quantum mechanics - What is the point of the reduced Planck's constant $hbar$ (h-bar)? - Why don't we just have Planck's Constant $h$?

I know that $\hbar$ is $h / 2\pi$ - and that $h$ is the Planck Constant ($6.62606957 × 10^{-34}\:\rm J\:s$). But why don't we just use $h$ - is it that $\hbar$ is used in angular momentum calculations?

Answer

Perhaps some additional information is in order to shed additional light...

The whole discussion begs the question: If $\hbar$ is so convenient, why do we have $h$ around?

As usual, "historical reasons".

Planck originally invented $h$ as a proportionality constant. The problem he was solving was blackbody radiation, for which the experimental data came from spectroscopy people. And spectroscopy people used $\nu$ (for frequency, for that or wavelengths were what they measured). So the data was tabulated in frequency. So, when he formulated his postulate, he used $E = nh\nu$ for his quantization.

In modern theory, we prefer working with $\omega$ rather than $\nu$, because it is annoying to write $\sin (2\pi\nu t)$ rather that $\sin (\omega t)$. With angular frequencies, the quantization postulate becomes:

$E = n \frac{h}{2 \pi} \omega$

Now life sucks. So we invented the shorthand:

$E = n \hbar \omega$

We are happy (almost) everywhere. If Planck had spectroscopy data in $\omega$, we probably would not have a bar on the $h$ now...