🏠 Home 📘 Track 1: Quantum Basics L21 — The CNOT Gate L22 — Your First Circuit L23 — Reading Results
L22 §4 · Build Something Real ~18 min

Your First Circuit —
Make an Entangled Pair

You know the gates. You know the rules. Now you put them together and run something real. Step by step, wire by wire — this is the Bell pair circuit, and the moment entanglement stops being theory and becomes data.

✦ One Idea Running the Bell pair circuit gives you $|00\rangle$ and $|11\rangle$ with equal probability — and never $|01\rangle$ or $|10\rangle$. That perfect, structurally-enforced correlation across every shot is the measurement signature of entanglement. The statistics are the proof.
Bell pair circuit H gate CNOT gate quantum circuit Bell state Φ⁺ measurement statistics entanglement signature
Section 01
① Hook

This Is the Moment

⚛️
Predict before you run
You've built the theory — now predict what the circuit will produce.

The Bell pair circuit applies H to Q0, then CNOT with Q0 as control and Q1 as target. Both qubits start in $|0\rangle$. If you run this circuit 100 times and measure both qubits each time, which outcome pattern do you expect?

Think about what you now know. In L20 you met the Hadamard gate — the one gate that turns a definite $|0\rangle$ into a genuine superposition. In L21 you understood CNOT — the gate that flips a target qubit only when the control is $|1\rangle$, and that creates entanglement when the control is in superposition.

This lesson is where both converge. You are going to run your first real quantum circuit. Not a thought experiment. Not an analogy. A two-gate sequence — H followed by CNOT — that produces an entangled pair. You will step through it gate by gate, watch the state transform at each step, then run it many times and observe the measurement statistics that make entanglement visible in data.

Every quantum computing platform in the world — IBM Quantum, Google Cirq, IonQ, Quantinuum — has this circuit in its beginner tutorials. It is the canonical first circuit. This is your first run.

Section 02
③ Framework

The Circuit — What You're About to Build

Here is the complete circuit, drawn exactly as you would see it in any textbook or quantum platform:

Bell pair circuit — the canonical two-qubit entanglement circuit
Q0 |0⟩ Q1 |0⟩ H Step 1 Step 2 |Φ⁺⟩ = (|00⟩+|11⟩)/√2 ← Bell state. Entangled. Step 3 0/1 0/1

Two qubits. Three steps. The simplest circuit that produces entanglement.

Both qubits start in $|0\rangle$ at the left — the default ground state for every qubit before a computation. Q0 passes through an H gate (the cyan box). Then a CNOT connects Q0 and Q1, with Q0 as the control (the filled dot ●) and Q1 as the target (the circled plus ⊕). Then both qubits are measured — the double lines after measurement represent classical information flowing to output registers.

📐
Why no H gate on Q1?
Q1 does not need a Hadamard gate before the CNOT. The H on Q0 creates superposition in the control qubit. When CNOT then acts, Q1 inherits a superposition — not from its own H gate, but from its correlation with Q0. Q1 starts boring and definite; it becomes entangled through its interaction with Q0. That is the whole point of CNOT: it spreads quantum-ness from one qubit to another.
Section 03
④ Theory

Step-by-Step Walkthrough

Let's trace every qubit through every gate. No steps skipped, no hand-waving.

