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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?

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

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