The Bell States

The state \tfrac{1}{\sqrt 2}(|00\rangle + |11\rangle) has appeared in almost every chapter in this part of the wiki. It is the canonical "this is what entanglement looks like" example — two qubits whose outcomes are perfectly correlated even though each qubit on its own is a 50/50 coin. Whenever we needed a concrete entangled state to illustrate a definition, a theorem, or a circuit, that one showed up.

It is time to tell you: that state has three siblings. Together they form the Bell states, named after John Bell, whose 1964 paper on Bell inequalities made these four states the headline objects of quantum foundations and, decades later, the standard currency of quantum communication. The four of them together span the two-qubit Hilbert space the same way |00\rangle, |01\rangle, |10\rangle, |11\rangle do — you can write any two-qubit state as a linear combination of Bell states instead of computational basis states, and for some problems the Bell-basis description is exactly the one you want.

By the end of this chapter you should be able to: write all four Bell states from memory, show they are pairwise orthogonal, derive |\Phi^+\rangle from |00\rangle by applying H and CNOT, construct the other three by adding X or Z at the start, and measure an unknown two-qubit state in the Bell basis by running the construction circuit in reverse.

The four states, written out

Here are all four, with their Dirac-notation form, their amplitude vector in the computational basis, and a short name you will see in papers.

Convention note. The column vector convention is (a_{00}, a_{01}, a_{10}, a_{11})^T — amplitude of |00\rangle first, then |01\rangle, then |10\rangle, then |11\rangle.

Why Phi vs Psi: the convention is that \Phi states are built from |00\rangle and |11\rangle (both qubits agree), and \Psi states are built from |01\rangle and |10\rangle (the qubits disagree). The + and - superscripts indicate the relative sign between the two terms.

Orthonormality — a one-pass check

Orthonormal means two things: each Bell state has unit norm (self-inner product 1), and every pair has inner product 0. Both are one-line checks using the column-vector forms above.

Self-inner products. Take |\Phi^+\rangle = \tfrac{1}{\sqrt 2}(|00\rangle + |11\rangle). Then

Why the cross terms vanish: \langle 00 | 11\rangle = 0 because the computational basis is orthonormal. Only \langle 00 | 00\rangle and \langle 11 | 11\rangle — both equal to 1 — survive.

The same calculation works for |\Phi^-\rangle, |\Psi^+\rangle, |\Psi^-\rangle. Each has self-inner product 1.

Pairwise orthogonality — \langle \Phi^+ | \Phi^-\rangle.

So |\Phi^+\rangle and |\Phi^-\rangle, which look like they differ "only" by a minus sign, are actually as orthogonal as |0\rangle and |1\rangle. This matters: the minus sign is a relative phase, which is a real, observable, and full-rank feature of a quantum state. States that differ by a global phase are the same physical state; states that differ by a relative phase (like |\Phi^+\rangle and |\Phi^-\rangle) are completely distinguishable by the right measurement.

Pairwise orthogonality — \langle \Phi^+ | \Psi^+\rangle.

Why every term is 0: no term in |\Phi^+\rangle shares a basis state with any term in |\Psi^+\rangle. The \Phi states live on \{|00\rangle, |11\rangle\} and the \Psi states live on \{|01\rangle, |10\rangle\} — orthogonal subspaces.

The remaining four pairs — \langle \Phi^+|\Psi^-\rangle, \langle \Phi^-|\Psi^+\rangle, \langle \Phi^-|\Psi^-\rangle, \langle \Psi^+|\Psi^-\rangle — all work out to 0 by the same arithmetic. You can verify them as a quick exercise or trust the pattern. Conclusion: the four Bell states form an orthonormal basis of the two-qubit Hilbert space.

Constructing |\Phi^+\rangle from |00\rangle

Preparing a Bell pair on real hardware is a two-gate routine. Start with both qubits in |0\rangle — the default init state of every real quantum computer. Apply a Hadamard to qubit 0 (the left qubit). Then apply a CNOT with qubit 0 as control and qubit 1 as target. The output is |\Phi^+\rangle.

The step-by-step derivation

Step 1 — start in |00\rangle. Both qubits in |0\rangle, joint state is |00\rangle, which as a column vector is (1, 0, 0, 0)^T.

