process matrix - physical interpretation I have a (probably) advanced question, concerning quantum process tomography.
Let's say I have made a measurement with a single qubit, and calculated a $\chi$-matrix which looks like $$\begin{bmatrix}0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}.$$
What's the physical interpretation for this, and why do I need a 4$\times$4 matrix for describing a 1 qubit system?
 A: I took a quantum information course last semester and we only touched quantum process tomography very briefly, so I'm no expert, but from what I understand from Nielsen & Chuang, I mighty still be able to give an answer to your questions.
The goal of process tomography is to find a set of operators $\{E_i\}$ which describes the effect of a superoperator $\mathcal E$ on density matrices,
$\mathcal E(\rho)=\sum_iE_i\rho E_i^\dagger$. For a qubit, the density matrices are from $\mathbb C^{2\times 2}$ and so are the operators $E_i$. Now, as the $E_i$s are initially unknown, we have to express them in some (arbitrarily chosen) basis of $\mathbb C^{2\times 2}$. The standard choice is $(I, X, Y, Z)$, i.e. the identity and the Pauli matrices. The particular choice of this basis (and the order of the basis matrices) will be important in the interpretation of $\chi$. Now, referring to those basis matrices as $\tilde E_m$, we may write
$E_i = \sum_m e_{im}\tilde E_m$
with coefficients $e_{im}$. Substituting the above, one gets for the superoperator
$\mathcal E(\rho) = \sum_{m,n}\left(\sum_ie_{im}e_{in}^*\right)\tilde E_m\rho\tilde E_n^\dagger = \sum_{m,n}\chi_{mn}\tilde E_m\rho\tilde E_n^\dagger$.
So, the reason why the $\chi$ matrix is $4\times 4$ for a single qubit is that the vector space for the operators on single qubits is four-dimensional. In general, a superoperator on a quantum system with a $d$-dimensional Hilbert space will be representable in terms of a $d^2\times d^2$ $\chi$ matrix.
The physical interpretation of the $\chi$ matrix you gave above depends on the choice of basis, as I mentioned before. In the conventional identity+Paulis basis, it means that $\mathcal E(\rho) = X\rho X$, i.e. the quantum process in simply a quantum NOT gate.
