Entangled States

Build the Bell pair and its n-qubit cousin the GHZ state, and see why their correlations have no product-state factorization.

Background

An n-qubit pure state $ |\psi\rangle $ is separable across a bipartition $ A|B $ if it factors as $ |\psi_A\rangle \otimes |\psi_B\rangle $. Otherwise it is entangled across that cut. A state that is entangled across every non-trivial bipartition is called genuinely multipartite entangled. Entanglement is the resource that separates quantum computing from a probabilistic classical register: classical correlations always factor once you condition on enough shared randomness; quantum correlations do not.

The two-qubit Bell pair $ |\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt{2} $ is the oldest worked example. Einstein, Podolsky, and Rosen used it in 1935 to argue quantum mechanics had to be incomplete¹. Bell turned that argument around in 1964 by deriving inequalities that any local-hidden-variable theory must satisfy and that $ |\Phi^+\rangle $ violates². Greenberger, Horne, and Zeilinger generalized the argument in 1989 to three qubits with a state that violates local realism with certainty rather than statistically³. The n-qubit generalization of their state — the GHZ state — is the natural “fan” circuit built on top of a Bell pair.

Nielsen and Chuang cover this material in §1.3.6 and §2.3⁴.

The math

Write $ |0\rangle = \binom{1}{0} $ and $ |1\rangle = \binom{0}{1} $. The Hadamard and controlled-NOT matrices we need are

$$ H = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad \text{CNOT}_{0 \to 1}|q_1 q_0\rangle = |q_1 \oplus q_0,\ q_0\rangle. $$

Here qubit 0 is the control and qubit 1 the target; $ \oplus $ is XOR.

Step 1. Start in $ |00\rangle $ and apply $ H $ to qubit 0:

$$ H_0|00\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |10\rangle). $$

Step 2. Apply $ \text{CNOT}_{0 \to 1} $. It leaves $ |00\rangle $ alone and sends $ |10\rangle \mapsto |11\rangle $:

$$ |\Phi^+\rangle ;=; \text{CNOT}_{0 \to 1} H_0 |00\rangle ;=; \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle). $$

The four Bell states $ |\Phi^\pm\rangle = (|00\rangle \pm |11\rangle)/\sqrt{2} $, $ |\Psi^\pm\rangle = (|01\rangle \pm |10\rangle)/\sqrt{2} $ span the two-qubit Hilbert space and are all maximally entangled across the $ 0|1 $ cut. They can be reached from $ |\Phi^+\rangle $ by local Pauli operations: $ Z_0, X_0, X_0 Z_0 $ respectively.

No product factorization. Suppose $ |\Phi^+\rangle = (a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle) $. Expanding, the amplitudes on $ |00\rangle, |01\rangle, |10\rangle, |11\rangle $ are $ ac, ad, bc, bd $. Matching to $ (1/\sqrt{2}, 0, 0, 1/\sqrt{2}) $ forces $ ad = bc = 0 $ while $ ac = bd = 1/\sqrt{2} $. The first pair forces one of $ {a,d} $ and one of $ {b,c} $ to vanish; any such choice kills one of the non-zero amplitudes. Contradiction, so no factorization exists.

Partial trace. Trace qubit 1 out of $ \rho = |\Phi^+\rangle\langle\Phi^+| $:

$$ \rho_0 = \operatorname{Tr}_1 \rho = \tfrac{1}{2}\bigl(|0\rangle\langle 0| + |1\rangle\langle 1|\bigr) = \tfrac{I}{2}. $$

A single qubit described by $ I/2 $ is maximally mixed — the observer with access only to qubit 0 sees uniform random bits. Maximum local disorder plus a pure global state is the signature of maximal entanglement.

GHZ generalization. On n qubits, apply $ H $ to qubit 0 and then $ \text{CNOT}_{0 \to k} $ for $ k = 1, \dots, n-1 $. Each CNOT copies the branch on qubit 0 onto qubit $ k $ in the computational basis, so

$$ |\text{GHZ}_n\rangle ;=; \tfrac{1}{\sqrt{2}}\bigl(|0\rangle^{\otimes n} + |1\rangle^{\otimes n}\bigr). $$

Tracing out any $ n-1 $ qubits of $ |\text{GHZ}_n\rangle $ again gives $ I/2 $ on the remaining qubit, so every single qubit is maximally entangled with the rest. The same no-factorization argument as above rules out any $ A|B $ product form.

The circuit

Bell circuit

Two elements. The circuit JSON follows the Circuit JSON Conventions:

{
  "num_qubits": 2,
  "elements": [
    {"type": "gate", "gate": "H", "targets": [0]},
    {"type": "gate", "gate": "X", "targets": [1], "controls": [0], "control_configs": [true]}
  ]
}

The CNOT is encoded as an X gate with a control on qubit 0; control_configs: [true] means the control fires on $ |1\rangle $ (active-high). Full Bell circuit JSON.

GHZ-4 circuit

GHZ-4 reuses the same pattern with three controlled-X gates fanning out from qubit 0:

{
  "num_qubits": 4,
  "elements": [
    {"type": "gate", "gate": "H", "targets": [0]},
    {"type": "gate", "gate": "X", "targets": [1], "controls": [0], "control_configs": [true]},
    {"type": "gate", "gate": "X", "targets": [2], "controls": [0], "control_configs": [true]},
    {"type": "gate", "gate": "X", "targets": [3], "controls": [0], "control_configs": [true]}
  ]
}

Full GHZ-4 circuit JSON. The fan topology is visible in the SVG above: one Hadamard, then three controls all rooted at qubit 0.

