1. Superposition

A qubit starts in a definite state, $|0\rangle$ . One gate is enough to put it into superposition: a state with two amplitudes at once. Measurement then returns 0 or 1 at random, with probabilities fixed by those amplitudes.

1. The idea

The Hadamard gate sends $|0\rangle$ to an equal mixture of $|0\rangle$ and $|1\rangle$ . Nothing in the circuit prefers one outcome. About half the shots are 0 and half are 1, and the split gets closer to even as you take more shots.

One Hadamard puts qubit 0 into superposition.

2. The code


import numpy as np
from fqkit import QuantumCircuit, Hadamard, run, measure_all
 
np.set_printoptions(precision=4, suppress=True)
 
qc = QuantumCircuit(1)
qc.add_gate(Hadamard(), [0])
 
state = run(qc)
print("Amplitudes   :", np.round(state, 4))
print("Probabilities:", np.round(np.abs(state) ** 2, 4))
print("Counts       :", measure_all(state, shots=1024))

3. The output


Amplitudes   : [0.7071+0.j 0.7071+0.j]
Probabilities: [0.5 0.5]
Counts       : {'0': 509, '1': 515}

The probabilities are exact. The counts are one sample of 1024 shots; yours will differ by a few tens.

4. The explanation

Before the gate, the state vector is [1, 0]: outcome 0 with certainty.
Hadamard writes the same amplitude, $1/\sqrt{2} \approx 0.707$ , onto both entries.
A probability is an amplitude’s squared length. $0.707^2 = 0.5$ , so each outcome has probability one half.
measure_all draws bitstrings from that distribution. It does not change the state vector you already printed.

5. The math

|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

H|0\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{|0\rangle + |1\rangle}{\sqrt{2}}

P(0) = P(1) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}

💡

Key insight: superposition is a pair of amplitudes, not a hidden bit. The simulator stores both numbers. A measurement throws one of them away and reports the bit that remains.

Try it yourself

Delete the Hadamard and run the circuit again. Which probability becomes 1?
Change shots to 10000. How much closer is the split to 5120 / 5120?
Continue with Interference, where a second Hadamard makes those two amplitudes cancel.