Quantum Country exercise 1

This is the first in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 1. For the Hadamard matrix H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1& -1 \end{bmatrix}, verify that HH = I where I is the identity matrix \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Answer: We have

HH = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1& -1 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1& -1 \end{bmatrix}

= \frac{1}{2} \begin{bmatrix} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 1 + 1 \cdot (-1) \\ 1 \cdot 1 + (-1) \cdot 1 & 1 \cdot 1 + (-1) \cdot (-1) \end{bmatrix}

= \frac{1}{2} \begin{bmatrix} 1 + 1 & 1 - 1 \\ 1 - 1 & 1 + 1 \end{bmatrix}

= \frac{1}{2} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

So we have shown that HH = I.

This entry was posted in quantum country. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s