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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}

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    $\begingroup$ Have you tried taking the conjugate transpose of the H matrix (which is equal to itself) and multiplying it by H to see what the result looks like? I'm not sure what exactly the question is... $\endgroup$ – Mariia Mykhailova Jan 3 at 8:36
  • $\begingroup$ sorry i am trying to learn. when i look at the matrix, i see a probability on the left (0.707), which can treated be a scalar and multiplied across each value in the matrix on the right. The first column would be .707, .707, and the second would be .707, -.707. So in my head i am picturing something other than the I matrix. I appreciate your help :) $\endgroup$ – VP9 Jan 3 at 8:42
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    $\begingroup$ @VP9 but what you're describing in your previous comment is just the matrix, not the matrix multiplied by itself. $\endgroup$ – DaftWullie Jan 3 at 10:17
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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):

  1. $h_{11} = 1\cdot1 + 1\cdot1 = 2$
  2. $h_{12} = 1\cdot 1 + 1\cdot (-1) = 0$
  3. $h_{21} = 1 \cdot1 + (-1) \cdot 1= 0$
  4. $h_{21} = 1 \cdot1 + (-1)\cdot (-1) = 2$

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}

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  • $\begingroup$ mind blown. poof. $\endgroup$ – VP9 Jan 3 at 18:44
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    $\begingroup$ Is the answer satisfactory for you? If so, would you mind to accept it? $\endgroup$ – Martin Vesely Jan 3 at 18:47
  • $\begingroup$ @VP9 What is it about this answer that blows your mind? $\endgroup$ – tparker Jan 5 at 5:51

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