# 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.
Build it yourself. The circuit below starts as this derivation's $H$ then $\mathrm{CNOT}$. Step the
cursor back to see the product state $|+\rangle|0\rangle$ after step 1, click the input kets to feed in
the other three basis states, or drag in gates of your own. The purity $\operatorname{Tr}\rho_0^2$ of
qubit 0 reads $1$ for a product state and $\tfrac12$ for a maximally entangled one (Term 1.5):
```qm-demo circuit-builder
{ "circuit": [["H", null], ["C", "X"]] }
```
### 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$^2 = \tfrac12(0+1+1+0)=1$. ✓
Solution
$I\otimes X$: it flips qubit 1, sending $|00\rangle\to|01\rangle$ and $|11\rangle\to|10\rangle$, so
$(I\otimes X)|\Phi^+\rangle=\tfrac1{\sqrt2}(|01\rangle+|10\rangle)=|\Psi^+\rangle$. ✓
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 $Z$.
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 $(Y\otimes Y)|\Phi^+\rangle=-|\Phi^+\rangle
\Rightarrow\langle Y\otimes Y\rangle=-1$.
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. ∎
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. ∎
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.