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

WHAT: As $\Delta x$ shrinks, the floor on $\Delta p$ rises. WHY: In $\Delta x\Delta p\ge \hslash /2$, the right side $\hslash /2$ is a fixed positive number. If $\Delta x\to$ small, then to keep the product $\ge \hslash /2$, $\Delta p$ must grow at least as fast as $1/\Delta x$. $\Delta p\ge \frac{\hslash }{2\Delta x}.$ Squeeze one, the other springs up — the balloon picture.

WHAT: Use the energy–time relation $\Delta E\Delta t\ge \hslash /2$. WHY: The unknown is an energy spread; the given is a time the state exists. The lifetime $\tau$ is $\Delta t$ — the timescale over which the state changes appreciably (here, by decaying). So $\Delta E\ge \hslash /(2\tau )$. (We compute the number in Q6.)

Momentum: minimum means equality. $\Delta p=\frac{\hslash }{2\Delta x}=\frac{1.055\times 10^{-34}}{2(5.0\times 10^{-11})}=1.055\times 10^{-24}\text{kg⋅m/s}.$ Speed: divide by the electron mass. $\Delta v=\frac{\Delta p}{m_e}=\frac{1.055\times 10^{-24}}{9.11\times 10^{-31}}\approx 1.16\times 10^6\text{m/s}.$ A million metres per second of irreducible velocity spread — this is why electrons can't sit still inside atoms.

$\Delta v=\frac{\hslash }{2m\Delta x}=\frac{1.055\times 10^{-34}}{2(0.16)(1.0\times 10^{-6})}\approx 3.30\times 10^{-28}\text{m/s}.$ Comment: $\hslash$ is tiny, $m$ is enormous — the floor is $10^{-28}\text{m/s}$, utterly unmeasurable. Quantum fuzziness is there, just cosmically irrelevant for balls.

$\Delta p$: $\Delta p_{\min }=\hslash /(2\Delta x)$ depends only on $\Delta x$, not on mass. So equal $\Delta p$ for both. $\Delta v$: $\Delta v=\Delta p/m$. Same $\Delta p$, bigger mass ⇒ smaller $\Delta v$. The electron has the larger speed spread, by the factor $1836$. Lesson: confinement dictates momentum spread; mass then decides how much velocity that is.

$\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(2.0\times 10^{-9})}\approx 2.64\times 10^{-26}\text{J}.$ Convert: $\frac{2.64\times 10^{-26}}{1.602\times 10^{-19}}\approx 1.65\times 10^{-7}\text{eV}.$ This tiny spread is the state's natural linewidth — see Natural linewidth and spectral broadening.

Set-up (WHAT & WHY): Confinement means the particle is somewhere within the box, so its position spread is about the box size: $\Delta x\sim a$ (the width of the cyan bump in the figure). By the relation, $\Delta p\sim \hslash /(2a)$. Since the box is symmetric, the average momentum is zero: $⟨p⟩=0$. Then the typical momentum-squared is just the spread squared: $⟨p^2⟩\sim (\Delta p{)}^2$. Energy: kinetic energy is $p^2/(2m)$, so $E_{\min }\sim \frac{(\Delta p{)}^2}{2m}\sim \frac{{\hslash }^2}{8m{a}^2}.$ Why it can't be zero: $E=0$ would need $p=0$ exactly, i.e. $\Delta p=0$. But then $\Delta x\ge \hslash /(2\Delta p)\to \infty$ — the particle would have to be spread over all space, contradicting "trapped in the box." So the ground-state energy is forced above zero: this is the zero-point energy (compare Particle in a box). Number (electron, $a=10^{-10}$ m): $E_{\min }\sim \frac{(1.055\times 10^{-34}{)}^2}{8(9.11\times 10^{-31})(10^{-10}{)}^2}\approx 1.53\times 10^{-19}\text{J}\approx 0.95\text{eV}.$ Order-of-magnitude right for atomic energies — good.

Analysis: The relevant relation is $\Delta E\Delta t\ge \hslash /2$ with $\Delta t=\tau$, the lifetime. The velocity $0.99c$ is a red herring — it tells you how far the muon flies, not its energy blur. $\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(2.2\times 10^{-6})}\approx 2.40\times 10^{-29}\text{J}.$ In eV: $\frac{2.40\times 10^{-29}}{1.602\times 10^{-19}}\approx 1.50\times 10^{-10}\text{eV}.$ A famously razor-sharp energy — because the muon lives comparatively "long."

