Checkpoint: Uncertainty Calculations

This checkpoint brings together the two key computational skills of Module 10: computing the standard deviation of an observable from its eigenvalue probability distribution, and evaluating the uncertainty product for two observables. Both quantities appear directly in the Robertson inequality — the theorem that underlies every form of the uncertainty principle.

Review: computing a single uncertainty

Recall the two-step recipe. For an observable $A$ in state $|\psi\rangle$ with eigenvalues $a_i$ and measurement probabilities $p_i$ :

\langle A^2 \rangle = \sum_i p_i\, a_i^2,$$ $$\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}.$$ The quantity $\Delta A$ is the root-mean-square spread of outcomes around their mean — zero only if the state is an eigenstate of $A$. ## The uncertainty product and the Robertson bound For two observables $A$ and $B$ the **Robertson inequality** states $$\Delta A \cdot \Delta B \;\geq\; \frac{1}{2}\,|\langle [A, B] \rangle|,$$ where $[A, B] = AB - BA$ is the commutator. The lower bound is set by the expectation value $\langle [A, B] \rangle$ in the given state, so it can depend on the state and may even vanish in particular states despite $[A, B] \neq 0$ as an operator — this state-dependent bound on the uncertainty product is the abstract core of every uncertainty relation in quantum mechanics. The Heisenberg relation $\Delta x \cdot \Delta p \geq \hbar/2$ is the most famous case, following from $[x, p] = i\hbar$. It is special because the commutator is $i\hbar$ times the identity, so $\langle [x, p] \rangle = i\hbar$ in every state and the bound $\hbar/2$ is fixed. ## Worked checkpoint example **Setup.** A quantum state yields the following measurement statistics: | Observable | Eigenvalue | Probability | |---|---|---| | $A$ | $+2$ | $3/4$ | | $A$ | $-2$ | $1/4$ | | $B$ | $+1$ | $1/4$ | | $B$ | $-1$ | $3/4$ | **Step 1 — uncertainty of $A$.** $$\langle A \rangle = \tfrac{3}{4}(+2) + \tfrac{1}{4}(-2) = \tfrac{3}{2} - \tfrac{1}{2} = 1.$$ $$\langle A^2 \rangle = \tfrac{3}{4}(4) + \tfrac{1}{4}(4) = 3 + 1 = 4.$$ $$(\Delta A)^2 = \langle A^2 \rangle - \langle A \rangle^2 = 4 - 1 = 3, \qquad \Delta A = \sqrt{3}.$$ **Step 2 — uncertainty of $B$.** $$\langle B \rangle = \tfrac{1}{4}(+1) + \tfrac{3}{4}(-1) = \tfrac{1}{4} - \tfrac{3}{4} = -\tfrac{1}{2}.$$ $$\langle B^2 \rangle = \tfrac{1}{4}(1) + \tfrac{3}{4}(1) = 1.$$ $$(\Delta B)^2 = 1 - \tfrac{1}{4} = \tfrac{3}{4}, \qquad \Delta B = \frac{\sqrt{3}}{2}.$$ **Step 3 — uncertainty product.** $$\Delta A \cdot \Delta B = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} = 1.5.$$ <Callout type="tip"> Notice that $\langle A^2 \rangle = 4$ regardless of the probabilities because both eigenvalues square to $4$. The uncertainty is entirely determined by the spread of the mean: a state with $\langle A \rangle$ far from the eigenvalues has large $\Delta A$, while an eigenstate has $\langle A \rangle = a_i$ and $\Delta A = 0$. </Callout> ## Try it This is a numerical exercise — your code should `return` a number, not a circuit. A quantum state is measured on two observables. Observable $A$ has eigenvalues $+2$ and $-2$ with probabilities $3/4$ and $1/4$. Observable $B$ has eigenvalues $+1$ and $-1$ with probabilities $1/4$ and $3/4$. Compute the uncertainty product $\Delta A \cdot \Delta B$. <RunnableExample />