Programming your first quantum circuits·Lesson 3 of 6

Two qubits and the Bell pair

The CNOT gate flips one qubit when another is 1. Use it to build a Bell pair, and learn how Qiskit orders a two-bit string.

Nicolaas SpijkerNicolaas SpijkerCommunity manager, Qollab
·9 min read
The Playground's Circuit panel: an H gate on q0, a CNOT from q0 down to q1, and a measurement on each qubit.

A Bell pair is two qubits whose answers are each uncertain and still agree with each other, on every shot of a perfect simulator. The circuit below builds one with a Hadamard and a CNOT, a gate that acts on two qubits at once.

The answer now has two characters, and Qiskit writes them with qubit 0 on the right; the gate row on two-character answers below explains the order.

A gate that acts on two qubits

Every gate so far acted on one qubit. The CNOT acts on two: a control and a target. The CNOT flips the target only if the control is 1; with the control at 0 it does nothing.

On its own the rule is a plain conditional, the kind an ordinary program writes with an if. When the control is in superposition, the CNOT applies its rule to each of the control's two options.

cx

The CNOT gate

cx(control, target). Flips the target if the control is 1, leaves it alone if the control is 0. Applied to a target at 0 while the control is in superposition, it makes the two qubits agree: measure them and, with no noise, you get 00 or 11 and never 01 or 10.

00

Two-character answers

With two ordinary bits, classical bits in Qiskit's words, every shot reads two characters, one per bit. Qiskit writes bit 0, which here holds qubit 0, on the right: the string 10 means qubit 1 read 1 and qubit 0 read 0. Read the string from the right to read it in qubit order.

The circuit, then the code

Two wires, q0 and q1, both starting at 0. Qubit 0 passes through an H gate, then a CNOT joins the wires with a dot on q0 and a crossed circle on q1. Two meters, one per wire, drop into two bit wires. The answer is 00 or 11, half the time each.
The CNOT is drawn as a dot on the control and a crossed circle on the target, joined by a line. After it, 00 and 11 are the only options the two wires hold.

The circuit is two gates and two measurements:

qc = QuantumCircuit(2, 2)     # two qubits, two ordinary bits
qc.h(0)                       # qubit 0 into superposition
qc.cx(0, 1)                   # CNOT: control qubit 0, target qubit 1
qc.measure([0, 1], [0, 1])    # qubit 0 into bit 0, qubit 1 into bit 1

The plumbing is lesson 2's, unchanged. The keys of the counts dict are two characters now, and the bar loop prints one line per string that was read.

two_qubits.py Python · PlaygroundOpen in Playground ↗
# 'backend' already exists when this runs: it is whatever you pick in the Run on Qollab dialog.
from qiskit import QuantumCircuit
from qiskit.providers.jobstatus import JobStatus
import time
​
shots = 1000
​
1qc = QuantumCircuit(2, 2)
2qc.h(0)
3qc.cx(0, 1)
4qc.measure([0, 1], [0, 1])
print(qc)
​
job = backend.run(qc, shots=shots)
while job.status() not in (JobStatus.DONE, JobStatus.ERROR, JobStatus.CANCELLED):
    time.sleep(5)
​
counts = job.result().get_counts()
print(counts)
5for bits, n in sorted(counts.items()):
    print(f"{bits}: {n:4d}  {100 * n / shots:5.1f}%  {'█' * (50 * n // shots)}")

  1. 1two qubits, two ordinary bits
  2. 2qubit 0 into superposition, as in lesson 2
  3. 3the CNOT: flip qubit 1 if qubit 0 is 1, and it acts inside the superposition
  4. 4read both: qubit 0 into bit 0, qubit 1 into bit 1
  5. 5each key is now two characters: bit 1 on the left, bit 0 on the right
▶ Run on Qollab

Run the circuit

Predict the tally first. Four strings are possible, 00, 01, 10 and 11. Which of them will you see, and in what shares? Then open the project, press Run on Qollab, choose IonQ Aria 1, and press Run. Then fork it, the way lesson 1 showed, so the edits below land in your own copy.

