Lesson: The Bell State

In this lesson you will create entanglement, the famously “spooky” connection between qubits, using only two gates, and see exactly why the math produces it.

1. The idea

A single qubit can be put into superposition: a state where it is genuinely both 0 and 1 at the same time, until measured. Two qubits can be entangled: their measurement results are perfectly correlated, no matter how far apart they are. The simplest entangled pair is called a Bell state.

Bell state. Hadamard on qubit 0, then CNOT onto qubit 1.

2. The code


import numpy as np
from fqkit import QuantumCircuit, Hadamard, CNOT, run, measure_all
 
np.set_printoptions(precision=4, suppress=True)
 
qc = QuantumCircuit(2)
qc.add_gate(Hadamard(), [0])   # put qubit 0 into superposition
qc.add_gate(CNOT(), [0, 1])    # copy qubit 0's value onto qubit 1
 
state = run(qc)
print("State :", np.round(state, 4))
print("Counts:", measure_all(state, shots=1024))

3. The output


State : [0.7071+0.j 0.+0.j 0.+0.j 0.7071+0.j]
Counts: {'00': 505, '11': 519}

(Your counts will vary a little: they are random samples.)

4. The explanation

The Hadamard gate turns qubit 0 from a definite 0 into a 50/50 mix of 0 and 1.
The CNOT gate flips qubit 1 only when qubit 0 is 1. Because qubit 0 is in a mix, the CNOT links the two qubits together: if qubit 0 is 0, qubit 1 is 0; if qubit 0 is 1, qubit 1 is 1.
After the CNOT there are no “01” or “10” possibilities left. Measuring the pair always gives matching bits: 00 or 11, each about half the time. That perfect correlation is entanglement.

5. The math

A qubit’s state is a vector. The two basis states are:

|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}

The Hadamard gate is the matrix:

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

Applying it to qubit 0 of the two-qubit system $|00\rangle$ gives superposition on the first qubit:

(H \otimes I)\,|00\rangle = \frac{|00\rangle + |10\rangle}{\sqrt{2}}

Then CNOT (control = qubit 0, target = qubit 1) flips the second qubit only in the $|10\rangle$ branch:

\text{CNOT}\,\frac{|00\rangle + |10\rangle}{\sqrt{2}} = \frac{|00\rangle + |11\rangle}{\sqrt{2}}

That final vector is the Bell state. In fqkit’s state-vector (big-endian ordering: index = $2q\_0 + q\_1$ ) it is:

\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \approx \begin{pmatrix} 0.707 \\ 0 \\ 0 \\ 0.707 \end{pmatrix}

which matches the printed output. The probability of each outcome is the squared magnitude of its amplitude:

P(00) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = 0.5, \qquad P(11) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = 0.5, \qquad P(01) = P(10) = 0

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Key insight: entanglement is not a force or a signal. It is simply a joint state that cannot be written as one qubit’s state times the other’s. $\frac{|00\rangle + |11\rangle}{\sqrt{2}}$ has no such factorization, so neither qubit has a definite value on its own.

Try it yourself

Remove the Hadamard and re-run: what state do you get, and why?
Swap the CNOT targets ([1, 0] instead of [0, 1]): which qubit is the control now, and how does the state vector change?
Export the circuit with qc.to_qasm() and run it on a real IBM quantum computer via the free tier: see OpenQASM & Hardware.

Continue with The GHZ state.