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 $\mathcal{E}(\rho)=\sum_i E_i\rho E_i^\dagger$.


1 Answer 1


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 $\lambda$.

  • $\begingroup$ 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? $\endgroup$
    – Sooraj S
    Nov 24, 2022 at 11:22
  • 1
    $\begingroup$ 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 $\endgroup$
    – glS
    Nov 24, 2022 at 11:25
  • $\begingroup$ sorry that was a typo, fixed my comment. $\endgroup$
    – Sooraj S
    Nov 24, 2022 at 11:27

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