Physics Asked on January 11, 2021
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.
$${cal L} =frac{1}{2} (partial^mu phi partial_muphi -left(frac{mc}{hbar}right)^2phi^2)$$
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.
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 $(nablaphi)^2 / 2$, 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 $(E/L^{D-2})^{1/2}$.
Correct answer by Chiral Anomaly on January 11, 2021
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 $L^{-1}$.
Answered by C Tong on January 11, 2021
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