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Derivation

Generalized Uncertainty Relation

D-214 Home PU-301 Threads chance · waves Depends on Spectral Theorem for Hermitian Observables, commutator-algebra-canonical
Statement

For any two observables (self-adjoint operators) \(\hat{A}\) and \(\hat{B}\) and any normalized state \(|\psi\rangle\) in their common domain, the product of the variances \(\sigma_A^2=\langle \hat{A}^2\rangle-\langle \hat{A}\rangle^2\) and \(\sigma_B^2=\langle \hat{B}^2\rangle-\langle \hat{B}\rangle^2\) obeys the Robertson–Schrödinger bound \(\displaystyle \sigma_A^2\,\sigma_B^2 \ge \left(\tfrac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle\right)^2+\left(\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2\), whose commutator part alone gives the Robertson relation \(\sigma_A\sigma_B\ge \tfrac{1}{2}\big|\langle[\hat{A},\hat{B}]\rangle\big|\).

Why it matters

This is the operator-theoretic heart of quantum indeterminacy: it shows that the impossibility of jointly sharp values is not a statement about clumsy measurement but a structural fact about non-commuting observables in a Hilbert space. Setting \(\hat{A}=\hat{x}\), \(\hat{B}=\hat{p}\) with \([\hat{x},\hat{p}]=i\hbar\) recovers Heisenberg's \(\sigma_x\sigma_p\ge\hbar/2\) as a special case, but the same one line covers spin components, angular momentum, quadratures of a field mode, and any other pair.

The Schrödinger refinement — keeping the symmetrized covariance term that Robertson discards — is what makes the bound tight for correlated and squeezed states, and it is the version that underlies modern discussions of squeezing, entanglement witnesses, and the geometry of quantum states.

