Week 6 - my notes

• X - 180 rotation around the X-axis
• Z - 180 rotation around the Z-axis
• H - creates a superposition

• \(|0\rangle\), \(|1\rangle\), \(|+\rangle\), \(|-\rangle\) are ways to represent them.
• But in an unequal superposition, we need a better way to represent contributions.
• That is why we will use two numbers in the form of a vector.
• 0 state is represented as \[\begin{bmatrix}1\\0\end{bmatrix}\]
• 1 state is represented as \[\begin{bmatrix}0\\1\end{bmatrix}\]
• It is worth mentioning that the top number represents the component of \(|0\rangle\) and the bottom, the component of \(|1\rangle\).

\begin{bmatrix} |0\rangle\ component\\|1\rangle\ component\end{bmatrix}

• To represent equal superpositions, we just need to add the two vectors: \[\begin{bmatrix}1\\0\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix}\]
• To get to an unequal superposition, we need to apply some gates to get there and this is the goal of a field called quantum algorithm design

• The vectors representing quantum states must be normalized, which means they have a lenght of 1
• The radius of the Bloch Sphere is 1, so as all quantum states lie on the Bloch sphere, they must all have a length of 1.
• [ ] Step 1: Write the state in the vector form \[\begin{bmatrix}a\\b\end{bmatrix}\]
• [ ] Step 2: Find the length of the state as \[\sqrt{a^2 + b^2}\]
• [ ] Step 3: If the length is 1, the state is normalized. If not, divide the state by its length to get the normalized state.
  • Normalized form of the state = \[\frac{1}{\sqrt{a^2 + b^2}}\begin{bmatrix}a\\b\end{bmatrix}\]

• Remember that all of this comes from linear algebra
• The length of the state \[\begin{bmatrix}1\\0\end{bmatrix}\]
  • A: 1
• Find the normalized state of \[\begin{bmatrix}1\\1\end{bmatrix}\]
  • Finding the length: \[\sqrt{1^2 + 1^2} = \sqrt{2}\]
  • As the length is not 1, we need to divide take the state and divide it by its length, then we get the normalized state:

\[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\]

• \(|+\rangle\) = \[\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\] same as \[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\]
• \(|-\rangle\) = \[\\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)\] same as \[\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}\]
• These two states above are both equal superpositions but they differ in phase
• Therefore, the sign between \(|0\rangle\) and \(|1\rangle\) is called phase. Is it positive or negative?
• The phase diferences lead to interference (more coming soon)
• There are more relations in phase that won’t be addressed in this course because they require complex numbers