GHZ vs W States

The two genuinely-tripartite SLOCC classes of three qubits — GHZ and W — entangle the same three particles in profoundly different ways. They differ in how robust their entanglement is to loss, in their bipartite reductions, and in the value of the tripartite invariant called the three-tangle.

The two states

|\mathrm{GHZ}\rangle = \tfrac{1}{\sqrt2}\big(|000\rangle + |111\rangle\big), \qquad |\mathrm{W}\rangle = \tfrac{1}{\sqrt3}\big(|001\rangle + |010\rangle + |100\rangle\big).

GHZ is a coherent superposition of "all up" and "all down." W is the equal superposition of all single-excitation states — exactly one qubit is $|1\rangle$ , but which one is undetermined.

Robustness to particle loss

The sharpest contrast appears when one party is traced out (lost). Discard qubit $C$ and look at the reduced state $\rho_{AB}$ .

For GHZ, tracing out $C$ gives a classically correlated mixture:

\rho_{AB}^{\mathrm{GHZ}} = \tfrac12\,|00\rangle\langle00| + \tfrac12\,|11\rangle\langle11|.

This is separable — its concurrence is $0$ . Losing one qubit of a GHZ state destroys all remaining entanglement. GHZ entanglement is maximally fragile and maximally "global."

For W, tracing out $C$ leaves an entangled two-qubit state:

\rho_{AB}^{\mathrm{W}} = \tfrac13\big(|01\rangle + |10\rangle\big)\big(\langle01| + \langle10|\big) + \tfrac13\,|00\rangle\langle00|,

whose concurrence is $C_{AB} = \tfrac23 \neq 0$ . W entanglement survives the loss of a party — it is distributed pairwise and robust.

The three-tangle

The three-tangle $\tau_{ABC}$ is a polynomial SLOCC invariant that quantifies the genuinely tripartite entanglement — the part shared by all three at once and not reducible to pairs. Via the Coffman–Kundu–Wootters relation it equals the residual in the monogamy inequality (next lesson):

\tau_{ABC} = C_{A(BC)}^2 - C_{AB}^2 - C_{AC}^2 \ \ge 0.

Its values cleanly separate the classes:

\tau_{\mathrm{GHZ}} = 1, \qquad \tau_{\mathrm{W}} = 0.

W has zero three-tangle: all of its entanglement is pairwise, none is irreducibly tripartite. GHZ saturates it: all of its entanglement is tripartite, with no pairwise concurrence ( $C_{AB}^{\mathrm{GHZ}} = 0$ , consistent with the separable reduction above).

Why no SLOCC bridge exists

Because $\tau$ is invariant (up to a positive factor) under invertible local operations, a state with $\tau = 1$ cannot be mapped to a state with $\tau = 0$ by any SLOCC protocol. The three-tangle is the witness of the GHZ/W dichotomy: it is the conserved quantity that forbids interconversion. This is the quantitative meaning of "three qubits can be entangled in two inequivalent ways."

Try it

Prepare the three-qubit GHZ state $\tfrac{1}{\sqrt2}(|000\rangle + |111\rangle)$ from $|000\rangle$ using a Hadamard on qubit 0 and two CNOTs (control 0, targets 1 and 2). Return the circuit.

Run your code to see the quantum state.