# Why is $\chi$ not uniquely determined by $\sum_{mn}\beta_{jk}^{mn}\chi_{mn}=\lambda_{jk}$?

The mathematical construct of the Quantum process tomography is given in Page 391, 392, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, as follows

Let a fixed set of operators $$\tilde{E}_i$$, which form a basis for the set of operators on the state space so that $$E_i=\sum_m e_{im}\tilde{E}_m$$, then the quantum operation can be represented as, \begin{align} \mathcal{E}(\rho)&=\sum_i E_i\rho E_i^\dagger \tag{8.150}\label{eq0}\\ &=\sum_{m,n}\tilde{E}_m\rho\tilde{E}^\dagger_n \chi_{mn} \tag{8.152}\label{eq2}\\ \end{align} where $$\chi_{mn}=\sum_i e_{im}e^*_{in}$$ are the entries of a matrix that is positive hermitian since $$\chi_{mn}^*=(\sum_i e_{im}e^*_{in})^*=\sum_i e_{in}e^*_{im}=\chi_{nm}$$ and $$\chi_{mm}=\sum_i e_{im}e^*_{im}=\sum_i |e_{im}|^2\geq 0$$.

Let $$\rho_j,1\leq j\leq d^2$$ be a fixed, linearly independent basis for the space of $$d\times d$$ matrices,i.e., any $$d\times d$$ matrix can be written as a unique linear combination of the $$\rho_j$$ and it is possible to determine $$E(\rho_j)$$ by state tomography, for each $$\rho_j$$.

Each $$E(\rho_j)$$ may be expressed as a linear combination of the basis states, $$\mathcal{E}(\rho_j)=\sum_k \lambda_{jk}\rho_k \tag{8.155}\label{eq5}$$

And, since $$\mathcal{E}(\rho_j)$$ is known from the state tomography, $$\lambda_jk$$ can be determined by standard linear algebraic algorithms. we can also write $$\tilde{E}_m\rho_j\tilde{E}_n^\dagger=\sum_k\beta_{jk}^{mn}\rho_k \tag{8.156}\label{eq6}$$

where $$\beta_{jk}^{mn}$$ are complex numbers that can be determined by standard algorithms from linear algebra given the $$\tilde{E}_m$$ operators and the $$\rho_j$$ operators.

Combining both obtain, \begin{align} \sum_k\sum_{mn}\chi_{mn}\beta_{jk}^{mn}\rho_k&=\sum_{k}\lambda_{jk}\rho_k\tag{8.157}\label{eq7}\\ \implies \sum_{mn}\beta_{jk}^{mn}\chi_{mn}&=\lambda_{jk} \tag{8.158}\label{eq8}\\ \implies \beta\vec{\chi}&=\vec{\lambda} \tag{8.161}\label{eq61} \end{align} Let $$\kappa$$ be the generalized inverse such that $$\beta\kappa\beta=\beta\Leftrightarrow \beta_{jk}^{mn}=\sum_{st,xy}\beta_{jk}^{st}\kappa_{st}^{xy}\beta_{xy}^{mn}$$

We can prove that $$\chi$$ defined by $$\chi_{mn}\equiv\sum_{jk}\kappa_{jk}^{mn}\lambda_{mn}\Leftrightarrow\vec{\chi}=\kappa\vec{\lambda}\tag{8.160}\label{eq60}$$ satisfies the equation $$\sum_{mn}\beta_{jk}^{mn}\chi_{mn}=\lambda_{jk}\Leftrightarrow\beta\vec{\chi}=\vec{\lambda}$$

But now it is stated that,

The difficulty in verifying that $$\chi$$ defined by \ref{eq0} satisfies \ref{eq8} is that, in general, $$\chi$$ is not uniquely determined by Equation \ref{eq8}

What is the logic in saying that $$\chi$$ is not uniquely defined by equation \ref{eq8} ?

In the next step, it says

From the construction that led to Equation \ref{eq2}, we know there exists at least onesolution to Equation \ref{eq61}

How do I make sense of this ?

In deriving equation \ref{eq2} we used the substitution $$E_i=\sum_m e_{im}\tilde{E}_m$$ into the expression $$\mathcal{E}(\rho)=\sum_i E_i\rho E_i^\dagger$$.

Because it's a linear system, and a linear system can in general have (infinitely) many solutions.

If you want to find $$x$$ such that $$Ax=b$$ for some matrix $$A$$ and vector $$b$$, then the set of solutions is the set of vectors of the form $$x=A^+ b + y$$, with $$A^+\equiv A^\dagger(AA^\dagger)^{-1}$$ the pseudoinverse of $$A$$ (this assumes that $$AA^\dagger$$ is invertible; if instead $$A^\dagger A$$ is invertible then a slightly different formula for $$A^+$$ applies), and $$y$$ any vector in the ker of $$A$$.

In other words, there are infinitely many solutions iff $$A$$ has nontrivial kernel.

To more explicitly connect with the $$\chi$$ matrix, imagine that $$x$$ becomes the "vector" of all components of $$\chi$$, and $$A$$ the matrix of coefficients $$\beta$$, and $$b$$ the vector $$\lambda$$.

• My understanding is that $Ax=b$ has a solution iff $b\in Range(A)$, else we will have to find the least square solution of the linear system, and that's where the pseudo inverse comes in. So how can we be sure that $\beta\vec{\chi}=\vec{\lambda}$ must have at least one solution? Nov 24, 2022 at 11:22
• has a solution for $x$ iff $b\in\operatorname{range}(A)$, is what you meant to write. It's the same thing for $\chi$. You'll have a solution for $\chi$ iff $\vec\lambda$ is in the range of the operator $\beta$. But you already know that there is a solution here, because you defined $\beta$ to make this system satisfied. You define $\beta$ so that there is such a solution. What might be less obvious is that there is still in general an entire space of solutions, whenever $\beta$ has nontrivial ker
– glS
Nov 24, 2022 at 11:25
• sorry that was a typo, fixed my comment. Nov 24, 2022 at 11:27