Bell States

The four standard Bell states are maximally entangled pure states of two qubits. They form an orthonormal basis of $\mathbb C^2\otimes\mathbb C^2$, and they are the standard two-qubit examples behind EPR pairs, Bell-basis measurements, teleportation, superdense coding, and many discussions of nonclassical correlation.

Helpful background. Product bases supplies the two-qubit coordinate order; entangled states explains why mixed marginals identify entanglement of a pure joint state.

The four Bell vectors

With the computational product basis ordered as

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle,$$

the Bell states are

$$\begin{aligned} \lvert\Phi^+\rangle &= \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),\\ \lvert\Phi^-\rangle &= \frac{1}{\sqrt2} \bigl( \lvert00\rangle-\lvert11\rangle \bigr),\\ \lvert\Psi^+\rangle &= \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr),\\ \lvert\Psi^-\rangle &= \frac{1}{\sqrt2} \bigl( \lvert01\rangle-\lvert10\rangle \bigr). \end{aligned}$$

The symbols $\Phi$ and $\Psi$ separate whether the two computational-basis bits are equal or different. The superscript records the relative sign.

Each Bell state is entangled across the first-qubit versus second-qubit split. None can be written as $\lvert\alpha\rangle_A\otimes\lvert\beta\rangle_B$.

Orthonormal Bell Basis

The Bell states are normalized because each is a sum of two orthogonal product-basis states with coefficients of magnitude $1/\sqrt2$.

They are mutually orthogonal. For example,

$$\begin{aligned} \langle\Phi^+\vert\Phi^-\rangle &= \frac12 \bigl( \langle00\vert+\langle11\vert \bigr) \bigl( \lvert00\rangle-\lvert11\rangle \bigr)\\ &= \frac12(1-1)\\ &=0. \end{aligned}$$

Similarly, $\Phi$ states are orthogonal to $\Psi$ states because their computational-basis supports are disjoint.

Since there are four orthonormal vectors in a four-dimensional two-qubit Hilbert space, they form a basis. This is the Bell basis.

The computational basis and Bell basis are related by a unitary change of basis. In one direction,

$$\begin{aligned} \lvert00\rangle &= \frac{1}{\sqrt2} \bigl( \lvert\Phi^+\rangle+\lvert\Phi^-\rangle \bigr),\\ \lvert11\rangle &= \frac{1}{\sqrt2} \bigl( \lvert\Phi^+\rangle-\lvert\Phi^-\rangle \bigr),\\ \lvert01\rangle &= \frac{1}{\sqrt2} \bigl( \lvert\Psi^+\rangle+\lvert\Psi^-\rangle \bigr),\\ \lvert10\rangle &= \frac{1}{\sqrt2} \bigl( \lvert\Psi^+\rangle-\lvert\Psi^-\rangle \bigr). \end{aligned}$$

Reduced States

All four Bell states have maximally mixed one-qubit reduced states:

$$\operatorname{Tr}_B \bigl( \lvert\beta\rangle\langle\beta\rvert \bigr) = \frac12 I_A, \qquad \operatorname{Tr}_A \bigl( \lvert\beta\rangle\langle\beta\rvert \bigr) = \frac12 I_B,$$

where $\lvert\beta\rangle$ is any Bell state.

For $\lvert\Phi^+\rangle$,

$$\rho_{AB} = \frac12 \Bigl( \lvert00\rangle\langle00\rvert +\lvert00\rangle\langle11\rvert +\lvert11\rangle\langle00\rvert +\lvert11\rangle\langle11\rvert \Bigr).$$

Tracing over $B$ gives

$$\rho_A = \frac12 \bigl( \lvert0\rangle\langle0\rvert +\lvert1\rangle\langle1\rvert \bigr) = \frac12 I_A.$$

The cross terms vanish in the partial trace because $\langle0\vert1\rangle=0$.

The operational consequence is that no one-qubit measurement can distinguish which Bell state was shared. The distinction lives in joint statistics, as developed in Local Measurement Statistics. If one qubit is measured and its outcome is known, the other qubit is described by a Conditional State.

Maximal Entanglement

A pure two-qubit state is maximally entangled when its two Schmidt coefficients are equal. There are infinitely many such states; the four Bell states are a particularly useful orthonormal basis among them. Each has the form

$$\lvert\beta\rangle = \frac{1}{\sqrt2} \lvert e_0\rangle_A\lvert f_0\rangle_B + \frac{e^{i\theta}}{\sqrt2} \lvert e_1\rangle_A\lvert f_1\rangle_B$$

for suitable local orthonormal bases. Absorbing $e^{i\theta}$ into $|f_1\rangle$ gives the Schmidt convention with nonnegative coefficients. Therefore each one-qubit reduced state has eigenvalues $1/2$ and $1/2$.

