2. Interference

Superposition is reversible. A second Hadamard brings the qubit back to $|0\rangle$ , because the two paths cancel. Insert a phase between the Hadamards and the same cancellation produces $|1\rangle$ instead.

1. The idea

Each amplitude is a complex number, so it has a size and a phase. The second Hadamard adds the paths together. Equal phases reinforce $|0\rangle$ and cancel $|1\rangle$ . A phase of $\pi$ on one path swaps which outcome survives.

Two Hadamards cancel and return 0.

A phase of pi between the Hadamards returns 1.

2. The code


import math
import numpy as np
from fqkit import QuantumCircuit, Hadamard, RZ, run
 
np.set_printoptions(precision=4, suppress=True)
 
plain = QuantumCircuit(1)
plain.add_gate(Hadamard(), [0])
plain.add_gate(Hadamard(), [0])
 
phased = QuantumCircuit(1)
phased.add_gate(Hadamard(), [0])
phased.add_gate(RZ(math.pi), [0])   # a relative phase of pi
phased.add_gate(Hadamard(), [0])
 
print("H H        :", np.round(run(plain), 4))
print("H RZ(pi) H :", np.round(run(phased), 4))

3. The output


H H        : [1.+0.j 0.+0.j]
H RZ(pi) H : [0.+0.j 0.-1.j]

The first circuit is definitely 0. The second is definitely 1. The -1j is a global phase: it does not change the measurement, which is 1 on every shot.

4. The explanation

The first Hadamard creates $(|0\rangle + |1\rangle)/\sqrt{2}$ .
Hadamard applied to $|0\rangle$ and to $|1\rangle$ produces opposite signs on $|1\rangle$ . Adding them cancels $|1\rangle$ and restores $|0\rangle$ .
RZ(π) multiplies $|0\rangle$ by $-i$ and $|1\rangle$ by $i$ . That relative minus sign reverses the cancellation, so $|0\rangle$ cancels and $|1\rangle$ remains.
No measurement noise is involved. Both results have probability 1.

5. The math

H|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \qquad H|1\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}

Adding those two expansions, the $|1\rangle$ terms cancel:

H\,H|0\rangle = \frac{1}{2}\big(|0\rangle + |1\rangle + |0\rangle - |1\rangle\big) = |0\rangle

RZ(π) is a phase gate,

R_Z(\pi) = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}

and the full sequence leaves $-i|1\rangle$ . Because $|-i|^2 = 1$ , every measurement returns 1.

💡

Key insight: a phase is invisible on a single amplitude and decisive when two amplitudes are added. Interference is that addition.

Try it yourself

Replace math.pi with math.pi / 2 and print the probabilities. Are they still 0 and 1?
Apply RZ before the first Hadamard. Does the final state change?
Continue with The Bell state, which uses the same Hadamard and then shares it with a second qubit.