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

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?

• Please note that English is this site's standard. It would be great if you could translate the Chinese (?) part. Thanks! Dec 5, 2021 at 20:39
• 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! Dec 5, 2021 at 21:00

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

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

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

$$^\dagger$$ 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.