The entanglement entropy is

$$S(\rho_A) = -\operatorname{Tr}(\rho_A\log_2\rho_A) = 1.$$

Thus a Bell state contains one ebit of bipartite pure-state entanglement.

Correlations

Bell states have no local bias in any Pauli direction because each one-qubit reduced state is $I/2$. Their information is in joint correlations.

For $\lvert\Phi^+\rangle$,

$$\langle\sigma_z\otimes\sigma_z\rangle=+1, \qquad \langle\sigma_x\otimes\sigma_x\rangle=+1, \qquad \langle\sigma_y\otimes\sigma_y\rangle=-1.$$

The first equality says that computational-basis measurements agree perfectly. The second says that $x$-basis measurements also agree perfectly. The third says that outcomes are anticorrelated along the usual $y$ axes: $Y|0\rangle=i|1\rangle$ and $Y|1\rangle=-i|0\rangle$. Thus $|\Phi^+\rangle$ is not equally correlated along every common direction.

For the singlet Bell state $\lvert\Psi^-\rangle$,

$$\langle\sigma_i\otimes\sigma_j\rangle = -\delta_{ij}, \qquad i,j\in\{x,y,z\}.$$

This rotationally invariant anticorrelation is why the singlet is central in spin-correlation experiments. The angular-momentum derivation begins in Two Spin-1/2 Particles, the rotational-invariance viewpoint belongs to Symmetry Singlet and Triplet States, and the composite-systems interpretation belongs to Singlet and Triplet States.

Stabilizer View

The Bell basis can also be labeled by eigenvalues of two commuting observables:

$$Z\otimes Z, \qquad X\otimes X.$$

These operators commute because the two minus signs from swapping $X$ past $Z$ cancel:

$$(Z\otimes Z)(X\otimes X) = (X\otimes X)(Z\otimes Z).$$

Their eigenvalue labels distinguish the four Bell states:

• $\lvert\Phi^+\rangle$ has $Z\otimes Z=+1$ and $X\otimes X=+1$.
• $\lvert\Phi^-\rangle$ has $Z\otimes Z=+1$ and $X\otimes X=-1$.
• $\lvert\Psi^+\rangle$ has $Z\otimes Z=-1$ and $X\otimes X=+1$.
• $\lvert\Psi^-\rangle$ has $Z\otimes Z=-1$ and $X\otimes X=-1$.

This is often the cleanest way to remember the basis: one bit tells whether the computational-basis outcomes are the same or different; the other bit tells the relative phase.

Local Pauli Transformations

All Bell states can be generated from $\lvert\Phi^+\rangle$ by applying a Pauli operator to one qubit:

$$\begin{aligned} (I\otimes I)\lvert\Phi^+\rangle &= \lvert\Phi^+\rangle,\\ (I\otimes Z)\lvert\Phi^+\rangle &= \lvert\Phi^-\rangle,\\ (I\otimes X)\lvert\Phi^+\rangle &= \lvert\Psi^+\rangle,\\ (I\otimes XZ)\lvert\Phi^+\rangle &= \lvert\Psi^-\rangle. \end{aligned}$$

Operators act right to left: $XZ$ applies $Z$ first. Reversing the order to $ZX$ gives $-|\Psi^-\rangle$, the same physical ray but a different vector identity. This local-Pauli labeling is the algebraic core of dense coding and teleportation circuits.

Bell-Basis Measurements

A Bell-basis measurement has four rank-one effects $P_\beta=|\beta\rangle\langle\beta|$, one for each Bell vector, with $\sum_\beta P_\beta=I_4$. It resolves both commuting labels $ZZ$ and $XX$.

One implementation reverses a Bell-preparation circuit. Let

$$B=\mathrm{CNOT}_{A\to B}(H_A\otimes I_B), \qquad H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

The input $|00\rangle$ becomes $(|00\rangle+|10\rangle)/\sqrt2$ after the Hadamard, then $|\Phi^+\rangle$ after CNOT. More generally,

Input to $B$	Output
$\lvert00\rangle$	$\lvert\Phi^+\rangle$
$\lvert01\rangle$	$\lvert\Psi^+\rangle$
$\lvert10\rangle$	$\lvert\Phi^-\rangle$
$\lvert11\rangle$	$\lvert\Psi^-\rangle$

Apply $B^\dagger=(H_A\otimes I_B)\mathrm{CNOT}_{A\to B}$ and then measure both qubits in the computational basis to distinguish the Bell states. CNOT acts first in this inverse circuit.

Measuring each qubit directly in $Z$ is different: it reveals only equal versus unequal bits and destroys the phase distinction. Nor can one resolve both Bell labels by measuring local $Z$ and then local $X$ on the same pair; the first measurement disturbs the second correlation. The parity measurement example explains this difference in state change.

Protocol Previews

In Quantum Teleportation, a Bell pair is shared between two parties. A Bell-basis measurement on the input qubit and one half of the pair produces two classical bits. The receiver uses those bits to choose a Pauli correction. The protocol transmits an unknown quantum state using shared entanglement plus classical communication; it does not transmit information faster than light.

