= Uncentered second-moment commutator inequality
{title2=$\langle A^2\rangle\langle B^2\rangle\geq\langle C\rangle^2/4,\quad[A,B]=iC$}
For <self-adjoint operators> $A,B$ and a normalized <quantum state> in their relevant product <operator domains>, put $u=A\psi$, $v=B\psi$, $z=\langle u,v\rangle$. With the <inner product> linear in its second entry, $\langle C\rangle=2\operatorname{Im}z$. The <Cauchy-Schwarz inequality> gives
$$
\langle A^2\rangle\langle B^2\rangle=\|u\|^2\|v\|^2\geq|z|^2\geq(\operatorname{Im}z)^2=\frac14\langle C\rangle^2.
$$
Equality holds exactly when $u,v$ are <linearly dependent> and $\operatorname{Re}z=0$. This includes either zero vector; if both vectors are nonzero it is equivalent to $B\psi=i\lambda A\psi$ for real nonzero $\lambda$. Applying this inequality to the centred <observables> $A-\langle A\rangle I$ and $B-\langle B\rangle I$ gives the <Robertson uncertainty principle>.
Back to article page
