# Negative powers of fields in a QFT Lagrangian

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?

• Presumably of interest: see the last paragraph in QMechanic's answer here. – AccidentalFourierTransform Jan 4 '18 at 10:54
• define "correct" please. – Rho Phi Jan 4 '18 at 16:43
• What's the rest of the Lagrangian? – Qmechanic Jan 4 '18 at 17:23
• @RhoPhi +1 .... – Kite.Y Jan 4 '18 at 18:59

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