How to derive the number of independent parameters in the $\chi$ matrix from the Choi matrix?

Question

In the section on Quantum process tomography, Page 391, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang. it is given that

In general, $\chi$ will contain $d^4−d^2$ independent real parameters, because a general linear map of $d$ by $d$ complex matrices to $d$ by $d$ matrices is described by $d^4$ independent parameters, but there are $d^2$ additional constraints due to the fact that $\rho$ remains Hermitian with trace one; that is, the completeness relation $\sum_i E_i^\dagger E_i=I$ is satisfied, giving $d^2$ real constraints.

---

A post, Why does the $\chi$ matrix have $d^4-d^2$ independent parameters?, has been asked on this, which contains an answer by @DaftWullie, which make use of the Choi matrix,

where it seems like $\chi$ is used to represent the Choi matrix, not the $\chi$ matrix.

---

$\chi$ is a hermitian $d^2\times d^2$ matrix which must have $d^2$ diagonal real (independent) terms, and the number of terms below the diagonal is $\dfrac{(d^2-1)d^2}{2}$ that contains a total of $2\times\dfrac{(d^2-1)d^2}{2}=d^4-d^2$ of real terms. Therefore, the total number of independent real parameters considering only that $\chi$ is hermitian, is $(d^4-d^2)+d^2=d^4$.

---

The Choi matrix is given by, $\sigma=(I_R\otimes\mathcal{E})(|\alpha\rangle\langle\alpha |)$ where $|\alpha\rangle=\dfrac{1}{\sqrt{d}}\sum_i |i_R\rangle\otimes|i_Q\rangle$ is a maximally entangled state of the systems $R$ and $Q$.

$$ \sigma=(I_R\otimes\mathcal{E})(|\alpha\rangle\langle\alpha |)=\dfrac{1}{d}\sum_{i,j}|i_R\rangle\langle j_R|\otimes\mathcal{E}(|i_Q\rangle\langle j_Q|) $$ which can be interpreted as the block matrix with $\frac{1}{d}\mathcal{E}(|i_Q\rangle\langle j_Q|)$ as the $(i,j)^{th}$ block.

The $\chi$ matrix is defined by setting $E_i=\sum_m e_{im}\tilde{E}_m$, with $\{\tilde{E}_m\}$ being any orthonormal basis for the set of operators on the state space, such that $$ \mathcal{E}(\rho)=\sum_i E_i\rho E_i^\dagger=\sum_{m,n}\chi_{mn}\tilde{E}_m\rho\tilde{E}_n^\dagger $$ where the $(m,n)^{th}$ element of $\chi$ is $\chi_{mn}=\sum_i e_{im}e_{in}^*$ such that $$ \chi=\sum_{m,n}\chi_{mn}|m\rangle\langle n| $$ $$ \sigma=(I_R\otimes\mathcal{E})(|\alpha\rangle\langle\alpha |)=\sum_{m,n}\chi_{mn}(I\otimes \tilde{E}_m)|\alpha\rangle\langle\alpha |(I\otimes \tilde{E}_n^\dagger)=\sum_{m,n}\chi_{mn}|\tilde{E}_m\rangle\langle\tilde{E}_n| $$ where \begin{align}|\tilde{E}_m\rangle&=(I\otimes \tilde{E}_m)|\alpha\rangle\\ &=(I\otimes \tilde{E}_m)\dfrac{1}{\sqrt{d}}\sum_i |i_R\rangle\otimes|i_Q\rangle\\ &=\dfrac{1}{\sqrt{d}}\sum_i |i_R\rangle\otimes\tilde{E}_m|i_Q\rangle \end{align} and therefore $$ \chi_{mn}=\langle\tilde{E}_m|\sigma|\tilde{E}_n\rangle $$ Now, the Choi matrix can be written as, \begin{align} \sigma&=\sum_{m}(I\otimes {E}_m)(\dfrac{1}{d}\sum_{i,j}|i_R\rangle\langle j_R|\otimes|i_Q\rangle\langle j_Q|)(I\otimes {E}_m^\dagger)\\ &=\sum_{i,j}|i\rangle\langle j|\otimes\dfrac{1}{d}\sum_m{E}_m|i\rangle\langle j|{E}_m^\dagger\\ \end{align} Therefore, the $(i,j)^{th}$ block of the Choi matrix $\sigma$ is $\dfrac{1}{d}\sum_m{E}_m|i\rangle\langle j|{E}_m^\dagger$.

