Entanglement and Protocols

Entanglement is a property of a composite quantum state that cannot be represented as a mixture of product states. For a pure entangled state, the joint state can be pure while each subsystem has a mixed reduced state. This page defines pure-state entanglement, catalogues standard entangled states, and works through teleportation, superdense coding and the Bell test.

Each circuit here is small enough to build in the Quantum Simulator .

Product states and entangled states

A two‑qubit state is a product state when it factorises,

|\psi\rangle = |\psi_A\rangle \otimes |\psi_B\rangle ,

and entangled when no such factorisation exists. Writing the state as $\alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$ , it factorises exactly when

\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = 0 .

For $|\Phi^+\rangle = (|00\rangle+|11\rangle)/\sqrt{2}$ this determinant is $1/2$ , not zero, so no pair of single‑qubit states can reproduce it.

For a bipartite pure state, tracing out one subsystem gives a useful test: the state is entangled exactly when either reduced state is mixed. For mixed joint states, mixed marginals do not by themselves establish entanglement, and separability requires a more general test.

The Bell basis

The four Bell states are standard maximally entangled two‑qubit states, and they form an orthonormal basis. Other maximally entangled bases can be obtained from them with local unitaries.

|\Phi^{\pm}\rangle = \tfrac{1}{\sqrt2}\big(|00\rangle \pm |11\rangle\big), \qquad |\Psi^{\pm}\rangle = \tfrac{1}{\sqrt2}\big(|01\rangle \pm |10\rangle\big).

One circuit produces all four: a Hadamard on the first qubit followed by a CNOT, with the input basis state selecting which one:

Input	Output
$\lvert 00\rangle$	$\lvert \Phi^+\rangle$
$\lvert 01\rangle$	$\lvert \Psi^+\rangle$
$\lvert 10\rangle$	$\lvert \Phi^-\rangle$
$\lvert 11\rangle$	$\lvert \Psi^-\rangle$

Reading the table backwards is just as useful. Running the circuit in reverse, CNOT and then Hadamard, maps the Bell basis onto the computational basis. A subsequent computational-basis measurement therefore implements a Bell measurement. Teleportation and superdense coding both use this transformation.

For every one of the four, tracing out either qubit gives $I/2$ . Each half alone is an unbiased random bit; all the structure is in the correlation.

Three qubits: GHZ and W

With three qubits entanglement stops having a single form. Under invertible local operations, there are two inequivalent classes of genuine tripartite pure-state entanglement, represented by GHZ and W. Product and biseparable states form additional, nongenuine classes.

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

The GHZ state is the natural extension of the Bell circuit: apply H to qubit 0, CNOT from qubit 0 to qubit 1, and CNOT from qubit 1 to qubit 2. Preparing the W state requires a different, asymmetric circuit with carefully chosen rotations; unlike the GHZ state, it cannot be produced by simply extending the Bell-state CNOT chain.

They behave differently when a qubit is lost. Tracing out one qubit of a GHZ state leaves the other pair in the separable mixture $(|00\rangle\langle00|+|11\rangle\langle11|)/2$ . Tracing out one qubit of a W state leaves the other pair entangled. GHZ and W represent inequivalent classes of genuine three-qubit entanglement under invertible local operations.

The no-cloning theorem

\langle\psi|\varphi\rangle = \langle\psi|\varphi\rangle^2 ,

which forces $\langle\psi|\varphi\rangle$ to be $0$ or $1$ . Thus two distinct states cloned perfectly by the same operation must be orthogonal. A device can clone states drawn from a known orthogonal set, but no device can perfectly clone every unknown quantum state.

This theorem rules out copying an unknown state for repeated inspection and explains why quantum error correction cannot encode by making independent copies. The disturbance caused by measuring nonorthogonal states, together with no-cloning and other protocol assumptions, also underlies the security of quantum key distribution.

