The Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle is one of the most profound and frequently misunderstood results in all of physics. In its most general form — the Robertson uncertainty relation — it states that for any two observables $\hat{A}$ and $\hat{B}$, the product of their standard deviations in any quantum state satisfies $\Delta A \cdot \Delta B \geq |\langle[\hat{A},\hat{B}]\rangle|/2$. For position and momentum, where $[\hat{x},\hat{p}] = i\hbar$, this gives the iconic inequality $\Delta x \cdot \Delta p \geq \hbar/2$. This is not a statement about experimental imprecision, the clumsiness of instruments, or the disturbance caused by measurement — it is a fundamental property of quantum states themselves, reflecting the incompatibility of conjugate observables at the kinematic level.
The naive "measurement disturbs the system" interpretation, while not entirely wrong as a heuristic, is incomplete and can be misleading. The uncertainty principle holds even for quantum states that have never been measured. It is a consequence of the non-commutativity of operators in quantum mechanics, which in turn reflects the non-classical structure of the algebra of observables. A state with $\Delta x = 0$ (definite position) must have $\Delta p = \infty$ (completely indefinite momentum) — not because measuring position disturbs momentum, but because no single quantum state can simultaneously assign definite values to both. This is a kinematic fact about the state space of quantum mechanics.
The mathematical derivation proceeds via the Cauchy-Schwarz inequality applied to the Hilbert space inner product. For operators $\hat{A}$ and $\hat{B}$ with $\delta\hat{A} = \hat{A} - \langle\hat{A}\rangle$ and $\delta\hat{B} = \hat{B} - \langle\hat{B}\rangle$, one considers the inner product $\langle\delta\hat{A}\psi|\delta\hat{A}\psi\rangle\langle\delta\hat{B}\psi|\delta\hat{B}\psi\rangle \geq |\langle\delta\hat{A}\psi|\delta\hat{B}\psi\rangle|^2$. Decomposing the cross term into symmetric and antisymmetric parts yields the Robertson relation, and the stronger Schrödinger uncertainty relation $\Delta A^2\Delta B^2 \geq (\frac{1}{2}\langle\{\delta\hat{A},\delta\hat{B}\}\rangle)^2 + (\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle)^2$, which includes both the commutator and anticommutator contributions.
Gaussian wave packets occupy a privileged role in this theory: they are the minimum-uncertainty states (coherent states, or squeezed states for the quadratures of the harmonic oscillator). A Gaussian in position space $\psi(x) \propto e^{-x^2/(4\sigma_x^2)}e^{ip_0x/\hbar}$ has $\Delta x = \sigma_x$ and $\Delta p = \hbar/(2\sigma_x)$, saturating the bound $\Delta x\Delta p = \hbar/2$. Any other state shape — square waves, triangular profiles, superpositions — gives a larger product. This characterization of Gaussians as optimal states for simultaneous approximate localization in phase space has deep implications for quantum information theory, where it underlies the theory of optimal state discrimination and quantum teleportation.
The energy-time uncertainty relation $\Delta E \cdot \Delta t \geq \hbar/2$ is qualitatively similar but fundamentally different in character. Time is not an operator in standard quantum mechanics (by Pauli's theorem: there is no self-adjoint time operator canonically conjugate to the Hamiltonian for systems bounded below). Instead, $\Delta t$ is interpreted as the time scale over which the system's state changes appreciably, or the lifetime of a state. The precise form $\Delta E \cdot \tau \geq \hbar/2$ (where $\tau$ is the coherence or lifetime) follows from the time-energy form of the Mandelstam-Tamm inequality: $\Delta E \cdot \Delta_\tau\hat{A} \geq \hbar|d\langle\hat{A}\rangle/dt|/2$, where $\Delta_\tau = |\Delta A/(d\langle A\rangle/dt)|$ is the time for the observable $\hat{A}$ to change by one standard deviation.
