Hadamard gate

The Hadamard gate maps

∣0 ⟩ \mapsto \frac{∣0 ⟩ + ∣1 ⟩}{2}, ∣1 ⟩ \mapsto \frac{∣0 ⟩ - ∣1 ⟩}{2} .

Table of contents
- Walsh-Hadamard transform Hadamard gate
- Uniform quantum superposition Hadamard gate

Walsh-Hadamard transform ( $H^{\otimes n}$ )

On

n

qubits, the Walsh-Hadamard transform satisfies

H^{\otimes n} ∣ x ⟩ = 2^{- n /2} \sum_{y \in {0, 1}^{n}} (- 1)^{x \cdot y} ∣ y ⟩,

where the dot product is evaluated modulo two.

Uniform quantum superposition

The uniform superposition of all

n

-bit strings is

\frac{1}{2 ^{n}} \sum_{x \in {0, 1}^{n}} ∣ x ⟩ = H^{\otimes n} ∣0 ⟩^{\otimes n} .

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Hadamard gate by

Ciro Santilli  40 Updated 2025-07-16

The Hadamard gate takes

∣ 0 ⟩

or

∣ 1 ⟩

(quantum states with probability 1.0 of measuring either 0 or 1), and produces states that have equal probability of 0 or 1.

H = \frac{1}{2} [11 1 - 1]

Hadamard gate matrix

.

Figure 1.
Hadamard gate symbol
. Source.

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