How to write an arbitrary qubit density operator?

Question

Physics noob here: I am reading the Wikipedia on Density Matrices (https://en.wikipedia.org/wiki/Density_matrix), and in the section labeled "Pure and mixed states", it states

"An arbitrary state for a qubit can be written as a linear combination of the Pauli matrices, which provide a basis for $2\times2$ self-adjoint matrices: $$\rho = \frac{1}{2}(I +r_x\sigma_x + r_y\sigma_y + r_z\sigma_z)."$$

My question is, how is this arbitrary? Couldn't I cook up some density operator that you can't write using this equation? i.e something like ~$\begin{pmatrix}i&i\\i&i\end{pmatrix}$ (not including normalization). What am I missing here?

Possible duplicate: How to prove that every mixed one-qubit state admits a Bloch-sphere representation? — Qmechanic, May 06 '22 at 12:21
Possibly related: Necessary and sufficient conditions for operator on $\mathbb C^2$ to be a density matrix — Tobias Fünke, May 06 '22 at 13:37

score 2 · Accepted Answer · answered May 06 '22 at 12:01

The density matrix has to be self-adjoint by definition and your example is not self-adjoint, or respectively, hermitian, i.e. $\rho \neq \rho^\dagger$.
As you wrote, the Pauli matrices provide a basis for self-adjoint matrices, so therefore you can write any self-adjoint matrix as a linear combination of the Pauli-matrices, similarly to how you can write any vector in $\mathbb{R}^3$ by a linear combination of the canonical basis vectors $e_1,e_2,e_3$.

This self-adjointness is the point I missed. Thanks! – physics_fan_123 May 06 '22 at 12:51 — physics_fan_123, May 06 '22 at 12:51

How to write an arbitrary qubit density operator?

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