Number-phase uncertainty is another important conjugate pair, especially in quantum optics. For a radiation field mode, the photon number $\hat{n}$ and the phase $\hat{\phi}$ satisfy $\Delta n \cdot \Delta\phi \geq 1/2$. A coherent state (Poissonian photon statistics with $\Delta n = \sqrt{\bar{n}}$) has phase uncertainty $\Delta\phi = 1/(2\sqrt{\bar{n}})$, while a Fock state (definite $n$, $\Delta n = 0$) has completely undefined phase. This trade-off is central to quantum metrology and the fundamental limits of interferometric phase measurement, where squeezed light can be used to surpass the shot-noise limit while respecting the uncertainty principle.
The quantum Zeno effect is an unexpected consequence of the uncertainty principle and the measurement postulate. Frequent measurements of whether a system has decayed (a yes/no observable with time resolution $\Delta t$) effectively halt the decay process when $\Delta t \to 0$. The survival probability after $N$ measurements in time $T$ goes as $P(T) = [1-(T/N)^2/\tau_Z^2]^N \to 1$ as $N \to \infty$, where $\tau_Z = \hbar/\Delta E$ is the Zeno time set by the energy uncertainty of the initial state. This is the quantum Zeno effect, experimentally confirmed in systems ranging from trapped ions to unstable atoms. The complementary quantum anti-Zeno effect (acceleration by frequent measurement) can occur for broader spectral densities.
Modern treatments distinguish between preparation uncertainty (the uncertainty in a quantum state as defined above) and measurement uncertainty (the imprecision-disturbance trade-off in a joint measurement of non-commuting observables). The Ozawa inequality and Busch-Lahti-Werner framework give rigorous bounds on the joint measurability of non-commuting observables, showing that the naive Heisenberg microscope argument, while capturing the right order of magnitude, is not quantitatively correct as a bound on measurement error. These distinctions matter for quantum information processing, where measurement back-action must be carefully controlled.
For any two observables $\hat{A}$ and $\hat{B}$ and any state $|\psi\rangle$, define the standard deviations $\Delta A = \sqrt{\langle\hat{A}^2\rangle - \langle\hat{A}\rangle^2}$ and $\Delta B = \sqrt{\langle\hat{B}^2\rangle - \langle\hat{B}\rangle^2}$. Then: $$\Delta A \cdot \Delta B \geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|.$$ The stronger Schrödinger uncertainty relation is: $$\Delta A^2 \cdot \Delta B^2 \geq \left(\frac{1}{2}\langle\{\delta\hat{A},\delta\hat{B}\}\rangle\right)^2 + \left(\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2,$$ where $\{\cdot,\cdot\}$ is the anticommutator. For $\hat{A} = \hat{x}$ and $\hat{B} = \hat{p}$: $[\hat{x},\hat{p}] = i\hbar$, giving $\Delta x \cdot \Delta p \geq \hbar/2$. The bound is saturated by coherent states (Gaussians).
For any observable $\hat{A}$ with time-dependent expectation value, the energy spread $\Delta E$ of the state and the time $\tau_A = \Delta A / |d\langle\hat{A}\rangle/dt|$ (the time for $\langle\hat{A}\rangle$ to change by $\Delta A$) satisfy: $$\Delta E \cdot \tau_A \geq \frac{\hbar}{2}.$$ The minimum time for any observable to change significantly (the "orthogonalization time") is $\tau_\perp = \pi\hbar/(2\Delta E)$. For an unstable state with lifetime $\tau_{\rm life}$: $\Delta E \cdot \tau_{\rm life} \geq \hbar/2$, giving a natural linewidth $\Gamma = \hbar/\tau_{\rm life}$. This is the fundamental origin of spectral line broadening in atomic physics. Note: $\Delta t$ is NOT the uncertainty in the eigenvalue of a time operator (none exists); it is a time scale characterizing state evolution.