Exact value: $E_1=\frac{{\pi }^2(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})(10^{-10}{)}^2}\approx 6.03\times 10^{-18}\text{J}\approx 37.6\text{eV}.$ Ratio: $E_1/E_{\min }=\frac{{\pi }^2{\hslash }^2/(2m{a}^2)}{{\hslash }^2/(8m{a}^2)}=4{\pi }^2\approx \mathrm{39.5.}$ Why the estimate is smaller: The uncertainty argument used the loosest bound $\Delta x\sim a$ and equality in $\Delta x\Delta p=\hslash /2$. The real confined wavefunction has a somewhat larger effective $\Delta p$ (it must vanish at both walls), so the true energy sits well above the crude floor. The uncertainty principle correctly predicts "nonzero and roughly atomic-scale," not the exact number.

(i) $p=\frac{h}{\lambda }=\frac{2\pi \hslash }{\lambda }=\frac{2\pi (1.055\times 10^{-34})}{1.0\times 10^{-10}}\approx 6.63\times 10^{-24}\text{kg⋅m/s}.$ (ii) Set $\Delta p=p$ and use $\Delta x\ge \hslash /(2\Delta p)$: $\Delta x=\frac{\hslash }{2p}=\frac{1.055\times 10^{-34}}{2(6.63\times 10^{-24})}\approx 7.96\times 10^{-12}\text{m}\approx 0.080\text{A˚}.$ Meaning: to make the momentum spread as big as a whole de Broglie momentum, you must localize the electron to under a tenth of an ångström. Momentum spread and de Broglie momentum are two faces of the same wave nature (de Broglie wavelength).

(i) $\Delta E=\frac{\hslash }{2\tau }=\frac{1.055\times 10^{-34}}{2(1.6\times 10^{-8})}\approx 3.30\times 10^{-27}\text{J}.$ (ii) Rearrange $|\Delta E|=\frac{hc}{{\lambda }^2}\Delta \lambda \Rightarrow \Delta \lambda =\frac{{\lambda }^2\Delta E}{hc}$: $\Delta \lambda =\frac{(589\times 10^{-9}{)}^2(3.30\times 10^{-27})}{(6.626\times 10^{-34})(3.0\times 10^8)}\approx 5.76\times 10^{-15}\text{m}\approx 5.8\times 10^{-6}\text{nm}.$ The spectral line is intrinsically about $6\times 10^{-6}$ nm wide — the natural width, before any Doppler or collision broadening (Natural linewidth and spectral broadening).

Verdict: Wrong. Energy is conserved exactly in quantum mechanics. What the symbols mean: $\Delta E$ is the statistical spread of energy values you would measure on identically prepared copies of a state; $\Delta t$ is the timescale over which the state changes appreciably (its lifetime). Neither is a "loan." Where it breaks: A state that lasts only $\Delta t$ is not a single sharp energy — it is a superposition whose energies are genuinely spread by $\Delta E\ge \hslash /(2\Delta t)$. Measuring gives one value with that spread; the average is conserved and no single measurement ever shows "extra" energy appearing then vanishing. The "borrowing" phrase is a heuristic for virtual particles, not a real violation.

Plug in: $E_{\min }\sim \frac{{\hslash }^2}{8m_p{a}^2}=\frac{(1.055\times 10^{-34}{)}^2}{8(1.67\times 10^{-27})(5.0\times 10^{-15}{)}^2}\approx 3.33\times 10^{-14}\text{J}.$ Convert to MeV: $E_{\min }\approx \frac{3.33\times 10^{-14}}{1.602\times 10^{-13}}\approx 0.21\text{MeV}.$ Comment: $E_{\min }\propto 1/a^2$. A nucleus is $\sim 10^5$ times smaller than an atom, so confinement energies are $\sim 10^{10}$ times larger — MeV instead of eV. This is why nuclear processes dwarf chemical ones: squeezing a proton into a nucleus costs an enormous zero-point energy, of the same MeV order as measured nuclear binding energies.