Bloch representation. Why Pauli operators?

Why do I know that a general qubit state can be written as $$\rho = \frac 1 2 \big(\mathbb 1 +\vec r \vec \sigma\big)\;\text ?$$

It is clear that the factor of $1/2$ comes from $\text{tr}\rho=1$. But is the set $\{\mathbb 1 , \sigma_x,\sigma_y,\sigma_z\}$ a unique basis for positive hermitian operators? If so, how can I see that there is no other?

• There are infinitely many bases for positive hermitian operators, just like any normal vector space. This is just the decomposition of $\rho$ in one particular complete, orthogonal basis. Commented Apr 11, 2015 at 11:34
• @MarkMitchison: That looks like an answer to me :P Commented Apr 11, 2015 at 12:44
• What is still unclear to me is the fact that the coefficients are just real: $\vec r=(r_x,r_y,r_z)\in\mathbb R^3$. Why is it still general? Or is $\rho = \rho^\dagger$ enough to prove that they have to be real? Commented Apr 13, 2015 at 15:37
• @thyme: yes that's enough. This is also true because the Hermitian matrices are a real vector space - not a complex one. Commented Apr 13, 2015 at 20:28
• Just want to add that these Pauli operators give us the intuition of orientation too. The r vector you have in front of these matrices could be any garbage vector if we take arbitrary matrices. Commented Jul 22, 2022 at 8:45

First note that the space of Hermitian matrices is a vector space, but a real one. Real linear combinations of Hermitian matrices $A$ and $B$ are also Hermitian, i.e. $\alpha A+B$ is Hermitian for $\alpha \in \mathbb{R}$. However, it is not a complex vector space, since $iA$ is anti-Hermitian for Hermitian $A$ as $(iA)^{\dagger}=-iA^{\dagger}=-iA$.
Now, we need to find a basis for the Hermitian operators. General considerations tell us that the real dimension of the vector space is four, i.e. our basis must consist of four linear independent Hermitian matrices. There are infinitely many bases of such kind, but since the Pauli matrices are very familiar to us, we'll stick to those. Adding unity, we have four matrices that are clearly linear independent and Hermitian: $\{1,\sigma_x,\sigma_y,\sigma_z\}$ hence any Hermitian $2\times 2$ matrix $A$ can be represented as a linear combination $A=x_11+x_2\sigma_x+x_3\sigma_y+x_4\sigma_z$ for some $x_i\in \mathbb{R}$.
Since we are talking about states, as you say, the normalization $\operatorname{tr}(A)=1$ forces us to fix $x_1$ and therfore, we obtain the formula you quote. The last constraint on $\vec{r}$ which is not mentioned in the question is obtained when we make sure that the state is a positive semi-definite matrix.