27 May 2025
Heisenberg Uncertainty Principle and Its Consequences
For a normalized state and a Hermitian observable $\hat A$, define the centred operator and standard deviation by
\[\delta\hat A=\hat A-\langle A\rangle, \qquad (\Delta A)^2=\langle(\delta\hat A)^2\rangle.\]Apply the Cauchy-Schwarz inequality to the vectors $\delta\hat A\lvert\psi\rangle$ and $\delta\hat B\lvert\psi\rangle$:
\[(\Delta A)^2(\Delta B)^2 \geq\left\lvert\langle\delta\hat A\,\delta\hat B\rangle\right\rvert^2.\]Split the product into Hermitian and anti-Hermitian parts,
\[\delta\hat A\,\delta\hat B =\frac12\{\delta\hat A,\delta\hat B\} +\frac12[\hat A,\hat B].\]The expectation of the anticommutator is real, while the expectation of the commutator is purely imaginary. Therefore the squared modulus is at least the square of its imaginary part:
\[(\Delta A)^2(\Delta B)^2 \geq\frac14\left\lvert\langle[\hat A,\hat B]\rangle\right\rvert^2.\]Taking the non-negative square root gives Robertson’s relation,
\[\boxed{\Delta A\,\Delta B \geq\frac12\left\lvert\langle[\hat A,\hat B]\rangle\right\rvert}.\]Since $[\hat x,\hat p_x]=i\hbar$,
\[\boxed{\Delta x\,\Delta p_x\geq\frac{\hbar}{2}}.\]Both sides have units of action: $\mathrm{m\,kg\,m\,s^{-1}}=\mathrm{J\,s}$. This spread is a property of identically prepared states, not merely a disturbance caused by an imperfect measuring instrument.
The path of an object
A classical path requires a definite position and momentum at every instant. If $\Delta x\to0$, the uncertainty relation requires $\Delta p_x\to\infty$; if $\Delta p_x\to0$, it requires $\Delta x\to\infty$. Thus an exact microscopic trajectory cannot be assigned. A wave-packet centre can approximate a classical path only when $\Delta x$ and $\Delta p$ are both negligible compared with the macroscopic length and momentum scales of the motion.
Zero-point energy
For the one-dimensional oscillator,
\[\hat H=\frac{\hat p^2}{2m}+\frac12m\omega^2\hat x^2.\]The minimum-energy state is centred at the equilibrium point with $\langle x\rangle=\langle p\rangle=0$. Hence
\[\langle H\rangle =\frac{(\Delta p)^2}{2m}+\frac12m\omega^2(\Delta x)^2.\]Using $\Delta p\geq\hbar/(2\Delta x)$ gives the lower bound
\[E\geq\frac{\hbar^2}{8m(\Delta x)^2} +\frac12m\omega^2(\Delta x)^2.\]Differentiate the right-hand side with respect to the positive width $\Delta x$:
\[-\frac{\hbar^2}{4m(\Delta x)^3} +m\omega^2\Delta x=0,\]so
\[(\Delta x)^2=\frac{\hbar}{2m\omega},\qquad (\Delta p)^2=\frac{m\hbar\omega}{2}.\]Substitution yields
\[\boxed{E_{\min}=\frac12\hbar\omega}.\]The two positive terms each contribute $\hbar\omega/4$. The oscillator therefore cannot have both $x=0$ and $p=0$, even at its lowest energy.
Size of an atom
Localizing the electron in hydrogen to a scale $r$ implies a momentum scale $\Delta p\sim\hbar/r$. This order-of-magnitude estimate deliberately suppresses numerical factors that depend on the three-dimensional trial state. The competing kinetic and Coulomb energies are
\[E(r)\sim \frac{\hbar^2}{2m_er^2} -\frac{e^2}{4\pi\varepsilon_0r}.\]The kinetic term is positive and varies as $r^{-2}$; the attractive potential is negative and varies as $r^{-1}$. Setting $dE/dr=0$ gives
\[-\frac{\hbar^2}{m_er^3} +\frac{e^2}{4\pi\varepsilon_0r^2}=0,\]and hence the atomic length scale
\[\boxed{r\sim a_0 =\frac{4\pi\varepsilon_0\hbar^2}{m_e e^2}}.\]The expression has units of length, and its value is $a_0\simeq5.29\times10^{-11}\,\mathrm m$. As $r\to0$, the $r^{-2}$ localization energy dominates the $-r^{-1}$ attraction, preventing classical collapse into the proton.
