# Where does $\pi$ come from in the Heisenberg equation?

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 $$4\pi$$?

• It's an interesting question as to how $\pi$ comes up everywhere in physics but I think in the case of the Heisenberg principle, it is just because of the convention. The quantity that actually shows up, physically, is $\hbar$ with the $\pi$ fully absorbed in it. For some reason, Planck isolated the $\pi$ out of it and defined an artificial number as $h$. That's why if you express things in terms of this $h$, $\pi$ shows up because $\pi$ was separated by construction. (I would have added this as an answer if I knew the "some reason" :P) – Dvij D.C. Nov 12 '19 at 23:32
• @DonielF Haha, again, the "natural" quantity to use here is $\omega$ and not $\nu$. In particular, this relationship comes (essentially) from the fact that any two adjecent energy levels in the energy spectrum of a harmonic oscillator are separated by $\hbar\omega$. Here, $\omega$ is simply coming from the harmonic potential $\frac{1}{2}m\omega^2x^2$. It would be unnatural to define the harmonic potential as $2\pi^2m\nu^2x^2$. – Dvij D.C. Nov 13 '19 at 0:17
• Ok but thanks for your last comment. Of course, this is how Planck introduced $h$, as the coefficient of $\nu$ in the quanta of energy of light. But, as I argue, the natural thing to have used would have been $\omega$ and then the coefficient would have been $\hbar$ all along. But, this wisdom that the natural thing to use would have been $\omega$ became available much later when we actually derived Planck's ansatz as a result from QED. I will compile all my comments into an answer. :) – Dvij D.C. Nov 13 '19 at 0:27
• Just saw your actual last comment. $\omega$ is the usual angular frequency, related to the frequency $\nu$ by a factor of $2\pi$. How the spectrum of a harmonic oscillator is related to the energy of the light quanta is an elaborate story. I will elaborate on it in my answer. Your professor is not talking about the spectrum of the harmonic oscillator in relation to photons is because this relation comes from Quantum Electrodynamics and not from non-relalativistic Quantum Mechanics. Strictly speaking, we should never speak of a photon in non-relativistic Quantum Mechanics in the first place. – Dvij D.C. Nov 13 '19 at 0:32
• I am on my phone so can't write a proper answer right now but I am too tempted to respond so I will leave one more comment :P Photon is a quanta of the EM field. That is true. But it is also the most relativistic object out there. It literally travels at the speed of light. So, one must use relativistic version of quantum mechanics to consistently describe it. This version is called quantum field theory. And the specific quantum field theory that describes electromagnetism is Quantum Electrodynamics. That's where a consistent quantum description of a photon lies. – Dvij D.C. Nov 13 '19 at 0:37

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 $$\hbar = 1$$!

• Really nice answer. – Árpád Szendrei Feb 25 at 16:45

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.

• And what does r represent? Distance from the nucleus, and therefore subsumed in the $\Delta x$ term? – DonielF Nov 12 '19 at 22:04
• But you have the same uncertainty relation, with the same $\pi$ for any pair of conjugate variables, so it seems that an explanation that explicitly invokes momentum can't be the right story. – WillO Nov 13 '19 at 0:45