Some usual calculations of the Bohr's radius (see 2.5 at Feynman's here, or this text here) starts by defining a radius $a$ ( most probable radius ? average radius ? ). Next, from Heisenberg's uncertainty principle:

$$\Delta p \Delta a \ge h/2$$

it is said:

$$p = h/a$$.

where $p$ is the electron momentum.

Remainder of the calculation is expression of the total electron energy and minimization.

This step from uncertainty to momentum is confusing to me: we pass from inequality to equality, from margins ($\Delta$) to concrete values of $p$ and $a$.

Any hint to understand this step? Thanks.


This is another similar example, from "Quantum Mechanics", Nouredine Zettili, example 1.6:

Estimate the uncertainty in the position of (a) a neutron moving at $5 \cdot 10^6 m/s$ ... Solution: Using (1.57), we can write the position uncertainty as $$\Delta x = \frac{\hbar}{2 \Delta p} = \frac{\hbar}{2 m v } = ... $$

again, $\Delta p$, an interval, is converted to the absolute value $p=mv$.

If something is in an interval $p \in [p_{min},p_{max}]$ we can say $p=p_{avg}\pm \Delta p$ where $p_{avg}=\frac{p_{max}+p_{min}}{2}$ and $\Delta p=\frac{p_{max}-p_{min}}{2}$, but these texts seems to assume $p_{min}=0$ and $\frac{p_{max}}{2}=mv=\Delta p$ or something similar.

In other words, when in this example we give a known value for the speed, we are fixing the momentum without any indetermination, thus, uncertainly in position should be infinite.

  • 2
    $\begingroup$ From the Feynman source: "We now consider another application of the uncertainty relation, Eq. (2.3). It must not be taken too seriously; the idea is right but the analysis is not very accurate." $\endgroup$
    – jacob1729
    Jan 27, 2019 at 19:57

1 Answer 1


The usual way this calculation is demonstrated that you complain about isn't a solid argument, it si more like a play with symbols that is repeated and taught for its ability to provide a shortcut procedure to remind us of the correct result for Bohr radius. It isn't solid for two reasons:

  • if we are using the uncertainty relation, $x,p$ do not have single simultaneous values and so it makes no sense to define $p=h/a$;

  • the uncertainty relation and the virial principle themselves do not actually fix the size of the atom, the atom can be as large as we want - great size $a$ corresponds to a high excitation number $n$.

A more solid way to derive some sort of estimate of minimal atom size would work only with expected averages and using some formal equivalent of that minimum requirement.

For example, we can assume that minimum size of atom is achieved when the expected average of its energy is the minimal possible value, and then try to use HUP and the virial principle to derive some estimate of atom size (defined as $\sqrt{\langle x^2\rangle}$). Similar procedure is known in the general theory of electron shells of an atom, where it is able to provide and estimation of the ground state energy and the corresponding psi function.

  • $\begingroup$ Maybe it comes from the fact that $p=h/\lambda$ and $\lambda=2\pi a$ so we would have $p=h/2\pi a$ or $pa=\hbar$ and its bigger then the limit given by the uncertanity principle. $\endgroup$
    – seVenVo1d
    Jan 28, 2019 at 4:37
  • $\begingroup$ That is very different argument, because it uses de Broglie's relation for particle with definite momentum, not the uncertainty relation. $\endgroup$ Jan 28, 2019 at 14:16

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