Step 2 — apply H to qubit 0. The Hadamard acts only on qubit 0; qubit 1 is untouched. This is H \otimes I on the two-qubit system. Using H|0\rangle = \tfrac{1}{\sqrt 2}(|0\rangle + |1\rangle):

Why the tensor product expands this way: when an operator acts only on one qubit, you distribute it into that slot and leave the other alone. Then you expand the sum: (\alpha + \beta) \otimes \gamma = \alpha \otimes \gamma + \beta \otimes \gamma.

The intermediate state is a product state — qubit 0 is in |+\rangle and qubit 1 is in |0\rangle, and the joint state factors. It is not yet entangled.

Step 3 — apply CNOT. The CNOT flips qubit 1 whenever qubit 0 is |1\rangle. Applied term by term to the superposition \tfrac{1}{\sqrt 2}(|00\rangle + |10\rangle):

• The |00\rangle term: qubit 0 is 0, so nothing happens. Result: |00\rangle.
• The |10\rangle term: qubit 0 is 1, so qubit 1 flips from 0 to 1. Result: |11\rangle.

So

Why this state is now entangled: you cannot factor \tfrac{1}{\sqrt 2}(|00\rangle + |11\rangle) as a product (a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle) = ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle. You would need ac = bd = \tfrac{1}{\sqrt 2} and ad = bc = 0. The second pair forces one of a, d to be 0 and one of b, c to be 0, which then breaks the first pair. No solution exists. This is the two-line proof that |\Phi^+\rangle is entangled.

You now have a Bell pair. On an IBM Quantum machine, this exact two-gate sequence is the first genuine quantum-computing "hello world" every student runs, and the resulting fidelity of the prepared |\Phi^+\rangle (usually 95-99% on current hardware) is one of the standard benchmark numbers reported for every new chip.

The other three Bell states

The same construction, preceded by a single Pauli on one qubit, gives each of the other three Bell states. Four input choices — identity, X, Z, or XZ — give four Bell states.

Deriving the four variants

For |\Phi^-\rangle — add Z on qubit 0 before H. Start at |00\rangle. Apply Z on qubit 0: Z|0\rangle = |0\rangle, so the state is unchanged at |00\rangle.

Wait — that's not quite how to get it at the start. Apply Z after H instead. Better: you can obtain |\Phi^-\rangle from |\Phi^+\rangle by applying Z to either qubit afterwards: Z \otimes I \cdot |\Phi^+\rangle = \tfrac{1}{\sqrt 2}(|00\rangle - |11\rangle) = |\Phi^-\rangle, using Z|0\rangle = |0\rangle and Z|1\rangle = -|1\rangle. The Z on the first qubit flips the sign of the |11\rangle term (where qubit 0 is 1) and leaves the |00\rangle term alone.

Alternatively, apply X on qubit 0 before H (so the initial state becomes |10\rangle, then H takes qubit 0 to \tfrac{1}{\sqrt 2}(|0\rangle - |1\rangle) = |-\rangle), then CNOT. Let's trace: after X, state is |10\rangle. After H on qubit 0, state is \tfrac{1}{\sqrt 2}(|00\rangle - |10\rangle). After CNOT, the |00\rangle is unchanged and |10\rangle flips to |11\rangle, giving \tfrac{1}{\sqrt 2}(|00\rangle - |11\rangle) = |\Phi^-\rangle.

Why both recipes work: X before H on qubit 0 is equivalent to H \cdot X = Z \cdot H (using the circuit identity HXH = Z), so pre-X plus H is the same as post-Z on the superposition side. Either way, you flip the sign on the |11\rangle amplitude of |\Phi^+\rangle.

For |\Psi^+\rangle — add X on qubit 1 before H+CNOT. Start at |00\rangle. Apply X on qubit 1: the state becomes |01\rangle. Apply H on qubit 0: \tfrac{1}{\sqrt 2}(|0\rangle + |1\rangle) \otimes |1\rangle = \tfrac{1}{\sqrt 2}(|01\rangle + |11\rangle). Apply CNOT: |01\rangle leaves qubit 1 alone (since qubit 0 is 0), and |11\rangle flips qubit 1 to |10\rangle. Result: \tfrac{1}{\sqrt 2}(|01\rangle + |10\rangle) = |\Psi^+\rangle.

