Position and momentum together
In the world of big things we can say exactly where a car is and how fast it goes. For electrons it is different, because an electron behaves like a wave.
A wave that is very spread out has a clear wavelength (so a clear momentum, p = h/λ) but no clear position. A wave squeezed into a small spot has a clear position but is a mixture of many wavelengths, so the momentum is uncertain.
The rule: Δx · Δp ≥ ħ/2
Werner Heisenberg wrote this as a rule:
Δx · Δp ≥ ħ/2, where ħ = h/2π ≈ 1.05 × 10⁻³⁴ J s. Some books write Δx · Δp ≥ h/4π, which is the same.
Δx is the uncertainty in position and Δp is the uncertainty in momentum (mass times velocity, so Δp = m Δv). The rule is not about bad instruments. It stays true even with perfect instruments.
For big objects ħ is so tiny that the uncertainty is not noticeable. For an electron it is huge.
Why Bohr orbits fail and orbitals appear
Bohr drew electrons on neat circular paths. That needs exact position and exact speed together, which the uncertainty principle forbids.
The modern model says: we cannot give a path. We can give a probability of finding the electron at each place. A region where this probability is high is an orbital. For the lowest state (1s), the orbital is a round, fuzzy ball around the nucleus, densest near the centre. Other orbitals (p, d) have other shapes.
Shrinking the cloud to the nucleus would make the electron move extremely fast (Δp large), which costs a lot of energy. The atom settles at a size that balances these.
Size of the uncertainty: two quick estimates
An electron in an atom: Δx ≈ 10⁻¹⁰ m. Then Δp ≥ ħ/(2Δx) ≈ 5.3 × 10⁻²⁵ kg m/s, and with m = 9.1 × 10⁻³¹ kg, Δv ≈ 5.8 × 10⁵ m/s. That is huge.
A cricket ball (m = 0.15 kg) known to Δx = 10⁻³ m: Δp ≥ 5.3 × 10⁻³² kg m/s, Δv ≈ 3.5 × 10⁻³¹ m/s. This is far too tiny to ever see.
Try it
In the 3D, put the slider at 3. Read Δx, Δp and the product. Move the slider to 1. Read again. Write both products. Predict first: will the product change?
At home: make a loose rope wave with a friend by shaking one end slowly: a long wave. Shake fast and short: you get a short bump. Which one has a clearer wavelength?
Key formulas and definitions
- Δx · Δp ≥ ħ/2 (ħ = h/2π)
- Also written Δx · Δp ≥ h/4π
- Δp = m Δv
- ħ ≈ 1.05 × 10⁻³⁴ J s; h ≈ 6.63 × 10⁻³⁴ J s
- Smaller Δx → larger Δp
Worked examples
1. If you squeeze the position spread Δx to half, what happens to the smallest possible Δp?
Δp ≥ ħ/(2Δx). Half Δx gives double the smallest Δp.
2. Δx = 10⁻¹⁰ m for an electron. Find the minimum Δp. (ħ = 1.05 × 10⁻³⁴ J s)
Δp ≥ ħ/(2Δx) = 1.05 × 10⁻³⁴ ÷ (2 × 10⁻¹⁰) = 5.3 × 10⁻²⁵ kg m/s.
3. Using the Δp above, find Δv for the electron (m = 9.1 × 10⁻³¹ kg).
Δv = Δp/m = 5.3 × 10⁻²⁵ ÷ 9.1 × 10⁻³¹ ≈ 5.8 × 10⁵ m/s.
4. A 0.15 kg ball has Δx = 10⁻³ m. Find the minimum Δv.
Δp ≥ 1.05 × 10⁻³⁴ ÷ (2 × 10⁻³) = 5.3 × 10⁻³² kg m/s. Δv = 5.3 × 10⁻³² ÷ 0.15 ≈ 3.5 × 10⁻³¹ m/s. Far too small to notice.
5. In units where ħ = 1, a particle has Δx = 0.5. Find the smallest Δp.
Δp ≥ 1/(2 × 0.5) = 1.
6. In the same units, Δp is measured as exactly 0.25. What is the smallest Δx?
Δx ≥ 1/(2 × 0.25) = 2.
Common mistakes
- Thinking the principle is about clumsy instruments. It is a feature of waves in nature.
- Thinking it says nothing can be known. We can know either one sharply; only the pair has a limit.
- Forgetting to use ħ = h/2π in Δx · Δp ≥ ħ/2.
- Thinking a heavy ball is "more uncertain". It is less noticeable because ħ is tiny compared with its mass.