🏠 Home 📘 Track 1: Quantum Basics L12 — Entanglement L13 — Creating Entanglement L14 — Entanglement Recap
L13 §3 · Three Superpowers ~20 min

Creating Entanglement — The Bell Pair

L12 showed what entanglement is. A phenomenon so strange Einstein spent 30 years refusing to believe it. Now you learn how to build it. The recipe has exactly two ingredients, four steps, and produces the most important two-qubit state in all of quantum computing.

✦ One Idea Two gates — a Hadamard followed by a CNOT — are all you need to create a Bell pair from scratch. H creates superposition on one qubit. CNOT spreads that superposition across both qubits, creating correlations that cannot be separated. That inseparability is entanglement.
Hadamard gate CNOT gate Bell state EPR pair two-qubit circuits entanglement creation
Section 01
① Hook

A Recipe That Shouldn't Work

🌀
Test your intuition first
Think carefully — this is the most common mistake in quantum computing education.

You apply a Hadamard gate (H) to Q1, putting it into superposition: (|0⟩ + |1⟩)/√2. Q2 is still in |0⟩. Are the two qubits now entangled?

You have just learned that entanglement is nature's most baffling phenomenon — two particles sharing one fate, no matter how far apart. Einstein spent decades refusing to believe it. Physicists needed 40 years of experiments to confirm it. And yet the recipe to create it takes two lines of code.

🎯 The challenge — stated precisely
Start with two completely ordinary qubits. Each one is in state $|0\rangle$ — boring, classical, independent. Nothing quantum is happening. No superposition. No correlation.

By the end of this lesson, you will know exactly how to transform them into a maximally entangled Bell pair — two qubits that share a single quantum fate — using just two quantum gates applied in sequence.

The strangeness isn't in the complexity. It's in the fact that something this simple produces something this profound.

This lesson is about the how. Once you know how to build a Bell pair, you understand the foundation of quantum teleportation, quantum cryptography, quantum error correction, and the exponential correlated state space of entangled multi-qubit systems. Every one of those protocols begins with exactly the circuit you will build today.

⚡ The punchline — before we start
Two gates. That is the complete recipe for creating the most non-classical state in physics. The same state that broke classical probability theory, won a Nobel Prize in 2022, and powers every quantum network protocol ever designed. Two gates.
Section 02
② Intuition

The Two Gates You Need

Before any algebra, here is the plain-English picture. Two gates. Two completely different jobs. Neither one creates entanglement alone — but together, in the right order, they do.

🔑 Everyday analogy — the scrambler and the linker
Imagine Alice and Bob each hold a plain white card, face down. Neither knows what's on theirs.

Step 1: You take Alice's card and pass it through a "scrambler" machine. Now Alice's card is in a genuine 50/50 uncertain state — it could become RED or BLUE, but hasn't decided yet. It is genuinely undecided, not just unknown.

Step 2: You pass both cards through a "linker" machine. This machine says: "Whatever colour Alice's card eventually becomes, Bob's card must match it — guaranteed, no exceptions."

Neither card has a colour yet. Both are genuinely undecided. But the moment Alice flips hers over and sees RED — Bob's card instantly becomes RED too, no matter where in the world Bob is. Not because someone told it to. Because the cards were linked at the source.

In quantum computing: the scrambler is the Hadamard gate. The linker is the CNOT gate. The linked cards are a Bell pair.

Gate 1 — Hadamard (H): the superposition maker

Hadamard Gate (H)
🌀
Takes certainty and makes uncertainty
$|0\rangle \to \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ — a definite 0 becomes a perfect 50/50 superposition. Acts on one qubit at a time. Think of it as starting the coin spinning — it had heads, now it has no face until you look.
Why H alone isn't enough
⚠️
Superposition ≠ entanglement
After H on Q1, the joint state is $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle$. This still factorises — Q1 and Q2 are independent. Measuring Q1 tells you nothing about Q2. You need one more step.

Gate 2 — CNOT (Controlled-NOT): the entanglement maker

CNOT Gate
🔗
A conditional flip across two qubits
Acts on two qubits: a control and a target. Rule: if the control is $|1\rangle$, flip the target. If the control is $|0\rangle$, do nothing. Simple for definite states — extraordinary for superposed ones.
Why CNOT creates entanglement here
✨
Superposition + CNOT = entanglement
When the control is in superposition, CNOT cannot execute just one branch — it executes both simultaneously. The $|0\rangle$ branch: do nothing. The $|1\rangle$ branch: flip Q2. Both outcomes coexist — and now Q1 and Q2 are inextricably linked.

