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For a normalized state with the required position and momentum moments, $[x,p]=i\hslash$ and the Robertson uncertainty principle give $\Delta x\Delta p\ge \hslash /2$. In a quantum harmonic oscillator, the arithmetic-geometric mean inequality then bounds the variance contribution to energy below by $\hslash \omega /2$.

Take a normalized state for which the indicated variances and operator pairings exist. Set $A={\hat{L}}_3-⟨{\hat{L}}_3⟩$ and $B={\hat{x}}_1-⟨{\hat{x}}_1⟩$. For real $\lambda$, positivity of the squared norm givesThe canonical commutation relations yield $[{\hat{L}}_3,{\hat{x}}_1]=i\hslash {\hat{x}}_2$. Thus the quadratic polynomial in $\lambda$ is nonnegative everywhere only if its discriminant is nonpositive. This gives the Robertson uncertainty principleIf either variance is zero, the same polynomial argument forces the commutator expectation to vanish, so that edge case is included. Centering the operators is essential to obtain variances rather than uncentered second moments.

The expectation value is the mean outcome over many identically prepared measurements of the observable. For real $\lambda$, positivity of the norm in the Hilbert space givesThis real quadratic has nonpositive quadratic discriminant, soApply this to the centered observables $Q-⟨Q⟩$ and $P-⟨P⟩$ to obtain the Robertson uncertainty principle

For the quantum harmonic oscillator, $[x,p]=i\hslash$ andThe arithmetic-geometric mean inequality and $\Delta x\Delta p\ge \hslash /2$ now give

An operator $Q$ is Hermitian whenon its domain. Hermitian operators represent quantum observables because their expectation values and eigenvalues are real. In a normalized state,

Positivity of the norm of $(Q+i\lambda P)\psi$ for every real $\lambda$ givesThe middle expectation is real because $i[Q,P]$ is Hermitian. The discriminant of this quadratic is nonpositive, soApply this to the centered operators $Q-⟨Q{⟩}_{\psi }$ and $P-⟨P{⟩}_{\psi }$, whose commutator is still $[Q,P]$. This yields the Robertson uncertainty principleNoncommuting observables therefore cannot both have arbitrarily sharply concentrated measurement distributions in the same state.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

For a normalized state, writeThen ${\Delta }_{\psi }A=\Vert A^{\prime }\psi \Vert$ and ${\Delta }_{\psi }B=\Vert B^{\prime }\psi \Vert$. The Schwarz inequality givesSince $A^{\prime }$ and $B^{\prime }$ are Hermitian and their commutator equals $[\hat{A},\hat{B}]$,Thereforewhich is the Robertson uncertainty principle.

${\Delta }_{\psi }Q=\sqrt{⟨Q^2⟩-⟨Q{⟩}^2}$. The Robertson uncertainty principle is

For a normalized state with finite second moment of a quantum observable $Q$, its uncertainty is the standard deviation of the measurement outcome: $(\Delta Q{)}^2=⟨Q^2⟩-⟨Q{⟩}^2$. This equals $\Vert (Q-⟨Q⟩)\psi {\Vert }^2$. The Robertson uncertainty principle bounds products of uncertainties using the expectation of the commutator.