= Spin uncertainty state dependence
{title2=$\Delta\sigma_x\,\Delta\sigma_z\geq|\langle\sigma_y\rangle|$}
For a <qubit>, use the dimensionless <Pauli matrices> as <observables> $A=\sigma_x$, $B=\sigma_z$. Their <Pauli matrix multiplication law> gives
$$
[\sigma_x,\sigma_z]=-2i\sigma_y,\qquad \sigma_x^2=\sigma_y^2=\sigma_z^2=I,\qquad \Delta\sigma_x\,\Delta\sigma_z\geq|\langle\sigma_y\rangle|.
$$
These <observables> have <eigenvalues> $\pm1$; the physical <spin> components are $S_j=(\hbar/2)\sigma_j$, where $\hbar$ is the reduced <Planck constant>, so their <standard deviations> acquire a factor $\hbar/2$.
In the <computational basis>, prepare $|+y\rangle=(|0\rangle+i|1\rangle)/\sqrt2$, an <eigenstate> of $\sigma_y$ with <eigenvalue> $+1$. Direct multiplication gives
$$
\langle\sigma_x\rangle=\langle\sigma_z\rangle=0,\qquad \langle\sigma_y\rangle=1,\qquad \Delta\sigma_x=\Delta\sigma_z=1.
$$
Both <Pauli measurements> return $+1$ and $-1$ with equal <probabilities>, and the <Robertson uncertainty principle> is saturated: $1\cdot1=1$.
Instead prepare $|+z\rangle=|0\rangle$. Now the <expectation values> are $\langle\sigma_x\rangle=\langle\sigma_y\rangle=0$, $\langle\sigma_z\rangle=1$, so $\Delta\sigma_x=1$, $\Delta\sigma_z=0$. The <Pauli measurement> of $\sigma_z$ is certain, and both sides of the bound vanish. The <commutator> is still nonzero as an operator: noncommuting <observables> do not guarantee a positive lower bound independent of the <quantum state>. This example is useful for distinguishing the algebra of the <observables> from the <expectation values> in a particular preparation.
\Image[/spin-uncertainty-state-dependence.png]
{title=Pauli standard deviations, their product, and the Robertson bound in two preparations: the same noncommuting observables give bounds of one and zero}
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