Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.
Sign up to join this community
The Hadamard gate is a unitary gate, but how does the matrix times its own conjugate transpose actually result in the $I$ matrix? I am currently looking at it as a scalar, 0.707..., multiplied by the matrix of ones and negative one.
\begin{equation} H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}
The Hadamard gate is described by this matrix \begin{equation} H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation} Conjugate transpose of $H$ is again $H$. Hence we have to check if $HH$ is $I$.
Multiplication goes as follows ($h_{ij}$ denotes elements of resulting matrix):
So
\begin{equation} HH=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =I. \end{equation}
Just a side comment to this question.
In the real world nothing is so obvious.
Example: Try to define Hadamard gate in Ruby language. Even when trying to define gate by using rational numbers which helps us sometimes to keep good precision the answer is that Hadamard is not unitary.
m = (1 / Math.sqrt(2)).to_r * Matrix[[1, 1],
[1, -1]]
Then
m.unitary?
results false and result of m*m multiplication is as below:
Matrix[[2535301200456458353478625262249/2535301200456458802993406410752, 0/1],
[0/1, 2535301200456458353478625262249/2535301200456458802993406410752]]
That means if we really want to be sure that something is 'true' as it should be, we need to double check existing feature in language and maybe fix/write own code when needed.
m.unitary? give true if you omit the .to_r? Of course $1/\sqrt{2}$ isn't rational, so it isn't clear to me whether the fractional form your code uses would be more precise.
$\endgroup$