Particles have a wavelength, $\lambda = h/p$ (matter waves). A wave with one exact wavelength, a perfect sine, goes on forever: it has an exact momentum but no position at all. To make a particle that is somewhere, you must add waves of many wavelengths, so that they cancel everywhere except in one region. That trade-off is the uncertainty principle.
Key idea
The spreads in position and momentum of any quantum state obey $$\Delta x\,\Delta p \ge \frac{\hbar}{2},\qquad \hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34}\ \text{J·s}.$$ A similar relation links a state's lifetime and its energy spread: $\Delta E\,\Delta t \gtrsim \frac{\hbar}{2}$.
Why it works
Build a wave packet by adding waves with wave numbers spread over a range $\Delta k$. Near the middle they are in step and add up; farther out their phases drift apart and they cancel. Waves whose $k$ differ by $\Delta k$ drift a full cycle apart over a distance of about $2\pi/\Delta k$, so the packet's width is about $\Delta x \sim 1/\Delta k$; the careful statement, for the narrowest possible (Gaussian) packet, is $\Delta x\,\Delta k = \frac{1}{2}$. Since $p = \hbar k$, multiplying by $\hbar$ gives $\Delta x\,\Delta p \ge \frac{\hbar}{2}$. Nothing here is about measuring. It is a property of waves, the same reason a short musical note cannot have a perfectly sharp pitch.
Worked examples
Worked example: An electron in an atom
An electron is confined to a region about $0.10$ nm across. What is the smallest possible spread in its momentum, and in its velocity?
- $\Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2(1.0 \times 10^{-10})} = 5.3 \times 10^{-25}$ kg·m/s.
- $\Delta v = \frac{\Delta p}{m_e} = \frac{5.3 \times 10^{-25}}{9.11 \times 10^{-31}} = 5.8 \times 10^{5}$ m/s.
Answer: $\Delta v \gtrsim 6 \times 10^5$ m/s: an electron in an atom cannot sit still
Worked example: A baseball
A 0.145 kg baseball's position is known to within 1.0 μm. What is the smallest spread in its velocity?
- $\Delta v \ge \frac{\hbar}{2m\Delta x} = \frac{1.055 \times 10^{-34}}{2(0.145)(1.0 \times 10^{-6})} = 3.6 \times 10^{-28}$ m/s.
Answer: About $4 \times 10^{-28}$ m/s, far too small to ever notice
Worked example: A short-lived state
An excited atomic state lives about 10 ns. Estimate the smallest spread in the energy of the photons it emits, in eV.
- $\Delta E \gtrsim \frac{\hbar}{2\Delta t} = \frac{1.055 \times 10^{-34}}{2(1.0 \times 10^{-8})} = 5.3 \times 10^{-27}$ J.
- In eV: $\frac{5.3 \times 10^{-27}}{1.602 \times 10^{-19}} = 3.3 \times 10^{-8}$ eV.
Answer: About $3 \times 10^{-8}$ eV, the natural width of the spectral line
Common mistake
Thinking the uncertainty comes from clumsy measuring tools that better tools could fix. The spreads belong to the state itself: a particle confined to $\Delta x$ simply does not have a sharper momentum than $\frac{\hbar}{2\Delta x}$. Evidence: an electron confined to an atom has a measurable kinetic energy of several eV from that spread, and electrons sent through a narrow slit spread out more when the slit is narrower, whatever detector you use.