The Bell States

Two coins glued back-to-back would always land showing the same face — look at one and you instantly know the other. Entanglement gives a pair of qubits a correlation far stranger and stronger than any pair of coins, and there are four "purest" ways to entangle two qubits. These four states are so central that they carry a name: the Bell states (after physicist John Bell). Each one is a two-qubit state that cannot be pulled apart into "this qubit does x, that qubit does y" — the two qubits only have a joint identity.

Here they are, all four:

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

Notice the pattern in the names: the \Phi ("Phi") states are built from the matching basis states |00\rangle and |11\rangle; the \Psi ("Psi") states from the opposite ones |01\rangle and |10\rangle. The superscript \pm is simply the sign in the middle.

A perfect basis, maximally entangled

The four computational states |00\rangle, |01\rangle, |10\rangle, |11\rangle form the everyday basis of the two-qubit space — any two-qubit state is a combination of them. The four Bell states are another basis for exactly the same space: they are mutually orthogonal (any two are perpendicular, inner product 0), each is normalised (length 1), and together they span everything. This is the Bell basis.

What makes it special is that every basis vector is maximally entangled. In each Bell state the two possible outcomes are perfectly balanced (\big|\tfrac{1}{\sqrt2}\big|^2 = \tfrac12 each), and measuring one qubit fixes the other with total certainty. You cannot write any Bell state as a product |a\rangle \otimes |b\rangle of a state for qubit 1 and a state for qubit 2 — the entanglement is baked in as strongly as two qubits allow.

The Bell basis at a glance

Read the table down: the two \Phi states always give matching outcomes; the two \Psi states always give opposite ones. Reveal the rows one at a time.

How they are made: Hadamard, then CNOT

The Bell states are not exotic curiosities — they are what you get by running the two most basic entangling gates on ordinary inputs. Take any of the four computational inputs, apply a Hadamard gate to the first qubit (putting it into an equal superposition), then apply a CNOT gate (which flips the second qubit whenever the first is 1). The four inputs map to the four Bell states, one for one:

|00\rangle \mapsto |\Phi^+\rangle, \quad |01\rangle \mapsto |\Psi^+\rangle, \quad |10\rangle \mapsto |\Phi^-\rangle, \quad |11\rangle \mapsto |\Psi^-\rangle.

Because this "H then CNOT" circuit is reversible, running it backwards (CNOT, then Hadamard) turns any Bell state back into a plain computational state — that reverse move is how you measure in the Bell basis, the key step in the protocols below.

Correlation and phase: the two things that vary

Only two features separate the four states, and it helps to hold them apart:

Worked example: two Bell states are orthogonal

Let us verify that |\Phi^+\rangle and |\Phi^-\rangle are perpendicular. Their inner product expands as

\langle \Phi^+ | \Phi^- \rangle = \tfrac12\big(\langle 00| + \langle 11|\big)\big(|00\rangle - |11\rangle\big) = \tfrac12\big(\langle 00|00\rangle - \langle 00|11\rangle + \langle 11|00\rangle - \langle 11|11\rangle\big).

The computational states are orthonormal, so \langle 00|00\rangle = \langle 11|11\rangle = 1 while the cross terms \langle 00|11\rangle = \langle 11|00\rangle = 0. Hence

\langle \Phi^+ | \Phi^- \rangle = \tfrac12(1 - 0 + 0 - 1) = 0. \quad \checkmark

Perpendicular, as promised. The same bookkeeping gives 0 for every distinct pair and 1 for a state with itself — the definition of an orthonormal basis.

Worked example: reading off a correlation

Take |\Psi^+\rangle = \tfrac{1}{\sqrt2}(|01\rangle + |10\rangle) and measure both qubits in the computational basis. Only two outcomes have non-zero amplitude: 01 and 10, each with probability \big|\tfrac{1}{\sqrt2}\big|^2 = \tfrac12. The outcomes 00 and 11 can never happen.

So whenever the first qubit reads 0, the second must read 1, and vice versa — the results are always opposite. That is exactly what "anti-correlated" means, and it is why \Psi sits in the anti-correlated column of the table.

Why they matter

The Bell basis is not just tidy mathematics — it is the fuel for the two headline tricks of quantum communication. In quantum teleportation, a shared Bell pair lets you transmit an unknown qubit's exact state using only two classical bits. In superdense coding, the mirror image, a shared Bell pair lets you send two classical bits by physically transmitting just one qubit — because nudging your half of a Bell pair with one of four simple operations rotates the shared state into one of the four Bell states, and the receiver, measuring in the Bell basis, can tell which. Both protocols hinge on the fact that four perfectly distinguishable, maximally entangled states exist at all.

It is worth pausing on how special this is. For a single qubit you can find an orthonormal basis (say |0\rangle, |1\rangle), but its members are the least entangled states imaginable — they are not entangled at all. For two qubits, nature hands you a basis whose every member is entangled as hard as possible. The four Bell states are simultaneously the most correlated two-qubit states and a complete, mutually distinguishable set. That rare combination — total entanglement plus perfect distinguishability — is precisely what makes them the workhorses of quantum information.

Do not blur the two labels together. The letter (\Phi vs \Psi) tells you the correlation: \Phi is correlated (00/11), \Psi is anti-correlated (01/10). The sign (+ vs -) tells you the relative phase, which a single joint measurement in the computational basis cannot even see — |\Phi^+\rangle and |\Phi^-\rangle give identical 00/11 statistics. Correlation and phase are independent knobs; changing one never changes the other.