For |\Psi^-\rangle — add both X (qubit 1) and Z (qubit 0) before H+CNOT. You can also get |\Psi^-\rangle by applying Z to one qubit of |\Psi^+\rangle: Z \otimes I \cdot \tfrac{1}{\sqrt 2}(|01\rangle + |10\rangle) = \tfrac{1}{\sqrt 2}(|01\rangle - |10\rangle) — the Z flips the sign on whichever terms have qubit 0 = 1, which in this case is just |10\rangle.

The Pauli-to-Bell-state correspondence

There is a compact way to say all of this at once. Write the four Bell states as

With (i, j) = (0, 0) you get |\Phi^+\rangle; (1, 0) gives |\Phi^-\rangle; (0, 1) gives |\Psi^+\rangle; (1, 1) gives |\Psi^-\rangle (up to an overall sign). A single Pauli on one qubit moves you between the four Bell states. This two-bit labelling (i, j) is exactly what Bell measurement will produce as its classical output — the measurement reads off i and j and tells you which of the four Bell states was detected.

Bell measurement — running the circuit backwards

Suppose someone hands you a two-qubit system in an unknown Bell state and asks you to identify which one it is. You cannot just measure in the computational basis — that tells you about |00\rangle, |01\rangle, |10\rangle, |11\rangle, which are not Bell states. You have to measure in the Bell basis.

The trick: run the Bell-construction circuit in reverse, then measure in the computational basis. The reverse of "H on qubit 0, then CNOT" is "CNOT, then H on qubit 0" — because both H and CNOT are their own inverses, so running them in reverse order undoes what they did.

The outcome table

Apply CNOT then H to each of the four Bell states and see what computational-basis state you land on. The construction was |00\rangle \to H \to \text{CNOT} \to |\Phi^+\rangle. Running backwards: |\Phi^+\rangle \to \text{CNOT} \to H \to |00\rangle. Analogously for the others:

Why each Bell state maps to a unique computational-basis state: the circuit CNOT then H is unitary and maps the Bell basis (four orthonormal states) to the computational basis (four orthonormal states). A unitary between two orthonormal bases is a one-to-one correspondence.

So a Bell measurement yields two classical bits (m_0, m_1) that uniquely identify which of the four Bell states was present. Because Bell states are orthogonal, the measurement is deterministic on pure Bell-state inputs — the same input always gives the same output pair.

Why Bell measurement is "non-local"

Here is the wild part. The two qubits of a Bell pair can be anywhere in the universe — qubit 0 in Mumbai and qubit 1 in Chennai. Local measurements on each qubit alone give you random classical bits with no correlation to the other side. To extract the full Bell-state identity you need to bring the two qubits together (or use a protocol that effectively does so, like entanglement swapping) so that the CNOT can fire across them. The Bell measurement is an entangling operation, and it cannot be done with separate local gates on the two qubits.

This fact — that Bell-basis information is globally distributed across a Bell pair and cannot be accessed locally — is the structural reason entanglement is a resource. It is also why you have met |\Psi^-\rangle (the singlet) as the state that maximally violates the Bell inequality: its correlations between local measurements are stronger than any classical model can reproduce, and the only way to see the violation is to collect local outcomes from both sides and compare them afterwards.

Applications — a preview

The Bell states are not academic exotica. They are the working currency of quantum information:

• Quantum teleportation (forthcoming in Part 7). You want to send a qubit's state from Alice to Bob without physically transporting the qubit. The protocol: share a Bell pair between Alice and Bob, have Alice do a Bell measurement on her qubit plus the state to teleport, and send the two classical bits over a classical channel. Bob applies a Pauli correction keyed by those bits and ends up with the original state. Bell states are the fuel; Bell measurement is the engine.

• Superdense coding (ch.50-ish). Alice and Bob share a Bell pair. Alice does one of four local operations on her qubit, sends it to Bob, and Bob does a Bell measurement on the pair. Bob extracts two classical bits from Alice's single qubit. The entanglement is a resource that doubles the classical capacity of a quantum channel. ISRO announced a 2022 demonstration of satellite-based Bell-pair generation over a free-space link between Bangalore and the PRL Mt. Abu station, as part of India's broader quantum-communication programme — a real-world setting where Bell-state preparation and measurement mattered.

