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

1. May 22, 2010

### diagopod

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.

Last edited: May 22, 2010
2. May 22, 2010

### 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.

3. May 22, 2010

### diagopod

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?

4. May 23, 2010

### clem

It just turns out that h/2pi appears in so many equations that hbar is a convenient constant. Then it turns out that 1 is even more convenient.

5. May 24, 2010

### diagopod

Thanks Clem. So in that sense, the fine structure constant is just one of several dimensionless numbers that could be produced by combining numbers lie e,h and so on?

6. May 24, 2010

### LostConjugate

Think of it as the value h per oscillation.

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