A state $|\psi\rangle$ saturates the Robertson bound $\Delta A \Delta B = |\langle[\hat{A},\hat{B}]\rangle|/2$ if and only if: (1) $\delta\hat{B}|\psi\rangle = i\lambda\,\delta\hat{A}|\psi\rangle$ for some real $\lambda > 0$, and (2) $\langle\{\delta\hat{A},\delta\hat{B}\}\rangle = 0$. For $\hat{x}$ and $\hat{p}$, condition (1) gives $(\hat{p} - \langle p\rangle)|\psi\rangle = i\lambda(\hat{x}-\langle x\rangle)|\psi\rangle$, which in position representation is a first-order ODE with solution $\psi(x) \propto \exp(ip_0x/\hbar - (x-x_0)^2/(4\sigma_x^2))$ — a Gaussian. The parameter $\lambda = \hbar/(2\sigma_x^2)$, giving $\Delta p = \lambda\sigma_x = \hbar/(2\sigma_x)$, thus $\Delta x \Delta p = \sigma_x \cdot \hbar/(2\sigma_x) = \hbar/2$. ✓
Let $|f\rangle = \delta\hat{A}|\psi\rangle$ and $|g\rangle = \delta\hat{B}|\psi\rangle$ where $\delta\hat{A} = \hat{A}-\langle A\rangle$. By Cauchy-Schwarz: $\langle f|f\rangle\langle g|g\rangle \geq |\langle f|g\rangle|^2$. Note $\langle f|f\rangle = \langle(\delta\hat{A})^2\rangle = (\Delta A)^2$ and $\langle g|g\rangle = (\Delta B)^2$. Decompose $\langle f|g\rangle = \frac{1}{2}\langle\{\delta\hat{A},\delta\hat{B}\}\rangle + \frac{1}{2}\langle[\delta\hat{A},\delta\hat{B}]\rangle = \frac{1}{2}\langle\{\delta\hat{A},\delta\hat{B}\}\rangle + \frac{1}{2}\langle[\hat{A},\hat{B}]\rangle$. The first term is real (anticommutator of Hermitian operators has real expectation value), the second is purely imaginary (commutator of Hermitian operators has purely imaginary expectation value). Thus $|\langle f|g\rangle|^2 = \frac{1}{4}\langle\{\delta\hat{A},\delta\hat{B}\}\rangle^2 + \frac{1}{4}|\langle[\hat{A},\hat{B}]\rangle|^2 \geq \frac{1}{4}|\langle[\hat{A},\hat{B}]\rangle|^2$. Taking square roots: $\Delta A\Delta B \geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|$. $\square$
Consider a system initially in state $|\psi_0\rangle$ with energy spread $\Delta E$. Define the survival probability $P_s(t) = |\langle\psi_0|e^{-i\hat{H}t/\hbar}|\psi_0\rangle|^2$. For short times: $P_s(t) = 1 - t^2/\tau_Z^2 + O(t^4)$ where the Zeno time $\tau_Z = \hbar/\Delta E$. If $N$ measurements are performed at times $T/N, 2T/N, \ldots, T$, the survival probability is $P_s^{(N)}(T) = P_s(T/N)^N = [1-(T/N)^2/\tau_Z^2]^N \to \exp(-(T/\tau_Z)^2/N) \to 1$ as $N \to \infty$ for fixed $T$. Conversely, for a system exhibiting exponential decay $P_s(t) \approx e^{-\Gamma t}$ (Lorentzian spectrum), $N$ measurements give $P_s^{(N)} = e^{-\Gamma T}$ independent of $N$ — the exponential regime is immune to the Zeno effect.
Prove the Robertson uncertainty relation for $\hat{L}_x$ and $\hat{L}_y$ (angular momentum components). Evaluate the bound for a state $|l,m\rangle$ with definite $L^2 = \hbar^2 l(l+1)$ and $L_z = m\hbar$. Show the bound vanishes for $|0,0\rangle$ but is non-trivial for $|1,0\rangle$.
