The Bell States

A Bell state is one of four specific two-qubit quantum states that are maximally entangled: knowing everything about the joint state tells you nothing about either qubit individually, yet the two qubits' measurement outcomes are perfectly correlated. The four Bell states are

Each is a superposition of exactly two of the four computational basis states $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, with equal magnitude amplitudes $1/\sqrt{2}$ and a relative sign or bit pattern distinguishing the four cases. Together they form the Bell basis, an orthonormal basis for the two-qubit space just as valid a coordinate system as the computational basis $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$. The defining property is entanglement: none of the four states can be written as $|\psi_1\rangle \otimes |\psi_2\rangle$ for any single-qubit states $|\psi_1\rangle, |\psi_2\rangle$. This is not a technicality — it is the entire reason these states are interesting. A separable state lets you describe each qubit on its own; a Bell state does not. Formally, a two-qubit pure state is entangled exactly when its Schmidt rank — the number of nonzero terms in its Schmidt decomposition — exceeds one. Every Bell state has Schmidt rank two, the maximum possible for two qubits, which is why they are called maximally entangled.

The standard circuit for producing $|\Phi^+\rangle$ from the fixed input $|00\rangle$ uses exactly two gates from Two-Qubit Gates and the CNOT: a Hadamard on the first qubit followed by a CNOT controlled by the first qubit acting on the second. Trace the state through the circuit. Starting from $|00\rangle$, applying $H$ to qubit 1 gives

This intermediate state is still separable — qubit 1 is in superposition, qubit 2 is definitely $|0\rangle$, and they are simply independent. The entangling step is the CNOT: it flips qubit 2 whenever qubit 1 is $|1\rangle$, and leaves qubit 2 alone otherwise. Applied to the superposition above,

The CNOT correlates the two branches of the superposition: the branch where qubit 1 is $|0\rangle$ now has qubit 2 also $|0\rangle$, and the branch where qubit 1 is $|1\rangle$ now has qubit 2 also $|1\rangle$. No local operation on qubit 1 alone, and no local operation on qubit 2 alone, could have produced this correlation — it requires a genuine two-qubit interaction, which is exactly what CNOT provides and a single-qubit gate cannot. The other three Bell states are reached by preparing the input differently or by applying a local Pauli gate after producing $|\Phi^+\rangle$: ``` procedure MAKE_BELL_STATE(a, b): // a, b ∈ {0,1} select which of the four Bell states to build input: two qubits initialized to |0⟩ ⊗ |0⟩ if a = 1: apply X to qubit 1 // pre-flip: changes |00⟩ input to |10⟩ apply H to qubit 1 // create local superposition apply CNOT(control=1, target=2) // entangle if b = 1: apply Z to qubit 1 // flip relative sign of the superposition branches return the two-qubit state // this is |Φ+⟩, |Ψ+⟩, |Φ-⟩, or |Ψ-⟩ depending on (a,b) ``` Running this with $(a,b) = (0,0)$ reproduces the derivation above and yields $|\Phi^+\rangle$. Setting $a=1$ starts from $|10\rangle$ instead of $|00\rangle$; the Hadamard and CNOT then correlate qubit 2 with the flipped value of qubit 1, producing $|\Psi^+\rangle = (|01\rangle + |10\rangle)/\sqrt{2}$. Setting $b=1$ instead inserts a $Z$ gate after the Hadamard, which flips the sign of the $|1\rangle$ branch of qubit 1's superposition before the CNOT correlates it, producing the states with a minus sign, $|\Phi^-\rangle$ and $|\Psi^-\rangle$. An equivalent and often more useful way to reach all four states is to fix the circuit (Hadamard then CNOT, unchanged) and instead apply one of the four single-qubit Pauli operators $\{I, X, Z, XZ\}$ to qubit 1 of an already-made $|\Phi^+\rangle$:

This fact — that any Bell state is reachable from any other by a local single-qubit operation on just one half of the pair — is the basis of both quantum teleportation and superdense coding, and it is the reason the Bell states are said to differ only by a "local Pauli frame."

Circuit cost. Preparing any Bell state costs exactly one single-qubit gate ($H$, and optionally a Pauli) plus one two-qubit gate (CNOT): $O(1)$ gates, independent of any problem size, since a Bell state is a fixed object on two qubits. This is the cheapest possible entangling circuit — you cannot produce an entangled two-qubit state with zero two-qubit gates, since any circuit built entirely from single-qubit gates applied to a product input state must return a product state (a tensor product of single-qubit unitaries maps $|\psi_1\rangle \otimes |\psi_2\rangle$ to $(U_1|\psi_1\rangle) \otimes (U_2|\psi_2\rangle)$, which is separable by construction). So one two-qubit gate is not just sufficient but necessary.