Quantum teleportation

Teleportation transfers an unknown qubit state from Alice’s system to Bob’s using one shared Bell pair and two classical bits. It does not move matter and does not permit faster-than-light communication. Before Bob receives the classical bits, his reduced state is $I/2$ , independent of the teleported state.

Alice holds the unknown $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ on qubit 0 and half of a Bell pair on qubit 1; Bob holds the other half on qubit 2.

Prepare $|\Phi^+\rangle$ on qubits 1 and 2 (H then CNOT).
Alice performs a Bell measurement on qubits 0 and 1: CNOT from qubit 0 to qubit 1, then H on qubit 0, then measure both. Call the results $m_0$ and $m_1$ .
Alice sends the two bits to Bob over a classical channel.
Bob applies $X^{m_1}$ followed by $Z^{m_0}$ to his qubit.

Bob’s qubit is then in the state $|\psi\rangle$ exactly. Each of Alice’s four outcomes occurs with probability $1/4$ regardless of $|\psi\rangle$ . Her two-bit outcome therefore reveals no information about the input state, and Bob cannot recover that state until the classical bits arrive.

Alice measures $m_0 m_1$	Bob’s qubit before correction	Bob applies
`00`	$\alpha\lvert 0\rangle + \beta\lvert 1\rangle$	$I$
`01`	$\alpha\lvert 1\rangle + \beta\lvert 0\rangle$	$X$
`10`	$\alpha\lvert 0\rangle - \beta\lvert 1\rangle$	$Z$
`11`	$\alpha\lvert 1\rangle - \beta\lvert 0\rangle$	$ZX$

Note what is not happening: no copy is made. Alice’s measurement destroys her copy of $|\psi\rangle$ in the same step that creates Bob’s, exactly as no‑cloning requires.

Superdense coding

Superdense coding is teleportation’s mirror image: two classical bits sent by transmitting one qubit, again using a shared Bell pair.

Alice and Bob start with $|\Phi^+\rangle$ . To send two bits, Alice applies one gate to her half and mails that single qubit to Bob:

Bits to send	Alice applies	Resulting pair	Bob measures
`00`	$I$	$\lvert \Phi^+\rangle$	`00`
`01`	$X$	$\lvert \Psi^+\rangle$	`01`
`10`	$Z$	$\lvert \Phi^-\rangle$	`10`
`11`	$XZ$	$\lvert \Psi^-\rangle$	`11`

Bob now holds both qubits, applies CNOT and then H to the first qubit, and measures both. Because the four Bell states are orthogonal, he recovers the two bits with certainty. This is an entanglement-assisted communication protocol: the pre-shared Bell pair is an additional resource, so transmitting two classical bits by sending Alice’s qubit does not contradict unassisted classical-capacity bounds.

Bell inequalities

Perfect correlation in one basis proves nothing; a shared coin flip does that. The quantum signature is that correlations persist across incompatible bases. For $|\Phi^+\rangle$ ,

\langle Z\otimes Z\rangle = +1, \qquad \langle X\otimes X\rangle = +1, \qquad \langle Y\otimes Y\rangle = -1,

all at once. The CHSH inequality turns this into a testable bound. Alice picks between measurement axes $a_0, a_1$ and Bob between $b_0, b_1$ , and one forms

S = \langle a_0 b_0\rangle + \langle a_0 b_1\rangle + \langle a_1 b_0\rangle - \langle a_1 b_1\rangle .

Any local hidden-variable theory, whether deterministic or stochastic, obeys $|S| \le 2$ . In the XZ plane, measure $|\Phi^+\rangle$ along $a_0=Z$ and $a_1=X$ for Alice and $b_0=(Z+X)/\sqrt2$ and $b_1=(Z-X)/\sqrt2$ for Bob. These choices give

S = 2\sqrt{2} \approx 2.828,

the maximum quantum value for CHSH, known as Tsirelson’s bound. Loophole-free Bell tests violate the local hidden-variable bound under standard assumptions such as independent measurement choices. They do not rule out every possible classical or deterministic description without those assumptions.