The $(k,l)^{th}$ term of the $(i,j)^{th}$ block of the Choi matrix is, $$ \sigma_{ij,kl}=\langle k|(\dfrac{1}{d}\sum_m{E}_m|i\rangle\langle j|{E}_m^\dagger)|l\rangle\\ =\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle\\ $$ $$ \sigma_{ij,kl}=\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle $$ We are free to choose $\tilde{E}_m=\sqrt{d}|t\rangle\langle q|$ such that $m=qd+t$ equates the Choi matrix and the $\chi$ matrix, therefore $$ \boxed{\chi_{ij,kl}=\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle} $$

---

Applying the trace preserving condition, $\sum_m E_m^\dagger E_m=I\implies\sum_m \langle k|E_m^\dagger E_m|i\rangle=\delta_{ik}$

This much is clear!

Then it says,

This directly influences the $\chi$: $$ \sum_j\chi_{ij,kj}=\delta_{ik} $$ There are (complex) constraints for all $d^2$ values of $i,k$. However, since everything is Hermitian, again this corresponds to $d^2$ real constraints. Thus, the total freedom remaining is $d^4-d^2$.

How do we obtain $\sum_j\chi_{ij,kj}=\delta_{ik}$ ?

How does imposing the trace-preserving condition $\sum_m E_m^\dagger E_m=I$ on the Choi matrix obtains the number of constraints on the $\chi$ matrix ?

My Attempt

Thanks @JSdJ for the hint.

$\chi_{mn}=\langle\tilde{E}_m|\sigma|\tilde{E}_n\rangle\implies \sigma=\sum_{m,n}\chi_{mn}|\tilde{E}_m\rangle\langle\tilde{E}_n|$

Therefore, in order for $\sigma=\chi$, we need $|\tilde{E}_m\rangle=|m\rangle$, where $m:0\to d^2-1$

$(I\otimes\tilde{E}_m)|\alpha\rangle=|\tilde{E}_m\rangle=\dfrac{1}{\sqrt{d}}\sum_i |i\rangle\otimes\tilde{E}_m|i\rangle=\dfrac{1}{\sqrt{d}}[|0\rangle\otimes\tilde{E}_m|0\rangle+|1\rangle\otimes\tilde{E}_m|1\rangle+\cdots+|d^2-1\rangle\otimes\tilde{E}_m|d^2-1\rangle]$

Let's consider the case when $d=2\implies d^2-1=3$,

Then,

where $\tilde{E}_m|i\rangle$ is the $i^{th}$ column of $\tilde{E}_m$, ie., $|i\rangle\otimes\tilde{E}_m|i\rangle$ is a column vector with the $i^{th}$ column of $\tilde{E}_m$ as it's $i^{th}$ block, all other elements are zero. So, $\sum_i |i\rangle\otimes\tilde{E}_m|i\rangle$ is the column vector with the columns of $\tilde{E}_m$ stacked on top of each other in order.

We can divide $|m\rangle$ into $d$ blocks of dimension $d$ such that let's take $m=qd+t$, where $q,t:0\to d-1$.

So, for $|\tilde{E}_m\rangle=|m\rangle$ we need to choose $\tilde{E}_m=\sqrt{d}|t\rangle\langle q|$, such that $\sigma=\sum_{m,n}\chi_{mn}|\tilde{E}_m\rangle\langle\tilde{E}_n|=\sum_{m,n}\chi_{mn}|m\rangle\langle n|=\chi$

$\therefore \chi_{ij,kl}=\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle$

Applying the trace preserving condition, $\sum_m E_m^\dagger E_m=I\implies\sum_m \langle k|E_m^\dagger E_m|i\rangle=\delta_{ik}$, how do we reach\begin{align} \sum_j\chi_{ij,kj}&=\delta_{ik} \end{align}

Answer 1

Counting the number of free parameters of Choi operators

The number of independent parameters in the $\chi$ matrix is identical to the number of independent parameters in the Choi matrix, or in the channel itself, because these are all isomorphic representations of the same objects.

A map $\Phi$ acting on (the set of linear operators defined on) a Hilbert $H$ is a channel (meaning, it's completely positive and trace-preserving) iff its Choi representations, defined as $J(\Phi)\equiv\sum_{ij}\Phi(|i\rangle\!\langle j|)\otimes|i\rangle\!\langle j|$, is positive semidefinite and such that $\operatorname{tr}_1(J(\Phi))=I$.

The set of Hermitian operators on a Hilbert space $H$ is a vector space of (real) dimension $\operatorname{dim}(H)^2$. The set of positive semidefinite operators on a Hilbert space $H$ is a (compact, convex) subset of this space, and thus has the same number of independent parameters.

