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Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrödinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

Centre the position operator and momentum operator by their real expectation values:Scalar multiples of the identity operator commute with every linear operator, so the canonical commutation relation givesThe second moments of the centred linear operators are the variances, ${\mathbb{E}}_{\psi }(A^2)={\Delta }_{\psi }(X{)}^2$ and ${\mathbb{E}}_{\psi }(B^2)={\Delta }_{\psi }(P{)}^2$. Applying the preceding inequality and taking nonnegative square roots proves the Heisenberg uncertainty relation:Subtracting the means is essential: the uncentred second moments alone do not equal the variances. The equality case of the Heisenberg uncertainty relation requires the centred vectors to be imaginary multiples, as in the preceding equality conditions.

The Robertson-Schrödinger uncertainty relation keeps the real part discarded in the proof of the Robertson uncertainty principle. Write $A^{\prime }=A-⟨A⟩I$ and $B^{\prime }=B-⟨B⟩I$. Their quantum covariance is $C_{AB}=\mathrm{Re}⟨A^{\prime }\psi |B^{\prime }\psi ⟩$. On the same finite-moment and product operator domains used above, the Cauchy-Schwarz inequality givesFor position and momentum, form the real, symmetrized covariance matrixA larger coordinatewise product can therefore encode a tilted covariance ellipse, rather than a larger determinant. This explains why the equality case of the Heisenberg uncertainty relation is more restrictive than equality in the Robertson-Schrödinger uncertainty relation.

The Robertson uncertainty principle bounds the product of the standard deviations of two observables in the same quantum state. For a normalized pure state $|\psi ⟩$ and self-adjoint operators $A,B$, write their expectation values and variances asand similarly for $B$, where $I$ is the identity operator. The quantum uncertainties $\Delta A,\Delta B$ then obeyHere $[A,B]$ is the commutator, and every expectation value is taken in the same $|\psi ⟩$. These standard deviations describe the outcome spreads when $A$ and $B$ are measured on separate ensembles prepared in that same quantum state. They are not instrument errors or the change caused by a preceding measurement in quantum mechanics. The definition and bound are developed in Barton Zwiebach, Uncertainty Principle and Compatible Observables, 2013, §§1–2.

For unbounded self-adjoint operators, the operator domains matter. Assume $\psi \in D(A)\cap D(B)$, so that both second moments are finite; $⟨A^2⟩$ here means the second moment of the measured probability distribution, equal to $\Vert A\psi {\Vert }^2$, even if $A^2\psi$ is undefined. To use the displayed operator commutator, also assume $\psi \in D(AB)\cap D(BA)$. Without these product-domain assumptions, the always meaningful version on $D(A)\cap D(B)$ uses $|\mathrm{Im}⟨A\psi |B\psi ⟩|$ as its right-hand side. These qualifications prevent a formal commutator identity from being applied to an inadmissible quantum state.

The brief proof is Cauchy-Schwarz inequality applied to $u=(A-⟨A⟩I)\psi$ and $v=(B-⟨B⟩I)\psi$:The final equality uses self-adjointness and the product-domain assumptions. Thus a geometric statement about inner products becomes a constraint on experimental spreads. For nonzero $\Delta A,\Delta B$, equality requires $u,v$ to be proportional with a purely imaginary proportionality factor; the equality case of the Heisenberg uncertainty relation turns this condition into a Gaussian wave packet.

The bound is useful because computing a single commutator and its expectation value can constrain two entire outcome distributions. Its right-hand side depends on the quantum state: the spin uncertainty state dependence example gives both a positive saturated bound and a zero bound for the same noncommuting observables. Robertson's paper, titled "The Uncertainty Principle", appeared on 1 July 1929 in Physical Review, volume 34, starting at page 163. ^[ref].

The squeezed vacuum is the zero-displacement squeezed coherent state $|\gamma ⟩=S(\gamma )|0⟩$. Its position and momentum variances are rescaled in opposite directions,so it remains an minimum-uncertainty state with $\Delta X\Delta P=\hslash /2$.