Assumptions
\(\hat{A},\hat{B}\) are self-adjoint.If either is merely symmetric or non-Hermitian, its expectation values and variances need not be real and the norm identity \(\sigma_A^2=\|\hat{A}'\psi\|^2\) fails, so the whole Cauchy–Schwarz route collapses. \(|\psi\rangle\) is normalized and lies in the domain of \(\hat{A}\), \(\hat{B}\), and their products.If \(|\psi\rangle\) is not in \(\mathrm{dom}(\hat{A}\hat{B})\cap\mathrm{dom}(\hat{B}\hat{A})\) the vectors \(\hat{A}'|\psi\rangle\), \(\hat{B}'|\psi\rangle\) or the commutator matrix element may be ill-defined; for unbounded operators the bound can be formally satisfied yet vacuous on states with infinite variance. Variances are finite.If \(\sigma_A\) or \(\sigma_B\) diverges (e.g. a momentum eigenstate has infinite \(\sigma_x\)) the inequality holds trivially but carries no information. The state is pure.For a mixed state \(\hat{\rho}\) one replaces \(\langle\cdot\rangle\) by \(\mathrm{Tr}(\hat{\rho}\,\cdot)\); the Robertson bound still holds but need not be tight, and stronger mixed-state (e.g. entropic or Fisher-information) bounds become the sharper statements.
Derivation
1
\[ \hat{A}'\equiv \hat{A}-\langle \hat{A}\rangle\,\hat{\mathbb{1}},\qquad \hat{B}'\equiv \hat{B}-\langle \hat{B}\rangle\,\hat{\mathbb{1}} \]
Define the deviation operators. Since \(\langle \hat{A}\rangle,\langle \hat{B}\rangle\in\mathbb{R}\) (self-adjointness), \(\hat{A}',\hat{B}'\) are again self-adjoint, and \(\sigma_A^2=\langle \hat{A}'^2\rangle\), \(\sigma_B^2=\langle \hat{B}'^2\rangle\). A
2
\[ \sigma_A^2=\langle\psi|\hat{A}'^2|\psi\rangle=\langle \hat{A}'\psi|\hat{A}'\psi\rangle=\langle f|f\rangle,\qquad |f\rangle\equiv \hat{A}'|\psi\rangle,\ \ |g\rangle\equiv \hat{B}'|\psi\rangle \]
Rewrite each variance as a squared norm. Moving one factor of \(\hat{A}'\) across the inner product uses \(\hat{A}'^\dagger=\hat{A}'\); the same gives \(\sigma_B^2=\langle g|g\rangle\). B
3
\[ \langle f|f\rangle\,\langle g|g\rangle \;\ge\; \big|\langle f|g\rangle\big|^2 \]
Cauchy–Schwarz inequality in the Hilbert space, valid for any two vectors, with equality iff \(|g\rangle=\lambda|f\rangle\) for some \(\lambda\in\mathbb{C}\). This is the single analytic input; everything after is algebra. C
4
\[ \langle f|g\rangle=\langle \hat{A}'\psi|\hat{B}'\psi\rangle=\langle\psi|\hat{A}'\hat{B}'|\psi\rangle=\langle \hat{A}'\hat{B}'\rangle\equiv z \]
Fold both deviation operators back onto the same side using \(\hat{A}'^\dagger=\hat{A}'\). The cross term is a single complex number \(z\). B
5
\[ \hat{A}'\hat{B}'=\tfrac{1}{2}\{\hat{A}',\hat{B}'\}+\tfrac{1}{2}[\hat{A}',\hat{B}'],\qquad [\hat{A}',\hat{B}']=[\hat{A},\hat{B}] \]
Split the product into its symmetric (anticommutator) and antisymmetric (commutator) parts. The constant shifts \(\langle \hat{A}\rangle,\langle \hat{B}\rangle\) commute with everything, so the commutator is unchanged by the subtraction. A
6
\[ \{\hat{A}',\hat{B}'\}^\dagger=\{\hat{A}',\hat{B}'\}\ \Rightarrow\ \langle\{\hat{A}',\hat{B}'\}\rangle\in\mathbb{R},\qquad [\hat{A},\hat{B}]^\dagger=-[\hat{A},\hat{B}]\ \Rightarrow\ \langle[\hat{A},\hat{B}]\rangle\in i\mathbb{R} \]
The anticommutator of two self-adjoint operators is self-adjoint (real expectation); the commutator is anti-self-adjoint (purely imaginary expectation). Hence \(\operatorname{Re}z=\tfrac12\langle\{\hat{A}',\hat{B}'\}\rangle\) and \(\operatorname{Im}z=\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\), both real. C
7
\[ |z|^2=(\operatorname{Re}z)^2+(\operatorname{Im}z)^2=\left(\tfrac{1}{2}\langle\{\hat{A}',\hat{B}'\}\rangle\right)^2+\left(\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2 \]
A complex modulus squared is the sum of squares of its (real and imaginary) parts. The two contributions are separately non-negative. A
8
\[ \sigma_A^2\sigma_B^2\;\ge\;\left(\tfrac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle\right)^2+\left(\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2 \]
Combine steps 2, 3, 7, and use \(\tfrac12\langle\{\hat{A}',\hat{B}'\}\rangle=\tfrac12\langle\{\hat{A},\hat{B}\}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle\) (the symmetrized covariance). Discarding the non-negative covariance term leaves the Robertson bound \(\sigma_A\sigma_B\ge\tfrac12|\langle[\hat{A},\hat{B}]\rangle|\). B
Result
\[ \sigma_A^2\,\sigma_B^2 \;\ge\; \underbrace{\left(\tfrac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle\right)^2}_{\text{covariance}^2}+\underbrace{\left(\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2}_{\text{commutator}}\;\ge\;\frac{1}{4}\big|\langle[\hat{A},\hat{B}]\rangle\big|^2 \]

Reading. The spread of \(A\) and the spread of \(B\) cannot both shrink freely: their product is floored by two things a state cannot hide — how much the two quantities co-fluctuate (the covariance) and how badly the observables fail to commute (the expected commutator). Only when \([\hat{A},\hat{B}]\) has zero expectation and the observables are uncorrelated in the state can the product of variances reach zero.

Units check. Every term carries units \([A]^2[B]^2\): a variance \(\sigma_A^2\) has units \([A]^2\); the covariance and \(\langle\{\hat{A},\hat{B}\}\rangle\) have units \([A][B]\); the commutator \([\hat{A},\hat{B}]\) also has units \([A][B]\), so \(|\langle[\hat{A},\hat{B}]\rangle|^2\) matches \([A]^2[B]^2\). For \(\hat{x},\hat{p}\): \(\big([\hat{x},\hat{p}]=i\hbar\big)\Rightarrow\) both sides in \((\mathrm{J\,s})^2=(\mathrm{kg\,m^2\,s^{-1}})^2\), and \(\sigma_x\sigma_p\ge\hbar/2\) reads \(\mathrm{m\cdot kg\,m\,s^{-1}}=\mathrm{J\,s}\). Consistent.

