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In class today we were taught about Heisenberg’s equation, $$\Delta x\Delta p\ge\frac{h}{4\pi}. $$

Experience tells me that any time an equation involves pi, circles aren’t far behind. Obviously this is true in geometry, but even pure number theory equations, such as $\Sigma_{n=1}^{\infty} \frac1{n^2}=\frac{\pi^2}6$, you can always find a way to construct the problem such that circles are involved and the solution, including pi, naturally jumps out.

The natural question, then, is: what do circles have to do with Heisenberg? Why is Planck’s constant divided by a multiple of pi, and why specifically $4\pi$?

Qmechanic
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DonielF
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2 Answers2

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There are two different conventions for the constant used in the uncertainty principle, which are written as $h$ and $\hbar$. They are related by

$$\hbar = \frac{h}{2\pi}$$

and the reason for the $2\pi$ is because both appear in the expression for the energy of a photon:

$$E = hf = \hbar \omega$$

and the relation between frequency $f$ and angular frequency $\omega$ is,

$$\omega = 2\pi f$$

from whence the relation to circles is more obvious. Indeed, I'm more familiar with writing the HUP as

$$\Delta x \Delta p \ge \frac{1}{2}\hbar$$

with no $\pi$ in it and $\hbar$ as the basic constant. And in theoretical papers on quantum theory, you'll much less often encounter $h$, and you amy not even see $\hbar$ because Planck units are used where $\hbar = 1$!

The_Sympathizer
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Very sketchy, but $\Delta x \Delta p$ has a unit of angular momentum. Angular momentum is quantized (Bohr's condition) which can be interpreted as standing wave condition on "circular" orbit of electron $$n \lambda = 2 \pi r,$$ for which the $\pi$ come from.

mollusca96
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