CNOT truth table — all four definite inputs

Control Q1 Target In Q2 Target Out Q2 What happened
$|0\rangle$$|0\rangle$$|0\rangle$Control is 0 — target unchanged
$|0\rangle$$|1\rangle$$|1\rangle$Control is 0 — target unchanged
$|1\rangle$$|0\rangle$$|1\rangle$Control is 1 — target flipped 0→1
$|1\rangle$$|1\rangle$$|0\rangle$Control is 1 — target flipped 1→0
✦
The key insight — memorise this
CNOT applied to two definite states produces two definite states. No superposition, no entanglement. But CNOT applied to a superposed control qubit — one that is simultaneously $|0\rangle$ and $|1\rangle$ — produces a state that cannot be factored. That unfactorable state is entanglement. The order matters: H first, then CNOT.
Section 03
③ Framework

The Bell Pair Recipe — Step by Step

Here is the complete recipe. Four steps. Two gates. One entangled pair. Watch the state transform at each step.

0
Start — two qubits, both in |0⟩
Starting state
You have two qubits: Q1 (the control) and Q2 (the target). Both start in the ground state $|0\rangle$. This is a completely definite, classical-like state. Nothing quantum is happening yet. No superposition. No entanglement.
Joint state of Q1 and Q2:
|Q1⟩ ⊗ |Q2⟩ = |00⟩
1
Apply H to Q1 only
H gate — Hadamard
Apply Hadamard to Q1. Q1 enters a perfect 50/50 superposition — it is simultaneously $|0\rangle$ and $|1\rangle$. Q2 is untouched, still $|0\rangle$. The two qubits are still independent. The joint state can be factored: $(|0\rangle + |1\rangle)/\sqrt{2}$ times $|0\rangle$. No entanglement yet — but the precondition for it has been set.
Q1 after H: (|0⟩ + |1⟩)/√2
Joint state: (|00⟩ + |10⟩)/√2
2
Apply CNOT — Q1 as control, Q2 as target
CNOT gate
Now apply CNOT. The rule is "if Q1 is $|1\rangle$, flip Q2." But Q1 is in superposition — it is simultaneously $|0\rangle$ and $|1\rangle$. So CNOT executes both branches at once. The $|00\rangle$ part stays $|00\rangle$ (Q1 was 0, do nothing). The $|10\rangle$ part becomes $|11\rangle$ (Q1 was 1, flip Q2). This is the moment entanglement is created. The state can no longer be factored.
|00⟩ → |00⟩ (Q1 was 0 — Q2 unchanged)
|10⟩ → |11⟩ (Q1 was 1 — Q2 flipped)
Φ⁺
Result — the Bell state |Φ⁺⟩
Entangled!
The final state is a superposition of $|00\rangle$ and $|11\rangle$ with equal probability. Notice what is missing: there is no $|01\rangle$ and no $|10\rangle$. The qubits will never disagree. Measure Q1 and get 0? Q2 is 0. Get 1? Q2 is 1. Every time. The two-gate circuit has created the most important state in quantum computing.
Final state: |Φ⁺⟩ = (|00⟩ + |11⟩) / √2
The Bell State — Φ⁺
The most important two-qubit state in quantum computing
$$|\Phi^+\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$$
Equal superposition of both-zeros and both-ones. Neither qubit has a definite value until measurement. But their outcomes are perfectly correlated — always. No signal travels between them. No hidden pre-arrangement. Just one shared quantum fate, created by two gates.
Bell pair circuit — how it looks written on paper
Q1 Q2 |0⟩ |0⟩ H |Φ⁺⟩ = (|00⟩+|11⟩) / √2 P(00) = 50% P(11) = 50% H on Q1 CNOT(Q1→Q2) entangled
Section 04
④ Theory

Full Derivation — Every Step Shown

Now let's see the algebra behind what just happened. We track the two-qubit state through each gate, one line at a time. No steps skipped. No "it can be shown that."