• Bell theorem and CHSH (already published). Bell's 1964 inequalities, and the later CHSH refinement, use |\Psi^-\rangle to show that quantum correlations cannot be reproduced by any local hidden-variable theory. The Nobel Prize in Physics 2022 went to Aspect, Clauser, and Zeilinger for experiments that confirmed these violations to high precision — using Bell-state preparation and measurement protocols close to what is described in this chapter.

• Quantum key distribution. The BB84 and Ekert91 protocols use pairs of entangled qubits as the source of shared randomness that becomes the cryptographic key. The Raman Research Institute in Bangalore runs an active experimental programme on entanglement-based QKD.

Common confusions

• "|\Phi^+\rangle and |\Phi^-\rangle differ only by a minus sign, so they're basically the same state." No. They are orthogonal — the inner product \langle\Phi^+|\Phi^-\rangle is exactly 0, as computed above. The minus sign is a relative phase between the two terms, which is observable (a Bell measurement distinguishes them deterministically) and hence physically real. A global phase would be unobservable; a relative phase is a first-class feature of the state.

• "Bell measurement is just a standard measurement in the computational basis." No. A computational-basis measurement on a Bell state gives a random outcome (|\Phi^+\rangle would give (0, 0) or (1, 1) with equal probability, destroying information about which Bell state it was). A Bell measurement first applies the inverse of the preparation circuit (CNOT then H) and then measures in the computational basis. The pre-rotation is essential — without it, you cannot distinguish the four Bell states by a local measurement.

• "The Bell states are the only entangled two-qubit states." No. Any non-product two-qubit state is entangled, and there are infinitely many of them. The Bell states are special because they are maximally entangled (each qubit individually is maximally mixed — zero information carried locally) and because they form an orthonormal basis of the two-qubit space. Other entangled states exist and show up in algorithms, but they typically decompose into a superposition over the Bell basis.

• "Bell-state preparation needs perfect gates; any noise makes it fail." Real devices prepare imperfect Bell states. A typical IBM superconducting chip prepares |\Phi^+\rangle with around 95-99% fidelity. The result is a density matrix \rho \approx 0.98\,|\Phi^+\rangle\langle\Phi^+| + 0.02\,\text{(other states)} — close enough for teaching demos and for many NISQ algorithms, but not perfect. State tomography on the prepared pair is a standard calibration routine.

• "The minus sign on |\Psi^-\rangle makes it a weirder or more special state." It is special in a specific sense: |\Psi^-\rangle is the only Bell state that is antisymmetric under exchange of the two qubits — swapping them sends |\Psi^-\rangle to -|\Psi^-\rangle. The other three are symmetric. This antisymmetry is the mathematical signature of fermionic particles (electrons, protons) in physics, which is why |\Psi^-\rangle is called the singlet. It also achieves the maximum CHSH Bell-inequality violation — a fact that connects quantum information to foundational physics.

• "You can use one Bell pair to send a message faster than light." No. The no-communication theorem proves that no local operation on one qubit of a Bell pair can convey any information to the holder of the other qubit. Measuring your qubit collapses the pair, but the statistics of the other side's measurement do not change — only the joint correlations, which require classical communication to detect. Teleportation is not faster-than-light: it needs the two classical bits sent from Alice to Bob, which travel at normal speeds.

Going deeper

The Pauli-Bell correspondence — full form

Every Bell state can be written as a Pauli applied to |\Phi^+\rangle on one qubit. Concretely:

(Up to an overall sign in the last case — (XZ \otimes I)|\Phi^+\rangle equals i|\Psi^-\rangle if you use XZ = iY, depending on convention.)

This means: once Alice and Bob share a Bell pair |\Phi^+\rangle, Alice can encode two classical bits into the pair by applying one of \{I, Z, X, XZ\} to her qubit. The receiver who gets both qubits can do a Bell measurement and read off which Pauli Alice applied — this is the superdense coding protocol. The Pauli-Bell correspondence is the algebraic heart of why two classical bits fit into one quantum bit (given pre-shared entanglement).

