Visual walkthrough — Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2
Step 1 — What does a "perfectly known momentum" look like?
WHAT. We draw one pure, endless wave: .
WHY. Before we can talk about uncertainty, we need to see the extreme case where one thing is known perfectly. A wave with a single, unchanging wavelength is that case for momentum. Here is the wavenumber — literally "how many radians of wiggle per metre." A single fixed means a single fixed wavelength , and by de Broglie a single fixed momentum .
PICTURE. Look at s01. The blue wave repeats forever, identical crest after identical crest.
Now ask: "Where is the particle?" Point anywhere on the blue curve — the next crest looks exactly the same. The wave gives no clue about location. Perfect momentum has cost us all knowledge of position.
Step 2 — What does a "perfectly known position" look like?
WHAT. We draw the opposite extreme: a single sharp spike at one point.
WHY. To pin position perfectly we want everything concentrated at one and zero everywhere else. This is the mirror image of Step 1 — and we'll see it costs us all knowledge of momentum.
PICTURE. s02 shows a tall yellow spike. It screams "the particle is HERE."
But now ask: "What is its wavelength?" A single spike has no repeating pattern at all — you cannot read off any . So momentum is completely unknown. Perfect position has cost us all knowledge of momentum.
Step 3 — Adding waves builds a bump (the wave packet)
WHAT. We add a few pure waves of slightly different together and watch a localized bump appear.
WHY. Real particles are neither everywhere (Step 1) nor infinitely sharp (Step 2) — they're somewhere in between: a wave packet. The only way to build a localized bump out of endless waves is to superpose many of them. This is the heart of the whole principle, so we draw it explicitly.
PICTURE. In s03 the thin coloured waves are individual . Where their crests happen to line up (centre) they reinforce → big bump. Where they fall out of step (edges) they cancel → flat. The thick green curve is their sum: a packet.
The key observation for later: to build this bump we needed a range of values, not just one.
Step 4 — The trade-off you can see: narrow bump ⇒ wide range of
WHAT. We compare two packets: a wide gentle bump vs. a narrow sharp bump, and count how many different 's each needed.
WHY. This is the mechanism. We want to see that squeezing the bump narrower forces us to throw in more widely-spread wavelengths. Once seen, the inequality is just this fact written in symbols.
PICTURE. s04 has two rows.
- Top: a wide position bump (small spread in , call it... actually a large ) built from waves whose 's are all bunched together → small spread in .
- Bottom: a narrow position bump (small ) — to make the edges cancel sharply you must add waves with widely different → large spread in .
The maths of Fourier analysis makes this precise: for the best possible (Gaussian, i.e. bell-shaped) bump, Every symbol here we have already drawn: = width of the green bump, = width of the coloured-wave spread. The is the smallest the product can ever be.
Step 5 — Turning into momentum (de Broglie enters)
WHAT. We convert the -language of Step 4 into -language using .
WHY. Step 4's result is about waves. Chemistry cares about momentum. The bridge is de Broglie's relation, which in wavenumber form reads . We use it because it is the only thing that links a wave's wiggle-rate to a particle's momentum — exactly the question we need answered.
PICTURE. s05 shows the two axes side by side: the same spread, once labelled in (waves), once relabelled in (momentum). Nothing about the picture changes — only the ruler on the axis.
Here is Planck's constant "per radian." Because is just multiplication by a constant, a spread in becomes a spread in the same way: Every symbol: = momentum spread, = the fixed conversion number, = wavenumber spread from Step 4.
Step 6 — Multiply and read off the principle
WHAT. Substitute into .
WHY. This is the final algebra that lands us on the boxed result — nothing new, just replacing by .
