INGENIA

QNT-05

Heisenberg uncertainty

Δx Δp ≥ ħ/2; given Δx, the bound on Δp.

Reading speed
UncertaintyHeisenberg 1927

Governing equation

Δx Δp≥ℏ2\Delta x\,\Delta p\ge\dfrac{\hbar}{2}

Δx∈[0.001,10]\Delta x \in \left[0.001, 10\right]

where

\Delta x
Position σ (nm)
\Delta p_{\min}
Min momentum σ (keV/c)
\Delta v_{\min}
Electron Δv min (km/s)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

QNT-05 · quantum
00:0 / 00:08
①

Governing law — red

Δx Δp≥ℏ2\Delta x\,\Delta p\ge\dfrac{\hbar}{2}

Δx∈[0.001,10]\Delta x \in \left[0.001, 10\right]

Δx=0.001\Delta x = 0.001Δpmin⁡=98.66\Delta p_{\min} = 98.66

②

Domain of the variable — blue

Δxmin⁡=0.001,Δxmax⁡=10\Delta x_{\min} = 0.001,\quad \Delta x_{\max} = 10

Δx∈[0.001,10]\Delta x \in \left[0.001, 10\right]

Δx=0.001\Delta x = 0.001Δx=0.001\Delta x = 0.001

③

Response curve — green

Δx Δp≥ℏ2\Delta x\,\Delta p\ge\dfrac{\hbar}{2}

Δx∈[0.001,10]\Delta x \in \left[0.001, 10\right]

Δx=0.001\Delta x = 0.001Δpmin⁡=98.66\Delta p_{\min} = 98.66

④

Rate of change — amber

dΔpmin⁡dΔx\frac{\mathrm{d}\Delta p_{\min}}{\mathrm{d}\Delta x}

Δx∈[0.001,10]\Delta x \in \left[0.001, 10\right]

Δx=0.064\Delta x = 0.064Δpmin⁡′=−778.3\Delta p_{\min}^{\prime} = -778.3

Lecture brief

Historical brief

Planck (1900), Einstein’s photoelectric law, Bohr, de Broglie, Heisenberg and Schrödinger’s 1926 equation rebuilt matter as amplitude. These sheets are the first solvable models: wells, spin, tunneling, uncertainty. This sheet (QNT-05 — Heisenberg uncertainty) is the form associated with Heisenberg 1927. Working symbols: Δx\Delta x →\rightarrow Δpmin⁡\Delta p_{\min}, Δvmin⁡\Delta v_{\min}. Non-commuting operators x and p imply a product of standard deviations bounded by ħ/2, the Robertson–Schrödinger form of Heisenberg's principle.

Purpose

Purpose: compute Δpmin⁡\Delta p_{\min}, Δvmin⁡\Delta v_{\min} from Δx\Delta x in Quantum mechanics via Δx Δp≥ℏ2\Delta x\,\Delta p\ge\dfrac{\hbar}{2} Δx Δp ≥ ħ/2; given Δx, the bound on Δp. Use it when a real quantum mechanics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given Δx=0.100 nm\Delta x = 0.100\,\mathrm{nm}, the governing relation Δx Δp≥ℏ2\Delta x\,\Delta p\ge\dfrac{\hbar}{2} yields Δpmin⁡=0.9866 keV/c\Delta p_{\min} = 0.9866\,\mathrm{keV/c}, Δvmin⁡=578.838 km/s\Delta v_{\min} = 578.838\,\mathrm{km/s}. Minimum-uncertainty bound, 1-D. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Min momentum σ \Delta p_{\min}0.9866 keV/c
  • Electron Δv min \Delta v_{\min}578.838 km/s
Reading speed

Watch on YouTube

Free library

Full library

Free PDF / open book

YouTube channels

QNT-05 · quantum
00:0 / 00:08

Narration of this film

Minimum-uncertainty bound, 1-D.

Non-commuting operators x and p imply a product of standard deviations bounded by ħ/2, the Robertson–Schrödinger form of Heisenberg's principle.

Reading speed

Watch on YouTube