Bell States

Four two-qubit states sit at the very heart of quantum information: the Bell states. They are the maximally entangled states of two qubits, an orthonormal basis of $\mathbb{C}^4$, and the fuel for teleportation, superdense coding, and the Bell/CHSH tests of Term 1.4.3. In this lesson we derive how a Hadamard followed by a CNOT manufactures one from scratch, prove the four form a basis, and expose the eerie perfect correlations that hold in every measurement basis.

Learning Objectives

After this lesson you will be able to:

1. Write the four Bell states and verify they form an orthonormal basis (the Bell basis) of the two-qubit space.
2. Derive $|\Phi^+\rangle$ step by step from $|00\rangle$ via $H$ on qubit 0 then $\mathrm{CNOT}$.
3. Prove the perfect correlation of $|\Phi^+\rangle$ in the computational basis and in the $X$-basis, and explain why measuring one qubit determines the other.
4. Argue why the Bell states are maximally entangled (forward-referencing the Schmidt spectrum).
5. Construct all four Bell states in NumPy and verify orthonormality; run a Bell-pair circuit on Braket's free local simulator.

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Intuition

Start with two qubits, both definitely $|0\rangle$. Put the first into an equal superposition with a Hadamard, so the pair is "$|0\rangle$ and $|0\rangle$" plus "$|1\rangle$ and $|0\rangle$" at once. Now apply a CNOT that flips the second qubit exactly when the first is $|1\rangle$. The two branches of the superposition become "$|00\rangle$" and "$|11\rangle$" — the second qubit has been chained to the first. The result, $\tfrac1{\sqrt2}(|00\rangle+|11\rangle)$, is a state where neither qubit has a definite value, yet they are perfectly correlated: measure either and you instantly know the other, in any basis you choose. That correlation-without-local-definiteness is entanglement at its purest, and it is what makes the Bell states the workhorses of quantum information.

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Theory

The four Bell states

The Bell states (also EPR pairs) are

$$\begin{aligned} |\Phi^+\rangle &= \tfrac{1}{\sqrt2}\big(|00\rangle + |11\rangle\big), &\quad |\Phi^-\rangle &= \tfrac{1}{\sqrt2}\big(|00\rangle - |11\rangle\big),\\[2pt] |\Psi^+\rangle &= \tfrac{1}{\sqrt2}\big(|01\rangle + |10\rangle\big), &\quad |\Psi^-\rangle &= \tfrac{1}{\sqrt2}\big(|01\rangle - |10\rangle\big). \end{aligned}$$

A compact label is $|\beta_{xy}\rangle = \tfrac1{\sqrt2}\big(|0,y\rangle + (-1)^x|1,\bar y\rangle\big)$, $x,y\in\{0,1\}$, where $\bar y = 1\oplus y$: $|\beta_{00}\rangle=|\Phi^+\rangle$, $|\beta_{01}\rangle=|\Psi^+\rangle$, $|\beta_{10}\rangle=|\Phi^-\rangle$, $|\beta_{11}\rangle=|\Psi^-\rangle$. Each is got from $|\Phi^+\rangle$ by a local Pauli on qubit 1:

$$|\Phi^-\rangle = (I\otimes Z)|\Phi^+\rangle,\quad |\Psi^+\rangle = (I\otimes X)|\Phi^+\rangle,\quad |\Psi^-\rangle = (I\otimes XZ)|\Phi^+\rangle = -i\,(I\otimes Y)|\Phi^+\rangle .$$

The Bell basis is orthonormal

Each Bell state is normalized: e.g. $\langle\Phi^+|\Phi^+\rangle = \tfrac12(\langle00|00\rangle + \langle11|11\rangle) = \tfrac12(1+1)=1$(the cross terms vanish since$\langle00|11\rangle=0$). For orthogonality, the $\Phi$'s live in the $\operatorname{span}\{|00\rangle,|11\rangle\}$ sector and the $\Psi$'s in $\operatorname{span}\{|01\rangle,|10\rangle\}$, which are orthogonal sectors — so any $\Phi$ is orthogonal to any $\Psi$. Within a sector,

$$\langle\Phi^+|\Phi^-\rangle = \tfrac12\big(\langle00|00\rangle - \langle11|11\rangle\big) = \tfrac12(1-1)=0, \qquad \langle\Psi^+|\Psi^-\rangle = \tfrac12(1-1)=0 .$$

