# 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. --- ## 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. --- ## 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](../../appendix/E-math-identities.md)): $H=\tfrac1{\sqrt2}\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix}$ 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](04-schmidt-decomposition.md)). 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). --- ## 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. --- ## Hands-on (Python) ```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] ``` ```python # 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] ``` ```python # 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). ```python # 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). --- ## 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. ∎

--- ## 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.

--- ## 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](01-multi-qubit-states.md) — the separability test used here. --- ← Prev: [Multi-Qubit States](01-multi-qubit-states.md) · Up: [Term 1](../README.md) · Next: [Nonlocality & the CHSH Inequality](03-nonlocality-chsh.md) →