Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2: Exercises

Definition What

$Δ x$ and $Δ p$ precisely are (the missing definition) When we say "standard deviation of the particle's position," here is the exact formula. Let $⟨ x ⟩$ mean the average position you'd get by measuring the same state many times, and $⟨ x^{2} ⟩$ the average of the squared position. Then $Δ x \equiv ⟨ x^{2} ⟩ - ⟨ x ⟩^{2}, Δ p \equiv ⟨ p^{2} ⟩ - ⟨ p ⟩^{2} .$

$⟨ x ⟩$ = the centre of the position bump; $⟨ x^{2} ⟩ - ⟨ x ⟩^{2}$ = how far, on average, the particle strays from that centre (squared). The square root converts it back to a length — that is the width of the blob in the phase-space picture below.
The same reading applies to $Δ p$ with momentum. This is exactly the "width" the double-headed arrows measure in every figure on this page, and it is what connects the intuitive blob to the rigorous quantum formalism (where $⟨ x^{2} ⟩$ etc. come from operators acting on the wavefunction $ψ$ ).

The picture behind the whole page (read before solving)

Figure — Heisenberg uncertainty principle — Δx Δp ≥ ℏ - 2

The top row shows a localized particle: a narrow position bump (small $Δ x$ ) forces a wide momentum bump (large $Δ p$ ). The bottom row flips it: a broad position bump (large $Δ x$ ) gives a sharp momentum bump (small $Δ p$ ). Both rows have the same shaded product area $Δ x Δ p$ — the principle says that area can never shrink below $ℏ/2$ , and equals $ℏ/2$ only for the Gaussian bell shape drawn here.

Level 1 — Recognition

L1.1

Which of these products is bounded below by $ℏ/2$ ? (a) $Δ x Δ p_{x}$ (b) $Δ x Δ p_{y}$ (c) $Δ y Δ p_{x}$ (d) $Δ E Δ t$

Recall Solution

Bounded: (a) and (d). Not bounded: (b), (c). Why? The principle constrains only conjugate pairs — quantities that live on the same phase-space plane (position and momentum along the same axis), or energy paired with a state's lifetime. Position on $x$ and momentum on $y$ are independent directions with independent planes, so you may pin both down freely.

L1.2

A textbook writes $Δ x Δ p \geq \frac{h}{4 π}$ ; your lecture wrote $\geq \frac{ℏ}{2}$ . Are these different rules? Compute both numbers.

Recall Solution

$\frac{ℏ}{2} = \frac{1.055 \times 1 0 ^{- 34}}{2} = 5.27 \times 1 0^{- 35} J\cdots$ . $\frac{h}{4 π} = \frac{6.626 \times 1 0 ^{- 34}}{4 π} = 5.27 \times 1 0^{- 35} J\cdots$ . Identical. Since $ℏ = h /2 π$ , dividing by $2$ gives $h /4 π$ . Same rule, two costumes.

L1.3

True or false: " $Δ x$ is the error made by my ruler."

Recall Solution

False. $Δ x$ is the standard deviation of the particle's position in its quantum state — $Δ x = ⟨ x^{2} ⟩ - ⟨ x ⟩^{2}$ , an intrinsic spread that exists before any measurement, with a perfect instrument. It is a property of the wave packet, not of your apparatus.

Level 2 — Application

L2.1

An electron is confined to a region of width $Δ x = 0.50 \overset{˚}{A}$ . Find the minimum uncertainty in its momentum, $Δ p$ .

Recall Solution

Use the bound at equality (minimum — the Gaussian best case): $Δ p = \frac{ℏ}{2 Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 0.50 \times 1 0 ^{- 10} )} = \frac{1.055 \times 1 0 ^{- 34}}{1.0 \times 1 0 ^{- 10}} = 1.055 \times 1 0^{- 24} kg\cdotm/s .$ What we did: rearranged $Δ x Δ p \geq ℏ/2$ to isolate $Δ p$ , using "=" for the smallest allowed value. A real (non-Gaussian) electron would have a larger $Δ p$ .

L2.2

From L2.1, find the minimum uncertainty in the electron's velocity.

Recall Solution

$Δ v = \frac{Δ p}{m _{e}} = \frac{1.055 \times 1 0 ^{- 24}}{9.11 \times 1 0 ^{- 31}} \approx 1.16 \times 1 0^{6} m/s .$ Why divide by mass: $p = m v$ so $Δ p = m Δ v$ , giving $Δ v = Δ p / m$ . Over a million metres per second of unavoidable spread — huge for an atom-sized box.

L2.3

A proton is localized to $Δ x = 1.0 \overset{˚}{A}$ . Find $Δ v$ and compare with the electron case.

Recall Solution

$Δ p = \frac{ℏ}{2Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 10} )} = 5.27 \times 1 0^{- 25} kg\cdotm/s .$ $Δ v = \frac{5.27 \times 1 0 ^{- 25}}{1.67 \times 1 0 ^{- 27}} \approx 316 m/s .$ Interpretation: Same box size, but the proton is ~1836× heavier, so its velocity spread is ~1836× smaller than an electron's. Heavier ⇒ more classical.

