# Representations of Pauli matrices involving outer product of qubit states

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.

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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}
 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?) – Nimza Apr 3 '12 at 14:10 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 ) – Frédéric Grosshans Apr 3 '12 at 14:57 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. – Frédéric Grosshans Apr 3 '12 at 15:01 Thank you for explication – Nimza Apr 3 '12 at 15:39