📐
Notation reminder
We write a two-qubit state as $|Q1\,Q2\rangle$ — left digit is Q1, right digit is Q2. So $|01\rangle$ means Q1 = $|0\rangle$, Q2 = $|1\rangle$. The symbol $\otimes$ is the tensor product — it combines two single-qubit states into one joint state: $|0\rangle \otimes |0\rangle = |00\rangle$. Quantum gates are linear — they distribute across sums exactly like multiplication distributes across addition.
Step 0 — Initial state
$$|\psi_0\rangle = |0\rangle \otimes |0\rangle = |00\rangle$$
Both qubits start in $|0\rangle$. The combined state is $|00\rangle$. Completely definite. No superposition. No entanglement. This is our starting point.
Step 1 — Apply Hadamard to Q1 (Q2 unchanged)
$$H|0\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr)$$ $$|\psi_1\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr) \otimes |0\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |10\rangle\bigr)$$
Hadamard sends $|0\rangle$ to $(|0\rangle+|1\rangle)/\sqrt{2}$. We distribute the tensor product across the sum. Q2 is unchanged. The system is in a superposition of $|00\rangle$ and $|10\rangle$ — but this state can still be factored as a product of two independent qubit states: $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle$. No entanglement yet.
Step 2 — Apply CNOT (Q1 control, Q2 target)
$$\text{CNOT rules:}\quad \text{CNOT}|00\rangle = |00\rangle,\quad \text{CNOT}|10\rangle = |11\rangle$$ $$|\psi_2\rangle = \text{CNOT}\left[\frac{1}{\sqrt{2}}\bigl(|00\rangle + |10\rangle\bigr)\right] = \frac{1}{\sqrt{2}}\bigl(\text{CNOT}|00\rangle + \text{CNOT}|10\rangle\bigr)$$ $$= \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$$
Because quantum gates are linear, CNOT acts on each term of the superposition separately. $|00\rangle$: Q1=0 so Q2 is untouched, $|00\rangle \to |00\rangle$. $|10\rangle$: Q1=1 so Q2 is flipped, $|10\rangle \to |11\rangle$. The $\frac{1}{\sqrt{2}}$ factor is preserved throughout. The result is the Bell state $|\Phi^+\rangle$.
Result — The Bell State Φ⁺ (boxed)
$$\boxed{|\Phi^+\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)}$$
The Born Rule gives $P(00) = |1/\sqrt{2}|^2 = 1/2$ and $P(11) = |1/\sqrt{2}|^2 = 1/2$. There is no $|01\rangle$ or $|10\rangle$ term — the probability of finding the qubits in opposite states is exactly zero. This is the canonical Bell state (also called the EPR pair). Source: Nielsen & Chuang (2000), §1.3.6.
✦
Why this state cannot be factored — the mathematical proof of entanglement
A separable (non-entangled) state can always be written as $|\alpha\rangle \otimes |\beta\rangle$ for some single-qubit states. Try to factor $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$: you'd need $(\alpha_0|0\rangle + \alpha_1|1\rangle)\otimes(\beta_0|0\rangle + \beta_1|1\rangle) = \alpha_0\beta_0|00\rangle + \alpha_0\beta_1|01\rangle + \alpha_1\beta_0|10\rangle + \alpha_1\beta_1|11\rangle$. For this to equal the Bell state: $\alpha_0\beta_1 = 0$ and $\alpha_1\beta_0 = 0$ — but also $\alpha_0\beta_0 \neq 0$ and $\alpha_1\beta_1 \neq 0$. These four conditions are mutually contradictory. The state cannot be factored. It is fundamentally, irreducibly entangled — not as a philosophical claim but as a provable algebraic fact.