$[\hat{L}_x,\hat{L}_y] = i\hbar\hat{L}_z$, so $\Delta L_x\Delta L_y \geq \frac{1}{2}|\langle\hat{L}_z\rangle| = \frac{\hbar}{2}|m|$. For $|0,0\rangle$: $m=0$, bound $= 0$ — trivially satisfied with $\Delta L_x = \Delta L_y = 0$ (since $L^2 = 0$ implies all components are zero). For $|1,0\rangle$: $m=0$, bound $= 0$. But $\langle L_x^2\rangle = \langle L_y^2\rangle = (\langle L^2\rangle - \langle L_z^2\rangle)/2 = \hbar^2(1\cdot2-0)/2 = \hbar^2$. So $\Delta L_x = \Delta L_y = \hbar$ and $\Delta L_x\Delta L_y = \hbar^2$, while the bound is 0 — the bound is non-tight here. For $|1,1\rangle$: bound $= \hbar^2/2$. $\langle L_x^2\rangle = (\hbar^2\cdot2 - \hbar^2)/2 = \hbar^2/2$. $\Delta L_x = \hbar/\sqrt{2}$, $\Delta L_y = \hbar/\sqrt{2}$. Product $= \hbar^2/2$. The bound is saturated! The state $|1,1\rangle$ is a minimum-uncertainty state for $L_x, L_y$ with respect to the $L_z$ commutator.
A Gaussian wave packet $\psi(x,0) = (2\pi\sigma_x^2)^{-1/4}e^{-x^2/(4\sigma_x^2)}$ (centered at origin, zero mean momentum) is evolved in time for a free particle. Calculate $\Delta x(t)$, $\Delta p(t)$, and $\Delta x(t)\Delta p(t)$. Show the product exceeds $\hbar/2$ for $t > 0$.
Momentum wave function: $\tilde{\psi}(p) = (2\pi\sigma_p^2)^{-1/4}e^{-p^2/(4\sigma_p^2)}$ with $\sigma_p = \hbar/(2\sigma_x)$. Since $\hat{p}$ commutes with $e^{-i\hat{p}^2t/(2m\hbar)}$, $\tilde{\psi}(p,t) = \tilde{\psi}(p,0)e^{-ip^2t/(2m\hbar)}$. So $|\tilde{\psi}(p,t)|^2 = |\tilde{\psi}(p,0)|^2$ — the momentum distribution is time-independent! $\Delta p(t) = \sigma_p = \hbar/(2\sigma_x)$ for all $t$. In position space: $\sigma_x(t)^2 = \sigma_x^2 + (\hbar t/(2m\sigma_x))^2 = \sigma_x^2(1+t^2/\tau^2)$ where $\tau = 2m\sigma_x^2/\hbar$. Thus $\Delta x(t) = \sigma_x\sqrt{1+t^2/\tau^2}$. Product: $\Delta x(t)\Delta p(t) = \frac{\hbar}{2}\sqrt{1+t^2/\tau^2} \geq \hbar/2$, with equality only at $t=0$. The minimum-uncertainty property is only instantaneous for a freely spreading Gaussian — dispersion immediately increases the position-momentum product above the minimum. This illustrates that the uncertainty principle sets a floor that quantum states are not obligated to remain at.
Use the uncertainty principle to estimate: (a) the ground state energy of the hydrogen atom, (b) the zero-point kinetic energy of a proton confined to a nucleus of radius $R = 5$ fm, (c) the minimum size of a white dwarf star (electron degeneracy pressure vs gravity).