Even so, entanglement alone sends no signal. Alice’s local statistics are $I/2$ no matter what Bob does, so her measurement outcomes are unbiased random bits either way. Every protocol on this page needs its classical channel.

Next: Quantum Error Correction builds on entanglement to protect information, and Quantum Algorithms shows what it buys computationally.

Simulator Lab: Build and Test Entanglement Protocols

The Quantum Simulator is most effective here when the circuit is tested in stages. Save or export JSON at each checkpoint so that a wiring error in a later protocol step does not require rebuilding the shared resource.

Bell and GHZ checkpoints

Start with the Bell (2) preset. Run and record the two nonzero probabilities and both Bloch radii. Add H to q0 after the preset’s CNOT and run again. This final H alone does not undo the Bell preparation because the inverse Bell transform requires CNOT first and then H.

Reset, reload Bell (2), append CNOT on 0,1, then append H on q0. The circuit now reverses its own preparation and returns |00⟩. This is the same Bell-to-computational-basis conversion used in Bell measurement.

Next load GHZ (3). The expected distribution has only |000⟩ and |111⟩, each with probability $1/2$ . All three reduced Bloch radii vanish. That local picture resembles three maximally mixed qubits, so it is the restricted global probability support, together with knowledge that the circuit prepared a pure state, that establishes the GHZ structure.

Check Bell correlations in three bases

Computational-basis correlation alone can be reproduced by shared classical randomness. Use copies of the exported Bell circuit to test more than one basis:

Z basis: Run the Bell preset unchanged. Only 00 and 11 occur, so $\langle Z\otimes Z\rangle=+1$ .
X basis: Append H to both qubits. Again only 00 and 11 occur, now representing equal X-basis outcomes, so $\langle X\otimes X\rangle=+1$ .
Y basis: Append RZ(-\pi/2) and then H to each qubit. Only 01 and 10 occur, representing opposite Y-basis outcomes, so $\langle Y\otimes Y\rangle=-1$ .

For any two-qubit result table, calculate a same-minus-different correlator as

E = P(00)+P(11)-P(01)-P(10).

Use QSim’s probabilities rather than counts when verifying the ideal identities. Use counts only when studying the statistical uncertainty that a finite number of measurements would introduce. Reload the same Bell JSON before adding each basis rotation so the three tests differ only in their measurement basis.

Teleport a family of test states

Use three qubits and construct the protocol in the order below:

Prepare the input on q0. For a useful nontrivial test, add RY with angle $1.1$ and RZ with angle $0.7$ .
Prepare a Bell pair on q1,q2 with H on q1 and CNOT targets 1,2.
Apply CNOT targets 0,1, then H on q0.
Add conditional X on q2 with condition q1 = 1.
Add conditional Z on q2 with condition q0 = 1.

Run the circuit and compare the final q2 Bloch values with the values obtained by running only the input preparation on a single qubit. They should agree up to numerical precision. Comparing a generic state is stronger than testing only |0⟩, because a missing correction can accidentally pass a basis-state test.

The conditional controls are classical feed-forward and cause measurement branches. The simulator averages over those branches in the reported final state. Before the correction bits are used, Bob’s reduced state is maximally mixed; after both corrections, every branch gives the same state on q2.

Superdense coding as a four-case regression test

Load Bell (2), apply one of I, X, Z or X followed by Z to q0, and then append CNOT targets 0,1 followed by H on q0. Run all four cases. The displayed outputs should be 00, 10, 01 and 11, respectively. QSim displays q0 as the rightmost bit of a ket, so the displayed strings reverse the q0 q1 ordering used in the protocol table above.

Export each version as JSON with a descriptive filename. This produces a small regression suite: if a control/target convention is misunderstood, at least two of the four cases will decode incorrectly.