Orthonormality. The four Bell states are pairwise orthogonal and each has norm 1, which is checked directly by computing inner products term by term; since the computational basis states are orthonormal, e.g. $\langle\Phi^+|\Phi^-\rangle = \frac{1}{2}(\langle00|-\langle11|)$ paired against $\frac{1}{2}(|00\rangle+|11\rangle)$ gives $\frac12(1 - 1) = 0$, and all other cross terms vanish similarly because the two states involved never share a nonzero overlap in more than one computational basis term with matching sign structure. Since they are four mutually orthogonal unit vectors in a four-dimensional space, they form a complete orthonormal basis — the Bell basis — and any two-qubit state can be written as a unique linear combination of them.

Entanglement is maximal, and provably so. For a bipartite pure state, entanglement is measured by the Schmidt decomposition: writing the state as $\sum_i \lambda_i |a_i\rangle \otimes |b_i\rangle$ with $\{|a_i\rangle\}, \{|b_i\rangle\}$ orthonormal and $\lambda_i \geq 0$. For $|\Phi^+\rangle = \frac{1}{\sqrt2}|00\rangle + \frac{1}{\sqrt2}|11\rangle$, this expansion is already in Schmidt form with $\lambda_1 = \lambda_2 = 1/\sqrt2$. A state is separable exactly when it has Schmidt rank 1 (a single nonzero $\lambda_i$); here there are two equal, nonzero Schmidt coefficients, which is the signature of maximal entanglement for a two-level system — no other normalized two-qubit state has more spread-out Schmidt coefficients than $(1/\sqrt2, 1/\sqrt2)$. The direct proof that $|\Phi^+\rangle$ is not separable: suppose for contradiction $|\Phi^+\rangle = (\alpha|0\rangle + \beta|1\rangle) \otimes (\gamma|0\rangle + \delta|1\rangle)$ for some complex $\alpha,\beta,\gamma,\delta$. Expanding gives $\alpha\gamma\,|00\rangle + \alpha\delta\,|01\rangle + \beta\gamma\,|10\rangle + \beta\delta\,|11\rangle$. Matching coefficients against $|\Phi^+\rangle = \frac{1}{\sqrt2}|00\rangle + 0\cdot|01\rangle + 0\cdot|10\rangle + \frac{1}{\sqrt2}|11\rangle$ requires $\alpha\delta = 0$ and $\beta\gamma = 0$ while $\alpha\gamma \neq 0$ and $\beta\delta \neq 0$. But $\alpha\gamma \neq 0$ forces both $\alpha \neq 0$ and $\gamma \neq 0$; then $\alpha\delta = 0$ forces $\delta = 0$, which contradicts $\beta\delta \neq 0$. No solution exists, so $|\Phi^+\rangle$ cannot be factored — it is entangled. The same argument applies to the other three Bell states by symmetry.

Measurement correlations. Measuring both qubits of $|\Phi^+\rangle$ in the computational basis: the state has amplitude $1/\sqrt2$ on $|00\rangle$ and $1/\sqrt2$ on $|11\rangle$ and zero amplitude elsewhere, so by the Born rule the outcome is $00$ with probability $1/2$ and $11$ with probability $1/2$, and the outcomes $01$ or $10$ never occur. The two measured bits are therefore always equal — perfectly correlated — even though each qubit alone, if measured first and in isolation, gives a uniformly random bit. This is the precise, provable content of "entangled but locally featureless": the marginal outcome distribution on either qubit is uniform (maximum uncertainty), yet the joint distribution is perfectly deterministic once you know one outcome. The same reasoning for $|\Psi^+\rangle = (|01\rangle+|10\rangle)/\sqrt2$ gives outcomes $01$ or $10$ each with probability $1/2$, and never $00$ or $11$ — the two bits are always opposite. The states with the minus sign, $|\Phi^-\rangle$ and $|\Psi^-\rangle$, give identical computational-basis statistics to their plus-sign counterparts, because the relative sign only affects the phase of the superposition, and the Born rule depends on amplitude magnitudes, not phases, when measuring in the basis that the state is already diagonal-ish in. The sign becomes visible only when you measure in a different basis (for example the Hadamard basis), which is exactly how these four otherwise similar-looking states are told apart in a real protocol.

