- #1
Ackbach
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I am beginning a study I have long wanted to engage in: quantum computing. This is a field lying at the intersection of mathematics, physics, computer science, and electrical engineering - all topics I studied, to varying levels. From time to time, I plan on posting notes and summaries that might prove useful to others studying the same thing. Without further ado:
$$\begin{array}{|c|c|c|c|c|c|} \hline
\textbf{Name} &\textbf{Matrix} &A^{\dagger}A=I? &A=A^{\dagger}? &\textbf{E-values} &\textbf{Norm. E-vectors} \\
\hline
\text{Hadamard} &H=\dfrac{1}{\sqrt{2}}\begin{bmatrix}1 &1\\1 &-1\end{bmatrix} &\text{Yes} &\text{Yes}
&1,\; -1 &\dfrac{1}{\sqrt{4-2\sqrt{2}}}\begin{bmatrix}1 \\ \sqrt{2}-1\end{bmatrix}, \;
\dfrac{1}{\sqrt{4+2\sqrt{2}}}\begin{bmatrix}1 \\ -\sqrt{2}-1\end{bmatrix} \\ \hline
\text{Pauli }X &X=\begin{bmatrix}0 &1\\1 &0\end{bmatrix} &\text{Yes} &\text{Yes} &1, \; -1
&\dfrac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}, \; \dfrac{1}{\sqrt{2}}
\begin{bmatrix}1\\-1\end{bmatrix} \\ \hline
\text{Pauli }Y &Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix} &\text{Yes} &\text{Yes}
&1, \; -1 &\dfrac{1}{\sqrt{2}}\begin{bmatrix}1\\i\end{bmatrix}, \; \dfrac{1}{\sqrt{2}}
\begin{bmatrix}1\\-i\end{bmatrix} \\ \hline
\text{Pauli }Z &Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix} &\text{Yes} &\text{Yes}
&1, \; -1 &\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\text{Phase} &S=\begin{bmatrix}1&0\\0&i\end{bmatrix} &\text{Yes} &\text{No} &1,\;i
&\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\pi/8 &T=\begin{bmatrix}1&0\\0&e^{i\pi/4}\end{bmatrix} &\text{Yes} &\text{No} &1, \; e^{i\pi/4}
&\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|} \hline
\textbf{Name} &\textbf{Matrix} &A^{\dagger}A=I? &A=A^{\dagger}? &\textbf{E-values} &\textbf{Norm. E-vectors} \\
\hline
\text{Hadamard} &H=\dfrac{1}{\sqrt{2}}\begin{bmatrix}1 &1\\1 &-1\end{bmatrix} &\text{Yes} &\text{Yes}
&1,\; -1 &\dfrac{1}{\sqrt{4-2\sqrt{2}}}\begin{bmatrix}1 \\ \sqrt{2}-1\end{bmatrix}, \;
\dfrac{1}{\sqrt{4+2\sqrt{2}}}\begin{bmatrix}1 \\ -\sqrt{2}-1\end{bmatrix} \\ \hline
\text{Pauli }X &X=\begin{bmatrix}0 &1\\1 &0\end{bmatrix} &\text{Yes} &\text{Yes} &1, \; -1
&\dfrac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}, \; \dfrac{1}{\sqrt{2}}
\begin{bmatrix}1\\-1\end{bmatrix} \\ \hline
\text{Pauli }Y &Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix} &\text{Yes} &\text{Yes}
&1, \; -1 &\dfrac{1}{\sqrt{2}}\begin{bmatrix}1\\i\end{bmatrix}, \; \dfrac{1}{\sqrt{2}}
\begin{bmatrix}1\\-i\end{bmatrix} \\ \hline
\text{Pauli }Z &Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix} &\text{Yes} &\text{Yes}
&1, \; -1 &\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\text{Phase} &S=\begin{bmatrix}1&0\\0&i\end{bmatrix} &\text{Yes} &\text{No} &1,\;i
&\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\pi/8 &T=\begin{bmatrix}1&0\\0&e^{i\pi/4}\end{bmatrix} &\text{Yes} &\text{No} &1, \; e^{i\pi/4}
&\begin{bmatrix}1\\0\end{bmatrix}, \; \begin{bmatrix}0\\1\end{bmatrix} \\ \hline
\end{array}$$
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