Why is the identity not considered when expanding a $2 \times 2$ matrix in the Pauli basis? [closed]

I am aware of the expansion of a two dimensional matrix $$M$$ in Pauli basis given by

$$M = \sum_{\mu=0,1,2,3} c_\mu \sigma_\mu$$

with $$\sigma_0 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$, the Identity matrix and $$\sigma_{1,2,3}$$ the three Pauli matrices.

However, in this published article on page-14, below equation 28, one finds the following:

Parameterising the qubit operators as $$Q = a \cdot \hat{\sigma}$$ , with $$\hat{\sigma}$$ the vector of Pauli matrices and $$a$$, a unit vector (I have dropped the subscripts $$i$$ on $$Q$$ and $$a$$ here which merely labels the time in this case and is irrelevant here)

My question: Why is the Identity matrix not taken into account as per this article?

• Presumably they are allowed to assume $Q$ to be traceless for whatever reason (possibly related to assuming that observables have mean 0?). – jacob1729 Apr 7 at 12:34
• ... or "vector of Pauli matrices" includes the identity as $\sigma_0$. – Norbert Schuch Apr 7 at 14:08

The set of $$2\times 2$$ complex matrices can be understood as a real vector space of dimension $$2\times 2^2$$, or equivalently, as a complex vector space of dimension $$2^2$$.
You can easily verify that a basis for the vector space is given by the Pauli matrices and the identity matrix. More precisely, we can write $$\mathfrak{gl}(2,\mathbb C) = \mathrm{span}_{\mathbb R}(\{I,iI,\sigma_x,i\sigma_x,\sigma_y,i\sigma_y,\sigma_z,i\sigma_z\}) = \mathrm{span}_{\mathbb C}(\{I,\sigma_x,\sigma_y,\sigma_z\}),$$ where I'm denoting with $$\mathfrak{gl}(2,\mathbb C)$$ the space of $$2\times2$$ complex matrices.
You clearly need the identity in the basis, as $$I$$ cannot be written as a linear combination of Pauli matrices.