In quantum process tomography, how does $\chi$ characterize a quantum process?

Question

I'm working through Nielsen and Chuang and I'm pretty confused by the discussion of quantum process tomography. I'm trying to work through an example of 1-qubit state tomography given by N&C (box 8.5), which provides an algorithm for determining $\chi$ in terms of block matrices and density matrices (determined by state tomography). The process seems pretty straight forward, but how does $\chi$ characterize a quantum process?

Answer 1

The linear map $\mathcal{E}$ is what characterizes a quantum process,

$$\rho \rightarrow \mathcal{E}(\rho),$$

but $\mathcal{E}$ can be determined by $\chi$. Using the operator-sum representation,

$$\mathcal{E}(\rho) = \sum_i A_i \rho A^\dagger_i = \sum_{i}\sum_{m}\sum_{n}a_{im}\tilde{A}_m \rho a^*_{in}\tilde{A}^\dagger_n,$$

where the $a_{ij}$ are some set of complex numbers that allow a fixed set of operators $\tilde{A}_{i}$ to form a basis for the unknown set of operators $A_i$ on the state space. Remember, if we can determine the operators $A_i$, then we can completely describe $\mathcal{E}$. Rearranging the above,

$$\mathcal{E}(\rho) = \sum_m \sum_n \tilde{A}_m \rho \tilde{A}^\dagger_n \sum_i a_{im} a^*_{in} = \sum_{mn} \tilde{A}_m \rho \tilde{A}^\dagger_n \chi_{mn},$$

where $\chi_{mn}$ is a "classical" error correlation matrix, which is positive Hermitian by definition. So, once the set of operators $\tilde{A}_i$ has been fixed, $\mathcal{E}$ can be determined completely by the complex number matrix $\chi$.$^{\text{[1]}}$

Note: The adjective "fixed" does not apply to any of the operators themselves, rather, it applies to the set of operators. The point is that the operators $A_i$ are unknown, so we are writing the $A_i$ in terms of a basis of known operators $\tilde{A}_i$ that we have chosen. The problem of determining the $A_i$ thus reduces to the problem of determining the coefficients $a_{ij}$ in this basis. This is not any different from determining an unknown vector by writing the vector as a linear combination of a fixed set of basis vectors, then finding the coefficients in this basis.

[1] Prescription for experimental determination of the dynamics of a quantum black box, Isaac L. Chuang, M. A. Nielson, 2008. https://arxiv.org/abs/quant-ph/9610001

Answer 2

TL;DR

Given a process matrix $\chi$ expressed in a normalized, ordered Pauli basis $\{P^{n}_{k}\}$, find its eigendecomposition $\{\lambda_{i},\mathbf{v_{i}}\}$, where $0 \leq \lambda_{i} \leq 1$ and $\mathbf{v_{i}} = (v^{i}_{P^{n}_{k}})$ is the eigenvector for eigenvalue $\lambda_{i}$ and contains 'coefficients' for every Pauli $P^{n}_{k}$. $\chi$ is Hermitian, so the $\lambda_{i}$'s are real and nonnegative, and the eigenvectors from different eigenvalues are orthogonal.

If the map represented by the process matrix is unitary there will be only one nonzero eigenvalue, let's call it $\lambda_{0}$ (then $\lambda_{0} = 1$). The corresponding eigenvector is then just the unitary $U$ expressed in the Pauli basis. We can readily compute $U$:

$$ U = \sum_{k} v^{0}_{P^{n}_{k}}P^{n}_{k} $$

For a single qubit, the $\chi$ matrix is $4 \times 4$ matrix. If again it represents a unitary channel, we get a single nonzero eigenvalue equal to one, and its corresponding eigenvector $\mathbf{v} = (v_{I},v_{X},v_{Y},v_{Z})$. Then, the single-qubit unitary represented by the map is:

$$ U = \sum_{P} v_{P}P = v_{I}I + v_{X}X + v_{Y}Y + v_{Z}Z. $$ Note that the $v_{P}$ are in general complex numbers.

If the map is supposed to be unitary, but it has a bit of noise in it, this will be reflected by one eigenvalue being close to $1$, corresponding to the unitary operation, and some (or all) other eigenvalues being small but nonzero - these correspond to the noise operators.

Density matrices and maps?

In its most general form, a quantum state is described not by a single ket state (e.g. $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, but by a statistical mixture of kets (e.g. $\{p_{i},|\psi_{i}\rangle\}$ described the situation that we have state $|\psi_{i}\rangle$ with probability $p_{i}$). If there is just one nonzero $p_{i}$, we get a single ket state back, and we call this state a pure state. Any other state is also referred to as a mixed state.

It turns out that we can - and it is extremely useful - to write this as a sum of outer products $\sum_{i}p_{i} |\psi_{i}\rangle\langle \psi_{i}|$. We call this the density matrix $\rho$ and, as mentioned before, this is the most general description that we can have of a quantum system. As a general rule of thumb, in quantum computing we always want

For pure states, the operations that we perform on them are gates, represented by unitary matrices. We can also apply these to density matrices, but there are also different valid operations that we can apply to density matrices. Any such operation we call a quantum operation or quantum channel, which is a linear map $\rho_{in} \rightarrow \Lambda(\rho_{in}) = \rho_{out}$ which is subject to some constraints to make sure that $\rho_{out}$ is a valid density matrix - I won't go into them here.

To fully characterize this $\Lambda$, there are multiple different representations - one of which is the process matrix $\chi$. In QPT, it is the goal to characterize a unknown map $\Lambda$ by finding one if its representations, where often the $\chi$ matrix is used. These can also represent unitary channels, but they can represent more.