The Standard Quantum Limit

When we average $N$ independent classical measurements, the central limit theorem shrinks the standard deviation of the mean by $1/\sqrt{N}$ . The quantum version of this rule is the standard quantum limit (SQL), and it sets the precision floor for any sensing protocol that uses $N$ independent (unentangled) probes. This lesson derives the $1/\sqrt{N}$ scaling from the Ramsey phase probe.

A single Ramsey probe

The workhorse is the Ramsey sequence on one qubit. Sandwiching the phase gate between two Hadamards gives the state just before measurement, and a direct computation yields the outcome probabilities

p_1(\theta) = \frac{1 - \cos\theta}{2}, \qquad p_0(\theta) = \frac{1 + \cos\theta}{2}.

The signal — the quantity that carries information about $\theta$ — is the probability $p_1$ . Its slope with respect to $\theta$ ,

\frac{\partial p_1}{\partial\theta} = \frac{\sin\theta}{2},

is largest near $\theta = \pi/2$ and vanishes at $\theta = 0, \pi$ , where the probe is least sensitive.

Error propagation

Each shot is a Bernoulli trial with success probability $p_1$ . After $N$ shots the estimated probability $\hat p_1$ has binomial variance

(\Delta \hat p_1)^2 = \frac{p_1(1 - p_1)}{N}.

We do not want $p_1$ , we want $\theta$ . Propagating the uncertainty through the relation $\theta(\hat p_1)$ gives

(\Delta\theta)^2 = \frac{(\Delta \hat p_1)^2}{\left(\partial p_1/\partial\theta\right)^2} = \frac{p_1(1-p_1)}{N\,(\sin\theta/2)^2}.

Substituting $p_1 = (1-\cos\theta)/2$ so that $p_1(1-p_1) = \frac{1}{4}\sin^2\theta$ , the $\sin^2\theta$ factors cancel exactly:

(\Delta\theta)^2 = \frac{\tfrac14 \sin^2\theta}{N\,\tfrac14\sin^2\theta} = \frac{1}{N} \quad\Longrightarrow\quad \boxed{\;\Delta\theta_{\text{SQL}} = \frac{1}{\sqrt{N}}\;}

This is the standard quantum limit: with $N$ independent probes the phase uncertainty falls as $1/\sqrt{N}$ , and — remarkably — the SQL is achieved at every working point $\theta$ , not only the most sensitive one.

Reading the limit off a histogram

On the simulator we cannot run $N \to \infty$ shots, but we can verify the per-shot statistics that the SQL is built on. For $\theta = \pi/3$ ,

p_1 = \frac{1 - \cos(\pi/3)}{2} = \frac{1 - \tfrac12}{2} = \frac{1}{4},

so a single probe yields outcome 1 with probability $0.25$ and outcome 0 with probability $0.75$ . Repeating it $N$ times and inverting $p_1(\theta)$ is exactly the procedure whose error scales as $1/\sqrt{N}$ .

Try it

Build the single-qubit Ramsey probe with $\theta = \pi/3$ and measure. The grader checks the exact analytic distribution: $0.75$ for outcome $0$ and $0.25$ for outcome $1$ .

Run your code to see the quantum state.