Existence of an electron inside the nucleus
If an electron were confined inside a nucleus of radius $R$, then
\[\Delta p\gtrsim\frac{\hbar}{2R},\qquad pc\gtrsim\frac{\hbar c}{2R}.\]Because $\hbar c\simeq197.3\,\mathrm{MeV\,fm}$, confinement to $R=1\,\mathrm{fm}$ requires $pc\gtrsim98.7\,\mathrm{MeV}$. This is far larger than the electron rest energy $m_ec^2=0.511\,\mathrm{MeV}$, so the non-relativistic expression $p^2/(2m_e)$ is invalid. The relativistic kinetic energy is
\[K=\sqrt{p^2c^2+m_e^2c^4}-m_ec^2 \gtrsim98\,\mathrm{MeV}.\]Such an energy is incompatible with a low-energy electron pre-existing inside an ordinary nucleus. Electrons emitted in beta decay are created in the decay process rather than released from nuclear confinement.
The stationary-width, zero-point-energy, and atomic-scale substitutions are verified in the Maxima worksheet.
Solved Problems
1. Minimum velocity spread caused by electron localization
An electron is prepared with position uncertainty $\Delta x=0.100\ \mathrm{nm}$. Find the minimum momentum spread, the corresponding velocity spread, and the minimum kinetic-energy contribution.
Solution. Standard deviations are non-negative, and equality gives the smallest allowed spread:
\[\Delta p_{\min}=\frac{\hbar}{2\Delta x} =\frac{1.05457\times10^{-34}} {2(1.00\times10^{-10})} =5.273\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}.\]No sign is assigned to a spread: momentum components occur on both sides of the mean. Since the result is non-relativistic,
\[\Delta v_{\min}=\frac{\Delta p_{\min}}{m_e} =\boxed{5.79\times10^5\ \mathrm{m\,s^{-1}}}.\]For a state with zero mean momentum, the least kinetic contribution compatible with this spread is
\[K_{\min}=\frac{(\Delta p_{\min})^2}{2m_e} =\boxed{0.952\ \mathrm{eV}}.\]The product $\Delta x\,\Delta p$ has units $\mathrm{J\,s}$, matching $\hbar$, and $\Delta v/c=1.93\times10^{-3}$ validates the non-relativistic step. As $\Delta x\to\infty$, both lower bounds approach zero; as $\Delta x\to0$, they diverge, which prevents simultaneous sharp position and momentum.
2. Diffraction spread and the loss of a sharp microscopic path
A beam of $100\ \mathrm{eV}$ electrons passes through a slit of width $b=1.00\ \mathrm{nm}$. Estimate the minimum angular spread of the emerging beam.
Solution. Take the transverse position uncertainty as $\Delta y\simeq b/2$. The uncertainty principle then gives
\[\Delta p_y\gtrsim\frac{\hbar}{2\Delta y} =\frac{\hbar}{b}.\]The incident longitudinal momentum is positive in the beam direction:
\[p_x=\sqrt{2m_eK}.\]For a small diffraction angle, $\Delta\theta\simeq\Delta p_y/p_x$, so
\[\begin{aligned} \Delta\theta_{\min} &\simeq\frac{\hbar}{b\sqrt{2m_eK}}\\ &=0.01952\ \mathrm{rad} =\boxed{1.118^\circ}. \end{aligned}\]The quotient is dimensionless because both numerator and denominator reduce to angular momentum. The spread occurs to both sides of the forward direction, not at a signed deflection of only $+1.118^\circ$. For $b\to\infty$ the lower bound tends to zero, while narrowing the slit broadens the angular distribution and destroys the notion of one exact path.
Descriptive Questions
- Why is the uncertainty principle a statement about a quantum state rather than only about disturbance by a measuring device?
- How does position-momentum uncertainty prevent an exact microscopic trajectory while still allowing a classical path approximation for macroscopic bodies?
- Explain why localization energy prevents atomic collapse and why its radial dependence dominates the Coulomb attraction at small radius.
- Why must a nuclear-confinement estimate for an electron be treated relativistically?
Numerical Problems
- A state lasts for $2.00\ \mathrm{ns}$. Using the textbook lifetime–linewidth estimate $\Delta E\,\Delta t\geq\hbar/2$, where $\Delta t$ is the characteristic lifetime rather than an operator uncertainty, find the lower-bound energy linewidth.
Final answer: $\boxed{\Delta E_{\min}=1.65\times10^{-7}\ \mathrm{eV}}$. - Find the uncertainty-principle zero-point energy of an oscillator with $\omega=5.00\times10^{14}\ \mathrm{rad\,s^{-1}}$.
Final answer: $\boxed{E_0=0.1646\ \mathrm{eV}}$. - Evaluate $4\pi\varepsilon_0\hbar^2/(m_e e^2)$, the atomic length scale obtained from the uncertainty estimate.
Final answer: $\boxed{a_0=0.0529\ \mathrm{nm}}$.
Every added uncertainty product, bound, and numerical value is checked in the MJ-11 problem-verification worksheet; every printed residual and check is zero.
Discussion