In Superdense Coding, one party applies one of four local Pauli operations to half of a shared Bell pair. Those four operations turn $\lvert\Phi^+\rangle$ into the four Bell states. A later Bell-basis measurement can distinguish them, allowing two classical bits to be encoded in one transmitted qubit when shared entanglement was prepared in advance.

Entanglement Swapping measures the middle halves of two Bell pairs in this basis, conditionally preparing a Bell state on the untouched endpoints. The outcome labels the endpoint Pauli frame.

E91 and Entanglement-Based QKD distributes singlet halves, uses same-direction anticorrelations for raw-key bits, and assigns other setting pairs to a Bell test.

The linked articles derive the protocols and state their communication requirements and noise assumptions.

Foundations Preview

Bell states are often used as the simplest examples of entangled pairs with strong correlations. The singlet state is especially important because its spin correlations are isotropic:

$$\langle (\boldsymbol\sigma\cdot\mathbf a) \otimes (\boldsymbol\sigma\cdot\mathbf b) \rangle_{\Psi^-} = -\mathbf a\cdot\mathbf b.$$

Bell’s theorem is not the statement that Bell states are entangled. It is a theorem about constraints satisfied by local hidden-variable models and violated by quantum predictions for suitable measurements. The states on this page provide standard examples, but the theorem itself requires a separate assumptions-and-inequalities analysis.

Common pitfalls

Four Bell states do not exhaust maximal entanglement. Applying arbitrary local unitaries produces infinitely many maximally entangled states. These four are a standard orthonormal choice.

A Bell measurement needs joint phase information. Separate computational-basis measurements cannot distinguish $\Phi^+$ from $\Phi^-$ or $\Psi^+$ from $\Psi^-$.

Pauli order changes vector signs. $XZ=-ZX$; the resulting global sign is irrelevant for a single state ray but must be retained in exact circuit identities.

References

• John S. Bell, “On the Einstein Podolsky Rosen paradox,” Physics Physique Fizika 1, 195–200 (1964). DOI.
• Charles H. Bennett, Gilles Brassard, Claude Crépeau, Richard Jozsa, Asher Peres, and William K. Wootters, “Teleporting an unknown quantum state via dual classical and Einstein–Podolsky–Rosen channels,” Physical Review Letters 70, 1895–1899 (1993). DOI.
• Charles H. Bennett and Stephen J. Wiesner, “Communication via one- and two-particle operators on Einstein–Podolsky–Rosen states,” Physical Review Letters 69, 2881–2884 (1992). DOI.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, sections 1.3 and 2.3–2.5.

Exercises

1. Apply the inverse Bell circuit to $|\Psi^-\rangle$ and find the computational-basis output.

Solution

CNOT sends $(|01\rangle-|10\rangle)/\sqrt2$ to $(|01\rangle-|11\rangle)/\sqrt2=|-\rangle|1\rangle$. The Hadamard on the first qubit sends $|-\rangle$ to $|1\rangle$, so the output is $|11\rangle$, as required by the table.

2. Show that $\lvert\Phi^+\rangle$ has reduced state $I/2$ on either qubit.

Solution

For

$$\rho_{AB} = \lvert\Phi^+\rangle\langle\Phi^+\rvert,$$

expand:

$$\rho_{AB} = \frac12 \Bigl( \lvert00\rangle\langle00\rvert +\lvert00\rangle\langle11\rvert +\lvert11\rangle\langle00\rvert +\lvert11\rangle\langle11\rvert \Bigr).$$

Tracing over $B$ removes the terms with $\langle0\vert1\rangle$ or $\langle1\vert0\rangle$, so

$$\rho_A = \frac12 \bigl( \lvert0\rangle\langle0\rvert +\lvert1\rangle\langle1\rvert \bigr) = \frac12 I.$$

The same calculation with $A$ traced out gives $\rho_B=I/2$.

3. Which Bell state is obtained by applying $I\otimes X$ to $\lvert\Phi^+\rangle$?

Solution

Since $X\lvert0\rangle=\lvert1\rangle$ and $X\lvert1\rangle=\lvert0\rangle$,

$$\begin{aligned} (I\otimes X)\lvert\Phi^+\rangle &= \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr)\\ &= \lvert\Psi^+\rangle. \end{aligned}$$

4. Explain why measuring both qubits of $\lvert\Phi^+\rangle$ in the computational basis is not a Bell-basis measurement.

Solution

Computational-basis measurements distinguish outcomes such as $00$ and $11$. A Bell-basis measurement distinguishes coherent superpositions such as

$$\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2} \qquad \text{and} \qquad \frac{\lvert00\rangle-\lvert11\rangle}{\sqrt2}.$$

The relative sign is not accessible from local computational-basis outcome probabilities alone.