitution $u = \frac{1}{h}$ and the new $u$ field appears linear in the Action. What will Change is the measure factor in the path integral

$\mathcal{D}[u] = - \frac{1}{h^2}\mathcal{D}[h] = -u^2 \mathcal{D}[h]$.

When computing the path integral, there will appear e.g. Gamma functions.

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  • $\begingroup$ .Many Thanks. If this Lagrangian is defined in QCD,again is it ok?Is there any problem due to renormalization? $\endgroup$ – elahe Jan 4 '18 at 14:34
  • $\begingroup$ define "ok" please. $\endgroup$ – Rho Phi Jan 4 '18 at 16:44
  • $\begingroup$ I mean what about the case that this Lagrangian is defined in QCD. Is there any problem because of the renormalization? $\endgroup$ – elahe Jan 5 '18 at 14:34

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