Not to be confused with Walsh matrix.
The product of a Boolean function and a Walsh matrix is its Walsh spectrum:[1]
(1,0,1,0,0,1,1,0) × H(8) = (4,2,0,−2,0,2,0,2)
Fast Walsh–Hadamard transform, a faster way to calculate the Walsh spectrum of (1,0,1,0,0,1,1,0).
The original function can be expressed by means of its Walsh spectrum as an arithmetical polynomial.

The Hadamard transform (also known as the Walsh–Hadamard transform, Hadamard–Rademacher–Walsh transform, Walsh transform, or Walsh–Fourier transform) is an example of a generalized class of Fourier transforms. It performs an orthogonal, symmetric, involutive, linear operation on 2m real numbers (or complex numbers, although the Hadamard matrices themselves are purely real).

The Hadamard transform can be regarded as being built out of size-2 discrete Fourier transforms (DFTs), and is in fact equivalent to a multidimensional DFT of size 2 × 2 × ⋯ × 2 × 2.[2] It decomposes an arbitrary input vector into a superposition of Walsh functions.

The transform is named for the French mathematician Jacques Hadamard, the German-American mathematician Hans Rademacher, and the American mathematician Joseph L. Walsh.

## Definition

The Hadamard transform Hm is a 2m × 2m matrix, the Hadamard matrix (scaled by a normalization factor), that transforms 2m real numbers xn into 2m real numbers Xk. The Hadamard transform can be defined in two ways: recursively, or by using the binary (base-2) representation of the indices n and k.

Recursively, we define the 1 × 1 Hadamard transform H0 by the identity H0 = 1, and then define Hm for m > 0 by:

${\displaystyle H_{m}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}H_{m-1}&H_{m-1}\\H_{m-1}&-H_{m-1}\end{pmatrix}}}$

where the 1/√2 is a normalization that is sometimes omitted.

For m > 1, we can also define Hm by:

${\displaystyle H_{m}=H_{1}\otimes H_{m-1}}$

where ${\displaystyle \otimes }$ represents the Kronecker product. Thus, other than this normalization factor, the Hadamard matrices are made up entirely of 1 and −1.

Equivalently, we can define the Hadamard matrix by its (kn)-th entry by writing

${\displaystyle k=\sum _{i=0}^{m-1}{k_{i}2^{i}}=k_{m-1}2^{m-1}+k_{m-2}2^{m-2}+\cdots +k_{1}2+k_{0}}$

and

${\displaystyle n=\sum _{i=0}^{m-1}{n_{i}2^{i}}=n_{m-1}2^{m-1}+n_{m-2}2^{m-2}+\cdots +n_{1}2+n_{0}}$

where the kj and nj are the binary digits (0 or 1) of k and n, respectively. Note that for the element in the top left corner, we define: ${\displaystyle k=n=0}$. In this case, we have:

${\displaystyle \left(H_{m}\right)_{k,n}={\frac {1}{2^{\frac {m}{2}}}}(-1)^{\sum _{j}k_{j}n_{j}}}$

This is exactly the multidimensional ${\displaystyle \scriptstyle 2\,\times \,2\,\times \,\cdots \,\times \,2\,\times \,2}$ DFT, normalized to be unitary, if the inputs and outputs are regarded as multidimensional arrays indexed by the nj and kj, respectively.

Some examples of the Hadamard matrices follow.

{\displaystyle {\begin{aligned}H_{0}=&+1\\H_{1}={\frac {1}{\sqrt {2}}}&{\begin{pmatrix}{\begin{array}{rr}1&1\\1&-1\end{array}}\end{pmatrix}}\end{aligned}}}

(This H1 is precisely the size-2 DFT. It can also be regarded as the Fourier transform on the two-element additive group of Z/(2).)