Hence $\{|\Phi^+\rangle,|\Phi^-\rangle,|\Psi^+\rangle,|\Psi^-\rangle\}$ is an orthonormal basis of $\mathbb{C}^4$ — the Bell basis. It is the unitary image of the computational basis under $\mathrm{CNOT}\,(H\otimes I)$, and measuring in it is a Bell measurement (Term 2.3, teleportation).

Deriving $|\Phi^+\rangle$: $H$ then $\mathrm{CNOT}$, step by step

We use big-endian gates (Appendix E): H=\tfrac1{\sqrt2}\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix} and CNOT with control = qubit 0, $|a,b\rangle\mapsto|a,\,a\oplus b\rangle$. Start from $|00\rangle$.

Step 1 — Hadamard on qubit 0 (the operator $H\otimes I$):

$$(H\otimes I)\,|0\rangle\otimes|0\rangle = (H|0\rangle)\otimes|0\rangle = \tfrac1{\sqrt2}(|0\rangle+|1\rangle)\otimes|0\rangle = \tfrac1{\sqrt2}\big(|00\rangle + |10\rangle\big).$$

This is still a product state, $|+\rangle\otimes|0\rangle$ (coefficient matrix rank $1$ — Lesson 1.4.1).

Step 2 — CNOT (control 0, target 1). It acts term-by-term: $|00\rangle\mapsto|00\rangle$ (control $0$, nothing happens) and $|10\rangle\mapsto|11\rangle$ (control $1$, flip target):

$$\mathrm{CNOT}\;\tfrac1{\sqrt2}\big(|00\rangle + |10\rangle\big) = \tfrac1{\sqrt2}\big(|00\rangle + |11\rangle\big) = |\Phi^+\rangle .$$

In one matrix line, $|\Phi^+\rangle = \mathrm{CNOT}\,(H\otimes I)\,|00\rangle$. The input was separable; the output has coefficient matrix $\tfrac1{\sqrt2}I$ of rank $2$ — entangled. The CNOT is the global gate that did the entangling; the Hadamard alone could not (Lesson 1.4.1). Feeding the other three computational inputs $|01\rangle,|10\rangle,|11\rangle$ through the same circuit produces $|\Psi^+\rangle,|\Phi^-\rangle,|\Psi^-\rangle$ respectively — the circuit is the change of basis to the Bell basis.

Perfect correlations in any basis

Computational ($Z$) basis. For $|\Phi^+\rangle$ the Born rule gives outcomes only on $|00\rangle$ and $|11\rangle$, each with probability $\tfrac12$; $|01\rangle,|10\rangle$ have probability $0$. So the two qubits' $Z$-outcomes are always equal: measuring qubit 0 to be $0$ collapses the state to $|00\rangle$ and forces qubit 1 to $0$ (and likewise for $1$). This holds no matter which qubit is measured first and regardless of the distance between them — the EPR puzzle of Lesson 1.4.3.

$X$-basis. The striking fact is the correlation persists in a rotated basis. Recall $|0\rangle=\tfrac1{\sqrt2}(|+\rangle+|-\rangle)$, $|1\rangle=\tfrac1{\sqrt2}(|+\rangle-|-\rangle)$. Substitute into $|\Phi^+\rangle$:

$$\begin{aligned} |\Phi^+\rangle &= \tfrac{1}{\sqrt2}\Big(\tfrac12(|+\rangle+|-\rangle)(|+\rangle+|-\rangle) + \tfrac12(|+\rangle-|-\rangle)(|+\rangle-|-\rangle)\Big)\\ &= \tfrac{1}{2\sqrt2}\Big(2|{+}{+}\rangle + 2|{-}{-}\rangle\Big) = \tfrac{1}{\sqrt2}\big(|{+}{+}\rangle + |{-}{-}\rangle\big). \end{aligned}$$

