Why Pauli called the following matrix $\:\sigma_x\:$ and not $\:\sigma_y$? \begin{equation} \sigma_x\boldsymbol{=} \begin{bmatrix} 0 & 1 \vphantom{\tfrac{a}{b}}\\ 1 & 0 \vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{01}\label{01} \end{equation}
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$\begingroup$ The Pauli matrices must obey the same commutator algebra as $L_x,L_y,L_z$, and, given a diagonal $\sigma_z$, interchanging $\sigma_x$ with $\sigma_y$ will fail that commutator algebra. $\endgroup$– Cosmas ZachosCommented Jul 8, 2020 at 0:36
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2 Answers
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Any vector in $\mathbb{R}^3$ can be represented by a $2\times2$ hermitian traceless matrix and vice versa. So, there exists a bijection (one-to-one and onto correspondence) between $\mathbb{R}^3$ and the space of $2\times2$ hermitian traceless matrices, let it be $\mathbb{H}$ : \begin{equation} \mathbf{r}\boldsymbol{=}(x,y,z)\in \mathbb{R}^3\;\boldsymbol{\longleftrightarrow} \; \mathrm R= \begin{bmatrix} z & x\boldsymbol{-}iy \\ x\boldsymbol{+}iy & \boldsymbol{-}z \end{bmatrix} \in \mathbb{H} \tag{01} \end{equation} From the usual basis of $\mathbb{R}^3$ \begin{equation} \mathbf{e}_x\boldsymbol{=}\left(1,0,0\right),\quad \mathbf{e}_y\boldsymbol{=}\left(0,1,0\right),\quad \mathbf{e}_z\boldsymbol{=}\left(0,0,1\right) \tag{02} \end{equation} we construct a basis for $\mathbb{H}$ \begin{align} \mathbf{e}_x & \boldsymbol{=}(1,0,0)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_x\boldsymbol{=} \begin{bmatrix} \:\: 0 & \hphantom{\boldsymbol{-}}1\:\:\vphantom{\dfrac{a}{b}}\\ \:\: 1 & \hphantom{\boldsymbol{-}}0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03a}\\ \mathbf{e}_y & \boldsymbol{=}(0,1,0)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_y\boldsymbol{=} \begin{bmatrix} \:\: 0 & \boldsymbol{-}i\:\:\vphantom{\dfrac{a}{b}}\\ \:\: i & \hphantom{\boldsymbol{-}}0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03b}\\ \mathbf{e}_z & \boldsymbol{=}(0,0,1)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_z\boldsymbol{=} \begin{bmatrix} \:\: 1 & \hphantom{\boldsymbol{-}}0\:\:\vphantom{\dfrac{a}{b}}\\ \:\: 0 & \boldsymbol{-}1\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03c} \end{align} where $\:\boldsymbol{\sigma}\equiv(\sigma_x,\sigma_y,\sigma_z)\:$ the Pauli matrices.
Note also that the matrix \begin{equation} U\boldsymbol{=}\cos\tfrac{\theta}{2}\,\mathrm I\boldsymbol{-}i\sigma_x\sin\tfrac{\theta}{2} \boldsymbol{=} \begin{bmatrix} \:\: \cos\tfrac{\theta}{2} & \boldsymbol{-}i\sin\tfrac{\theta}{2}\:\:\vphantom{\dfrac{a}{b}}\\ \:\: \boldsymbol{-}i\sin\tfrac{\theta}{2} & \hphantom{\boldsymbol{-}} \cos\tfrac{\theta}{2}\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{04} \end{equation} is the unitary matrix representation of the rotation around the $x$-axis through an angle $\theta$.
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$\begingroup$ Thank you so much!! It's crystal clear! $\endgroup$– y ingCommented May 1, 2020 at 6:47
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$\begingroup$ @ying out of curiosity, why did $\sigma_y$ seem more natural to you? $\endgroup$ Commented May 1, 2020 at 7:08
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$\begingroup$ @user2723984 To be honest, they both seem strange to me... I think the present form is "a" correct one, but not "the" only solution. I mean, if Pauli switched the right-up and left-lower terms of the matrix shown in equation (01), the formula would be different. But the physical observable should be the same. Am I correct? $\endgroup$– y ingCommented May 1, 2020 at 11:06
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$\begingroup$ well yes of course, it's just a matter of choosing a basis for a space. Everything that matters should be basis independent anyway. The most compelling argument anyway to call that matrix $\sigma_x$ is that $e^{i\theta\sigma_x}$ is a rotation around the $x$ axis $\endgroup$ Commented May 1, 2020 at 11:57
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$\begingroup$ @Frobenius Can you explain what is the justification of this particular form of the Hermitian traceless matrix R? Why is that this expression in particular satisfies the bijection and not any other? $\endgroup$– pll04Commented Jun 21, 2023 at 11:01
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The matrix $$ \sigma_x=\left(\begin{array}{cc} 0&1 \\ 1&0\end{array}\right) $$ is just the (standard) permutation matrix interchanging two objects: if $$ \vert 1\rangle \to \left(\begin{array}{c} 1 \\ 0\end{array}\right)\, ,\qquad \vert 2\rangle \to \left(\begin{array}{c} 0 \\ 1\end{array}\right)\, , $$ then $\sigma_x\vert 1\rangle =\vert 2\rangle$ and $\sigma_x\vert 2\rangle=\vert 1\rangle$ without any phases. $\sigma_y$ introduces an additional phase to the permutation operation.
answered May 1, 2020 at 9:03