Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ - 2: Worked examples

This is the "roll up your sleeves" child of the Heisenberg parent note. The parent told you the why. Here we hit every kind of problem the two relations

$Δ x Δ p \geq \frac{ℏ}{2}, Δ E Δ t \geq \frac{ℏ}{2}$

can throw at you. Before any numbers, let us pin down what every symbol means so nothing sneaks in unexplained.

The scenario matrix

Every uncertainty problem is one of these cells. The examples below are labelled with the cell they cover, so by the end every row is filled.

Cell	What makes it distinct	Example
A. Solve for $Δ p$	given $Δ x$ , find momentum spread	Ex 1
B. Solve for $Δ x$	given $Δ p$ (or $Δ v$ ), find position spread	Ex 2
C. Macroscopic limit	huge mass ⇒ effect vanishes	Ex 3
D. Degenerate input $Δ x \to 0$	perfect position ⇒ what happens to $Δ p$ ?	Ex 4
E. Energy–time: lifetime ⇒ linewidth	given $τ$ , find $Δ E$	Ex 5
F. Energy–time: linewidth ⇒ lifetime	run it backwards	Ex 6
G. Zero-point energy (real-world word problem)	confinement ⇒ minimum energy $\neq = 0$	Ex 7
H. Unit / exam-twist trap	wrong constant, eV↔J, "disturbance" red herring	Ex 8

There are no "quadrants" or signs here — all four quantities are spreads, so they are $\geq 0$ by construction. The interesting extremes are instead the limits (mass $\to$ large, $Δ x \to 0$ , $τ \to 0$ ), which cells C, D, and F cover explicitly.

Example 1 — Cell A: find the momentum spread

Worked example Electron trapped in an atom,

$Δ x = 1 \overset{˚}{A} = 1.0 \times 1 0^{- 10} m$ . Smallest possible $Δ p$ ? Forecast: Guess before computing — will $Δ p$ be tiny or large for an electron? Jot a number.

1. Use the relation at its floor (equality), because "smallest possible" means we sit on the wall: $Δ x Δ p = \frac{ℏ}{2} \Rightarrow Δ p = \frac{ℏ}{2 Δ x} .$ Why this step? The word "minimum" turns $\geq$ into $=$ ; any larger $Δ p$ is allowed but not forced.

2. Plug in: $Δ p = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 10} )} \approx 5.3 \times 1 0^{- 25} kg\cdotm/s .$ Why this step? Just arithmetic — keep $ℏ$ in SI so the answer comes out in SI momentum units.

3. Convert to a speed spread using $Δ v = Δ p / m_{e}$ , $m_{e} = 9.11 \times 1 0^{- 31} kg$ : $Δ v \approx \frac{5.3 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 5.8 \times 1 0^{5} m/s .$ Why this step? Speed is more intuitive than momentum — this shows why electrons never "sit still".

Verify: Units: $\frac{J\cdots}{m} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{m} = kg\cdotm/s$ ✓. And $5.8 \times 1 0^{5} m/s$ is a fraction of light speed — big but sub-relativistic, physically sane.

Example 2 — Cell B: go the other way, find the position spread

Example 3 — Cell C: the macroscopic limit (why you never notice it)

Example 4 — Cell D: the degenerate input $Δ x \to 0$

The figure below makes the trade-off literal. It draws two cases stacked on top of each other. In the upper pair, the magenta position bump is squeezed narrow (small $Δ x$ ) and its orange momentum partner is forced wide (big $Δ p$ ). In the lower pair, the violet position bump is allowed to spread wide (large $Δ x$ ) and its navy momentum partner sharpens to a spike (small $Δ p$ ). Trace either colour across: whenever one curve narrows, its dashed partner flares — that is $Δ x Δ p \geq ℏ/2$ drawn as a picture. The $Δ x \to 0$ limit of this example is simply the upper case pushed to its extreme, where the orange curve becomes infinitely wide.

Figure — Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2, ΔE Δt ≥ ℏ - 2

Example 5 — Cell E: lifetime ⇒ linewidth (energy–time forward)

Worked example An excited atomic state lives for

$τ = 1.0 \times 1 0^{- 8} s$ . Find its minimum energy spread $Δ E$ , in joules and in eV. Forecast: Will this energy blur be big or small compared to a photon energy (~eV)?

1. Set $Δ t = τ$ (the lifetime is the timescale over which the state changes): $Δ E = \frac{ℏ}{2 τ} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 8} )} \approx 5.3 \times 1 0^{- 27} J .$ Why this step? $Δ t$ in the energy–time relation is not a stopwatch error — it is exactly the state's lifetime. That identification is the physics.

2. Convert to eV using $1 eV = 1.602 \times 1 0^{- 19} J$ : $Δ E = \frac{5.3 \times 1 0 ^{- 27}}{1.602 \times 1 0 ^{- 19}} \approx 3.3 \times 1 0^{- 8} eV .$ Why this step? Spectroscopists measure line widths in eV; the conversion makes the answer comparable to real data.

Verify: A $\sim$ eV transition blurred by only $3 \times 1 0^{- 8} eV$ is an extremely sharp line — consistent with the Natural linewidth and spectral broadening of allowed atomic transitions. Units: $J\cdots / s = J$ ✓.

Example 6 — Cell F: linewidth ⇒ lifetime (run it backwards)

Worked example A spectral line is measured to have a natural energy width

$Δ E = 4.0 \times 1 0^{- 6} eV$ . Estimate the lifetime of the emitting state. Forecast: Shorter or longer than the $1 0^{- 8}$ s of Example 5? (A wider line means what?)

1. Convert $Δ E$ to joules: $Δ E = (4.0 \times 1 0^{- 6}) (1.602 \times 1 0^{- 19}) = 6.41 \times 1 0^{- 25} J .$ Why this step? The relation is in SI; convert before dividing.

