Let $| 0 \rangle$ and $| 1 \rangle $ be the states of qubit. Let $\hat{\sigma_x}$, $\hat{\sigma_y}$, $\hat{\sigma_z}$ be Pauli matrices: $$ \hat{\sigma}_{x} = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \;\;\; \hat{\sigma}_{y} = \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right), \;\;\; \hat{\sigma}_{z} = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right). $$ Using the representation $\hat{I} = | 0 \rangle \langle 0 | + |1\rangle \langle 1| $ I tried to receive the similar representation for Pauli matrices: $$ \hat{\sigma}_x = | 1 \rangle \langle 0 | + |0\rangle \langle 1|, \\ \hat{\sigma}_y = -i| 1 \rangle \langle 0 | + i|0\rangle \langle 1|, \\ \hat{\sigma}_z = -| 0 \rangle \langle 0 | + |1\rangle \langle 1|, \\ $$ Is it correct? I'm not sure if it is possible but I assumed that $|0\rangle = (0,1)^{T}$ and $|1\rangle = (1,0)^{T}$ and I looked at $|0\rangle \langle 1|$ as on outer product $(0,1)^{T}(1,0)$. I'm afraid that I use matrices instead of operators etc. Help me please to understand the topic.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
$\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bra}[1]{\left<#1\right|}$ You actually got the notation slightly wrong and had a typo in $I$, but you're almost right.
The usual convention are $$ \begin{align} \ket0&=\begin{bmatrix}1 \\ 0\end{bmatrix}& \ket1&=\begin{bmatrix}0 \\ 1\end{bmatrix}\\ \text{if }\ket\psi&=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}& \text{then }\bra\psi&=\begin{bmatrix}\alpha^* & \beta^*\end{bmatrix} \end{align} $$ So we have $$ \begin{align} I&=\ket0\bra0+\ket1\bra1\\ \sigma_x&=\ket1\bra0+\ket0\bra1\\ \sigma_y&=i\ket1\bra0-i\ket0\bra1\\ \sigma_z&=\ket0\bra0-\ket1\bra1 \end{align} $$
answered Apr 3, 2012 at 12:53
-
$\begingroup$ Thank you, can you give me some book reference please (I got stuck in notation $|0 \rangle = (1,0)^{T}$: $|i\rangle$ corresponds to one of stationary states of psi function of qubit, how we can identify these states with two-dimensional vectors?) $\endgroup$– AppliquéApr 3, 2012 at 14:10
-
$\begingroup$ The qubit notation $\ket0$ $\ket1$ is a "quantum-information standard", so the best book reference is probably the standard textbook : the Nielsen & Chuang ( squint.org/qci ) $\endgroup$ Apr 3, 2012 at 14:57
-
$\begingroup$ About the identification of states with vectors: quantum mechanics tells you that any pure state is a vector in a Hilbert state. The dimension of the vector space depends on the system you look at, and is the number of distinguishable (orthogonal) states. By definition, a qubit has 2 two orthogonal states and lives in dimension 2. (A qutrit in dim. 3; a ququart in dim 4). There is nothing special in dimension 2, except that it's the simplest non-trivial case. $\endgroup$ Apr 3, 2012 at 15:01
-
1$\begingroup$ Is there a consistent way to get this answer? Or do we have to guess? $\endgroup$ Mar 23, 2021 at 19:02