The set of Choi operators $J(\Phi)$ of channels $\Phi$ is a subset of the set of positive semidefinite operators on $H\otimes H$. It thus has $\dim(H\otimes H)^2=\dim(H)^4$ independent real parameters. However, the trace-preserving condition removes some of these degrees of freedom. The trace-preserving condition, which reads $\operatorname{tr}_1(J(\Phi))=I$, amounts explicitly to the $\dim(H)^2$ conditions: $$\sum_i J(\Phi)_{ij,ik}=\delta_{jk}, \qquad \forall j,k=1,...,\dim(H). $$ You can verify that these conditions are linearly independent, and thus remove $\dim(H)^2$ free parameters. You conclude that the set of Choi operators is a subset of $\operatorname{Lin}(H\otimes H)$ with $\dim(H)^4-\dim(H)^2$ free parameters.

Reason directly from the $\chi$ representation

You can also just reason directly in terms of the $\chi$ matrix, because $\chi$ and Choi are related via $\chi_{mn}=\langle \tilde E_m|J(\Phi)|\tilde E_n\rangle$, $J(\Phi)=\sum_{mn} \chi_{mn} |\tilde E_m\rangle\!\langle \tilde E_n|$ for some choice of orthonormal operatorial basis $\{\tilde E_m\}$.

I think it's worth showing where some of the relations given in this answer come from though, because the notation can be tricky here. Namely, what exactly the kets and bras in $J(\Phi)=\sum_{mn} \chi_{mn} |\tilde E_m\rangle\!\langle \tilde E_n|$ mean.

I'll somewhat follow the notation in the linked answer, so that the relation between channel/map $\Phi$ and its Choi representation reads $$\Phi(\rho) = \sum_{m,n} \chi_{m,n} P_m \rho P_n^\dagger.$$ Each $P_m$ is here an operator acting on $H$, and $\{P_m\}$ is a basis for the space of such operators. The Choi representation derived from this decomposition is $$J(\Phi) = \sum_{ij} \Phi(E_{ij})\otimes E_{ij} \equiv \sum \chi_{m,n} (P_m E_{ij} P_n^\dagger)\otimes E_{ij}.$$ Consider the action of $J(\Phi)$ on some simple vector $|\alpha\rangle\otimes|\beta\rangle$, which gives $$J(\Phi)(|\alpha\rangle\otimes|\beta\rangle)= \sum_{m,n,i} \chi_{m,n} (P_m E_{i,\beta} P_n^\dagger |\alpha\rangle)\otimes |i\rangle \\ = \sum \chi_{m,n} (P_n^\dagger)_{\beta,\alpha} (P_m |i\rangle)\otimes |i\rangle = \sum \chi_{m,n} (\bar P_n)_{\alpha\beta} (P_m)_{ji} (|j\rangle\otimes |i\rangle).$$ In other words, the action of $J(\Phi)$ on a generic vector $v\in H\otimes H$ can be written as $$J(\Phi)v=\sum_{m,n} \chi_{m,n}\langle \operatorname{vec}(P_n), v \rangle \operatorname{vec}(P_m),$$ where $\operatorname{vec}(P)$ denotes the vectorisation of an operator $P$. Explicitly, if $P:H\to H$, then $\operatorname{vec}(P)\in H\otimes H$ with $\operatorname{vec}(P)_{ij}=P_{ij}$.

The above relation is what we mean when writing $J(\Phi)$ as a sum over ket-bras of a basis of operators. So, armed with this knowledge, how do we compute $\operatorname{tr}_1(J(\Phi))$ from the expression with the $\chi$ matrix, which is what we want to do to derive what trace-preserving translates into for $\chi$? At the cost of being pedantic, we can start observing that $\operatorname{tr}_1(J(\Phi))$ means explicitly $$\operatorname{tr}_1(J(\Phi)) = \sum_{\alpha} (\langle\alpha|\otimes I)J(\Phi)(|\alpha\rangle\otimes I) \equiv \sum_{\alpha\beta\gamma} (\langle\alpha|\otimes\langle\gamma|)J(\Phi)(|\alpha\rangle\otimes|\beta\rangle) \, |\gamma\rangle\!\langle \beta|,$$ and using the expression for $J(\Phi)$ in terms of $\chi_{m,n}$ we get $$(\langle\alpha|\otimes\langle\gamma|)J(\Phi)(|\alpha\rangle\otimes|\beta\rangle) = \sum_{m n\alpha\beta\gamma} \chi_{m,n} (\bar P_n)_{\alpha,\beta} (P_m)_{\alpha,\gamma},$$ and therefore $$\operatorname{tr}_1(J(\Phi)) = \sum \chi_{m,n} P_n^\dagger P_m,$$ hence the trace-preserving condition becoming $\sum_{m,n} \chi_{m,n} P_n^\dagger P_m = I$.