The four Bell states — a complete family

There are exactly four maximally entangled two-qubit states, collectively called the Bell basis. You just built $|\Phi^+\rangle$. Here are all four:

Φ⁺ (Phi-plus) — built today
$(|00\rangle + |11\rangle)/\sqrt{2}$
Q1 and Q2 always agree. 50% chance of |00⟩, 50% of |11⟩. Built with: H on Q1, then CNOT.
Φ⁻ (Phi-minus)
$(|00\rangle - |11\rangle)/\sqrt{2}$
Same correlations as Φ⁺ but with a relative phase of −1. The minus sign matters for interference, not for measurement outcomes alone.
Ψ⁺ (Psi-plus)
$(|01\rangle + |10\rangle)/\sqrt{2}$
Q1 and Q2 always disagree. 50% of |01⟩, 50% of |10⟩. Built with: X gate on Q2, then H on Q1, then CNOT.
Ψ⁻ (Psi-minus)
$(|01\rangle - |10\rangle)/\sqrt{2}$
Anti-correlated with a relative phase. The "singlet state" — important in quantum cryptography (E91 protocol) and quantum key distribution.
⚠️
Don't memorise all four yet
For Track 1, the essential Bell state is $|\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$ — the one you just built and derived. The other three appear naturally in quantum algorithms and cryptography. What matters now is understanding why the state is entangled and exactly how the H + CNOT circuit produces it.
Section 05
⑤ Interactive

Bell Pair Builder

Step through the Bell pair recipe yourself. Press Next Step to apply each gate and watch the quantum state update in real time. Once the Bell pair is built, press Measure repeatedly — observe the perfect correlation build up in the statistics.

🔗 Bell Pair Builder
|00⟩ → H → (|00⟩+|10⟩)/√2 → CNOT → |Φ⁺⟩ · measurements: 0 · correlation: —
INTERACTIVE
0
Initial state — both qubits in |0⟩
Q1 = |0⟩ · Q2 = |0⟩ · Joint = |00⟩
START
1
Apply H to Q1 → superposition, no entanglement yet
Waiting…
H
2
Apply CNOT → entanglement created
Waiting…
CNOT
✓
Bell state |Φ⁺⟩ — ready to measure
Waiting…
Φ⁺
Current quantum state
|00⟩
Press Next Step to begin building the Bell pair from |00⟩. Watch the state transform at each gate.
Measurement history — always |00⟩ or |11⟩, never |01⟩ or |10⟩
Result |00⟩
0% 0 shots
Result |11⟩
0% 0 shots
💡 After 20+ measurements, the |00⟩ and |11⟩ bars should approach 50% each — never |01⟩ or |10⟩. That mathematical absence is the direct consequence of having no $|01\rangle$ or $|10\rangle$ terms in the Bell state equation.
Quick Check
Lesson Summary

What You Now Know About Creating Entanglement

  • 🌀
    Hadamard creates superposition — but not entanglement alone
    $H|0\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$. After this step, Q1 is in superposition and Q2 is still $|0\rangle$. The joint state $(|00\rangle + |10\rangle)/\sqrt{2}$ can still be factored — no entanglement yet. Superposition is the precondition, not the result.
  • 🔗
    CNOT on a superposed control qubit creates entanglement
    CNOT says "flip Q2 if Q1 is 1." When Q1 is in superposition, CNOT propagates that superposition across both qubits: $|00\rangle \to |00\rangle$, $|10\rangle \to |11\rangle$. The resulting state $(|00\rangle + |11\rangle)/\sqrt{2}$ cannot be factored. This is entanglement — created by the gate, not by magic.
  • 📐
    The Bell state $|\Phi^+\rangle$ cannot be factored — provably
    Any attempt to write $(|00\rangle + |11\rangle)/\sqrt{2}$ as a product of two independent qubit states leads to a mathematical contradiction. That unfactorability is the formal definition of entanglement. The absence of $|01\rangle$ and $|10\rangle$ terms guarantees perfect correlation in every measurement.
  • 🎯
    There are four Bell states — you built the most important one
    $|\Phi^\pm\rangle$ and $|\Psi^\pm\rangle$ form the Bell basis — a complete set of maximally entangled two-qubit states. For Track 1, $|\Phi^+\rangle$ is the essential one. The others appear in quantum algorithms and cryptography later.
  • ⚛️
    This two-gate circuit is the foundation of everything
    Quantum teleportation, superdense coding, quantum key distribution, quantum error correction — all begin with a Bell pair. You now understand, at the circuit level, how entanglement is made. Not as a magical phenomenon, but as the direct result of two precise quantum operations applied in sequence.
How clearly does the H + CNOT recipe click?

You can now create a Bell pair.
You understand superposition, interference, and entanglement.
The three quantum superpowers are complete.
So — what do you do with them?

→ Three Superpowers Together — L14
Sources & Further Reading
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L12 — When qubits become inseparable
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