Circuit Bench 01: The Bell State

What this circuit does

Two qubits whose measurement results are correlated because they share one quantum state.

The circuit is tiny:

h q[0];
cx q[0], q[1];

After those two gates, the qubits are in the Bell state

\[ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}. \]

If you measure both qubits in the usual computational basis, you should see only 00 and 11, about half the time each. That is the first visible signature.

It is not the whole story. A classical shared random bit can also produce matching 00/11 outcomes. The quantum content of the Bell state is the coherence between the two branches, and you see that by changing basis before measurement. This circuit note shows both pieces carefully.

Circuit components

• 2 qubits
• 1 Hadamard gate (h)
• 1 CNOT gate (cx)
• Measurement
• Optional: 2 more Hadamard gates for the X-basis check
• A Quokka puck or app

If terms like gate, basis, and measurement are new, start with Circuit Bench 00: Reading a Quantum Circuit. Otherwise, this note is self-contained.

Files on the bench

Open the source directory on GitHub.

File	Purpose
bell.qasm	Prepare \(\lvert\Phi^+\rangle\) and measure it in the Z basis
expected.txt	Ideal probabilities and approximate 1024-shot counts
circuit.png	Circuit diagram

Step 1: Declare qubits and classical bits

OPENQASM 2.0;
include "qelib1.inc";

qreg q[2];
creg c[2];

The two qubits start in \(|00\rangle\).

Step 2: Put the first qubit in superposition

h q[0];

The Hadamard turns the first qubit into an equal superposition:

\[ |0\rangle \xrightarrow{H} \frac{|0\rangle + |1\rangle}{\sqrt{2}}. \]

The two-qubit state is now

\[ \frac{|00\rangle + |10\rangle}{\sqrt{2}}. \]

The first qubit is in superposition; the second is still just \(|0\rangle\).

Step 3: Entangle with CNOT

cx q[0], q[1];

CNOT flips the target qubit when the control qubit is 1:

|00> -> |00>
|10> -> |11>

Because the control qubit is in superposition, both branches are transformed coherently:

\[ \frac{|00\rangle + |10\rangle}{\sqrt{2}} \xrightarrow{\mathrm{CNOT}} \frac{|00\rangle + |11\rangle}{\sqrt{2}}. \]

This is \(|\Phi^+\rangle\).

Step 4: Measure in the Z basis

measure q[0] -> c[0];
measure q[1] -> c[1];

You should see:

00 about half the time
11 about half the time
01 never, ideally
10 never, ideally

This confirms correlation in the computational basis.

It does not, by itself, prove non-classicality. A classical source that flips one fair coin and copies the result would also produce only 00 and 11. To see the difference, we need to ask a second question.

Optional check: measure in the X basis

To measure in the X basis, apply a Hadamard to each qubit immediately before measurement:

h q[0];
h q[1];

measure q[0] -> c[0];
measure q[1] -> c[1];

For \(|\Phi^+\rangle\), the outcomes are again correlated: 00 and 11.

That is not what a classical 50/50 mixture of 00 and 11 would do. If you took the classical mixture and measured in the X basis, you would get all four outcomes roughly equally. The extra X-basis correlation is evidence that the Bell state is a coherent superposition, not just a hidden coin flip.

This is still not a full Bell test. A Bell test uses several measurement settings and checks a Bell inequality. But the X-basis check is the right next step after the simple 00/11 check.

The complete Z-basis circuit

The file bell.qasm contains the simple Z-basis version:

OPENQASM 2.0;
include "qelib1.inc";

qreg q[2];
creg c[2];

h q[0];
cx q[0], q[1];

measure q[0] -> c[0];
measure q[1] -> c[1];

Run it

Run bell.qasm on Quokka. You should see counts concentrated on 00 and 11:

{'00': 512, '11': 512}   # idealised example

Real shot counts vary, but 01 and 10 should be absent or very rare in a noiseless simulator.