(a) Hydrogen: $\Delta r \sim a$ (Bohr radius), $\Delta p \sim \hbar/a$ (uncertainty principle). Energy: $E = p^2/(2m_e) - e^2/(4\pi\epsilon_0 a) \sim \hbar^2/(2m_ea^2) - e^2/(4\pi\epsilon_0 a)$. Minimize over $a$: $dE/da = -\hbar^2/(m_ea^3) + e^2/(4\pi\epsilon_0 a^2) = 0 \Rightarrow a_0 = 4\pi\epsilon_0\hbar^2/(m_ee^2) = 0.529$ Å (Bohr radius). $E_{\min} = -m_ee^4/(2(4\pi\epsilon_0)^2\hbar^2) = -13.6$ eV. ✓ (b) $\Delta p \sim \hbar/R = 1.055\times10^{-34}/(5\times10^{-15}) = 2.11\times10^{-20}$ kg·m/s. $T = (\Delta p)^2/(2m_p) = (2.11\times10^{-20})^2/(2\times1.673\times10^{-27}) = 1.33\times10^{-13}$ J $= 830$ keV. Nuclear binding energies are MeV-scale, consistent with confinement at fm scales. (c) Star of $N$ electrons, mass $M = Nm_p$ (charge neutrality), radius $R$. Fermi momentum $p_F \sim \hbar N^{1/3}/R$. Kinetic energy $E_K \sim N^{5/3}\hbar^2/(m_eR^2)$. Gravitational energy $E_G \sim -GM^2/R \sim -GN^2m_p^2/R$. Total: $E = E_K + E_G$. Minimize: $R_{\min} \sim N^{-1/3}\hbar^2/(m_eGNm_p^2) = \hbar^2/(m_eGNm_p^2/N^{2/3}) \sim \hbar^2N^{-1/3}/(m_eGm_p^2)$. For $M = M_\odot$: $N \approx 10^{57}$, giving $R \sim 10^4$ km — the order of magnitude of white dwarf radii.
Analyze the Heisenberg microscope thought experiment. A photon of wavelength $\lambda$ is used to measure the position of an electron. Show that the photon recoil gives an uncertainty $\Delta p \geq h/\lambda$ while position resolution is $\Delta x \geq \lambda/(2\sin\alpha)$ (Rayleigh criterion, aperture $\alpha$). Verify $\Delta x\Delta p \geq \hbar/2$. Then explain why this "derivation" of the uncertainty principle is logically circular.
Position resolution: by the Rayleigh criterion, two points can be resolved if their angular separation exceeds $\lambda/D$ where $D = 2f\sin\alpha$ is the aperture ($f$ = focal length). Minimum resolvable position: $\Delta x \geq \lambda/(2\sin\alpha)$. Momentum uncertainty: the scattered photon has momentum $\hbar\omega/c = h/\lambda$, and its direction is uncertain within the acceptance cone $\pm\alpha$. The $x$-component of recoil momentum is uncertain by $\Delta p_x \geq (h/\lambda)\sin\alpha$. Product: $\Delta x \cdot \Delta p_x \geq \frac{\lambda}{2\sin\alpha}\cdot\frac{h\sin\alpha}{\lambda} = h/2 \geq \hbar/2$. ✓ However, this argument is logically circular: it uses classical optics (Rayleigh criterion) to derive a quantum result. The Rayleigh criterion itself implicitly assumes wave optics and diffraction, which encodes the Fourier uncertainty for classical waves $\Delta x \Delta k \geq 1/2$. The uncertainty principle has not been derived — it has merely been illustrated that classical wave optics and quantum momentum quantization are mutually consistent with the uncertainty principle. The true derivation requires only the Cauchy-Schwarz inequality in Hilbert space and the commutation relation $[\hat{x},\hat{p}] = i\hbar$.
Practice Problems
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1. Define $|f\rangle = \delta\hat{A}|\psi\rangle$, $|g\rangle = \delta\hat{B}|\psi\rangle$. Cauchy-Schwarz: $(\Delta A)^2(\Delta B)^2 = \langle f|f\rangle\langle g|g\rangle \geq |\langle f|g\rangle|^2$. Split $\langle f|g\rangle$ into real/imaginary parts; imaginary part is $\langle[\hat{A},\hat{B}]\rangle/(2i)$. So $(\Delta A\Delta B)^2 \geq |\langle[\hat{A},\hat{B}]\rangle|^2/4$. Equality when $|f\rangle = c|g\rangle$ (collinear) with purely imaginary $c$, i.e., $\delta\hat{A}|\psi\rangle = i\lambda\delta\hat{B}|\psi\rangle$, giving Gaussian $\psi(x)$.
