What is the significance of the quantity in the Heisenberg uncertainty principle? $Δp \times Δx = \frac{h}{4π}$ The focus of the question is to learn why the ratio of a circle's circumference to its diameter, universally known as $π$, is used in the denominator in the term $\frac{h}{4π}$. To put it another way, why is this geometric number relevant in Heisenberg's formula?

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    $\begingroup$ Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. $\endgroup$
    – Philipp
    Sep 17, 2014 at 8:41
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    $\begingroup$ Related Phys.SE posts: physics.stackexchange.com/q/69604/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 17, 2014 at 17:27
  • $\begingroup$ Aha! Now it is clear what are you asking for. For de Broglie plane waves $\pi$ appears naturally in a connection between $x$ and $p$ as I tried to show at the very end of my answer. For a general case connection between $x$ and $p$ is non-trivial, but it still involves $\pi$, as it is postulated by canonical commutation relation. $\endgroup$
    – Wildcat
    Sep 18, 2014 at 17:27

3 Answers 3


The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter) $$ \Delta x \Delta p \sim \hbar \, .$$ To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations. $$ \Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\ \Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\ $$ where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that $$ \Delta x \Delta p \geq \hbar / 2 \, . $$ Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity $$ \sigma_x \sigma_p \geq \hbar / 2 \, . $$

So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.

In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ $$ \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|. $$ Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to $$ \sigma_x \sigma_p \geq \hbar / 2 \, . $$

From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 noted, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.

Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation $$ p = \hbar k \, , $$ where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows $$ \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, , $$ and its derivative with respect to $x$ is $$ \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, . $$ If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get $$ \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, , $$ or $$ p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, , $$ which suggest that $$ \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, . $$ $\hat{x}$ is simple multiplicative operator in this picture $$ \hat{x} \psi(x) = x \psi(x) \, , $$ so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.

  • $\begingroup$ I am puzzled by the answer that has been posted. Rather than answering my question directly, it introduces a new dimension to my question which, although it may be correct, is not relevant to my specific question. To clarify this commenmt and my original question, I want to know what 4π means as the divisor in Heisenberg's formula. $\endgroup$ Sep 17, 2014 at 9:22
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    $\begingroup$ @TheStarVoyager $4 \pi$ is just a factor of proportionality. What kind of its significance beyound that you are looking for? $\endgroup$
    – Wildcat
    Sep 17, 2014 at 9:46

It seems that you are looking for a more mathematical answer.

Actually Heisenberg's inequality naturally comes from using Fourier Analysis. $2\pi$ comes from deriving a Fourier transform and the $2$ that remains comes from playing around with the terms.

For a detailed proof of the Heisenberg inequality that explains every term, you can check GB Folland's Lectures on Partial Differential Equations (pdf) http://www.math.tifr.res.in/~publ/ln/tifr70.pdf and go to page 11.


I will try to address your question very directly:

why is this geometric number relevant in Heisenberg's formula?

$\pi$ is not a geometric number. Among many things, $\pi$ the ratio of a circle's circumference to its diameter, but it is not its deepest meaning. That is, this geometrical relation can be viewed as a consequence of most fundamental properties of $\pi$. The inclusion of $\pi$ in the uncertainty relationships is due to those other more fundamental properties, and it is not a direct consequence the geometric properties mentioned by you.

$\pi$ arises naturally (after imposing some desirable restrictions) in the solution of the differential equation

$$f(z) = \frac{df(z)}{dz}$$

with $f:\mathbb{C} \rightarrow \mathbb{C}$ (and clearly $z \in \mathbb{C}$) which is found everywhere in physics, chemistry, etc.

In order to get a more detailed (case specific) answer of how it arises, please take a look at the Wildcat answer.

As I understand it, that is the deepest reason why $\pi$ is found in the uncertainty principle equation.


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