0
Start — both qubits initialised to |0⟩
Initial state
Q0 and Q1 are both in the ground state $|0\rangle$. Nothing quantum is happening. No superposition. No entanglement. Just two classical-looking qubits sitting there, ready.
Joint state: |Q0, Q1⟩ = |00⟩
Q0 = |0⟩ · Q1 = |0⟩ · Separable — no correlation
1
H gate applied to Q0 only
Hadamard gate
The Hadamard gate transforms Q0 from $|0\rangle$ to $(|0\rangle + |1\rangle)/\sqrt{2}$ — a perfect superposition. Q1 is completely untouched by this gate and remains in $|0\rangle$. The two qubits are still independent at this point — no entanglement yet, because neither qubit knows anything about the other.
Q0 after H: (|0⟩ + |1⟩) / √2
Joint state: (|00⟩ + |10⟩) / √2
Q1 unchanged at |0⟩ · Still separable · No entanglement yet
2
CNOT applied — Q0 is control, Q1 is target
CNOT gate
Here is where entanglement is born. CNOT says: "if Q0 is $|1\rangle$, flip Q1. Otherwise, do nothing." But Q0 is in superposition — it is simultaneously both $|0\rangle$ and $|1\rangle$. So CNOT acts on both branches simultaneously. The $|00\rangle$ branch has Q0=0, so Q1 stays: → $|00\rangle$. The $|10\rangle$ branch has Q0=1, so Q1 is flipped: → $|11\rangle$. The result cannot be separated into individual qubit states. The qubits are entangled.
|00⟩ branch: CNOT(|00⟩) = |00⟩ (Q0=0, do nothing)
|10⟩ branch: CNOT(|10⟩) = |11⟩ (Q0=1, flip Q1: 0→1)
Final state: |Φ⁺⟩ = (|00⟩ + |11⟩) / √2
3
Measure both qubits
Measurement
The Bell state collapses to a definite classical outcome. There are only two possible outcomes — $|00\rangle$ and $|11\rangle$ — each with probability $\frac{1}{2}$. The outcome is random: you cannot predict which one you will get on any single shot. But whatever you get, the two qubits always agree. You will never get $|01\rangle$ or $|10\rangle$. That guaranteed agreement — across every single run — is the measurement signature of entanglement.
P(00) = |1/√2|² = 50% ← qubits both measured as 0
P(11) = |1/√2|² = 50% ← qubits both measured as 1
P(01) = 0% · P(10) = 0% ← impossible. Amplitude is exactly zero.
The Bell State — Φ⁺
What this circuit produces
$$|\Phi^+\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$$
Neither qubit has a definite value before measurement. But their outcomes are perfectly correlated — guaranteed, every time, with no communication between them. This is entanglement made visible in statistics.
Section 04
∑ Mathematics

The Mathematics — Every Line Shown

Step 0 — Initial state
$$|\psi_0\rangle = |0\rangle \otimes |0\rangle = |00\rangle$$
Both qubits start in the ground state. $\otimes$ is the tensor product — it combines two single-qubit states into a joint two-qubit state. $|00\rangle$ is shorthand for $|0\rangle_0 \otimes |0\rangle_1$, where subscripts label Q0 and Q1.
Step 1 — Apply H to Q0, identity I to Q1
$$|\psi_1\rangle = (H \otimes I)\,|00\rangle = H|0\rangle \otimes I|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle$$ $$= \frac{1}{\sqrt{2}}(|0\rangle \otimes |0\rangle + |1\rangle \otimes |0\rangle) = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)$$
H acts on Q0: $H|0\rangle = (|0\rangle+|1\rangle)/\sqrt{2}$. The identity $I$ acts on Q1: $I|0\rangle=|0\rangle$. We distribute the tensor product over the sum. The $\frac{1}{\sqrt{2}}$ factor ensures the state is normalised: $|\frac{1}{\sqrt{2}}|^2 + |\frac{1}{\sqrt{2}}|^2 = 1$. Q1 is still untouched — no entanglement yet.
Step 2 — Apply CNOT (Q0=control, Q1=target)
$$|\psi_2\rangle = \text{CNOT}\,|\psi_1\rangle = \text{CNOT}\!\left[\frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)\right]$$ $$= \frac{1}{\sqrt{2}}\bigl(\text{CNOT}|00\rangle + \text{CNOT}|10\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr) = |\Phi^+\rangle$$
CNOT is linear: it acts on each term of the superposition separately. $\text{CNOT}|00\rangle = |00\rangle$ (control=0, target unchanged). $\text{CNOT}|10\rangle = |11\rangle$ (control=1, target flipped: $0\to1$). The $\frac{1}{\sqrt{2}}$ is preserved. The result $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ cannot be written as a product $|\alpha\rangle\otimes|\beta\rangle$ — it is genuinely entangled. Source: Nielsen & Chuang §1.3.6.
Step 3 — Measurement probabilities via Born Rule
$$P(00) = \left|\langle 00 | \Phi^+\rangle\right|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$ $$P(11) = \left|\langle 11 | \Phi^+\rangle\right|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$ $$P(01) = \left|\langle 01 | \Phi^+\rangle\right|^2 = |0|^2 = 0 \qquad P(10) = \left|\langle 10 | \Phi^+\rangle\right|^2 = |0|^2 = 0$$
The Born Rule: the probability of measuring $|x\rangle$ equals the squared magnitude of the amplitude $\langle x|\psi\rangle$. The Bell state has amplitude $\frac{1}{\sqrt{2}}$ on $|00\rangle$ and $|11\rangle$, and zero amplitude on $|01\rangle$ and $|10\rangle$. Those two outcomes are not just unlikely — they are provably, algebraically impossible. Their amplitudes are exactly zero.
Section 05
⑤ Interactive

Run the Bell Pair Circuit

Step through the circuit gate by gate and watch the quantum state change at each step. Once you reach the Bell state, run as many measurement shots as you want. Watch $|00\rangle$ and $|11\rangle$ appear in roughly equal numbers. Watch $|01\rangle$ and $|10\rangle$ never appear.

