Why is h-bar rather than h used in Planck units?

Main Question or Discussion Point

Learning a bit more about Planck units, it looks like a number of arguably arbitrary, or at least pragmatic, choices had to be made, regarding using G versus 4piG, using 1/4piEpsiolon0 versus Epsilon0, and so on, but in the reading I don't see any question that h/2pi, rather than h, is the best choice when it comes to h, and was just curious about that.

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alxm
Planck units are arbitrary, as are any set of units. The reason why $$\hbar$$ is set to 1 rather than $$h$$ is simply because you tend to use the former a lot more. (e.g. the momentum operator)

Personally i don't have any need for gravity, but do want a simpler expression for coulomb potentials, so I use Atomic Units instead.

The reason why $$\hbar$$ is set to 1 rather than $$h$$ is simply because you tend to use the former a lot more. (e.g. the momentum operator)
Thanks for the reply. I did notice that h-bar seems to be necessary to produce the dimensionless fine-structure constant. Without it, we would have to say 2 pi e^2 / hc (in cgs) right or am I missing something there?

clem