The Cramér–Rao Bound

The quantum Fisher information measured how much a probe knows about $\theta$ . The quantum Cramér–Rao bound (QCRB) converts that number into a hard floor on precision: no unbiased estimator, and no measurement whatsoever, can beat it.

The bound

For an estimator built from $M$ independent repetitions of a probe whose single-shot quantum Fisher information is $F_Q$ , the variance of any unbiased estimator satisfies

(\Delta\theta)^2 \;\ge\; \frac{1}{M\,F_Q} \quad\Longleftrightarrow\quad \Delta\theta \;\ge\; \frac{1}{\sqrt{M\,F_Q}}.

The factor of $M$ is the familiar statistical averaging: independent repetitions add their Fisher information. The factor $F_Q$ is everything quantum — it is set by the probe state and the generator, as derived in the previous lesson.

Two bounds in one

The QCRB sits at the end of a chain of inequalities. For a specific measurement with classical Fisher information $F(\theta)$ one has the classical Cramér–Rao bound $(\Delta\theta)^2 \ge 1/(M F)$ . Maximising over all measurements gives the quantum bound, because

F(\theta) \;\le\; F_Q,

with equality achievable by an optimal measurement. So the QCRB is attainable: there always exists a measurement (often a simple projective one) that saturates it asymptotically as $M \to \infty$ .

Recovering the standard quantum limit

Put the pieces together for $N$ independent single-qubit probes, each repeated so that $M = N$ probe uses contribute. Each equatorial probe has $F_Q = 1$ , and Fisher information is additive over independent probes, so the total is $F_Q^{\text{tot}} = N$ . The bound gives

\Delta\theta \;\ge\; \frac{1}{\sqrt{N \cdot 1}} = \frac{1}{\sqrt{N}},

which is precisely the standard quantum limit. The SQL is not a technological accident — it is the Cramér–Rao bound for the best unentangled probe. To do better we must raise $F_Q$ itself, which requires entanglement.

A worked number

Suppose we run $M = 100$ repetitions of the optimal single-qubit probe ( $F_Q = 1$ ). The precision floor is

\Delta\theta \ge \frac{1}{\sqrt{100 \cdot 1}} = \frac{1}{10} = 0.1\ \text{rad}.

No estimator can do better than a standard deviation of $0.1$ radians on this data. Doubling the precision would require quadrupling $M$ — the hallmark of $1/\sqrt{M}$ scaling.

Try it

Compute the Cramér–Rao precision floor for $M = 100$ repetitions of a probe with $F_Q = 1$ . Return the number $1/\sqrt{M\,F_Q}$ ; the grader checks it equals $0.1$ .

Run your code to see the quantum state.