# 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?

• $\hbar$ is a lot more common than $h$ is almost all (quantum-mechanical) calculations. It's simply laziness. – Danu Dec 17 '14 at 17:08
• So we can write, e.g, $E = h\nu = \hbar \omega$ instead of $E = h \nu = \frac{h}{2\pi}\omega$ – Alfred Centauri Dec 17 '14 at 17:22
• We do exactly the same thing with angular frequencies. It's much better in classical mechanics and electrodynamics (and EE) to deal with $\omega$ than with $2\pi f$. – CuriousOne Dec 17 '14 at 17:45
• @Danu - laziness, or efficiency? If everyone understands what you mean there is no need to waste time / ink. – Floris Dec 17 '14 at 20:00
• It looks cooler honestly – Connor Dolan Mar 13 '18 at 12:36

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...

• I'd add cultural differences. Electrical engineers like to state frequency in cycles per second (Hertz); physicists prefer radians per second. – Bert Barrois Mar 13 '18 at 14:51
• @BertBarrois but your talking about people who think $\sqrt{-1} = j$.... – JEB Mar 13 '18 at 15:24
• ... and this is physics.stackexchange.com :-) – safkan Mar 23 '18 at 21:34

To quote Stephen Gasciorowicz,

Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is $h/2\pi$ rather than $h$ that appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$

So basically it's just a matter of convenience.

The "quantities" in the quote are the energy and radius of the Bohr atom

Of course $ħ$ as the short form of $h/2\pi$ is more practical. This answer is simple but is not the answer to the question "what is the physical meaning (and convenience and difference) of ħ compared to h?" Let's consider the Bohm-Sommerfeld relationship $$\int_C\mathbf p \cdot \text{dx = nh}$$ For $n=1$ we see that the physical meaning of Planck constant is that of a complete rotation of a quantized vortex. This is normal if we consider quantum vacuum as a superfluid and fermions as quantum vortices in this superfluid as it happens in other superfluids as $^4\text{He}$. It is moreover interesting to observe that a vortex ring with healing distance, i.e. a vortex torus can perfectly express fermions spin$\frac{1}{2}$. Refer to chapters §3 and §3.1 in https://hal.archives-ouvertes.fr/hal-01312579 So, vacuum fluctuations $$\Delta E\Delta t \ge ħ$$ just means the spontaneous manifestation of quantum vortex-antivortex pairs (particle-antiparticle pairs) in the superfluid vacuum. A really modern view in quantum physics has indeed to consider quantum vacuum as a superfluid (Planck didn't know this, for this reason "h" is still "in circulation" (using a pun!)) which probably coincides with the ubiquitous scalar field of dark energy, whose mass density $\rho_0$ is expressed in the cosmological constant of Einstein field equations $\Lambda=\rho_0k$ and whose internal pressure causes the well-known repulsive action of dark energy. Indeed the question "Planck constant is a quantum of action. But what kind of action?" has answer: "a rotation". So we understand why we have to put $2\pi$, as it refers to a complete rotation.

## protected by Qmechanic♦Mar 13 '18 at 7:41

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