Bit ordering callout. In yao-rs, qubit 0 is the most significant bit of the probability-array index: probabilities[k] is the probability of the basis state whose binary label, read with enough leading zeros to fill $ n $ bits, has $ q_0 $ at the left. See bit ordering for the full convention.

Running it

Quick run (after cargo install --path yao-cli, so yao is on your PATH):

yao example bell | yao simulate - | yao probs -

Expected output:

{
  "locs": null,
  "num_qubits": 2,
  "probabilities": [
    0.5000000000000001,
    0.0,
    0.0,
    0.5000000000000001
  ]
}

For GHZ-4 the pipeline is the same with --nqubits 4:

yao example ghz --nqubits 4 | yao simulate - | yao probs -

Expected output (16 entries; only indices 0 and 15 are non-zero):

{
  "locs": null,
  "num_qubits": 4,
  "probabilities": [
    0.5000000000000001, 0.0, 0.0, 0.0,
    0.0, 0.0, 0.0, 0.0,
    0.0, 0.0, 0.0, 0.0,
    0.0, 0.0, 0.0, 0.5000000000000001
  ]
}

Regenerating this page’s artifacts (SVG diagrams and probability plots), run from the repo root:

cargo build -p yao-cli --no-default-features
target/debug/yao example bell --json --output docs/src/examples/generated/circuits/bell.json
target/debug/yao visualize docs/src/examples/generated/circuits/bell.json --output docs/src/examples/generated/svg/bell.svg
target/debug/yao simulate docs/src/examples/generated/circuits/bell.json | target/debug/yao probs - > docs/src/examples/generated/results/bell-probs.json
target/debug/yao example ghz --nqubits 4 --json --output docs/src/examples/generated/circuits/ghz4.json
target/debug/yao visualize docs/src/examples/generated/circuits/ghz4.json --output docs/src/examples/generated/svg/ghz4.svg
target/debug/yao simulate docs/src/examples/generated/circuits/ghz4.json | target/debug/yao probs - > docs/src/examples/generated/results/ghz4-probs.json
python3 scripts/plot_cli_results.py docs/src/examples/generated/results docs/src/examples/generated/plots

Interpreting the result

Bell probabilities

The Bell result is probabilities = [0.5, 0, 0, 0.5]. Index 0 is $ |00\rangle $, index 3 is $ |11\rangle $; indices 1 and 2 — the $ |01\rangle $ and $ |10\rangle $ outcomes where the two qubits disagree — have probability zero. Measurements on the two qubits are perfectly correlated: whichever outcome qubit 0 reports, qubit 1 reports the same bit.

GHZ-4 probabilities

GHZ-4 has probabilities[0] = probabilities[15] = 0.5 and every one of the fourteen middle entries is exactly zero. Under the yao-rs convention qubit 0 is the MSB, so index 15 is binary $ 1111 $ read as $ |q_0 q_1 q_2 q_3\rangle = |1111\rangle $ — i.e. all four qubits in $ |1\rangle $. So on measurement the four qubits collapse together: all zeros or all ones, with equal probability and no intermediate outcomes.

The non-classical content is this: the agreement is there before measurement, not imposed by it. Any local-hidden-variable story — each qubit carries a pre-assigned bit that the measurement reveals — must commit to a joint distribution over the four classical bit-strings. Such distributions cannot reproduce the full set of correlations $ |\Phi^+\rangle $ produces across different measurement bases; Bell’s 1964 inequality is the quantitative form of this impossibility², and $ |\Phi^+\rangle $ violates it. GHZ sharpens the statement to an all-or-nothing contradiction³.

Variations & next steps

Change --nqubits on yao example ghz to produce $ |\text{GHZ}_n\rangle $ for any $ n \geq 2 $. With $ n = 2 $ you recover the Bell state.
See Quantum Fourier Transform for the superposition+phase pattern that underlies phase estimation, and Bernstein–Vazirani for a single-shot algorithm that reads a hidden bitstring using the same H-fan structure.
The W state $ (|100\ldots 0\rangle + |010\ldots 0\rangle + \dots + |0\ldots 01\rangle)/\sqrt{n} $ is another genuinely multipartite entangled family with correlations that survive tracing out a qubit (unlike GHZ, whose entanglement collapses into a classical mixture after one loss). It is not covered in this catalogue.

References

A. Einstein, B. Podolsky, and N. Rosen, “Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?”, Phys. Rev. 47, 777 (1935). ↩
J. S. Bell, “On the Einstein–Podolsky–Rosen paradox”, Physics 1, 195 (1964). ↩ ↩2
D. M. Greenberger, M. A. Horne, and A. Zeilinger, “Going beyond Bell’s theorem”, in Bell’s Theorem, Quantum Theory, and Conceptions of the Universe, ed. M. Kafatos (Kluwer, 1989). ↩ ↩2
M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th Anniversary Edition (Cambridge University Press, 2010), §1.3.6 (Bell states) and §2.3 (entanglement and mixed states). ↩

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