2. Solve the floor relation for $Δ t = τ$ : $τ = \frac{ℏ}{2 Δ E} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 6.41 \times 1 0 ^{- 25} )} \approx 8.2 \times 1 0^{- 11} s .$ Why this step? Same equation, unknown swapped — this is the "measure the blur, deduce how fast it lives" direction used for short-lived resonances.

Verify: The line is $\sim$ 100× wider than Example 5's, and indeed $τ \approx 8 \times 1 0^{- 11}$ s is $\sim$ 100× shorter than $1 0^{- 8}$ s. Wider line ⇒ shorter life — matches the mnemonic "live fast, die fuzzy." ✓

Example 7 — Cell G: real-world word problem, zero-point energy

Worked example An electron is confined inside a 1D "box" of length

$a = 1.0 \times 1 0^{- 10} m$ (roughly an atom). Estimate its minimum kinetic energy — its zero-point energy — and argue it cannot be zero. Forecast: Can a truly trapped electron ever be perfectly at rest ( $E = 0$ )?

1. Confinement sets the position spread: $Δ x \approx a$ . Hence $Δ p \approx \frac{ℏ}{2 a} .$ Why this step? The walls limit where the electron can be, so $Δ x$ can't exceed the box size — that's what "confined" means.

2. The average momentum is zero (it bounces both ways), i.e. $⟨ p ⟩ = 0$ , so the typical momentum magnitude is the spread itself: $p_{m i n} \approx Δ p$ . Kinetic energy $E = p^{2} /2 m$ : $E_{m i n} \approx \frac{( Δ p ) ^{2}}{2 m _{e}} = \frac{ℏ ^{2}}{8 m _{e} a ^{2}} .$ Why this step? Since $⟨ p ⟩ = 0$ (defined above), all the "motion" lives in the spread; squaring it gives the smallest energy consistent with being trapped.

3. Plug numbers: $E_{m i n} \approx \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{- 10} ) ^{2}} \approx 1.5 \times 1 0^{- 19} J \approx 0.95 eV .$ Why this step? An eV-scale energy is exactly the order of atomic binding energies — the estimate is physically believable. Compare with the exact Particle in a box ground state.

4. Why not zero? $E = 0$ needs $p = 0$ exactly, i.e. $Δ p = 0$ , which forces $Δ x = \infty$ — the electron would escape the box. Contradiction. Why this step? This is the logical heart: confinement + uncertainty $\Rightarrow$ nonzero ground energy.

Verify: $E_{m i n} \approx 0.95 eV \neq = 0$ ✓. The exact particle-in-a-box ground state is $E_{1} = π^{2} ℏ^{2} / (2 m_{e} a^{2}) \approx 37.6 eV$ — our rough $Δ x \approx a$ estimate lands a factor $\sim$ 40 low, i.e. the right order-of-magnitude ballpark and correctly a lower-side guess, as a crude uncertainty estimate should be.

Example 8 — Cell H: the exam-twist / unit trap

Worked example A student writes: "For an electron confined to

$Δ x = 2.0 \times 1 0^{- 10} m$ , the minimum momentum spread is $Δ p = ℏ/Δ x$ ." Find the correct $Δ p$ and name the two traps. Forecast: By what factor is the student's answer wrong?

1. Spot Trap #1: the missing $\frac{1}{2}$ . The theorem is $Δ x Δ p \geq ℏ/2$ , not $\geq ℏ$ . So: $Δ p = \frac{ℏ}{2 Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 2.0 \times 1 0 ^{- 10} )} \approx 2.6 \times 1 0^{- 25} kg\cdotm/s .$ Why this step? The student's version overshoots by exactly a factor of 2 — a classic order-of-magnitude slip called out in the parent's "mistakes".

2. Spot Trap #2 (the red herring): if the question adds "...because the measuring photon kicks the electron," mark that reasoning wrong. The bound is intrinsic (Fourier/wave nature); disturbance is only a consequence. Why this step? Exams love pairing a correct number with a wrong cause. The relation holds even with a perfect, non-disturbing measurement, because it comes from the particle being a wave packet — Fourier math, present before you ever look.

Verify: Student's value $ℏ/Δ x = 5.3 \times 1 0^{- 25}$ ; correct value $2.6 \times 1 0^{- 25}$ ; ratio $= 2$ ✓. Units $J\cdots/m = kg\cdotm/s$ ✓.

Active recall

What turns "

\geq

" into "

=

" in a minimum problem?

Asking for the smallest possible value — you sit exactly on the floor

ℏ/2

In the energy–time relation, what is

Δ t

physically?

The lifetime of the state — the time over which it changes appreciably, not a clock error.

Δ x \to 0

, what happens to

Δ p

It diverges to infinity (

Δ p \geq ℏ/2Δ x \to \infty

Why is a confined particle's ground energy nonzero?

E = 0

needs

Δ p = 0

, which forces

Δ x = \infty

, contradicting confinement.

A wider spectral line implies what about lifetime?

A shorter lifetime, since

τ = ℏ/ (2Δ E)

What does

⟨ p ⟩

mean, and how does it differ from

Δ p

⟨ p ⟩

is the average (mean) momentum;

Δ p

is the spread. A box particle has

⟨ p ⟩ = 0

but

Δ p \neq = 0

Connections

Parent topic (Hinglish) — the theory these examples exercise.
Particle in a box — Example 7's exact counterpart.
Natural linewidth and spectral broadening — Examples 5 and 6 in the lab.
de Broglie wavelength — the $p = ℏ k$ bridge behind every momentum step.
Wave packets and Fourier analysis — why the trade-off in Example 4 exists at all.