Level 3 — Analysis

L3.1

A cricket ball, $m = 0.16 kg$ , is located to $Δ x = 1.0 μ m = 1 0^{- 6} m$ . Find $Δ v$ and explain why the classical world looks deterministic.

Recall Solution

Chain the rearrangement explicitly so no step is hidden: $Δ x Δ p \geq \frac{ℏ}{2} \Rightarrow Δ p \geq \frac{ℏ}{2 Δ x} \Rightarrow Δ v = \frac{Δ p}{m} \geq \frac{ℏ}{2 m Δ x} .$ Now substitute: $Δ v = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 0.16 ) ( 1 0 ^{- 6} )} = \frac{1.055 \times 1 0 ^{- 34}}{3.2 \times 1 0 ^{- 7}} \approx 3.3 \times 1 0^{- 28} m/s .$ Analysis: $3 \times 1 0^{- 28}$ m/s is unmeasurable — the ball would drift less than a proton's width over the age of the universe. Large $m$ crushes the bound, so classical mechanics is an excellent approximation. Uncertainty only bites when $m$ is tiny.

L3.2

An excited atomic state lives for $Δ t = 1.0 \times 1 0^{- 8} s$ before emitting. Find the minimum energy spread $Δ E$ (in J and in eV). This spread is the natural line width. (Recall: $Δ t$ here is the state's lifetime, not a clock error.)

Recall Solution

Use the energy–time form, where $Δ t$ is how long the state survives: $Δ E = \frac{ℏ}{2 Δ t} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 8} )} = 5.27 \times 1 0^{- 27} J .$ In eV: $\frac{5.27 \times 1 0 ^{- 27}}{1.602 \times 1 0 ^{- 19}} \approx 3.3 \times 1 0^{- 8} eV .$ Analysis: A state that exists only briefly cannot have a razor-sharp energy — so the emitted photon's energy (and thus the spectral line) has an intrinsic width. Shorter-lived states ⇒ broader lines.

L3.3

Estimate the minimum kinetic energy of an electron confined to a box of width $L = 1.0 \overset{˚}{A}$ . Take $Δ x \approx L$ and $Δ p \approx p_{m i n}$ .

Recall Solution

Confinement forces a minimum momentum: $p ≳ Δ p = \frac{ℏ}{2Δ x}$ . $Δ p = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 1.0 \times 1 0 ^{- 10} )} = 5.27 \times 1 0^{- 25} kg\cdotm/s .$ $E = \frac{p ^{2}}{2 m _{e}} = \frac{( 5.27 \times 1 0 ^{- 25} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = \frac{2.78 \times 1 0 ^{- 49}}{1.82 \times 1 0 ^{- 30}} \approx 1.53 \times 1 0^{- 19} J \approx 0.95 eV .$ Analysis: An electron cannot sit still in a small box — squeezing it (small $Δ x$ ) forces a nonzero momentum spread, hence a nonzero zero-point energy of order 1 eV. This is on the scale of chemical bond energies.

Level 4 — Synthesis

L4.1

An electron and a proton are each confined to $Δ x = 1.0 \overset{˚}{A}$ . Using $E \approx \frac{( Δ p ) ^{2}}{2 m}$ , find the ratio of their confinement energies. What does this say about which particle "spreads out" into an orbital?

Recall Solution

$Δ p$ is the same for both (it depends only on $Δ x$ ): $5.27 \times 1 0^{- 25}$ kg·m/s. The energy $E = (Δ p)^{2} /2 m$ scales as $1/ m$ , so $\frac{E _{e}}{E _{p}} = \frac{m _{p}}{m _{e}} = \frac{1.67 \times 1 0 ^{- 27}}{9.11 \times 1 0 ^{- 31}} \approx 1833.$ Synthesis: The electron pays ~1800× more confinement energy than the proton for the same squeeze. That is why the light electron cannot be pinned into a tiny orbit — it delocalizes into a large probability cloud (an orbital), while heavy nuclei stay compact. See Bohr's failure: fixed orbits demand exact $x$ and $p$ together.

L4.2

Combine de Broglie and Heisenberg. A particle with well-defined wavelength $λ$ has momentum $p = h / λ$ . If we localize it to $Δ x \approx λ$ (about one wavelength), show the fractional momentum uncertainty $Δ p / p$ is of order 1.

Recall Solution

With $Δ x \approx λ$ : $Δ p \geq \frac{ℏ}{2Δ x} = \frac{ℏ}{2 λ} = \frac{h}{4 π λ} .$ Since $p = h / λ$ : $\frac{Δ p}{p} \geq \frac{h / ( 4 π λ )}{h / λ} = \frac{1}{4 π} \approx 0.08.$ Synthesis: Even squeezing the packet to a single wavelength leaves at least an ~8% momentum spread. To make $Δ p / p$ small you must let $Δ x$ span many wavelengths — exactly the de Broglie / wave trade-off from the parent note: a sharp momentum needs a long, spread-out wave.

