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From here I learn that, if Pauli vector is defined as $\boldsymbol\sigma=\sigma_\alpha\hat x_\alpha$, and $\boldsymbol a$ denotes a vector, whose components are all numbers, not matrices, then $$\det(\boldsymbol{a}\cdot\boldsymbol{\sigma})=-\boldsymbol a\cdot\boldsymbol a$$ That website proves this by concretizing the form of Pauli matrices.

However, I want to find a way of proving independent of the explicit form of Pauli matrices in a representation, such as starting from the relation $$ \sigma _{\alpha}\sigma _{\beta}=\mathrm{\delta}_{\alpha \beta}I_2+\mathrm{i}\epsilon _{\alpha \beta \gamma}\sigma _{\gamma} $$ Could anyone help me? I have no idea how to deal with it or search it.

What's more, if the component of $\boldsymbol a$ is also a matrix, in which way the conclusion should be fixed?

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    $\begingroup$ Please note that English is this site's standard. It would be great if you could translate the Chinese (?) part. Thanks! $\endgroup$
    – Jonas
    Dec 5 '21 at 20:39
  • $\begingroup$ Sorry about that, I wrote the Chinese part as I am not certain if my translation (I want to find a way of proving independent with the concrete form of Pauli matrices in a representation) of it is precise... I try to edit it at once, thanks! $\endgroup$
    – SHBooKP
    Dec 5 '21 at 21:00
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A cute proof that involves your suggested relation goes as follows. First, note that by definition, $\mathbf a \cdot \boldsymbol \sigma$ is a $2\times 2$ traceless Hermitian matrix. As such, its eigenvalues are $\pm c$ for some $c\in \mathbb R$ (because its trace is just the sum of its eigenvalues) and its determinant is $-c^2$ (because its determinant is the product of its eigenvalues). Therefore, we have that $$\mathrm{det}\big((\mathbf a \cdot \boldsymbol \sigma)^2\big)= \mathrm{det}(\mathbf a \cdot \boldsymbol \sigma)^2 = c^4$$

However, note that $$(\mathbf a \cdot \boldsymbol \sigma)^2= a_i a_j \sigma_i \sigma_j = a_i a_j (\delta_{ij} I_2 + i\epsilon_{ijk} \sigma_k)= a^2 I_2$$ where we've used that, because $a_ia_j$ and $\epsilon_{ijk}$ are respectively symmetric and antisymmetric under the exchange $i\leftrightarrow j$, their contraction vanishes. However, $\mathrm{det}(a^2 I_2) = a^4$, and so comparison with the above yields that $\mathrm{det}(\mathbf a \cdot \boldsymbol \sigma)= - a^2$.

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For a coordinate-free proof, we physicists think about the problem's symmetries. The left-hand side is a rotationally invariant scalar function of $\boldsymbol a$ that multiplies by $\lambda^2$ under $\boldsymbol a\mapsto\lambda\boldsymbol a$, so some $c\in\Bbb R$ satisfies $\det(\boldsymbol a\cdot\boldsymbol\sigma)=c\boldsymbol a\cdot\boldsymbol a$ for all $\boldsymbol a\in\Bbb R^3$. We want to prove $c=-1$. If $\boldsymbol a$ is an arbitrary element of the standard basis of $\Bbb R^3$, say $a_\alpha=\delta_{\alpha\gamma}$ for some $\gamma\in\{1,\,2,\,3\}$,$$c=\det(\boldsymbol a\cdot\boldsymbol\sigma)=\det\sigma_\gamma=\pm\sqrt{\det\sigma_\gamma^2}=\pm\sqrt{\det I_2}=\pm1.$$Since all $\det\sigma_\gamma=c$,$$c^2=\det(\sigma_1\sigma_2)=\det(i\sigma_3)=-\det\sigma_3=-c\implies c=-1.$$

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