{\displaystyle {\begin{aligned}H_{2}={\frac {1}{2}}&{\begin{pmatrix}{\begin{array}{rrrr}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{array}}\end{pmatrix}}\\H_{3}={\frac {1}{2^{\frac {3}{2}}}}&{\begin{pmatrix}{\begin{array}{rrrrrrrr}1&1&1&1&1&1&1&1\\1&-1&1&-1&1&-1&1&-1\\1&1&-1&-1&1&1&-1&-1\\1&-1&-1&1&1&-1&-1&1\\1&1&1&1&-1&-1&-1&-1\\1&-1&1&-1&-1&1&-1&1\\1&1&-1&-1&-1&-1&1&1\\1&-1&-1&1&-1&1&1&-1\end{array}}\end{pmatrix}}\\(H_{n})_{i,j}={\frac {1}{2^{\frac {n}{2}}}}&(-1)^{i\cdot j}\end{aligned}}}

where ${\displaystyle i\cdot j}$ is the bitwise dot product of the binary representations of the numbers i and j. For example, if ${\displaystyle \scriptstyle n\geq 2}$, then ${\displaystyle \scriptstyle ({H_{n}})_{3,2}\;=\;(-1)^{3\cdot 2}\;=\;(-1)^{(1,1)\cdot (1,0)}\;=\;(-1)^{1+0}\;=\;(-1)^{1}\;=\;-1}$, agreeing with the above (ignoring the overall constant). Note that the first row, first column of the matrix is denoted by ${\displaystyle \scriptstyle ({H_{n}})_{0,0}}$.

The rows of the Hadamard matrices are the Walsh functions.

## Computational complexity

The Hadamard transform can be computed in n log n operations (n = 2m), using the fast Hadamard transform algorithm.

## Quantum computing applications

Further information: Quantum gate § Hadamard gate

In quantum information processing the Hadamard transformation, more often called Hadamard gate in this context (cf. quantum gate), is a one-qubit rotation, mapping the qubit-basis states ${\displaystyle |0\rangle }$ and ${\displaystyle |1\rangle }$ to two superposition states with equal weight of the computational basis states ${\displaystyle |0\rangle }$ and ${\displaystyle |1\rangle }$. Usually the phases are chosen so that we have

${\displaystyle H={\frac {|0\rangle +|1\rangle }{\sqrt {2}}}\langle 0|+{\frac {|0\rangle -|1\rangle }{\sqrt {2}}}\langle 1|}$

in Dirac notation. This corresponds to the transformation matrix

${\displaystyle H_{1}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}}$

in the ${\displaystyle |0\rangle ,|1\rangle }$ basis.

Many quantum algorithms use the Hadamard transform as an initial step, since it maps n qubits initialized with ${\displaystyle |0\rangle }$ to a superposition of all 2n orthogonal states in the ${\displaystyle |0\rangle ,|1\rangle }$ basis with equal weight.

It is useful to note that computing the quantum Hadamard transform is simply the application of a Hadamard gate to each qubit individually because of the tensor product structure of the Hadamard transform. This simple result means the quantum Hadamard transform requires n operations, compared to the classical case of n log n operations.

{\displaystyle {\begin{aligned}H(|0\rangle )&={\frac {1}{\sqrt {2}}}|0\rangle +{\frac {1}{\sqrt {2}}}|1\rangle \\H(|1\rangle )&={\frac {1}{\sqrt {2}}}|0\rangle -{\frac {1}{\sqrt {2}}}|1\rangle \\H\left({\frac {1}{\sqrt {2}}}|0\rangle +{\frac {1}{\sqrt {2}}}|1\rangle \right)&={\frac {1}{2}}(|0\rangle +|1\rangle )+{\frac {1}{2}}(|0\rangle -|1\rangle )=|0\rangle \\H\left({\frac {1}{\sqrt {2}}}|0\rangle -{\frac {1}{\sqrt {2}}}|1\rangle \right)&={\frac {1}{2}}(|0\rangle +|1\rangle )-{\frac {1}{2}}(|0\rangle -|1\rangle )=|1\rangle \end{aligned}}}
One application of the Hadamard gate to either a 0 or 1 qubit will produce a quantum state that, if observed, will be a 0 or 1 with equal probability (as seen in the first two operations). This is exactly like flipping a fair coin in the standard probabilistic model of computation. However, if the Hadamard gate is applied twice in succession (as is effectively being done in the last two operations), then the final state is always the same as the initial state. This would be like taking a fair coin that is showing heads, flipping it twice, and it always landing on heads after the second flip. The ket ${\displaystyle |0\rangle ={\begin{bmatrix}1\\0\\\end{bmatrix}}}$ and the ket ${\displaystyle |1\rangle ={\begin{bmatrix}0\\1\\\end{bmatrix}}}$.