L4.3

The parent note's Bohr-microscope argument gave $Δ x Δ p \approx h$ , but the exact bound is $ℏ/2 = h /4 π$ . By what factor does the crude estimate overshoot the true minimum?

Recall Solution

$\frac{h}{ℏ/2} = \frac{h}{h /4 π} = 4 π \approx 12.57.$ Synthesis: The order-of-magnitude " $\approx h$ " is about $4 π \approx 12.6$ times larger than the true Gaussian minimum $ℏ/2$ . The microscope story gets the scale right (an $h$ -sized product) but not the exact constant — the Fourier/Gaussian treatment supplies the precise $\frac{1}{2}$ .

Level 5 — Mastery

L5.1

A spectral line has measured width $Δ E = 4.1 \times 1 0^{- 24} J$ arising purely from the excited state's finite lifetime. Estimate that lifetime $τ \approx Δ t$ .

Recall Solution

Rearrange the energy–time bound at equality (here $Δ t$ is exactly the lifetime we want): $Δ t = \frac{ℏ}{2 Δ E} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 4.1 \times 1 0 ^{- 24} )} = \frac{1.055 \times 1 0 ^{- 34}}{8.2 \times 1 0 ^{- 24}} \approx 1.29 \times 1 0^{- 11} s .$ Mastery: We ran the line-width logic backwards — from an observed broadening to the underlying lifetime (~13 ps). This is how physicists measure fleeting states they can never watch directly.

L5.2

A nucleus has diameter $\approx 5.0 \times 1 0^{- 15} m$ . If an electron were confined inside it, estimate $Δ p$ , then its energy via the relativistic estimate $E \approx (Δ p) c$ (with $c = 3.0 \times 1 0^{8}$ m/s). Argue why electrons cannot live in nuclei.

Recall Solution

$Δ p = \frac{ℏ}{2Δ x} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 5.0 \times 1 0 ^{- 15} )} = 1.055 \times 1 0^{- 20} kg\cdotm/s .$ Because this momentum is enormous, use the ultra-relativistic estimate $E \approx p c$ : $E \approx (1.055 \times 1 0^{- 20}) (3.0 \times 1 0^{8}) = 3.17 \times 1 0^{- 12} J \approx 19.8 MeV .$ Mastery: An electron squeezed into a nucleus would need ~20 MeV of energy — far more than nuclear binding energies (~few MeV) can supply. So electrons cannot be confined in nuclei; the uncertainty principle forbids it. (Nuclear electrons in beta decay are created at emission, not pre-stored.)

L5.3

Decide, with one line of reasoning each, which form of the principle to use — and solve: (a) You know a molecule's transition energy is uncertain by $Δ E$ ; you want its state's lifetime. (b) You know an electron's box size $Δ x$ ; you want its speed spread. Given: (a) $Δ E = 2.1 \times 1 0^{- 25} J$ ; (b) $Δ x = 2.0 \overset{˚}{A}$ .

Recall Solution

(a) Energy is paired with time (the state's lifetime), so use $Δ E Δ t \geq ℏ/2$ : $Δ t = \frac{ℏ}{2Δ E} = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 2.1 \times 1 0 ^{- 25} )} \approx 2.51 \times 1 0^{- 10} s .$ (b) Position is paired with momentum; chain the full rearrangement, then divide by $m_{e}$ : $Δ x Δ p \geq \frac{ℏ}{2} \Rightarrow Δ p \geq \frac{ℏ}{2Δ x} \Rightarrow Δ v \geq \frac{ℏ}{2 m _{e} Δ x} .$ $Δ p = \frac{1.055 \times 1 0 ^{- 34}}{2 ( 2.0 \times 1 0 ^{- 10} )} = 2.64 \times 1 0^{- 25} kg\cdotm/s, Δ v = \frac{2.64 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 2.90 \times 1 0^{5} m/s .$ Mastery: The whole skill is matching the conjugate pair to the quantity you're given — energy↔lifetime, position↔momentum — then rearranging and, for velocity, dividing by mass.

Recall Quick self-check (reveal after attempting all)

L2.1 $Δ p$ ::: $1.06 \times 1 0^{- 24}$ kg·m/s L2.2 $Δ v$ (electron) ::: $1.16 \times 1 0^{6}$ m/s L3.1 $Δ v$ (cricket ball) ::: $3.3 \times 1 0^{- 28}$ m/s L3.2 $Δ E$ ::: $5.3 \times 1 0^{- 27}$ J $\approx 3.3 \times 1 0^{- 8}$ eV L4.1 ratio $E_{e} / E_{p}$ ::: $\approx 1833$ L5.2 electron-in-nucleus energy ::: $\approx 20$ MeV — forbidden L5.3 (a) lifetime ::: $\approx 2.5 \times 1 0^{- 10}$ s; (b) $Δ v \approx 2.9 \times 1 0^{5}$ m/s