The cross terms $|{+}{-}\rangle,|{-}{+}\rangle$ cancel exactly. So in the $X$-basis the qubits are again perfectly correlated: both $+$ or both $-$. In fact $|\Phi^+\rangle$ has the same form in any real-rotated basis $\{|\theta\rangle,|\theta^\perp\rangle\}$ — a basis-independence that is the formal signature of maximal entanglement. (This same algebra, applied at different angles on the two sides, is exactly what produces the CHSH violation in Lesson 1.4.3.)

Why "maximally" entangled

A bipartite state's entanglement is measured by the Schmidt spectrum — the singular values $\sigma_i$ of its coefficient matrix $C$, normalized so $\sum_i\sigma_i^2=1$ (Lesson 1.4.4). For $|\Phi^+\rangle$, $C=\tfrac1{\sqrt2}I$ has singular values $\sigma_0=\sigma_1=\tfrac1{\sqrt2}$ — a flat spectrum, the most spread-out possible for two qubits. Equivalently, tracing out either qubit leaves the maximally mixed state $\tfrac12 I$ (Term 1.5): the local state carries zero information, because all the information lives in the correlations. A flat Schmidt spectrum / maximally mixed marginals is precisely what "maximally entangled" means, and all four Bell states share it. The von Neumann entanglement entropy is then $S=-\sum_i\sigma_i^2\log_2\sigma_i^2 = 1$ bit — the maximum for a qubit (Lesson 1.4.4).

Forward links. Maximal entanglement is exactly what makes superdense coding (2 classical bits through 1 qubit) and teleportation work (Term 2.3), and what lets the Bell state saturate the Tsirelson bound $2\sqrt2$ in CHSH (Lesson 1.4.3).

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Worked Examples

Example 1 — Building $|\Psi^-\rangle$ from a computational input

Apply the same circuit $\mathrm{CNOT}\,(H\otimes I)$ to $|11\rangle$. Step 1: $(H\otimes I)|11\rangle=(H|1\rangle)\otimes|1\rangle=\tfrac1{\sqrt2}(|0\rangle-|1\rangle)\otimes|1\rangle =\tfrac1{\sqrt2}(|01\rangle-|11\rangle)$. Step 2: CNOT sends $|01\rangle\mapsto|01\rangle$ (control $0$) and $|11\rangle\mapsto|10\rangle$ (control $1$, flip target):

$$\mathrm{CNOT}\,\tfrac1{\sqrt2}(|01\rangle-|11\rangle)=\tfrac1{\sqrt2}(|01\rangle-|10\rangle)=|\Psi^-\rangle .$$

This is the singlet — the unique Bell state that is antisymmetric under swapping the two qubits and rotationally invariant (it has the same form $\tfrac1{\sqrt2}(|01\rangle-|10\rangle)$ in every basis, up to a global phase).

Example 2 — Decoding: $|\Phi^+\rangle$ back to $|00\rangle$

Bell-state preparation is unitary, hence reversible. Apply the inverse circuit $(H\otimes I)\,\mathrm{CNOT}$ to $|\Phi^+\rangle$ (CNOT and $H$ are each self-inverse):

$$\mathrm{CNOT}\,|\Phi^+\rangle = \tfrac1{\sqrt2}(|00\rangle+|10\rangle),\qquad (H\otimes I)\,\tfrac1{\sqrt2}(|00\rangle+|10\rangle) = (H|+\rangle)\otimes|0\rangle = |00\rangle .$$

Disentangling a Bell pair is the first move of a Bell measurement: rotate the Bell basis to the computational basis, then measure (Term 2.3).

