Quantum Fisher Information

How much information does a quantum probe carry about an unknown parameter $\theta$ ? The quantum Fisher information (QFI), $F_Q(\theta)$ , answers this with a single number. It is the quantum-mechanical ceiling on how sharply any measurement can resolve $\theta$ , and the next lesson turns it into a hard precision bound.

From states to distinguishability

Estimating $\theta$ is possible only because the state $\rho_\theta$ changes with $\theta$ . The faster nearby states $\rho_\theta$ and $\rho_{\theta + d\theta}$ pull apart — in the sense of being statistically distinguishable — the more precisely $\theta$ can be pinned down. Braunstein and Caves (1994) made this geometric idea exact: the rate of distinguishability is the QFI.

For a pure probe $|\psi_\theta\rangle$ there is a clean formula. Writing $|\partial_\theta\psi\rangle = \partial_\theta|\psi_\theta\rangle$ ,

F_Q(\theta) = 4\big(\langle\partial_\theta\psi|\partial_\theta\psi\rangle - |\langle\psi_\theta|\partial_\theta\psi\rangle|^2\big).

Unitary encoding: QFI is a variance

In sensing, $\theta$ enters through a unitary $U(\theta) = e^{-i\theta G}$ generated by a Hermitian operator $G$ , so $|\psi_\theta\rangle = e^{-i\theta G}|\psi_0\rangle$ . Substituting into the formula collapses it to a variance of the generator in the initial state:

F_Q = 4\,\mathrm{Var}_{\psi_0}(G) = 4\big(\langle G^2\rangle - \langle G\rangle^2\big).

The QFI does not depend on $\theta$ at all for unitary encoding — only on the probe state and the generator. The whole art of probe design is choosing $|\psi_0\rangle$ to maximise $\mathrm{Var}(G)$ .

The single-qubit case

Take the phase generator $G = Z/2$ , so $U(\theta) = e^{-i\theta Z/2} = R_Z(\theta)$ . Then

F_Q = 4\,\mathrm{Var}\!\left(\frac{Z}{2}\right) = \mathrm{Var}(Z) = \langle Z^2\rangle - \langle Z\rangle^2 = 1 - \langle Z\rangle^2,

using $Z^2 = I$ . The variance is maximal when $\langle Z\rangle = 0$ , which happens exactly on the equator of the Bloch sphere. The optimal probe is therefore $|+\rangle = H|0\rangle$ , whose Bloch vector is $(x, y, z) = (1, 0, 0)$ and which gives

F_Q^{\max} = 1.

Why the variance picture is powerful

The variance form $F_Q = 4\,\mathrm{Var}(G)$ generalises immediately to $N$ qubits. If the collective generator is $G = \tfrac12\sum_{k} Z_k$ , then for $N$ independent equatorial probes the variances add and $F_Q = N$ . But an entangled probe can make $\mathrm{Var}(G)$ as large as $N^2/4$ , giving $F_Q = N^2$ . That quadratic jump — visible already in this single formula — is the seed of the Heisenberg limit we reach later in the module.

Try it

Prepare the optimal single-qubit probe for the generator $G = Z/2$ . The grader checks the probe's Bloch vector: it must sit on the equator at $(1, 0, 0)$ , the configuration that maximises $\mathrm{Var}(Z)$ and hence the QFI.

Run your code to see the quantum state.