Limiting cases
  • Canonical pair: \([\hat{x},\hat{p}]=i\hbar\) gives \(\sigma_x\sigma_p\ge\hbar/2\), saturated by Gaussian (coherent/vacuum) states.
  • Commuting observables: \([\hat{A},\hat{B}]=0\) removes the floor from the commutator; a simultaneous eigenstate then makes both variances zero, so the bound is \(0\ge0\).
  • Uncorrelated state: when the symmetrized covariance vanishes, Robertson and Schrödinger coincide — the extra term buys nothing.
  • Squeezed / correlated states: a nonzero covariance raises the Schrödinger floor above Robertson's, so a state can saturate Schrödinger while sitting strictly above the naive \(\tfrac12|\langle[\hat{A},\hat{B}]\rangle|\).
  • Classical limit: for macroscopic actions \(\hbar\) is negligible against \(\sigma_x\sigma_p\), and the bound becomes operationally invisible.
Breaks when
  • Non-Hermitian or merely symmetric operators. If \(\hat{A}^\dagger\ne\hat{A}\), the step \(\langle \hat{A}'\psi|\hat{A}'\psi\rangle=\langle\psi|\hat{A}'^2|\psi\rangle\) fails, expectation values go complex, and there is no meaningful real variance to bound.
  • States outside the operator domain / infinite variance. For unbounded \(\hat{A},\hat{B}\) (position, momentum) acting on a state not in \(\mathrm{dom}(\hat{A}\hat{B})\), the matrix element \(\langle[\hat{A},\hat{B}]\rangle\) can be ambiguous; a state with \(\sigma_A=\infty\) satisfies the inequality vacuously. This is the technical loophole behind apparent "violations" on singular wavefunctions.
  • Commutator with state-dependent domain subtleties. When \([\hat{A},\hat{B}]\) is not a bounded operator (e.g. angular variables on a circle where \([\hat{\phi},\hat{L}_z]\) meets periodic boundary conditions), a naive \(\tfrac12|\langle[\hat{A},\hat{B}]\rangle|\) can exceed the true attainable spread and the bound must be replaced by a version respecting the topology.
  • Mixed states pushed past purity. The inequality still holds for \(\hat{\rho}\), but treating a mixed ensemble as if a single \(|\psi\rangle\) saturated it gives wrong (too tight) conclusions; classical mixing generally loosens, not tightens, the achievable product.
Failure modes
  • Dropping the modulus. Writing \(\sigma_A\sigma_B\ge\tfrac12\langle[\hat{A},\hat{B}]\rangle\) — but \(\langle[\hat{A},\hat{B}]\rangle\) is imaginary; you must take \(\tfrac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\) or \(\tfrac12|\langle[\hat{A},\hat{B}]\rangle|\).
  • Forgetting to subtract the means. Using \(\langle \hat{A}\hat{B}\rangle\) directly instead of the deviation operators \(\hat{A}',\hat{B}'\); the bound is about variances, not raw second moments.
  • Confusing operator uncertainty with measurement disturbance. Robertson bounds the state's statistical spreads, not the error–disturbance of a sequential measurement (that is the separate Ozawa/Branciard relation).
  • Assuming the bound is always tight. Robertson can be strictly loose (e.g. for angular momentum on many states); non-saturation is the rule, not the exception.
  • Treating \(\tfrac12|\langle[\hat{A},\hat{B}]\rangle|\) as a constant. For \([\hat{A},\hat{B}]\) an operator (like \(i\hbar\hat{S}_z\)), the floor is state-dependent and can even be zero on a state with \(\langle\hat{S}_z\rangle=0\).
  • Applying it to a single observable with itself. \([\hat{A},\hat{A}]=0\) trivially; there is no self-uncertainty.
Discussion

The derivation isolates exactly what physics enters and where. The only quantum-mechanical inputs are (i) states live in an inner-product space, giving Cauchy–Schwarz, and (ii) observables are self-adjoint, forcing the commutator's expectation to be imaginary and the anticommutator's to be real. The uncertainty principle is therefore not an extra postulate but a corollary of Hilbert-space kinematics plus non-commutativity — it would hold in any theory with the same algebraic skeleton.

