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?
Best Answer
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 $\mathrm{det}(\mathbf a \cdot \boldsymbol \sigma)= - a^2$.