How to prove the det of the dot product of a vector and a Pauli vector is minus the vector itself squared?

Question

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?

Answer 1

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 $\det\sigma_\gamma=c$,$$c^2=\det(\sigma_1\sigma_2)=\det(i\sigma_3)=-\det\sigma_3=-c\implies c=-1.$$

Answer 2

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

Answer 3

Let $a=|\mathbf a|$ and write $\mathbf a= a \mathbf n$ with $|\mathbf n|=1$. It suffices to show that $\det \mathbf n \cdot \boldsymbol \sigma = -1$. To this end, note that we can easily prove$^\dagger$ the following well-known relation:

$$e^{ia\mathbf n \cdot \boldsymbol \sigma} = \mathbb I \cos(a) + i \sin(a)\, \mathbf n \cdot \boldsymbol \sigma \quad . $$

This holds for all $a \in \mathbb R$, so choose $a=\pi/2$ to obtain

$$\det e^{i\frac{\pi}{2}\mathbf n \cdot \boldsymbol \sigma} = \det i\,\mathbf n \cdot \boldsymbol \sigma = - \det \mathbf n \cdot \boldsymbol \sigma \quad . $$

Finally, by using $\det e^A = e^{\mathrm{Tr}A}$, we find $$\det \mathbf n \cdot \boldsymbol \sigma = -1 $$

and hence $$\det \mathbf a \cdot \boldsymbol \sigma = \det a \mathbf n \cdot \boldsymbol \sigma = a^2 \det \mathbf n \cdot \boldsymbol \sigma = -a^2 \quad .$$

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$^\dagger$ This relation can be shown without using the explicit form of the Pauli matrices. In fact, it immediately follows from the relation the OP is suggesting.