Tangent, but there has to be a better way to measure population density than taking total area as the denominator and headcount as the numerator.
Something like 'median density': sort citizens by local density, find the median, and report that. My guess is that California would have a higher number than Florida by that measure. I don't know about Switzerland vs. California, but that's the question I'd like to answer here!
Another approach might be to start by subtracting land not available for most human uses for whatever reason, e.g. up on the Eiger, or in a large national park, or on a military reservation (places like Pendleton and Hunter Liggett immediately come to mind in California).
I'd guess those two approaches converge on a similar number, but using the median density is pure statistics, while removing unused land is more of a judgement call.
The median approach does risk minor artifacts based on the divisions used to calculate density (census tracts in the United States), but since we're looking for a median value, and census tracts are pretty small, it should be reasonably precise.
Something like 'median density': sort citizens by local density, find the median, and report that. My guess is that California would have a higher number than Florida by that measure. I don't know about Switzerland vs. California, but that's the question I'd like to answer here!