Heisenberg Uncertainty Principle

The uncertainty principle states that the position and momentum of a particle cannot both be determined with arbitrary precision at the same time. The product of the two uncertainties has a lower bound: $\Delta{}x\cdot{}\Delta{}p \ge{}\frac{ \hbar{}}{2}$, with the reduced Planck constant $\hbar{} \approx{} 1.055\times{}10^{-34} J\cdot{}s$. The more precisely the position is fixed, the larger the momentum uncertainty inevitably becomes and vice versa. This limit is not a matter of poor instruments but a fundamental property of the quantum nature. Δx and Δp are the standard deviations of the respective distributions. For everyday objects the limit is completely negligible because of the tiny ℏ, but for electrons and other quantum particles it is decisive.

The uncertainty is not an imprecision of the instruments but a fundamental property of the quantum states themselves. A particle does not simultaneously possess an exactly sharp position and an exactly sharp momentum; these quantities are not physically defined together with arbitrary precision at all. Δx and Δp are standard deviations of the probability distributions that describe the state. Even with perfect instruments the limit $\Delta{}x\cdot{}\Delta{}p \ge{}\frac{ \hbar{}}{2}$ would remain. The reason is that position and momentum are Fourier-conjugate quantities: a spatially tightly localized wave function needs many different wave numbers and thus a broad momentum spectrum. Treating this interpretation as a mere measurement error is a widespread misconception.

Rearrange the uncertainty relation for Δp: $\Delta{}p \ge{}\frac{ \hbar{}}{(2\cdot{}\Delta{}x)}$. The minimum momentum uncertainty is obtained when you take the equals sign. Insert the reduced Planck constant $\hbar{} \approx{} 1.055\times{}10^{-34} J\cdot{}s$ and the position uncertainty Δx in metres. Example: an electron localized to $\Delta{}x = 10^{-10} m$ has $\Delta{}p \ge{}\frac{ 1.055\times{}10^{-34}}{(2\cdot{}10^{-10})} \approx{}\frac{ 5.3\times{}10^{-25} kg\cdot{}m}{s}$. Through $\Delta{}p = m\cdot{}\Delta{}v$ this gives a speed uncertainty of about $\frac{5.8\times{}10^{5} m}{s}$. If you halve the position uncertainty, the minimum momentum uncertainty doubles, because the two quantities are inversely proportional to each other. Make sure to use ℏ and not h.

Because the reduced Planck constant ℏ, at about $1.055\times{}10^{-34} J\cdot{}s$, is unimaginably small. For macroscopic objects the limit set by $\Delta{}x\cdot{}\Delta{}p \ge{}\frac{ \hbar{}}{2}$ is so tiny that it vanishes completely against any practically achievable measurement precision. A car or a ball has such a large momentum that the associated momentum and position uncertainty lies billions of times below any detection threshold. Therefore everyday objects appear sharply localized and with a definite speed. Only for extremely light particles such as electrons, whose momentum is small and whose wavelength is large, does the uncertainty become significant and determine the behaviour, for example the size of atoms and the stability of the electron shell.

In the uncertainty relation $\Delta{}x\cdot{}\Delta{}p \ge{}\frac{ \hbar{}}{2}$ it is the reduced Planck constant ℏ, not the ordinary h. Both are linked through $\hbar{} =\frac{ h}{(2\pi{})}$, so they differ by the factor $2\pi{} \approx{} 6.28$. With $h = 6.626\times{}10^{-34} J\cdot{}s$ you get $\hbar{} \approx{} 1.055\times{}10^{-34} J\cdot{}s$. A common mistake is to accidentally insert h instead of ℏ; then the result is too large by exactly the factor 2π. The reduced ℏ appears wherever angular frequencies or angular momenta play a role, while h appears for example in $E = h\cdot{}f$ and in the de Broglie relation $\lambda{} =\frac{ h}{p}$. Check carefully in every formula which constant is required.