# Why is a Hadamard gate unitary?

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.

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

• 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... – Mariia Mykhailova Jan 3 at 8:36
• 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 :) – VP9 Jan 3 at 8:42
• @VP9 but what you're describing in your previous comment is just the matrix, not the matrix multiplied by itself. – DaftWullie Jan 3 at 10:17

The Hadamard gate is described by this matrix $$$$H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$$$ Conjugate transpose of $$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

$$$$HH=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =I.$$$$

• mind blown. poof. – VP9 Jan 3 at 18:44
• Is the answer satisfactory for you? If so, would you mind to accept it? – Martin Vesely Jan 3 at 18:47