The Schrödinger term is geometrically the covariance of the two observables; together with the two variances it forms a \(2\times2\) covariance matrix whose positive-semidefiniteness is the Robertson–Schrödinger inequality (the determinant condition). This viewpoint generalizes cleanly to \(n\) observables, where positivity of the full quantum covariance matrix reproduces multimode uncertainty relations and, for Gaussian states, the symplectic (Williamson) spectrum that certifies squeezing and entanglement.

Saturation deserves emphasis: equality in Cauchy–Schwarz demands \(\hat{B}'|\psi\rangle=\lambda\,\hat{A}'|\psi\rangle\), and killing the covariance term requires \(\lambda\) purely imaginary. For \(\hat{x},\hat{p}\) this eigenvalue equation is a first-order ODE whose solutions are exactly the Gaussians — which is why coherent and squeezed vacuum states, and no others, sit on the \(\sigma_x\sigma_p=\hbar/2\) floor.

A subtlety worth internalizing at degree level: the Robertson bound can be trivial even when observables fail to commute. Angular momentum components satisfy \([\hat{L}_x,\hat{L}_y]=i\hbar\hat{L}_z\), yet on a state with \(\langle\hat{L}_z\rangle=0\) the Robertson floor is zero although \(\sigma_{L_x}\sigma_{L_y}\) is generally positive. The bound is a lower bound, not a value; its weakness on such states is precisely what motivated stronger (sum, entropic, and Maccone–Pati) uncertainty relations that stay nontrivial whenever the observables genuinely disagree.

Common misconceptions. The relation does not say a measurement of \(A\) mechanically "kicks" \(B\); it constrains the statistics of an ensemble of identically prepared states, each measured once. Nor does it forbid knowing both \(\langle A\rangle\) and \(\langle B\rangle\) precisely — it constrains the spreads, not the means. And it is symmetric in \(A,B\): there is no "primary" quantity whose measurement disturbs a "secondary" one.

Worked examples
1
Position–momentum spread of a confined electron
Take \(\hat{A}=\hat{x}\), \(\hat{B}=\hat{p}\), with \([\hat{x},\hat{p}]=i\hbar\). Symbols first, then numbers. A
2
\[ \tfrac{1}{2i}\langle[\hat{x},\hat{p}]\rangle=\tfrac{1}{2i}(i\hbar)=\tfrac{\hbar}{2}\ \Rightarrow\ \sigma_x\sigma_p\ge\frac{\hbar}{2},\qquad \sigma_p\ge\frac{\hbar}{2\sigma_x} \]
The commutator is a c-number, so the floor is state-independent; solve for \(\sigma_p\). A
3
\[ \sigma_x=1.0\times10^{-10}\,\mathrm{m}\ \Rightarrow\ \sigma_p\ge\frac{1.055\times10^{-34}\,\mathrm{J\,s}}{2(1.0\times10^{-10}\,\mathrm{m})}=5.3\times10^{-25}\,\mathrm{kg\,m\,s^{-1}} \]
Insert \(\hbar=1.055\times10^{-34}\,\mathrm{J\,s}\) and an atomic-scale confinement. A
4
\[ \sigma_v=\frac{\sigma_p}{m_e}\ge\frac{5.3\times10^{-25}}{9.11\times10^{-31}}\ \mathrm{m\,s^{-1}}\approx 5.8\times10^{5}\,\mathrm{m\,s^{-1}} \]
Convert to a velocity spread using \(m_e=9.11\times10^{-31}\,\mathrm{kg}\). A
\[ \sigma_p\gtrsim5.3\times10^{-25}\,\mathrm{kg\,m\,s^{-1}},\qquad \sigma_v\gtrsim5.8\times10^{5}\,\mathrm{m\,s^{-1}} \]

Reading. Pinning an electron to one ångström forces an irreducible momentum spread corresponding to a velocity uncertainty near \(10^6\,\mathrm{m/s}\) — a large fraction of a percent of \(c\), which is why bound electrons are inherently "fast." Units check. \(\mathrm{J\,s}/\mathrm{m}=\mathrm{kg\,m\,s^{-1}}\); dividing by \(\mathrm{kg}\) gives \(\mathrm{m\,s^{-1}}\).

