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Quantum Mechanics

Position and Momentum Cannot Both Be Sharp

Squeeze position and momentum spreads; a narrow wave packet is a sum of many wavenumbers, so the product cannot fall below Δx·Δp ≥ ℏ/2.

The harder you pin a particle to a single point, the more its momentum blurs out. Call the spread in position Δx and the spread in momentum Δp; the two are a seesaw, where squeezing one stretches the other. For a Gaussian wave packet the product bottoms out at Δx·Δp = ℏ/2, and it can never go below that. We start with a single small bump in space in the first figure.

A particle with a sharp position is a single small bump in space. We plot the Gaussian packet's probability density |ψ(x)|² and let the slider change its width σ. Shrink σ and the bump grows narrow and tall; grow σ and it spreads wide and low. The area stays fixed at 1, so the narrower it gets, the taller it stands. If you shrink the width toward zero to make a perfect single point, the bump shoots up without bound and becomes physically impossible, which is the first hint that position cannot be narrowed without limit.

One and the same σ sets both the position spread Δx and the momentum spread Δp. For a Gaussian packet Δx is proportional to σ, while Δp goes as 1/σ, namely ℏ/(2σ). Slide σ down and the Δx bar shrinks, but the Δp bar rises by the same factor. The product of the two bars, Δx·Δp, always stays at the same value. That one σ fixes both spreads at once is because position and momentum are a pair sprung from a single root, and it is the very same trade-off as a shorter sound having a blurrier pitch.

Written in one line, this limit is the Heisenberg relation Δx·Δp ≥ ℏ/2, where ℏ = h/2π is the reduced Planck constant. Draw the boundary curve Δp = ℏ/(2Δx) and shade below it: the dot can sit on the curve or above it, but never inside the shaded band beneath. A Gaussian packet sits exactly on that boundary line. This limit is not due to clumsy measurement but a floor fixed by nature, and it is exactly why an electron cannot fall all the way into the nucleus and why atoms do not collapse.

Why does squeezing blur the momentum? A sharply localized bump is built by adding plane waves of many wavenumbers. Use the slider to raise the number of waves N. With just one, you have an endless repeating wave and no location; add more and they reinforce only near x = 0, so a sharp bump emerges. Confining it tightly takes that many more wavenumbers. That the summed wavenumbers are spread wide means the momentum is spread wide as well, so at the very moment you make a narrow bump the blur in momentum comes along with it.

The position picture and the momentum picture are two languages for the same particle, paired by a Fourier transform. The left panel is a Gaussian in position space, the right one a Gaussian in momentum space, and their widths are reciprocals. Shrink σx and the left narrows while the right widens to σp = 1/(2σx). This reciprocal link is exactly why sharpening one side blurs the other. This Fourier-pair relation of reciprocal widths is not unique to quantum but a property every wave carries, and because nature writes even particles as waves, uncertainty becomes a fundamental law of matter.

In PracticeThe more you narrow position (small Δx), the wider momentum spreads (large Δp), and for a Gaussian packet that product touches its floor ℏ/2. Pinning to a point takes many wavenumbers, and position and momentum are a Fourier pair with reciprocal widths. In one line: Δx·Δp ≥ ℏ/2, and both can never be sharp at once.
Quantum Mechanics
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