Can anyone explain the process required to make a Hadamard gate that acts on 1st, 2nd and 3rd qbits?

For Hgates acting on the first qubit i realise the matrix is $H=\begin{pmatrix} 1&1\\1&-1\end{pmatrix}$, but I am unsure how to formulate such a gate. As to acting on the second qubit I have read this web page 1 but I am not really sure if it is correct, because it is just a 4x4 matrix without any zeroes. I found this $\begin{pmatrix} 1 & 1 \\-1 & 1 \end{pmatrix} $ on another website which will obviously have zeroes when it is multiplied by $I$. Simply can anyone explain how Hadarmd gates can be formulated and built to n qubits?

  • $\begingroup$ The web page talks about a Hadamard transform and not a Hadamard gate. To perform a Hadamard transform to an $n$-qubit state, you apply a Hadamard gate to each qubit individually. So the website is correct, but it doesn't directly apply to your question. $\endgroup$ – Peter Shor Mar 11 '13 at 10:14

For two qubits, the answer is as follows.

The Hadamard gate acting on the second qubit is $I \otimes H$:

$$ A = \frac{1}{\sqrt{2}} \left( \begin{array}{rrrr} 1&1&0&0\\ 1&-1&0&0\\ 0&0&1&1\\ 0&0&1&-1 \end{array} \right). $$

The Hadamard gate acting on the first qubit is $H \otimes I$:

$$ B = \frac{1}{\sqrt{2}} \left( \begin{array}{rrrr} 1&0&1&0\\ 0&1&0&1\\ 1&0&-1&0\\ 0&1&0&-1 \end{array} \right). $$

For three qubits, the matrices for the Hadamard gate on the 3rd, 2nd, and 1st qubits are

$$ \left(\begin{array}{cc} A & 0 \\ 0 & A \end{array}\right), $$ $$ \left(\begin{array}{cc} B & 0 \\ 0 & B \end{array}\right), $$ $$ \frac{1}{\sqrt{2}} \left(\begin{array}{rr} I & I \\ I & -I \end{array}\right). $$

Hopefully, you can see the pattern.


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