Example 3 — Expectation of a correlation observable

Consider the observable $Z\otimes Z$ on $|\Phi^+\rangle$. Since $Z|0\rangle=|0\rangle$ and $Z|1\rangle=-|1\rangle$,

$$(Z\otimes Z)|\Phi^+\rangle=\tfrac1{\sqrt2}\big((+1)(+1)|00\rangle+(-1)(-1)|11\rangle\big)=|\Phi^+\rangle,$$

so $\langle Z\otimes Z\rangle=\langle\Phi^+|\Phi^+\rangle=+1$: the $Z$-outcomes are perfectly correlated (product of signs always $+1$), the quantitative face of "both qubits agree." Likewise $\langle X\otimes X\rangle=+1$ (from the $|{+}{+}\rangle+|{-}{-}\rangle$ form), while $\langle Y\otimes Y\rangle=-1$. These three numbers are the seeds of the CHSH calculation.

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Hands-on (Python)

import numpy as np
from functools import reduce

ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)
tensor = lambda *ks: reduce(np.kron, ks)   # big-endian: first arg = qubit 0

# Gates (big-endian, control = qubit 0):
H = (1/np.sqrt(2)) * np.array([[1, 1], [1, -1]], dtype=complex)
I = np.eye(2, dtype=complex)
CNOT = np.array([[1, 0, 0, 0],
                 [0, 1, 0, 0],
                 [0, 0, 0, 1],
                 [0, 0, 1, 0]], dtype=complex)

# Prepare |Φ+> = CNOT (H⊗I) |00> and verify the step-by-step derivation:
step1 = tensor(H, I) @ tensor(ket0, ket0)     # |+>|0> = (|00>+|10>)/√2
phi_plus = CNOT @ step1                        # (|00>+|11>)/√2
print(np.round(step1, 3))                      # [0.707 0    0.707 0   ]
print(np.round(phi_plus, 3))                   # [0.707 0    0     0.707]

# Build all four Bell states by feeding the four computational inputs
# through the SAME circuit U = CNOT (H⊗I).
U = CNOT @ tensor(H, I)
inputs = {'00': tensor(ket0, ket0), '01': tensor(ket0, ket1),
          '10': tensor(ket1, ket0), '11': tensor(ket1, ket1)}
bell = {name: U @ vec for name, vec in inputs.items()}

labels = {'00': 'Φ+', '01': 'Ψ+', '10': 'Φ-', '11': 'Ψ-'}
for k, v in bell.items():
    print(f"|{labels[k]}> =", np.round(v, 3))
# |Φ+> = [0.707 0 0 0.707] ; |Ψ+> = [0 0.707 0.707 0]
# |Φ-> = [0.707 0 0 -0.707]; |Ψ-> = [0 0.707 -0.707 0]

# Verify orthonormality: the Gram matrix ⟨βi|βj⟩ must be the 4x4 identity.
B = np.column_stack([bell['00'], bell['01'], bell['10'], bell['11']])  # columns = Bell states
gram = B.conj().T @ B
print(np.allclose(gram, np.eye(4)))            # True -> orthonormal basis

# Confirm maximal entanglement: every Bell state has Schmidt rank 2 (flat spectrum).
def schmidt_singular_values(state2q):
    return np.linalg.svd(state2q.reshape(2, 2), compute_uv=False)
print(np.round(schmidt_singular_values(bell['00']), 3))  # [0.707 0.707] -> flat, maximal

A brief Braket version. The local simulator runs on your machine and is free — no AWS charges. We build the Bell circuit and confirm the $\sim$50/50 split over 00 and 11 (the deep Braket introduction is Term 2).

# pip install amazon-braket-sdk
from braket.circuits import Circuit
from braket.devices import LocalSimulator

device = LocalSimulator()                      # free, runs locally
circ = Circuit().h(0).cnot(0, 1)               # H on qubit 0, then CNOT(0->1) = |Φ+>
result = device.run(circ, shots=1000).result()
print(result.measurement_counts)
# e.g. Counter({'00': 503, '11': 497})  -- only 00 and 11 appear (perfect Z-correlation)

Endianness check. Braket's bit-string keys read qubit 0 first (big-endian), matching our ket order, so '00' and '11' correspond exactly to the $|00\rangle,|11\rangle$ amplitudes of phi_plus. No bit reversal needed — unlike little-endian SDKs (Appendix C).