No signaling. A subtlety worth being explicit about: this perfect correlation does not allow communication. Whoever holds qubit 2 sees a uniformly random bit regardless of what basis, or whether at all, the distant party measures qubit 1 in — the marginal probability distribution on qubit 2 alone is unaffected by any operation performed only on qubit 1. Correlation is revealed only when the two measurement records are later compared, which requires an ordinary classical channel. This is why Bell states enable protocols like teleportation only together with a classical communication step, never instead of one.

Trace $|\Phi^+\rangle$ and $|\Psi^-\rangle$ through the same measurement scenario to see how the sign and bit pattern manifest concretely.

Setup. Alice and Bob share one qubit each of a Bell pair. Both measure in the computational basis $\{|0\rangle, |1\rangle\}$.

Now push further and distinguish $|\Phi^+\rangle$ from $|\Phi^-\rangle$, which the table shows are indistinguishable under computational-basis measurement alone. Apply a Hadamard to both qubits before measuring. Recall $H|0\rangle = (|0\rangle+|1\rangle)/\sqrt2$ and $H|1\rangle = (|0\rangle - |1\rangle)/\sqrt2$. Expanding $H \otimes H$ acting on $|\Phi^+\rangle$:

Expanding each product of superpositions and collecting terms, the cross terms with an odd number of $|1\rangle$s cancel, leaving

$|\Phi^+\rangle$ is unchanged — it is an eigenstate of $H \otimes H$. Repeating the same expansion for $|\Phi^-\rangle = (|00\rangle - |11\rangle)/\sqrt2$, the surviving terms now come with a relative minus sign that turns into the $|01\rangle, |10\rangle$ terms rather than canceling them:

So after applying $H$ to both qubits and measuring in the computational basis: if the state was $|\Phi^+\rangle$, Alice and Bob see matching bits ($00$ or $11$); if it was $|\Phi^-\rangle$, they see mismatched bits ($01$ or $10$). The change-of-basis measurement, unavailable in the first table, cleanly separates the two states that computational-basis measurement alone could not tell apart — exactly the mechanism the Bell measurement circuit (CNOT followed by $H$ on the control qubit, then measure both) exploits to identify which of the four Bell states a pair is in, a step at the heart of quantum teleportation.

Bell's 1964 theorem grew out of the 1935 Einstein–Podolsky–Rosen paper, which argued that quantum mechanics must be incomplete because it permitted such strange correlations at a distance. Bell showed the opposite conclusion follows from taking the correlations seriously: no theory obeying locality and realism as EPR assumed them can reproduce the statistics that entangled states like the Bell states actually predict. The states now bear his name because they are the simplest, most extreme examples that saturate this gap between quantum and classical correlation.

Quantum teleportation consumes exactly one Bell pair per qubit teleported: Alice performs a joint Bell measurement on her half of the pair and the qubit she wants to send, then transmits two classical bits telling Bob which of four Pauli corrections to apply to his half. Superdense coding runs the transaction in reverse, letting Alice send two classical bits by manipulating only her half of a pre-shared Bell pair and sending a single qubit. Entanglement swapping chains Bell pairs together across intermediate nodes, the basic primitive proposed for extending a Entanglement and Correlated Measurement-based quantum internet beyond direct line-of-sight fiber links, since it lets two parties who never shared a direct photon link end up holding an entangled pair. Experimental quantum key distribution protocols such as E91 use Bell-inequality violation itself as the security guarantee: an eavesdropper who tries to intercept and resend qubits necessarily degrades the correlation below the quantum-mechanical bound, and comparing the observed correlation against the CHSH bound catches the intrusion. Photonic Bell-pair sources (via spontaneous parametric down-conversion) and superconducting-qubit Bell-pair generation (via tunable couplers implementing an effective CNOT) are both standard calibration benchmarks for quantum hardware — a device's ability to produce and verify a high-fidelity Bell state is one of the first tests run on any new quantum processor. The Bell states are also the simplest system on which to introduce density matrices, since asking "what does Bob's qubit alone look like, given that he has no access to Alice's half" cannot be answered with a state vector at all — that question is exactly what motivates Density Matrices: An Introduction.

Ask the tutor: How does the Bell measurement circuit (CNOT then Hadamard, then measuring both qubits) actually distinguish all four Bell states in one shot?