# What are the units of a scalar field if I only impose $c=1$?

I know that a scalar field in 4d in natural units ($\hbar = c =1$) has mass dimension 1. We can see this by requiring that the kinetic term in the action $$\int \text{d}x^4 \: \partial_{\mu} \phi \: \partial^{\mu}\phi$$ be dimensionless. However, what are the dimensions of $\phi$ when I only let $c=1$ (i.e. time and distance are measured in the same units, say L)? To get his I've imposed that the Lagrangian have units of energy E $$[L] = \left[\int \text{d}x^3 \partial_{\mu} \phi \: \partial^{\mu}\phi \right] \stackrel{!}{=} \text{E}.\\ \implies \text{L}[\phi]^2 \stackrel{!}{=} E \quad \quad \implies [\phi] = \sqrt{\text{E}\:\text{L}^{-1}}$$ I don't think this is right. It seems strange to have non integer powers of units. Also because in QFT the expectation value of a field is something physical ($\phi$ is hermitian).

So one conclusion is that there are powers of $\hbar$ to be restored. If, for example the true kinetic term was $\hbar\:\partial_{\mu} \phi \: \partial^{\mu}\phi$ then $[\phi] = L^{-1}$ which is more reasonable. (Notice restoring any $\hbar^\text{n}$, $n$ odd, would also give a reasonable answer; so what is the correct $n$?).

PS: This however begs the question: how to write down a $\textit{classical}$ scalar field theory? i.e. one where $\hbar$ doesn't appear at all. Supposing we have some notion of mass, $m \: \partial_{\mu} \phi \: \partial^{\mu}\phi$ would give $[\phi] = \sqrt{L^{-1}}$; again, not quite right.

• I think the lagrangian should be dimensionless. – PeaBrane Mar 20 '18 at 23:27
• The Lagrangian is (schematically) kinetic energy - potential energy. It has units of energy. – Rudyard Mar 20 '18 at 23:44
• Do you mean a classical scalar field theory where $c=1$ but $\hbar \neq 1$ and doesn't appear? Or just scalar field theory in general sans $\hbar$? If the latter here is one (though of course their eventual aim is to describe quantum field theory) quantummechanics.ucsd.edu/ph130a/130_notes/node448.html . Indeed this page starts a nice non-quantum application of field theory to a classical scalar field. – Triatticus Mar 21 '18 at 2:29
• @Triatticus Thank you for the link. However notice the 'classical' Lagrangian that they write down would imply that the dimensions of of the scalar field are $\sqrt{E L^{-1}}$, as I showed above. Unless of course one introduces powers of $\hbar$... but this is something quantum mechanical! So this leads me to wonder: what is the correct theory for a relativistic $\textit{non-quantum-mechanical}$ scalar field? If the Lagrangian is still truly the same but without the factors of $\hbar$ then it seems the scalar field will have quite unnatural (unphysical?) units. – Rudyard Mar 22 '18 at 19:02
• For the sake of argument you will always have an implicit $\hbar$ since it has the same dimensions as the action does in non-natural units. The issue is there isn't a unique way to non-dimensionalize the action if you only let two of the three constants (remember that natural units is defined by $\hbar = c = k_b = 1$ as there are three linearly independent units we wish to remove, namely L, T, and M.). – Triatticus Mar 22 '18 at 20:45