2. (a) $\Delta p \geq \hbar/(2\Delta x) = 1.055\times10^{-34}/(2\times10^{-10}) = 5.28\times10^{-25}$ kg·m/s. (b) $\langle T\rangle_{\min} = (\Delta p)^2/(2m_e) = (5.28\times10^{-25})^2/(2\times9.109\times10^{-31}) = 1.53\times10^{-19}$ J $= 0.95$ eV $\approx 1$ eV. (c) $\lambda = h/\Delta p = 6.626\times10^{-34}/(5.28\times10^{-25}) = 1.25\times10^{-9}$ m $= 1.25$ nm $= 2\times\Delta x$. Consistent: $\lambda/(2\Delta x) = 1$ (approximately one half-wavelength fits in the confinement region).
3. $|+x\rangle = (|\uparrow\rangle+|\downarrow\rangle)/\sqrt{2}$. $\langle S_x\rangle = \hbar/2$, $\langle S_y\rangle = 0$, $\langle S_z\rangle = 0$. $\langle S_x^2\rangle = \hbar^2/4$ (all Pauli squares give $\hbar^2/4$). $\Delta S_x = 0$, $\Delta S_y = \hbar/2$, $\Delta S_z = \hbar/2$. Check $S_y$-$S_z$: $\Delta S_y\Delta S_z = \hbar^2/4 \geq \hbar|\langle S_x\rangle|/2 = \hbar^2/4$. ✓ Saturated! Check $S_x$-$S_y$: $0 \cdot \hbar/2 = 0 \geq \hbar|\langle S_z\rangle|/2 = 0$. ✓
4. Suppose $\hat{A}$ were bounded with all eigenvalues in $[-M, M]$. On a finite-dimensional subspace, $\text{Tr}([\hat{A},\hat{B}]) = \text{Tr}(\hat{A}\hat{B}) - \text{Tr}(\hat{B}\hat{A}) = 0$ (cyclic property). But $\text{Tr}(c\hat{\mathbf{1}}) = cn \neq 0$ for finite $n$. Contradiction. In infinite dimensions, the trace diverges, and neither $\hat{x}$ nor $\hat{p}$ has bounded spectrum — their eigenvalues range over all of $\mathbb{R}$.
5. $\Delta E = \hbar/\tau = 1.055\times10^{-34}/(1.6\times10^{-9}) = 6.6\times10^{-26}$ J. In frequency units: $\Delta\nu = \Delta E/h = 6.6\times10^{-26}/(6.626\times10^{-34}) \approx 10^8$ Hz $= 100$ MHz. The Lyman-alpha line at 121.6 nm has a natural linewidth of $\sim 100$ MHz — observable in precision spectroscopy.
6. $\langle n|\hat{x}|n\rangle = \sqrt{\hbar/(2m\omega)}\langle n|(\hat{a}+\hat{a}^\dagger)|n\rangle = 0$ (since $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$, $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$, off-diagonal). Similarly $\langle p\rangle = 0$. $\langle x^2\rangle = (n+1/2)\hbar/(m\omega)$, $\langle p^2\rangle = (n+1/2)m\hbar\omega$. Product $\langle x^2\rangle\langle p^2\rangle = (n+1/2)^2\hbar^2$. Since $\langle x\rangle=\langle p\rangle=0$, $\Delta x\Delta p = (n+1/2)\hbar \geq \hbar/2$. ✓ Photon number states have minimum phase uncertainty $\Delta\phi \sim 1/(2\Delta n) \to \infty$ for $|n\rangle$ ($\Delta n=0$).