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Exercises

E1 (easy). Write $|\Psi^+\rangle$ as a $4$-vector and confirm $\langle\Psi^+|\Psi^+\rangle=1$.

Solution

$|\Psi^+\rangle=\tfrac1{\sqrt2}(|01\rangle+|10\rangle)=\tfrac1{\sqrt2}(0,1,1,0)^T$. Norm$^2 = \tfrac12(0+1+1+0)=1$. ✓

E2 (easy). Which Pauli on qubit 1 turns $|\Phi^+\rangle$ into $|\Psi^+\rangle$? Verify by computing.

Solution

$I\otimes X$: it flips qubit 1, sending $|00\rangle\to|01\rangle$ and $|11\rangle\to|10\rangle$, so $(I\otimes X)|\Phi^+\rangle=\tfrac1{\sqrt2}(|01\rangle+|10\rangle)=|\Psi^+\rangle$. ✓

E3 (medium). Show that $|\Phi^-\rangle=\tfrac1{\sqrt2}(|00\rangle-|11\rangle)$ has the form $\tfrac1{\sqrt2}(|{+}{-}\rangle+|{-}{+}\rangle)$ in the $X$-basis. What correlation does this predict?

Solution

Substitute $|0\rangle=\tfrac1{\sqrt2}(|+\rangle+|-\rangle)$, $|1\rangle=\tfrac1{\sqrt2}(|+\rangle-|-\rangle)$: $|00\rangle=\tfrac12(|{+}{+}\rangle+|{+}{-}\rangle+|{-}{+}\rangle+|{-}{-}\rangle)$ and $|11\rangle=\tfrac12(|{+}{+}\rangle-|{+}{-}\rangle-|{-}{+}\rangle+|{-}{-}\rangle)$. Subtracting and multiplying by $\tfrac1{\sqrt2}$: the $|{+}{+}\rangle,|{-}{-}\rangle$ terms cancel, leaving $\tfrac1{\sqrt2}(|{+}{-}\rangle+|{-}{+}\rangle)$. Prediction: in the $X$-basis the qubits are anti-correlated (one $+$, one $-$), even though they were correlated in $Z$.

E4 (medium). Prove $\langle X\otimes X\rangle=+1$ and $\langle Y\otimes Y\rangle=-1$ for $|\Phi^+\rangle$.

Solution

$X\otimes X$ swaps both qubits' bits: $|00\rangle\mapsto|11\rangle$, $|11\rangle\mapsto|00\rangle$, so $(X\otimes X)|\Phi^+\rangle=\tfrac1{\sqrt2}(|11\rangle+|00\rangle)=|\Phi^+\rangle\Rightarrow\langle X\otimes X\rangle=+1$. For $Y$: $Y|0\rangle=i|1\rangle$, $Y|1\rangle=-i|0\rangle$, so $(Y\otimes Y)|00\rangle=(i)(i)|11\rangle=-|11\rangle$ and $(Y\otimes Y)|11\rangle=(-i)(-i)|00\rangle=-|00\rangle$; thus $(Y\otimes Y)|\Phi^+\rangle=-|\Phi^+\rangle \Rightarrow\langle Y\otimes Y\rangle=-1$.

E5 (hard). The singlet $|\Psi^-\rangle$ is rotationally invariant: show that $(U\otimes U)|\Psi^-\rangle = (\det U)\,|\Psi^-\rangle$ for any $U\in SU(2)$ (so up to a phase it is the same state in every basis). Conclude it is anti-correlated in every basis.

