Case Study: Diagnosing a GHZ State That Isn't

Executive Summary

A team reports preparing a 5-qubit GHZ state, $\frac{1}{\sqrt2}(|00000\rangle + |11111\rangle)$, and offers as evidence a histogram showing 47% 00000, 44% 11111, and 9% scattered elsewhere. It looks convincing.

It is not evidence of entanglement at all. A classical machine that flips one fair coin and copies the result to five output bits produces exactly that histogram. This case study shows why the computational-basis histogram is the single most over-interpreted plot in quantum computing, and builds the measurement that actually distinguishes a GHZ state from a correlated classical mixture: multiple-quantum coherence, or its cheaper cousin, the parity oscillation.

Skills applied

  • Distinguishing a coherent superposition from a classical mixture (§5.4, §5.7).
  • Designing a phase-sensitive measurement (§5.8).
  • Extracting GHZ fidelity from populations plus coherence (§5.12).
  • Reasoning about how entanglement fidelity degrades with qubit number.

Background

The imposter

Consider two states of five qubits:

$$|\text{GHZ}\rangle = \tfrac{1}{\sqrt2}\big(|00000\rangle + |11111\rangle\big) \qquad\text{vs.}\qquad \rho_{\text{mix}} = \tfrac12|00000\rangle\langle00000| + \tfrac12|11111\rangle\langle11111|$$

The first is maximally entangled. The second is a classical coin flip with no entanglement whatsoever — preparable with a random number generator and five $X$ gates.

In the computational basis they are statistically identical. Both give 50% 00000 and 50% 11111. Every measurement in the $Z$ basis, on any subset of qubits, returns the same distribution.

The difference is the relative phase between the two branches, which exists only in the superposition. As Chapter 2 established, phase is invisible to $Z$-basis measurement and visible in every other basis. So we need a different basis.

Phase 1: The parity oscillation

Apply a phase rotation $R_z(\phi)$ to every qubit, then rotate into the $X$ basis and measure the parity $\langle X^{\otimes 5}\rangle$ — the expectation of the product of all outcomes.

For the GHZ state, the two branches accumulate a relative phase of $5\phi$ (each of the five qubits contributes $\phi$), giving

$$P(\phi) = \langle \text{parity}\rangle = \cos(5\phi)$$

For the classical mixture, there is no relative phase to accumulate — each branch is a separate classical possibility, not a superposition — so

$$P(\phi) = 0 \quad \text{for all } \phi$$

An oscillating parity signal is the entanglement. Its amplitude is the coherence.

from qiskit import QuantumCircuit
import numpy as np

def ghz_parity(n, phi):
    qc = QuantumCircuit(n, n)
    qc.h(0)
    for q in range(n - 1):
        qc.cx(q, q + 1)                  # GHZ
    for q in range(n):
        qc.rz(phi, q)                    # phase kick
        qc.h(q)                          # rotate to X basis
    qc.measure(range(n), range(n))
    return qc

def parity(counts):
    total = sum(counts.values())
    return sum(((-1) ** bits.count('1')) * c for bits, c in counts.items()) / total

Note the frequency: the signal oscillates at $n\phi$, not $\phi$. A 5-qubit GHZ state accumulates phase five times faster than a single qubit — this is the metrological advantage behind quantum sensing, and it is also why GHZ states are five times more sensitive to dephasing noise.

Phase 2: The data

Sweeping $\phi$ from $0$ to $2\pi/5$ in 20 steps, 4,096 shots each, the team's device gives a clean cosine at the right frequency with amplitude

$$C = 0.62$$

against an ideal amplitude of 1.0.

Finding 1. The state is entangled — a classical mixture would have given a flat line at zero, and 0.62 is nowhere near zero. The original histogram simply could not have told them this.

Phase 3: Fidelity from populations and coherence

GHZ fidelity decomposes into exactly two measurable pieces:

$$F = \frac{1}{2}\big(P_{00000} + P_{11111}\big) + \frac{1}{2}C$$

where the first term is the population (from the $Z$-basis histogram) and $C$ is the coherence (from the parity oscillation amplitude).