7. From earlier: $\Delta x = L\sqrt{1/12-1/(2n^2\pi^2)}$, $\Delta p = n\pi\hbar/L$. $\Delta x\Delta p = n\pi\hbar\sqrt{1/12-1/(2n^2\pi^2)} = \hbar\sqrt{n^2\pi^2/12-1/2}$. For large $n$: $\Delta x\Delta p \approx n\pi\hbar/(2\sqrt{3}) = n\pi\hbar/2\sqrt{3}$. Ratio to $\hbar/2$: $n\pi/\sqrt{3} \to \infty$ as $n\to\infty$. The uncertainty product grows without bound for excited states of the box.
8. By Robertson applied to $\hat{H}$ and $\hat{A}$: $\Delta E \cdot \Delta A \geq \frac{1}{2}|\langle[\hat{H},\hat{A}]\rangle| = \frac{\hbar}{2}|d\langle\hat{A}\rangle/dt|$ (Heisenberg equation). Define $\tau_A = \Delta A / |d\langle\hat{A}\rangle/dt|$ (time for $\langle A\rangle$ to change by $\Delta A$). Then $\Delta E \cdot \tau_A = \Delta E \cdot \Delta A / |d\langle A\rangle/dt| \geq \hbar/2$. ✓
9. Without measurement: $P_s(T) = 1 - T^2/\tau_Z^2 = 1 - (10^{-6})^2/(10^{-5})^2 = 1 - 0.01 = 0.99$. With $N=100$ measurements: $P_s^{(N)} = [1-(T/N)^2/\tau_Z^2]^{100} = [1-(10^{-8}/10^{-5})^2]^{100} = [1-10^{-6}]^{100} \approx 1 - 10^{-4} \approx 0.9999$. The frequent measurements dramatically suppress the decay: from 99% to 99.99% survival.
10. $\Delta x = e^{-0.5}\Delta x_{\rm vac}$. Since $\Delta x\Delta p = \hbar/2$ (squeezed states saturate): $\Delta p = \hbar/(2\Delta x) = e^{0.5}\hbar/(2\Delta x_{\rm vac}) = e^{0.5}\Delta p_{\rm vac}$. Verification: $\Delta x\Delta p = e^{-0.5}e^{0.5}\Delta x_{\rm vac}\Delta p_{\rm vac} = \hbar/2$. ✓ Noise reduction: $\Delta x^2/\Delta x_{\rm vac}^2 = e^{-2r} = e^{-1} \approx 0.368$. In dB: $10\log_{10}(0.368) \approx -4.34$ dB. (About 4.3 dB squeezing.)
11. $\hat{p}_r = -i\hbar(\partial_r + 1/r)$ is the self-adjoint radial momentum. $[\hat{r}, \hat{p}_r] = i\hbar\hat{\mathbf{1}}$ on the appropriate domain. Robertson gives $\Delta r\Delta p_r \geq \hbar/2$. For bound states, $\psi \to 0$ as $r\to 0$ and $r\to\infty$; the current $j_r \propto \text{Re}(\psi^*\hat{p}_r\psi)$ integrates to zero over all space, so $\langle\hat{p}_r\rangle = 0$.
12. Chandrasekhar limit: for relativistic electrons ($v \to c$ at high density), energy per electron $E_e \sim p_Fc \sim \hbar cN^{1/3}/R$. Gravitational energy $E_G \sim -GN^2m_p^2/R$. Total energy: $E = \hbar cN^{4/3}/R - GN^2m_p^2/R$. This is independent of $R$ only if the coefficients balance: $\hbar cN^{4/3} = GN^2m_p^2 \Rightarrow N_C = (\hbar c/(Gm_p^2))^{3/2} \approx (137/(1.8\times10^{-38}))^{3/2} \approx 1.5\times10^{57}$. $M_C = N_Cm_p \approx 2.5\times10^{30}$ kg $\approx 1.3 M_\odot$. (Exact result: $1.44 M_\odot$, very close.)