Then try the X-basis version by inserting two Hadamards before measurement. You should again see matching outcomes. That second basis is what tells you there is more here than a copied classical bit.

Extend and experiment

1. Check coherence in the X basis. Copy bell.qasm and insert h q[0]; h q[1]; immediately before measurement. The outcomes should still match: only 00 and 11 ideally. A classical 50/50 mixture of 00 and 11 would instead produce all four outcomes in this test.
2. Remove the entangler. Delete the CNOT and run the original Z-basis measurement. One qubit remains fixed at 0 while the other varies, so the matching Bell correlation disappears.
3. Make the relative phase visible. Add z q[0]; after the CNOT. Z-basis counts remain 00/11, because that basis cannot see the sign change. Then add the two X-basis Hadamards: the outcomes become anti-correlated (01/10 ideally), distinguishing \(|\Phi^-\rangle\) from \(|\Phi^+\rangle\).

Analysis

Gate matrices and state evolution

The Hadamard gate is

\[ H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. \]

So

\[ H|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}. \]

The CNOT gate flips the target qubit when the control qubit is 1:

|00> -> |00>
|01> -> |01>
|10> -> |11>
|11> -> |10>

Therefore:

\[ |00\rangle \xrightarrow{H \otimes I} \frac{|00\rangle + |10\rangle}{\sqrt{2}} \xrightarrow{\mathrm{CNOT}} \frac{|00\rangle + |11\rangle}{\sqrt{2}}. \]

Bell state versus classical mixture

The Bell state is

\[ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}. \]

A classical mixture with the same Z-basis statistics is:

with probability 1/2: prepare 00
with probability 1/2: prepare 11

These look the same if you only measure in the Z basis.

They differ in the X basis. Applying \(H\) to both qubits maps:

\[ |\Phi^+\rangle \mapsto |\Phi^+\rangle, \]

so the outcomes are still correlated.

But the classical mixture becomes a mixture of \(|++\rangle\) and \(|--\rangle\) when written in the X basis, which gives no guaranteed matching bit after Z-basis measurement. Operationally, after the basis-change Hadamards it produces all four bit strings with equal probability.

That is why the second measurement basis matters.

The four Bell states

The four Bell states are:

\[ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}, \qquad |\Phi^-\rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}}, \]

\[ |\Psi^+\rangle = \frac{|01\rangle + |10\rangle}{\sqrt{2}}, \qquad |\Psi^-\rangle = \frac{|01\rangle - |10\rangle}{\sqrt{2}}. \]

In the Z basis, \(|\Phi^+\rangle\) and \(|\Phi^-\rangle\) both produce matching outcomes. The relative phase is invisible.

In the X basis, that phase becomes visible:

State	Z-basis pattern	X-basis pattern
\(\lvert\Phi^+\rangle\)	same	same
\(\lvert\Phi^-\rangle\)	same	different
\(\lvert\Psi^+\rangle\)	different	same
\(\lvert\Psi^-\rangle\)	different	different

This is why measuring only one basis is not enough to distinguish all four Bell states.

Bell tests

A full Bell test is stronger than the checks in this circuit note. It measures entangled qubits in several carefully chosen bases and evaluates an inequality, such as the CHSH inequality.

Classical local-hidden-variable theories obey a bound. Quantum mechanics predicts that Bell states can violate it. That violation is the sharp sense in which Bell-state correlations cannot be explained classically.

This note stops earlier: it prepares a Bell state, shows the obvious Z-basis correlation, and then uses an X-basis check to reveal coherence.

Practical notes

• Do not over-read the first histogram. Seeing only 00 and 11 is the start of the story, not the proof of entanglement.
• Basis matters. Relative phase is invisible in the computational basis but visible after a basis change.
• This is the first reusable primitive. Bell states are used in teleportation, entanglement swapping, superdense coding, and tests of non-classical correlation.
• Next natural circuit: Teleportation, which uses a Bell pair as a resource.