From the reported histogram: $P_{00000} + P_{11111} = 0.47 + 0.44 = 0.91$. From the sweep: $C = 0.62$.

$$F = \tfrac12(0.91) + \tfrac12(0.62) = 0.455 + 0.310 = 0.765$$

The genuine GHZ fidelity is about 76.5%, not the ~91% the histogram implied.

Finding 2. The histogram over-reports fidelity by 15 percentage points, because it measures only half of what fidelity is made of. The populations were nearly fine; the coherence was the problem.

The entanglement threshold

For an $n$-qubit GHZ state, genuine multipartite entanglement is certified when $F > 1/2$. At $F = 0.765$ the claim of a genuine 5-qubit GHZ state is sound — but note how much of the margin was consumed by coherence loss, and that a coherence of $C < 0.18$ would have dropped fidelity below the threshold entirely while leaving the histogram looking essentially unchanged.

Phase 4: Why coherence dies faster than population

Population errors require a bit flip. Coherence requires only a phase flip — and phase errors are both more common and more damaging in a GHZ state, because any one of the $n$ qubits dephasing destroys the global superposition.

If each qubit retains coherence $e^{-t/T_2}$, the GHZ state retains

$$C \approx e^{-nt/T_2}$$

The decay is $n$ times faster than for a single qubit. This is the central practical fact about large entangled states:

$n$ Coherence at $t = 0.1\,T_2$
2 0.82
5 0.61
10 0.37
20 0.14
50 0.007

Observed $C = 0.62$ at $n = 5$ corresponds to $t \approx 0.1\,T_2$ — entirely consistent with the circuit's duration. The device is behaving as expected; the ambition was simply larger than the coherence budget.

Phase 5: The recommendation

To the team: report $F = 0.765 \pm$ (error bar), stating both components. Publish the parity sweep, not the histogram — the histogram is compatible with a classical device and a reviewer will say so.

On scaling: the linear-in-$n$ coherence decay means the GHZ preparation circuit's depth matters as much as its width. Their circuit uses a linear CNOT chain of depth $n-1$; a balanced binary tree achieves depth $\lceil\log_2 n\rceil$, cutting the exposure time substantially and typically buying several percentage points of coherence at $n = 5$ and much more at $n = 20$.

Discussion Questions

  1. Why is the computational-basis histogram identical for the GHZ state and the classical mixture, and what general lesson does that carry for validating quantum states?
  2. The parity signal oscillates at $n\phi$. Explain both consequences — enhanced sensing precision and enhanced noise sensitivity — from the same fact.
  3. The fidelity formula splits into population and coherence. Construct a state with perfect population and zero coherence, and one with the reverse.
  4. Why does the entanglement threshold sit at $F > 1/2$ specifically?

Your Turn: Extensions

  • Implement the sweep for $n = 2, 3, 4, 5$ and plot $C$ against $n$; fit the exponential and extract an effective $T_2$.
  • Rebuild the GHZ preparation as a binary tree and compare measured coherence against the linear chain at equal $n$.
  • Simulate with a dephasing noise model and confirm the $e^{-nt/T_2}$ scaling.
  • Look up the multiple-quantum-coherence protocol used for larger GHZ certifications and explain how it generalizes the parity sweep.

Key Takeaways

  • A computational-basis histogram cannot distinguish a GHZ state from a classical coin flip. Publishing one as evidence of entanglement is a category error.
  • Fidelity = ½(population) + ½(coherence). Measuring only the population systematically over-reports.
  • Coherence is measured by a phase sweep; an $n$-qubit GHZ state oscillates $n$ times faster than a single qubit.
  • That same factor of $n$ makes GHZ coherence decay $n$ times faster — the reason large cat states are hard.
  • Shallower preparation circuits buy coherence; depth is as important as width for entangled state preparation.