That is entanglement — not as a concept, but as a pattern in data.

⚛ Bell Pair Circuit Runner
Step through H → CNOT → measure · shots: 0 · correlation: —
INTERACTIVE
Step 0 of 3 — Initialise
Current quantum state
|Q0, Q1⟩ = |00⟩
Press ▶ Next Step to apply the first gate. Watch the quantum state update with each operation.

Key observation: run 100+ shots and watch the correlation rate displayed in the header converge toward 100%. Every single shot, the qubits agree. The randomness (which outcome appears) and the correlation (that the qubits always agree) are both real, simultaneous, and both correct.

Section 06
③ Framework

What the Statistics Are Telling You

After running many shots, you have a histogram. Two bars at roughly equal heights — one for $|00\rangle$, one for $|11\rangle$ — and two bars permanently at zero. Take a moment to appreciate what that pattern means.

The correlation is not a coincidence

Every single shot: Q0 and Q1 always agree. Every time Q0 comes out 0, Q1 is also 0. Every time Q0 comes out 1, Q1 is also 1. Not most of the time. Every time, without exception.

Now ask yourself: could this correlation have been set up in advance? Could we have secretly programmed both qubits at the start to agree — like two people agreeing before a coin flip to both say "heads"? That was Einstein's idea — hidden variables. He thought quantum particles must carry pre-programmed outcomes we just can't see.

But Bell's theorem rules that out. The correlations in the Bell state are stronger than any pre-programming can explain. The qubits genuinely had no definite value before measurement. The measurement itself created the correlated outcome — and the other qubit's state was determined at the same instant, wherever it was.

What |01⟩ and |10⟩ never appearing means

The absence of $|01\rangle$ and $|10\rangle$ is not a feature of your simulator's randomness. It is a mathematical fact. Their amplitude in the Bell state is exactly zero — not small, not unlikely, but provably, algebraically zero. Run this circuit a trillion times on a real quantum computer and they will never appear.

Key Insight
The histogram you just generated is the measurement signature of a fundamentally non-classical correlation — one that no classical physics can produce. The statistics are the evidence. The pattern in the data is the proof of entanglement.
🔮
This is what every real Bell test measures
Since Alain Aspect's 1982 experiment, physicists have run Bell tests on real entangled particles — photons, electrons, atoms — and observed exactly this pattern: two outcomes that always match, with probabilities that cannot be explained by hidden variables. The 2022 Nobel Prize in Physics was awarded to Aspect, Clauser, and Zeilinger for confirming this beyond any doubt. What you just ran in simulation is the logical structure of those experiments.
Lesson Summary

What You Just Did — and Why It Matters

  • 🔗
    You ran your first real quantum circuit
    Two qubits, two gates, one measurement. H on Q0 created superposition. CNOT entangled Q0 and Q1. Measurement revealed the correlation. This is the actual Bell pair circuit used by every quantum platform worldwide — IBM Quantum, Google Cirq, IonQ, Quantinuum, all of them begin their beginner tutorials here.
  • 📊
    The statistics prove the entanglement
    $|00\rangle$ and $|11\rangle$ appear with ~50% each. $|01\rangle$ and $|10\rangle$ never appear — not rarely, but provably never. Their amplitudes are exactly zero. That perfect, structurally-enforced correlation across every shot is the measurement signature of a genuinely entangled quantum state.
  • 📐
    The mathematics is exact — no approximations
    $|\psi_0\rangle=|00\rangle$ → $|\psi_1\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|10\rangle)$ → $|\psi_2\rangle=|\Phi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. Every step follows from the linearity of quantum gates and the CNOT truth table. No hand-waving anywhere.
  • 🏆
    This circuit is the foundation of quantum information science
    Quantum teleportation, superdense coding, quantum cryptography (BB84, E91), and quantum error correction all begin with a Bell pair. You now understand, at the circuit level, exactly how entanglement is created and what its signatures look like in measurement data.
Quick Check
How clearly does the Bell pair circuit click for you?

You have built and run a real quantum circuit.
Now learn to read its output correctly —
what circuit depth means for real hardware,
and how repeated shots reconstruct a probability distribution.

→ Reading Results — L23
Sources & Further Reading
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The CNOT Gate
L21 — The entanglement maker
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