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.
My Understanding
We have the quantum operation $\mathcal{E}$ defined by the Kraus operators $\{E_i\}$ such that $\mathcal{E}(\rho)=\sum_i E_i\rho E_i^\dagger$.
Consider the set of operators $\{\tilde{E}_i\}$, which form a basis for the set of operators on the state space such that, $E_i=\sum_m e_{im}\tilde{E}_m$, for some set of complex numbers $e_{im}$.
\begin{align} \mathcal{E}(\rho)&=\sum_i E_i\rho E_i^\dagger\\ &=\sum_i \sum_m e_{im}\tilde{E}_m\rho \sum_n e^*_{in}\tilde{E}^\dagger_n\\ &=\sum_{m,n}\tilde{E}_m\rho\tilde{E}^\dagger_n \sum_i e_{im}e^*_{in}\\ &=\sum_{m,n}\tilde{E}_m\rho\tilde{E}^\dagger_n \chi_{mn}\\ \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 $\chi_{mm}=\sum_i e_{im}e^*_{im}=\sum_i |e_{im}|^2\geq 0$.
$E_i$ is a $d$ by $d$ matrix and therefore we have $d^2$ number of $\tilde{E}_m$, i.e., $m,n:1\to d^2$ and therefore there are $d^4$ elements in the $\chi$ matrix.
Since $\mathcal{E}$ is a trace preserving quantum operation, i.e., $\sum_i E_i^\dagger E_i=I$ \begin{align} \sum_i E_i^\dagger E_i&=\sum_i \sum_m e_{im}^*\tilde{E}_m^\dagger \sum_n e_{in}\tilde{E}_n\\ &=\sum_{m,n}\tilde{E}_m^\dagger\tilde{E}_n\sum_i e_{im}^*e_{in}\\ I&=\sum_{m,n}\tilde{E}_m^\dagger\tilde{E}_n\chi_{mn}\\ \end{align}
How does it say that "$\chi$ will contain $d^4−d^2$ independent real parameters" ?
My Attempt
Thanks @DaftWullie for the hint.
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 $Q$. \begin{align} |\alpha\rangle\langle\alpha |&=\dfrac{1}{d}(\sum_i |i_R\rangle\otimes|i_Q\rangle)(\sum_j \langle j_R|\otimes\langle j_Q|)\\ &=\dfrac{1}{d}(\sum_{i,j} |i_R\rangle\otimes|i_Q\rangle)(\langle j_R|\otimes\langle j_Q|)\\ &=\dfrac{1}{d}\sum_{i,j}|i_R\rangle\langle j_R|\otimes|i_Q\rangle\langle j_Q| \end{align} $$ \sigma=(I_R\otimes\mathcal{E})(|\alpha\rangle\langle\alpha |)=\dfrac{1}{d}\sum_{i,j}|i_R\rangle\langle j_R|\otimes\mathcal{E}(|i_Q\rangle\langle j_Q|) $$ 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 $|\tilde{E}_m\rangle=(I\otimes \tilde{E}_m)|\alpha\rangle$, and therefore $$ \chi_{mn}=\langle\tilde{E}_m|\sigma|\tilde{E}_n\rangle $$ Now, the Choi matrix can be written as, \begin{align} \sigma&=(I_R\otimes\mathcal{E})(|\alpha\rangle\langle\alpha |)\\ &=\sum_{m}(I\otimes {E}_m)|\alpha\rangle\langle\alpha |(I\otimes {E}_m^\dagger)\\ &=\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, $$ \sigma_{ij,kl}=\langle k|(\dfrac{1}{d}\sum_m{E}_m|i\rangle\langle j|{E}_m^\dagger)|l\rangle\\ =\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle\\ $$
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 $$ \color{blue}{\chi_{ij,kl}=\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|l\rangle} $$
Applying the trace preserving condition, $\sum_m E_m^\dagger E_m=I\implies\sum_m \langle k|E_m^\dagger E_m|i\rangle=\delta_{ik}$
\begin{align} \sum_j \chi_{ij,kj}&=\sum_j \dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\langle j|{E}_m^\dagger|j\rangle\\ &=\dfrac{1}{d}\sum_m\langle k|{E}_m|i\rangle\sum_j\langle j|{E}_m^\dagger|j\rangle\\ \end{align}
How do we reach the expression $\sum_j\chi_{ij,kj}=\delta_{ik}$?
How do I proceed further and obtain that there is a $d^2$ number of constraints?