The Commutator Bound

The commutator

Two observables $A$ and $B$ do not, in general, commute. Their commutator is the operator

$[A, B] \equiv AB - BA.$

If $[A, B] = 0$ the operators share a complete set of eigenstates and can in principle be measured simultaneously with arbitrary precision. When $[A, B] \neq 0$, a fundamental constraint kicks in.

Robertson's inequality

In 1929, H. P. Robertson derived a general lower bound on the product of standard deviations of any two observables. For a system in state $|\psi\rangle$, let $\sigma_A$ and $\sigma_B$ be the standard deviations of $A$ and $B$ respectively. Then

$\sigma_A \,\sigma_B \;\geq\; \frac{1}{2}\,\bigl|\langle [A, B] \rangle\bigr|,$

where $\langle [A, B] \rangle = \langle \psi | [A, B] | \psi \rangle$ is the expectation value of the commutator in that state.

The inequality follows from the Cauchy–Schwarz inequality applied to the vectors $|f\rangle = (A - \langle A\rangle)|\psi\rangle$ and $|g\rangle = (B - \langle B\rangle)|\psi\rangle$. One writes $\langle f | g \rangle = \text{Re}\langle f|g\rangle + i\,\text{Im}\langle f|g\rangle$ and notes that the imaginary part equals $\frac{1}{2i}\langle [A,B]\rangle$ (so that $|\text{Im}\langle f|g\rangle| = \frac{1}{2}|\langle [A,B]\rangle|$, recalling that $\langle[A,B]\rangle$ is purely imaginary for Hermitian $A,B$), so the Cauchy–Schwarz bound $\|f\|^2\|g\|^2 \geq |\langle f|g\rangle|^2 \geq |\text{Im}\langle f|g\rangle|^2$ gives exactly Robertson's result.

Position and momentum

The canonical commutation relation,

$[x, p] = i\hbar,$

is the foundational algebraic fact of non-relativistic quantum mechanics. Because $[x,p]$ equals $i\hbar$ times the identity, the expectation value $\langle [x, p]\rangle = i\hbar$ in every normalizable state, and Robertson's inequality immediately gives

$\sigma_x \,\sigma_p \;\geq\; \frac{1}{2}\,|i\hbar| = \frac{\hbar}{2}.$

This is the Heisenberg uncertainty principle, now understood as a theorem about operators rather than a statement about experimental disturbance. The numerical bound is

$\frac{\hbar}{2} = \frac{1.0545718 \times 10^{-34}\,\text{J·s}}{2} \approx 5.27 \times 10^{-35}\,\text{J·s}.$

States that saturate the bound — Gaussian wave packets for which $\sigma_x \sigma_p = \hbar/2$ — are called minimum-uncertainty states. The harmonic-oscillator coherent states are the canonical example (they keep equal, balanced spreads in $x$ and $p$); the squeezed states are minimum-uncertainty packets too, trading a smaller spread in one variable for a larger spread in the other. All non-Gaussian states have a strictly larger product.

Other pairs

The Robertson inequality applies to any two observables. For the components of angular momentum, $[L_x, L_y] = i\hbar L_z$, so

$\sigma_{L_x}\,\sigma_{L_y} \;\geq\; \frac{\hbar}{2}\,|\langle L_z \rangle|.$

Here the bound does depend on the state through $\langle L_z \rangle$. In an eigenstate of $L_z$ with quantum number $m=0$ the right-hand side vanishes, so Robertson's inequality imposes no constraint at all. A vanishing bound does not imply the two observables become sharp, however: for any $\ell \geq 1$ that same $m=0$ state has $\sigma_{L_x} = \sigma_{L_y} = \hbar\sqrt{\ell(\ell+1)/2} > 0$. The inequality is simply silent there.

Try it

This is a numerical exercise — return a number, not a circuit. Using $\hbar = 1.0545718 \times 10^{-34}\,\text{J·s}$, compute the lower bound on $\sigma_x \,\sigma_p$ implied by the Robertson inequality and the canonical commutation relation.

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