Solution

Write $|\Psi^-\rangle=\tfrac1{\sqrt2}(|01\rangle-|10\rangle)$. For U=\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}, compute $(U\otimes U)(|01\rangle-|10\rangle)$. The coefficient of $|ij\rangle$ in $(U\otimes U)|kl\rangle$ is $U_{ik}U_{jl}$; the antisymmetric combination $|01\rangle-|10\rangle$ maps to $\sum_{ij}(U_{i0}U_{j1}-U_{i1}U_{j0})|ij\rangle$. The bracket is the $2\times2$ determinant of columns $(0,1)$ of $U$ when $i\neq j$, and $0$ when $i=j$; explicitly it equals $(ad-bc)(|01\rangle-|10\rangle) =\det(U)(|01\rangle-|10\rangle)$. For$U\in SU(2)$,$\det U=1$, so$(U\otimes U)|\Psi^-\rangle=|\Psi^-\rangle$. Measuring both qubits in the same rotated basis $\{|u_0\rangle,|u_1\rangle\}=\{U|0\rangle,U|1\rangle\}$ therefore always gives opposite labels — perfect anti-correlation in every basis. ∎

E6 (hard). Show that $\mathrm{CNOT}\,(H\otimes I)$ maps the computational basis onto the Bell basis, and hence that $(H\otimes I)\,\mathrm{CNOT}$ is a Bell measurement circuit (it rotates the Bell basis to the computational basis before a standard measurement).

Solution

From the Hands-on, feeding $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ through $U=\mathrm{CNOT}(H\otimes I)$ yields $|\Phi^+\rangle,|\Psi^+\rangle,|\Phi^-\rangle,|\Psi^-\rangle$, a bijection between two orthonormal bases, so $U$ is unitary and maps one ONB onto the other. Its inverse $U^{-1}=(H\otimes I)\,\mathrm{CNOT}$ (using $H^2=I$, $\mathrm{CNOT}^2=I$) maps each Bell state back to a distinct computational basis ket. Applying $U^{-1}$ and then measuring in the computational basis thus reads out which Bell state the input was — that is a Bell measurement. ∎

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Checkpoint

1. Write all four Bell states. Which two share the $Z$-sector $\{|00\rangle,|11\rangle\}$?
2. Give the two-step derivation of $|\Phi^+\rangle$ from $|00\rangle$ and name the gate that does the entangling.
3. Why are the four Bell states an orthonormal basis?
4. State the $Z$-basis and $X$-basis correlations of $|\Phi^+\rangle$. Why is basis-independence the hallmark of maximal entanglement?
5. What is the Schmidt spectrum of a Bell state, and what does it imply about the reduced state of one qubit?

Answers

1. $|\Phi^\pm\rangle=\tfrac1{\sqrt2}(|00\rangle\pm|11\rangle)$ and $|\Psi^\pm\rangle=\tfrac1{\sqrt2}(|01\rangle\pm|10\rangle)$. The $\Phi$'s share the $\{|00\rangle,|11\rangle\}$ sector.
2. $(H\otimes I)|00\rangle=\tfrac1{\sqrt2}(|00\rangle+|10\rangle)$, then $\mathrm{CNOT}$ gives $\tfrac1{\sqrt2}(|00\rangle+|11\rangle)=|\Phi^+\rangle$. The CNOT (the global gate) does the entangling; $H$ alone keeps it separable.
3. Each is normalized; $\Phi$'s and $\Psi$'s live in orthogonal sectors, and within a sector the $\pm$ pair is orthogonal — four mutually orthogonal unit vectors in $\mathbb{C}^4$.
4. Perfectly correlated in $Z$ (both $0$ or both $1$) and in $X$ (both $+$ or both $-$). Basis independence means the local marginals carry no information; all information is in correlations — the definition of maximal entanglement.
5. Flat spectrum $\sigma_0=\sigma_1=\tfrac1{\sqrt2}$. Tracing out one qubit gives the maximally mixed state $\tfrac12 I$, entanglement entropy $1$ bit.

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Further Reading

• [NC] Nielsen & Chuang, §1.3.6 (Bell states) and §2.5 (Schmidt decomposition for "maximal").
• [EPR35] Einstein, Podolsky & Rosen — the original two-particle correlated state behind "EPR pairs."
• [Pre] Preskill, Ph219, Ch. 4 — Bell states, entanglement, and the Bell measurement.
• 1.4.1 Multi-Qubit States — the separability test used here.

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