1
Spin components in the \(z\)-up state (Robertson saturated)
Take \(\hat{A}=\hat{S}_x\), \(\hat{B}=\hat{S}_y\) for a spin-\(\tfrac12\) particle in \(|\psi\rangle=|{\uparrow}_z\rangle\), with \([\hat{S}_x,\hat{S}_y]=i\hbar\hat{S}_z\). B
2
\[ \sigma_{S_x}\sigma_{S_y}\ge\tfrac12\big|\langle i\hbar\hat{S}_z\rangle\big|=\tfrac{\hbar}{2}\big|\langle\hat{S}_z\rangle\big|=\tfrac{\hbar}{2}\cdot\frac{\hbar}{2}=\frac{\hbar^2}{4} \]
Insert the commutator; on \(|{\uparrow}_z\rangle\), \(\langle\hat{S}_z\rangle=+\hbar/2\). The floor is state-dependent here. B
3
\[ \langle\hat{S}_x\rangle=\langle\hat{S}_y\rangle=0,\qquad \hat{S}_x^2=\hat{S}_y^2=\tfrac{\hbar^2}{4}\hat{\mathbb{1}}\ \Rightarrow\ \sigma_{S_x}=\sigma_{S_y}=\frac{\hbar}{2} \]
For spin-\(\tfrac12\) each component squares to \((\hbar/2)^2\hat{\mathbb{1}}\); the means vanish by symmetry of \(|{\uparrow}_z\rangle\) under \(x\!\to\!-x\), \(y\!\to\!-y\). B
4
\[ \sigma_{S_x}\sigma_{S_y}=\frac{\hbar}{2}\cdot\frac{\hbar}{2}=\frac{\hbar^2}{4};\qquad \tfrac12\langle\{\hat{S}_x',\hat{S}_y'\}\rangle=\tfrac12\langle \hat{S}_x\hat{S}_y+\hat{S}_y\hat{S}_x\rangle=0 \]
The anticommutator \(\{\hat{S}_x,\hat{S}_y\}=0\) for spin-\(\tfrac12\), so the covariance term vanishes and Schrödinger reduces to Robertson — which the state then saturates exactly. C
\[ \sigma_{S_x}\sigma_{S_y}=\frac{\hbar^2}{4}=\tfrac12\big|\langle[\hat{S}_x,\hat{S}_y]\rangle\big| \]

Reading. A spin polarized along \(z\) has maximally uncertain transverse components, and it does so at the exact Robertson floor: the inequality is an equality. Because \(\langle\hat{S}_z\rangle\) sets the floor, a state with \(\langle\hat{S}_z\rangle=0\) (e.g. an \(x\)-eigenstate) would give bound zero even though \(\hat{S}_x,\hat{S}_y\) still fail to commute. Units check. \(\hat{S}\) has units of action (\(\mathrm{J\,s}\)); \(\sigma_{S_x}\sigma_{S_y}\) and \(\hbar^2\) are both \((\mathrm{J\,s})^2\).

