What type of mathematical object is the “Pauli vector”?

The three pauli matrices $\sigma_x$, $\sigma_y$, $\sigma_z$ are sometimes combined in the "Pauli vector", usually denoted $\boldsymbol{\sigma} = \sigma_{x} \boldsymbol{e_x} + \sigma_y \boldsymbol{e_y} + \sigma_z \boldsymbol{e_z}$, the intuitive meaning of this is somewhat clear, for example, if one multiplies, for instance, this object by $\frac{\hbar}{2}$ one gets the operator that represents the spin angular momentum "vector" of an electron. This is a "vector operator", angular momentum is a vector in classical mechanics, kinda makes sense. On the other hand, what is the formal meaning of this? Is this a vector belonging to a 3-dimensional vector space over a "pseudo-field" of hermitian matrices? Is this a rank 3 tensor, some sort of 3D grid of numbers? What type of mathematical object is the Pauli Vector?

Apologies if the question is obvious or trivial, something inside me tells me that it may be, but I don't recall working with "vectors of matrices" in any linear algebra course and my efforts to makes sense of this have been in vain so far. Perhaps I have seen this but with a different name or context and I'm just not seeing the connection when written in this notation?

• Well, WP's formal definition is evidently not to your liking. Why don't you luxuriate in the delights of what you may actually do with it, which, e.g., that article illustrates? matrices, may, indeed be manipulated analogously to, and with, numbers. Read up on the rules and do something with them. – Cosmas Zachos Nov 28 '16 at 20:23
• I had read wikipedia's article but not that remark, I guess perhaps you are right, what it does is more important than what it is. "An element of the tensor product of yabba yabba" doesn't say much, and it's more or less what I was expecting I guess. – Ignacio Nov 30 '16 at 1:45
• Indeed, you define the Pauli vector so, dotted by a plain (number) vector it gives you a scalar (matrix) satisfying eqn (1) in that article. From that point on, everything about SU(2) is easy, including the group composition law, following. You'll hardly find anything as nice for rank > 1 Lie groups. – Cosmas Zachos Nov 30 '16 at 1:55

This is best understood using group theory. The rotation algebra $so(3)$ contains both spinor and vector representations. Multiplying two copies of the spin 1/2 representation yields the vector and scalar representations, and the vector part is the Pauli vector.
In ordinary terms: a "vector" is defined as anything with 3 indices that transforms as a vector under rotations: $v^i \rightarrow \sum_j R^{ij}(\theta^1, \theta^2, \theta^3) v^j$, where $R$ is a rotation matrix. You can verify that the quantity $A^i \equiv \psi^\dagger \sigma^i \psi$ is a vector, where $\psi$ is a 2-component spinor and $\sigma^i$, $i=1,2,3$ are the Pauli matrices.
\begin{aligned} A^i=\psi^\dagger \sigma^i \psi \rightarrow \psi^\dagger e^{-i\sigma^3 \theta^3/2} \sigma^i e^{i\sigma^3 \theta^3/2} \psi &= \psi^\dagger (1-i\sigma^3 \theta^3/2) \sigma^i (1+i\sigma^3 \theta^3/2) \psi \\ &=\psi^\dagger \sigma^i \psi - i\theta^3/2\psi^\dagger\left[\sigma^3,\sigma^i\right]\psi \\ &=\psi^\dagger \sigma^i \psi + \theta^3\epsilon_{3ik}\psi^\dagger\sigma^k\psi \end{aligned}
so $A^3\rightarrow A^3$, $A^2 \rightarrow A^2 -\theta^3A^1$, and $A^1\rightarrow A^1+\theta^3 A^2$.