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

The normalized Hadamard transform maps a vector indexed by binary strings through entries $2^{-n/2}(-1{)}^{x\cdot y}$, where the dot product is evaluated modulo two. It is its own inverse and an orthogonal transformation, hence also a unitary operator on complex input vectors. The Hadamard gate is the $n=1$ quantum case, and the Hadamard basis is the image of the computational basis. The Walsh-Hadamard transform gives the same tensor product construction.

For two qubits, applying one Hadamard gate to each prepares the uniform quantum superpositionEach probability amplitude is $1/2$, so each computational-basis measurement outcome has probability $1/4$. Applying the Hadamard transform again reconstructs $|00⟩$: signed contributions cancel at the other three outputs. Its usefulness lies in combining simple local quantum gates into an operation on exponentially many amplitudes. Extracting useful information still requires an appropriate subsequent transformation and measurement.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

A flat reference over $k$ adjacent Fock states has overlap $r_k$ with a one-photon shift and $s_k$ with a two-photon shift. These overlaps, obtained by counting shared number states, determine the reduced qubit channel of a number-conserving Hadamard interaction. For $k\ge 2$, its squared fidelity with the desired pure output is $1-(1+xz)/(2k)$ for input Bloch vector $(x,y,z)$. Its worst-case infidelity is $3/(4k)$ because $|xz|\le 1/2$. At $k=1$ the two-shift overlap is zero, not $-1$, and the fidelity is $1/2+(y^2-xz)/4$. Broad coherent number support, rather than merely a large fixed photon number, suppresses distinguishability of the outgoing field states.

The uniform superposition of all $n$-bit strings is

On $n$ qubits, the Walsh-Hadamard transform satisfieswhere the dot product is evaluated modulo two.

This is the Hadamard transform acting on the computational basis of $n$ qubits.

The qubit Hadamard basis consists of $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$.