= Robertson-Schrödinger uncertainty relation
{c}
{title2=$(\Delta A)^2(\Delta B)^2-C_{AB}^2\geq|\langle[A,B]\rangle|^2/4$}
The <Robertson-Schrödinger uncertainty relation> keeps the real part discarded in the proof of the <Robertson uncertainty principle>. Write $A'=A-\langle A\rangle I$ and $B'=B-\langle B\rangle I$. Their <quantum covariance> is $C_{AB}=\operatorname{Re}\langle A'\psi|B'\psi\rangle$. On the same finite-moment and product <operator domains> used above, the <Cauchy-Schwarz inequality> gives
$$
(\Delta A)^2(\Delta B)^2\geq|\langle A'\psi|B'\psi\rangle|^2
=C_{AB}^2+\frac14|\langle[A,B]\rangle|^2.
$$
For <position> and <momentum>, form the real, symmetrized <covariance matrix>
$$
\Sigma=\begin{pmatrix}(\Delta x)^2&C_{xp}\\C_{xp}&(\Delta p)^2\end{pmatrix},
\qquad \det\Sigma\geq\frac{\hbar^2}{4}.
$$
A 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>.
Back to article page