The run console: the two-qubit circuit drawn in text, the counts dict, two long bars for 00 and 11, and lines for 01 and 10 that show a few shots and no bar
Two of the four strings carry almost all the shots. The 01 and 10 lines show their counts but no bar, because a bar is fifty characters for the full shot count and four shots in a thousand round to none.

Our press on 15 September 2026, on the Aria 1 noise model, came back {'00': 512, '01': 3, '10': 1, '11': 484}: 51.2% 00, 48.4% 11, and four shots in a thousand where the two qubits disagreed.

The few 01 and 10 shots come from the noise model; on a perfect simulator the pair never disagrees.

Why the pair always agrees

After the Hadamard, qubit 0 carries two options, 0 and 1, with no value chosen. The CNOT's rule is: flip qubit 1 wherever qubit 0 is 1.

The CNOT applies that rule to both options at once. Where qubit 0 is 0, qubit 1 stays 0; where qubit 0 is 1, qubit 1 becomes 1.

So the circuit never holds 01 or 10 as an option, and measuring cannot produce them. The circuit holds 00 or 11, weighted half and half, and the measurement returns one of those. Two qubits in that state are called a Bell pair, the smallest case of entanglement.

Assignment: make the CNOT copy a plain 1

  1. In your fork, replace qc.h(0) with qc.x(0), so qubit 0 is a definite 1 when the CNOT sees it. There is no superposition now, so the CNOT has one case to handle.
  2. Apply the gate row's rule, flip the target if the control is 1, and work out what each of the two measurements reads. Write the answer as a string with qubit 0 on the right.
  3. Press Run on Qollab and choose IonQ Aria 1.
  4. Compare the tally with your answer.
Solution
qc = QuantumCircuit(2, 2)
qc.x(0)
qc.cx(0, 1)
qc.measure([0, 1], [0, 1])

Qubit 0 is 1, so the CNOT flips qubit 1 to 1. Both read 1 and every shot is 11. With a definite control the CNOT is an ordinary conditional; the previous press was the same gate with a control in superposition.

Which character is which qubit

In the Bell pair 00 and 11 read the same in either bit order, so the order never shows. The next circuit produces a string where the order matters; predict the string before you run.

Assignment: find qubit 0 in the string

  1. Starting from the version with qc.x(0), change that line to qc.x(1) and leave the CNOT in.
  2. Work out whether the CNOT changes its target, now that its control, qubit 0, is never flipped.
  3. Work out which character the qubit that reads 1 is written in, by the gate row on two-character answers.
  4. Write down 01 or 10.
  5. Press Run on Qollab and choose IonQ Aria 1.
Solution

Qubit 0 stays 0, so the CNOT does nothing to qubit 1, which keeps the 1 the X gate gave it. Qubit 1 is 1 and qubit 0 is 0, and Qiskit writes qubit 0 on the right: the string is 10.

If you expected 01, you read the string left to right in qubit order. Read the tally from the right, qubit 0 first.

Measuring is per qubit

The pair agrees when both qubits are measured. Measuring only one of them is a valid circuit too, and the tally still shows two characters, because there are still two bits. A bit that no measurement writes to stays at 0.

Assignment: read only one qubit of the pair

  1. Change qc.x(1) back to qc.h(0), so the circuit is the Bell pair again: qc.h(0), then the CNOT.
  2. Replace the measurement line with qc.measure(0, 0), so only qubit 0 is read.
  3. Work out what qubit 0 alone reads over many shots, given it was put into superposition, and what bit 1 holds when nothing ever writes to it.
  4. Put the two together into two-character strings, qubit 0 on the right.
  5. Press Run on Qollab and choose IonQ Aria 1.
Solution
qc = QuantumCircuit(2, 2)
qc.h(0)
qc.cx(0, 1)
qc.measure(0, 0)

Qubit 0 reads 0 or 1, half each. Bit 1 stays 0. So the tally is 00 and 01, about half each. Qubit 1 was still in the pair; the circuit did not measure it. Measure both again and the tally goes back to 00 and 11.

Lesson 4's RY gate takes an angle and moves a qubit part of the way from 0 to 1.

Next up4 · Rotate a qubit by any angleRY sets the odds of reading 1 with an angle; RZ changes the phase and, on its own, leaves the tally unchanged.Continue →

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