Heisenberg uncertainty principle is mathematically given as

$$\sigma_x \cdot \sigma_p \ge {{\hbar} \over {2}}$$

The two terms on the left being the standard deviations of position and momentum.

But on many places the HUP is as

$$\Delta x \Delta p\geq \frac{h}{4π}$$ and used as in this example( as in beisers modern Physics):

If a particle can be anywhere in a sphere of radius $\Delta R$ and can have momentum in a range $\Delta p$ then we have have $\Delta R$. $\Delta p \geq \frac{\hbar}{2}$

How does this example follow from the definition given on the top?

  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/69604/2451 , physics.stackexchange.com/q/103208/2451 , physics.stackexchange.com/q/271059/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Dec 23, 2020 at 12:01
  • $\begingroup$ I'm glad you didn't close this question, although there are similar ones but the answers there imho don't answer the question. $\endgroup$
    – Kashmiri
    Commented Dec 23, 2020 at 12:16
  • 1
    $\begingroup$ I’m very sorry to say that Beiser is full of conceptual errors and should not be relied on for anything. We stopped using it 10-12 years ago because of these multiple issues. $\endgroup$ Commented Dec 23, 2020 at 13:41
  • $\begingroup$ Thank you for your advice, I'll replace it soon hopefully by some other standard texts. Thank you. $\endgroup$
    – Kashmiri
    Commented Dec 23, 2020 at 16:39

3 Answers 3


The idea is that if a particle can be anywhere in the sphere of radius $\Delta R$, then $\sigma_x\sim\Delta R$ (you can try to calculate the standard deviation of the position of a particle that can be anywhere on a sphere of radius $\Delta R$ and it will be proportional to $\Delta R$). It is a bit like the Fermi Problem (https://en.wikipedia.org/wiki/Fermi_problem), in which you are only interested in estimating the order-of-magnitude of something.

This kind of "estimates" are not rigorous but common. However, this is usually enough. Notice that, independently of the rigor of the inequality, the physical interpretation will remain the same: if the radius of this sphere to which the particle is confined decreases, the uncertainty of the momentum increases (i.e. it is very hard to confine particles to very small spaces).

  • $\begingroup$ Thank you Johny, one more request. Say we know that $\Delta P= 10^5 $ how can we use this to say that the average $P$ will be of the order of $10^5$? P is the momentum. $\endgroup$
    – Kashmiri
    Commented Dec 23, 2020 at 12:45
  • $\begingroup$ Consider a similar analysis but for the position. If $\Delta R=10^5$, can we conclude that the average position is of order $10^5$? That might be true, but there are several distributions that have the same value of $\Delta R$. These are characterized by the solutions of the Schrödinger's equation. In the link: en.wikipedia.org/wiki/…, section "Derivation of the radial equation", the say that if $\lim _{r \rightarrow 0} r^{2} V(r)=0$ then near R = 0, $R \sim r^{l}, l\geq 0$. This would somehow justify your claim. $\endgroup$
    – JGBM
    Commented Dec 23, 2020 at 17:53

The first formula is more precise. For a gaussian wave package the standard deviations in x and p are related by this expression.

A wave function that is uniform over a spherical volume and zero outside it has high momentum components due to its sharp edges. This causes he product of x and p standard deviations to be larger than $\hbar/2 = h/4\pi$.

  • $\begingroup$ A wave function like that wouldn't be a solution of Schrödinger's equation. $\endgroup$
    – kaylimekay
    Commented Dec 23, 2020 at 11:52
  • $\begingroup$ @kaylimekay Yes it would. The free Schrödinger equation allows solutions with arbitrary momentum. From these I can construct a wave function that is uniformly nonzero inside a sphere and zero outside, at 1 point in time. $\endgroup$
    – my2cts
    Commented Dec 23, 2020 at 12:19
  • $\begingroup$ Thank you my2cts, Beiser doesn't talk about the wave function being Gaussian yet proceeds to use the second form of HUP. $\endgroup$
    – Kashmiri
    Commented Dec 23, 2020 at 12:42
  • $\begingroup$ @my2cts Sorry, I said that wrong. What I should have said is, the wave function should be continuous, so as to not have infinite energy. It doesn't really matter for the question. Only mention it since you gave an extra example. $\endgroup$
    – kaylimekay
    Commented Dec 23, 2020 at 15:41
  • $\begingroup$ @kaylimekay Indeed, you cannot make the step function infinitely sharp. $\endgroup$
    – my2cts
    Commented Dec 23, 2020 at 20:12

This answer is a comment really.

The standard deviation has a strict statistical meaning

The root-mean-square deviation of x from its average is called the standard deviation. For a set of discrete measurements, the standard deviation takes the form


for continuous:

stdevcont ....

Determining the average or mean in the above expression involves the distribution function for the variable.


The distribution function is also statistically defined, for example:

distrfun , Poisson

That is why the $σ$ symbol is usually confined to standard deviation.

The Heisenberg uncertainty(HUP) is usually given as


with the $Δ$ symbol instead of the $σ$ to keep clear that the distribution function is not one of the statistical ones, but given by the quantum mechanical solution of the equations and boundary conditions of the problem, $Ψ^*Ψ$, a probability distribution, but not a statistical one.


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