Problems
  1. (A) Starting from the general bound, insert \([\hat{x},\hat{p}]=i\hbar\) and show carefully that the imaginary unit cancels to leave \(\sigma_x\sigma_p\ge\hbar/2\).
    Solution The commutator term is \(\left(\tfrac{1}{2i}\langle[\hat{x},\hat{p}]\rangle\right)^2=\left(\tfrac{1}{2i}\cdot i\hbar\right)^2=(\hbar/2)^2\), which is real and positive. Dropping the non-negative covariance term, \(\sigma_x^2\sigma_p^2\ge(\hbar/2)^2\). Taking positive square roots (variances are non-negative) gives \(\sigma_x\sigma_p\ge\hbar/2\). The factor \(1/i=-i\) makes \(\tfrac{1}{2i}(i\hbar)=\hbar/2\) real, which is why the physical spread is bounded by \(\hbar/2\), not by an imaginary number.
  2. (A) An electron is localized in a region of size \(\Delta x\approx1\,\mathrm{nm}\) (a quantum dot). Estimate the minimum kinetic-energy scale \(\sim\sigma_p^2/2m_e\) implied by the uncertainty bound.
    Solution With \(\sigma_x\approx\Delta x=1\times10^{-9}\,\mathrm{m}\), \(\sigma_p\ge\hbar/(2\sigma_x)=1.055\times10^{-34}/(2\times10^{-9})=5.3\times10^{-26}\,\mathrm{kg\,m\,s^{-1}}\). Then \(E\sim\sigma_p^2/2m_e=(5.3\times10^{-26})^2/(2\times9.11\times10^{-31})\approx1.5\times10^{-21}\,\mathrm{J}\approx9.6\,\mathrm{meV}\). So confinement to a nanometre costs on the order of \(10\,\mathrm{meV}\) of zero-point kinetic energy — comparable to thermal energy at room temperature (\(k_BT\approx25\,\mathrm{meV}\)), which is why quantum-dot level spacings are experimentally relevant.
  3. (B) For spin-\(\tfrac12\) in the state \(|{\uparrow}_z\rangle\), verify the Robertson bound for the pair \((\hat{S}_x,\hat{S}_z)\). Is it saturated?
    Solution \([\hat{S}_x,\hat{S}_z]=-i\hbar\hat{S}_y\), so the floor is \(\tfrac12|\langle-i\hbar\hat{S}_y\rangle|=\tfrac{\hbar}{2}|\langle\hat{S}_y\rangle|\). On \(|{\uparrow}_z\rangle\), \(\langle\hat{S}_y\rangle=0\), so the Robertson floor is \(0\). Meanwhile \(\sigma_{S_z}=0\) (it is a \(\hat{S}_z\) eigenstate) and \(\sigma_{S_x}=\hbar/2\), giving product \(0\). Thus \(0\ge0\): the bound holds and is saturated trivially, because one variance is exactly zero. This illustrates that a zero on the right does not require commuting observables.
  4. (B) Show that if \([\hat{A},\hat{B}]=0\) and \(|\psi\rangle\) is a common eigenstate of \(\hat{A}\) and \(\hat{B}\), both sides of the Robertson relation vanish. Then argue why commuting observables can be simultaneously sharp.
    Solution If \(\hat{A}|\psi\rangle=a|\psi\rangle\) then \(\hat{A}'|\psi\rangle=(a-\langle \hat{A}\rangle)|\psi\rangle=0\) since \(\langle \hat{A}\rangle=a\); hence \(\sigma_A^2=\|\hat{A}'\psi\|^2=0\), and likewise \(\sigma_B=0\), so the left side is \(0\). The right side has commutator term \(\tfrac12|\langle[\hat{A},\hat{B}]\rangle|=0\) by assumption, and covariance \(\tfrac12\langle\{\hat{A}',\hat{B}'\}\rangle=0\) because \(\hat{A}'|\psi\rangle=0\). By the spectral theorem, commuting self-adjoint operators share a complete orthonormal eigenbasis, so such a common eigenstate always exists — there is no obstruction to both being dispersion-free, exactly what the vanishing bound permits.
  5. (C) Derive the equality condition for the full Robertson–Schrödinger bound and use it to explain why only Gaussian wavefunctions saturate \(\sigma_x\sigma_p=\hbar/2\).
    Solution Equality in Cauchy–Schwarz (step 3) needs \(\hat{B}'|\psi\rangle=\lambda\,\hat{A}'|\psi\rangle\) for some \(\lambda\in\mathbb{C}\). Writing \(z=\langle f|g\rangle=\lambda\langle f|f\rangle=\lambda\sigma_A^2\), the covariance is \(\operatorname{Re}z=\sigma_A^2\operatorname{Re}\lambda\) and the commutator part is \(\operatorname{Im}z=\sigma_A^2\operatorname{Im}\lambda\). Saturating the Schrödinger bound needs only the eigenvector condition; the Robertson bound additionally needs the covariance to vanish, i.e. \(\operatorname{Re}\lambda=0\), so \(\lambda\) is purely imaginary. For \(\hat{A}=\hat{x}\), \(\hat{B}=\hat{p}=-i\hbar\,d/dx\), the condition \((\hat{p}-\langle p\rangle)\psi=\lambda(\hat{x}-\langle x\rangle)\psi\) is the ODE \(-i\hbar\psi'-\langle p\rangle\psi=\lambda(x-\langle x\rangle)\psi\), whose solution is \(\psi(x)\propto\exp\!\big[\tfrac{i\lambda}{2\hbar}(x-\langle x\rangle)^2+\tfrac{i\langle p\rangle}{\hbar}x\big]\). With \(\lambda\) purely imaginary, \(\lambda=-i\hbar/(2\sigma_x^2)\), this is a normalizable Gaussian of width \(\sigma_x\) — no other functional form solves the equality condition, so Gaussians (coherent and squeezed) are the unique minimum-uncertainty states.