The singlet and its fermionic flavour

The singlet |\Psi^-\rangle = \tfrac{1}{\sqrt 2}(|01\rangle - |10\rangle) is antisymmetric under qubit exchange — the swap operator maps |\Psi^-\rangle \to -|\Psi^-\rangle. Particles whose wavefunctions are antisymmetric under exchange are fermions; electrons, protons, and neutrons are fermions. The Pauli exclusion principle — "two electrons cannot occupy the same quantum state" — is exactly the statement that the two-electron wavefunction must be antisymmetric, and if they are in the same state the antisymmetric combination is 0.

The other three Bell states are symmetric under exchange and correspond (in a loose sense) to bosonic statistics. Photons, phonons, and the Higgs boson are bosons. S.N. Bose's 1924 paper — the foundational work on bosonic quantum statistics — is the Indian physics root of this split.

The singlet also has the property that it is rotationally invariant — applying the same single-qubit unitary U to both qubits, (U \otimes U)|\Psi^-\rangle = |\Psi^-\rangle (up to a phase). This uniqueness is not shared by the other Bell states and is why |\Psi^-\rangle shows up in any scenario with rotational symmetry, including NMR and some QKD protocols.

Maximal CHSH violation

The CHSH inequality (Bell chapter, already published) defines a quantity S that any local hidden variable theory must satisfy |S| \leq 2. Quantum mechanics can achieve |S| = 2\sqrt{2} \approx 2.828 — the Tsirelson bound. The state that achieves this maximum is exactly |\Psi^-\rangle, with specific choices of measurement angles on each side. Experiments verifying this violation are how we know, empirically, that nature is not locally realistic — the 2022 Physics Nobel recognised this exact lineage of experiments.

Bell states in quantum key distribution

The Ekert 1991 protocol (E91) uses Bell pairs directly: Alice and Bob share a stream of |\Phi^+\rangle pairs, each measures randomly in one of several bases, and the measurement correlations (when they happen to choose compatible bases) form a shared secret key. An eavesdropper cannot intercept without disturbing the CHSH statistics, and the CHSH inequality becomes a detector for eavesdropping. Modern satellite-based QKD — including ISRO's 2022 demonstration, and China's Micius satellite experiments — uses this kind of entanglement-based scheme or its BBM92 variant. The Raman Research Institute in Bangalore has a dedicated quantum-optics lab producing polarisation-entangled photon Bell pairs for QKD and foundational experiments.

Preview of multipartite entanglement

Two qubits give you four Bell states. Three qubits give you two inequivalent classes of entanglement: the GHZ state \tfrac{1}{\sqrt 2}(|000\rangle + |111\rangle) and the W state \tfrac{1}{\sqrt 3}(|001\rangle + |010\rangle + |100\rangle). These cannot be converted into each other by any local operation on the three qubits — they are genuinely different kinds of three-qubit entanglement. Chapter 39 covers GHZ and W in detail. The general story — that multipartite entanglement has a richer classification than bipartite — is one of the foundational surprises of quantum information.

Where this leads next

• Entanglement, defined — the formal test for entanglement; the Bell states are the standard examples.
• GHZ and W states — three-qubit entanglement, two inequivalent classes.
• Quantum teleportation — the canonical application of Bell pairs plus Bell measurement.
• Superdense coding — using a shared Bell pair to send two classical bits in one qubit.
• Bell theorem and CHSH — the maximal violation achieved by |\Psi^-\rangle, and the Nobel-winning experiments.
• Schmidt decomposition — the linear-algebraic structure theorem that explains why Bell states are "maximally entangled."

References

1. Wikipedia, Bell state — the four states, their preparation, and properties.
2. John Preskill, Lecture Notes on Quantum Computation, Ch. 4 — theory.caltech.edu/~preskill/ph229.
3. Nielsen and Chuang, Quantum Computation and Quantum Information (2010), §2.3 (Bell states and measurements) — Cambridge University Press.
4. Qiskit Textbook, Entanglement and Bell States — Qiskit code to prepare and measure Bell states.
5. Wikipedia, CHSH inequality — the quantitative Bell inequality, and why |\Psi^-\rangle maximally violates it.
6. IBM Quantum Learning, Entanglement in Action: Bell States — interactive tutorial for preparing and measuring Bell states on real hardware.