Entanglement

Every essay before this one fit on a single qubit. One Bloch sphere, two amplitudes, a handful of gates. That's been enough to cover superposition, phase, measurement, and rotation — most of what's surprising about a single quantum particle.

None of that prepares you for what happens when you have two.

Two qubits, calmly

A two-qubit register has four computational-basis states: |00⟩, |01⟩, |10⟩, |11⟩. We write the register |q1 q0⟩ — qubit 0 is the least-significant bit, matching Qiskit and the simulator under the hood here. So |10⟩ means "q₁ = 1, q₀ = 0", which is the only place that convention ever bites you and now we've named it.

A general two-qubit state is a unit vector with four complex components — one amplitude per basis state — instead of two. Most of the time, this is no big deal: if q₀ is in some state |ψ⟩ and q₁ is in some state |φ⟩, the joint state is just their tensor product |φ⟩ ⊗ |ψ⟩. You can read either qubit separately, you can act on either qubit separately, and the bars below behave exactly like two independent single-qubit bars pressed up against each other.

P(|0⟩) = …, P(|1⟩) = …

Try this: press H. Only q₀ moves — the H buttons here are wired to act on q₀. The bars split four ways into two equal pairs (because q₁ is still firmly |0⟩, only |00⟩ and |01⟩ light up). The left Bloch card sweeps to the equator. The right one stays at the north pole. Both cards still show r = 1: each qubit has a perfectly well-defined individual state. The joint state is a product, the two pictures add up to the whole story, life is fine.

Press Reset before reading on, so we're starting fresh from |00⟩.

Build the Bell pair

Now we apply two gates in sequence. First, H on q₀ — the move we just did, taking us to (|00⟩ + |01⟩)/√2. Then a CNOT with q₀ as the control and q₁ as the target. CNOT flips q₁ exactly when q₀ is |1⟩; it does nothing when q₀ is |0⟩. The result is the most famous two-qubit state in physics:

The bars now show exactly two outcomes, each at 50%: either both qubits read 0, or both read 1. Never one of each. That's already a striking correlation — but the bars alone don't tell you why it's quantum. The Bloch cards do.

Try this: reset, then press Apply H to q₀, then Apply CNOT(0, 1). Watch both Bloch arrows. They don't move to the equator. They don't pick up phase. They shrink to the origin, and the little amber "Entangled" badge lights up on both cards.

The arrow length r = √(x² + y² + z²) measures how much of the qubit's state can be captured by a single-qubit Bloch picture. r = 1 means a pure state — the arrow is the whole truth. r = 0 means the maximally mixed state I / 2: no preferred direction, no preferred basis. From the perspective of q₀ alone, after the H and the CNOT, every measurement outcome in every basis is 50/50. Q₀ looks like a fair coin.

That's not because the joint state is uncertain. The joint state (|00⟩ + |11⟩)/√2 is a perfectly pure, unit-norm vector — it's just not factorable. There is no |ψ⟩ and |φ⟩ such that |ψ⟩ ⊗ |φ⟩ equals the Bell pair. The information isn't distributed unevenly between q₀ and q₁; it's distributed in the correlations between them, in a place no per-qubit picture can reach.

Try this: with the Bell pair set up, press H on q₀ again. The bars rearrange — the individual outcomes shift — but watch the Bloch cards. Both arrows stay near the origin. You can rotate either qubit all you like; you can't unscramble what H + CNOT did. The only way to recover full single-qubit states is to either undo the CNOT, measure one qubit (which collapses the joint state and pulls the other into a definite outcome), or perform a joint operation that touches both qubits at once.

Why this matters

Entanglement is the resource that makes quantum computers interesting. Superdense coding, quantum teleportation, the speedup in Grover's and Shor's algorithms, the security argument behind BB84 — none of them work on product states. They all need at least one pair of qubits that can't be described separately.

The "Bloch arrow shrinks to the origin" picture is the cheapest, most honest single-qubit visualization of that fact, and it's the one the Sandbox uses on its results panel for any circuit you build. The math behind the shrinking arrow is the reduced density matrix: take the joint state, trace out every qubit except the one you care about, and you're left with a 2×2 matrix that captures everything a measurement on that single qubit could ever reveal. For the Bell pair, that matrix is I / 2 — maximally mixed. Hence the origin.

Self-test

1. After H on q₀ and CNOT(0, 1), what does the Bloch arrow for q₀ look like, and what does that not tell you about the joint state? (Press the buttons; then explain the picture to a friend.)
2. The probability bars after the Bell pair show only |00⟩ and |11⟩, each at 50%. If entanglement is a "shared fate," what would change if we'd built (|01⟩ + |10⟩)/√2 instead? (Hint: try X on q₁ first, then H + CNOT.)

Open the Bell pair in the sandbox →

Tensor products, partial traces, and why ρ_q₀ = I/2

Two single-qubit states |ψ⟩, |φ⟩ ∈ ℂ² combine into a two-qubit state by the tensor product:

$$|\psi\rangle \otimes |\varphi\rangle \;=\; \begin{bmatrix}\psi_0\varphi_0\\ \psi_0\varphi_1\\ \psi_1\varphi_0\\ \psi_1\varphi_1\end{bmatrix}$$

Every product state has this shape. A state that can't be written this way for any choice of |ψ⟩ and |φ⟩ is, by definition, entangled. The Bell pair is the canonical example — try to solve ψ₀φ₀ = ψ₁φ₁ = 1/√2 with ψ₀φ₁ = ψ₁φ₀ = 0 and you'll fail.

To get a single-qubit picture out of a joint state we take a partial trace. For qubit 0 of a two-qubit state |Ψ⟩:

$$\rho_{q_0} \;=\; \operatorname{Tr}_{q_1}\bigl(|\Psi\rangle\langle\Psi|\bigr) \;=\; \sum_{k \in \{0,1\}} \langle k|_{q_1}\, |\Psi\rangle\langle\Psi|\, |k\rangle_{q_1}$$

For the Bell pair, both terms contribute ½ |k⟩⟨k|, and the sum is the maximally mixed state:

$$\rho_{q_0} \;=\; \tfrac{1}{2}|0\rangle\langle 0| + \tfrac{1}{2}|1\rangle\langle 1| \;=\; \tfrac{1}{2} I$$

Bloch coordinates of a density matrix ρ = ½(I + r·σ) are x = 2 Re ρ₀₁, y = −2 Im ρ₀₁, z = ρ₀₀ − ρ₁₁. Plug in ρ = I/2 and you get (0, 0, 0) — the arrow at the origin.

One-liner the texts call Schmidt decomposition: every pure two-qubit state can be written as |Ψ⟩ = Σᵢ λᵢ |aᵢ⟩ ⊗ |bᵢ⟩ for some pair of orthonormal bases {|aᵢ⟩} on q₀ and {|bᵢ⟩} on q₁. The number of nonzero λᵢ is the Schmidt rank — 1 for product states, 2 for the Bell pair. That's the cleanest single number that tells you whether a state is entangled, and how much.