\quad\xrightarrow{\;\Delta k=\Delta p/\hbar\;}\quad \Delta x\left(\frac{\Delta p}{\hbar}\right)\ge \tfrac12$$ Multiply both sides by $\hbar$ (positive, so the "$\ge$" stays pointing the same way): $$\boxed{\;\Delta x\;\Delta p \;\ge\; \frac{\hbar}{2}\;}$$ > [!formula] Every symbol, one last time > $$\underbrace{\Delta x}_{\substack{\text{how blurry}\\\text{the position is}}}\;\underbrace{\Delta p}_{\substack{\text{how blurry}\\\text{the momentum is}}}\;\ge\;\underbrace{\frac{\hbar}{2}}_{\substack{\text{fixed floor}\\\approx 5.3\times10^{-35}\,\text{J·s}}}$$ > Since $\dfrac{\hbar}{2} = \dfrac{h}{4\pi}$, this is the *identical* statement chemistry books write as $\Delta x\,\Delta p \ge h/4\pi$. --- ## Step 7 — The degenerate cases (check the extremes) **WHAT.** Test the formula against Steps 1 and 2, the perfect-knowledge extremes. **WHY.** A law must survive its own edge cases. If letting one uncertainty go to zero breaks the picture sensibly, we trust it. **PICTURE.** s06 plots the boundary curve $\Delta x\,\Delta p = \hbar/2$ — a hyperbola. The forbidden zone (below it, tinted red) is where nature never goes; the allowed zone (above) is everywhere real particles live. ![[deepdives/dd-chemistry-2.1.04-d2-s06.png]] - **$\Delta x \to 0$ (Step 2 spike):** to keep the product $\ge \hbar/2$, $\Delta p$ must $\to \infty$. Momentum totally unknown — exactly what the spike showed. - **$\Delta p \to 0$ (Step 1 pure wave):** then $\Delta x \to \infty$. Position totally unknown — exactly what the endless wave showed. - **Big mass (cricket ball):** $\Delta v = \dfrac{\hbar}{2m\,\Delta x}$. With $m$ enormous, $\Delta v$ is absurdly tiny — the hyperbola hugs the axes so tightly the classical point-particle sits safely in the allowed zone. Uncertainty "switches off" for big things. > [!mistake] "Can't we just sit *on* the curve, or beneath it?" > **Why it feels right:** minimising things is usually good. **The fix:** you may sit *on* the hyperbola (Gaussian packet, best case) or *above* it (any messier state), **never below** it. The red region is physically empty. --- ## The one-picture summary ![[deepdives/dd-chemistry-2.1.04-d2-s07.png]] This final figure stacks the whole story: a pure wave (sharp $p$, no $x$) at one end, a spike (sharp $x$, no $p$) at the other, a balanced packet in the middle — and beneath them the hyperbola $\Delta x\,\Delta p = \hbar/2$ showing each case as a point that never crosses into the forbidden red. > [!recall]- Feynman retelling — say it in plain words > A particle isn't a dot; it's a **little heap of ripples** — a wave packet. If you make the heap out of ripples that all have the *same* wavelength, the heap spreads out forever: you know the wavelength (so the momentum) perfectly, but the particle is *everywhere*. If instead you pile ripples of *many* wavelengths on top of each other, they cancel out except in one small spot — now you know *where* it is, but you've thrown in so many different wavelengths that its momentum is a blur. There's no way to have both: a narrow heap **needs** a wide mix of wavelengths. Fourier maths says the best you can do is $\Delta x\,\Delta k = \tfrac12$; de Broglie ($p=\hbar k$) turns that into $\Delta x\,\Delta p = \hbar/2$. For a cricket ball the numbers are so tiny you'd never notice, which is why the everyday world looks solid and certain — but for an electron in an atom the blur is huge, which is why we draw **[[Atomic orbitals and probability density|orbital clouds]]** instead of neat **[[Bohr model and its failures|Bohr orbits]]**. > [!recall]- Quick self-check > Why does a narrow position bump need many wavelengths? ::: To cancel everywhere except one spot, you must superpose waves of many different $k$ — sharp edges require a broad $\Delta k$. > What single relation converts $\Delta k$ into $\Delta p$? ::: de Broglie's $p=\hbar k$, so $\Delta p = \hbar\,\Delta k$. > Where on the hyperbola does a Gaussian packet sit? ::: Exactly *on* it — the minimum-uncertainty (equality) case. > Why is the cricket ball unaffected? ::: Its huge mass makes $\Delta v=\hbar/(2m\Delta x)$ vanishingly small. --- **Related:** [[Wave-particle duality]] · [[Schrödinger equation and wavefunction ψ]] · [[Quantum numbers n, l, m, s]]