What is the unit of Klein Gordon field?

Question

Normally I don't care about units in the derivations on relativity or QM. Just set $\hbar = c = 1$.

But learning about the energy momentum tensor for the Klein Gordon equation, I couldn't make $T^{00}$ for example have units of energy density, that means energy per (spatial) volume.

Of course $T^{\mu \nu}$ comes from the Lagrangian density, that should also have units of energy per volume. So I tried to examine it.

In the expression below, the second term for example has units of $L^{-2}$ if the field is adimensional.

$${\cal L} =\frac{1}{2} (\partial^\mu \phi \partial_\mu\phi -\left(\frac{mc}{\hbar}\right)^2\phi^2)$$

It could be fixed if the field has units of $$\left(\frac{E}{L}\right)^{\frac{1}{2}}$$

But I don't see it mentioned anywhere, so I am not sure about it. Just to compare, both the Lagrangian density and energy density for electromagnetism have consistent units.

Answer 1

Assuming that spacetime is four-dimensional, your result is correct. When $\hbar=c=1$, it reduces to the statement that $\phi$ has mass dimension $1$, which is often stated in the literature about relativistic quantum field theory.

To see this more directly, start with the fact that the action $S$ should have the same units as Planck's constant $\hbar$. (We usually say it the other way around: Planck's constant has units of action!) The action is the integral of ${\cal L}$ over spacetime, which implies that ${\cal L}$ has units of energy density (energy per unit spatial volume). If we use the convention that the spatial-derivative part of kinetic term is $(\nabla\phi)^2 / 2$, where $\nabla$ is the spatial gradient, then we arrive at the result shown in the question: $\phi$ must have units $(E/L)^{1/2}$, where $E$ is energy and $L$ is length.

One virtue of this argument is that it doesn't assume anything about how the coefficient of the $\phi^2$ term is related to mass, and it works even if the $\phi^2$ term is absent.

The same argument can be generalized to $D$-dimensional space for arbitrary $D$. The conclusion is that $\phi$ has units $(E/L^{D-2})^{1/2}$.

Answer 2

As mentioned in the comments, with $c$ and $\hbar$ set to 1. Energy has the unit of $L^{-1}$ and the 4D integral of the Lagrangian density is dimensionless as $\hbar$. It means that the Lagrangian density is of the unit $L^{-4}$, so $\partial_\mu \phi$ is of the unit $L^{-2}$ and $\phi$ of the unit $L^{-1}$. It is consistent with your conclusion of $(E/L)^{1/2}$, but you could reduce it to just $L^{-1}$.