Entanglement-Enhanced Sensing

The standard quantum limit assumed independent probes. Entanglement breaks that assumption. By correlating $N$ qubits into a GHZ state, we make them accumulate phase collectively and beat the $1/\sqrt{N}$ scaling — the central result that motivates this whole module.

The GHZ probe

The Greenberger–Horne–Zeilinger state on $N$ qubits is the maximal superposition of "all zeros" and "all ones":

|\mathrm{GHZ}_N\rangle = \frac{|0\rangle^{\otimes N} + |1\rangle^{\otimes N}}{\sqrt 2}.

It is built with one Hadamard and $N-1$ CNOTs: the Hadamard creates a superposition on the first qubit, and each CNOT copies that bit's value into another qubit, locking them together.

Why it senses $N$ times faster

Apply the same phase $\theta$ to every qubit, $U(\theta) = \prod_k e^{-i\theta Z_k/2}$ . On the two GHZ branches the phases add up:

|\mathrm{GHZ}_N\rangle \xrightarrow{U(\theta)} \frac{e^{-iN\theta/2}|0\rangle^{\otimes N} + e^{+iN\theta/2}|1\rangle^{\otimes N}}{\sqrt 2}.

The relative phase between the two branches is $N\theta$ — $N$ times larger than the single-qubit phase. The GHZ state oscillates $N$ times faster as $\theta$ changes, so the same change in $\theta$ is far easier to detect. This is sometimes called phase super-resolution.

The Fisher information jumps to $N^2$

Recall $F_Q = 4\,\mathrm{Var}(G)$ with collective generator $G = \tfrac12\sum_k Z_k$ . For the GHZ state the two branches are eigenstates of $G$ with eigenvalues $\pm N/2$ , so $\langle G\rangle = 0$ and $\langle G^2\rangle = N^2/4$ , giving

F_Q^{\mathrm{GHZ}} = 4 \cdot \frac{N^2}{4} = N^2.

Compare with $F_Q = N$ for $N$ independent probes. Through the Cramér–Rao bound this turns into

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

a $1/N$ scaling that beats the SQL's $1/\sqrt N$ by a factor of $\sqrt N$ .

Reading out a GHZ phase

To convert the amplified phase into counts, the readout mirrors the preparation in reverse (disentangle with CNOTs, then a Hadamard) followed by a measurement. The parity of the all-zeros versus all-ones outcome then oscillates as $\cos(N\theta)$ — $N$ fringes for every one fringe of a single qubit.

Try it

Build the 3-qubit GHZ state $(|000\rangle + |111\rangle)/\sqrt2$ with a Hadamard and two CNOTs. The grader compares the full state vector, so only the correct entangled superposition will pass.

Run your code to see the quantum state.