Why does the $\chi$ matrix have $d^4-d^2$ independent parameters?

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.

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" ?

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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?

Answer 1

First, note that for a $k\times k$ Hermitian matrix, there are $k^2$ parameters. This is because there are $k$ diagonal elements, which must be real, and $\frac12 k(k-1)$ complex elements (i.e. 2 real parameters) below the diagonal.

Generically, the $\chi$ matrix is $d^2\times d^2$, so I'll index its elements by $\chi_{ij,kl}$, each index running over $d$ elements. Since it's Hermitian, without constraint there are $d^4$ real parameters.

Strictly, we have $$ \chi_{ij,kl}=\frac{1}{d}\sum_m\langle j|E_m|i\rangle\langle k|E_m^\dagger|l\rangle. \tag{1}\label{eq1} $$ Now the condition $$ \sum_mE_m^\dagger E_m=I $$ basically imposes that $$ \sum_m\langle k|E_m^\dagger E_m|i\rangle=\delta_{ik}.\tag{2}\label{eq2} $$ This directly influences the $\chi$: if I substitute into Eq. (\ref{eq1}) with $j=l$, then I have $$ \chi_{ij,kj}=\frac{1}{d}\sum_m\langle j|E_m|i\rangle\langle k|E_m^\dagger|j\rangle=\frac{1}{d}\sum_m\langle k|E_m^\dagger|j\rangle\langle j|E_m|i\rangle. $$ Summing over all $j$, I can use the completeness relation $$ \sum_j\chi_{ij,kj}=\frac{1}{d}\sum_m\langle k|E_m^\dagger\left(\sum_j|j\rangle\langle j|\right)E_m|i\rangle=\frac{1}{d}\sum_m\langle k|E_m^\dagger E_m|i\rangle. $$ Now use Eq. (\ref{eq2}) to get $$ \sum_j\chi_{ij,kj}=\frac{\delta_{ik}}{d}. $$ 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 $d^4-d^2$.

Answer 2

Consider the following observations:

1. Let $V,W$ be finite-dimensional vector spaces. The set of linear functions $A:V\to W$, which I'll denote with $\operatorname{Lin}(V,W)$, is in itself a vector space, and $\dim(\operatorname{Lin}(V,W))=\dim(V)\dim(W)$. To see it, you just need to note that linear maps are characterised by their coefficients with respect to a basis for $V$ and $W$ (or in other words, $\operatorname{Lin}(V,W)$ can be parametrised by the set of $\dim(W)\times\dim(V)$ matrices). I didn't specify the underlying field here; "dimension" can refer to the number of complex free parameters if $V,W$ are complex vector spaces, or to the number of real free parameters, if $V,W$ are real vector spaces (and same goes in any other field).

2. Let $V$ be a finite-dimensional complex vector space. The set of Hermitian operators defined on $V$, denoted with $\operatorname{Herm}(V)$, is a real vector space of dimension $\dim(V)^2$. It's a real vector space because you multiplying a Hermitian matrix with something like $i$ makes it non-Hermitian. Note the difference between this and the vector space of all linear functions $\operatorname{Lin}(V)$, which as complex dimension $\dim(V)^2$, or equivalently, real dimension $2\dim(V)^2$.

3. The set of traceless Hermitian operators on $V$, which I'll write with $\operatorname{Herm}_0(V)$ is also a vector space, and has dimension $\dim(V)^2-1$. To see it, just observe that you can find a basis of orthonormal operators for $\operatorname{Herm}(V)$ of the form $\{\sigma_1,\sigma_2,...,\sigma_{d^2}\}$ with $\sigma_0=I/\sqrt{\dim(V)}$ and $\operatorname{tr}(\sigma_\alpha)=0$ for $\alpha\neq1$. Removing the identity term $\sigma_1$ then gives you a basis for $\operatorname{Herm}_0(V)$ containing $d^2-1$ elements.

4. The set of linear maps between Hermitian operators, $\operatorname{Lin}(\operatorname{Herm}(V),\operatorname{Herm}(W))$, is thus a real vector space of dimension $\dim(V)^2\dim(W)^2$. This can be restated by saying that the set of (not necessarily trace-preserving or positive) Hermitian-preserving quantum maps sending Hermitians in $V$ to Hermitians in $W$ has dimension $\dim(V)^2\dim(W)^2$. Or if $V=W$ and $\dim(V)=d$, it has dimension $d^4$.

5. The set of trace-preserving, Hermitian-preserving quantum maps between $V$ and $W$ is a subset of $\operatorname{Lin}(\operatorname{Herm}(V),\operatorname{Herm}(W))$. This is not a vector space: as multiplying by a scalar or adding together multiple such maps breaks the trace-preservingness. This means that "dimension" in this case has a slightly different meaning than it had before. Nonetheless, this set forms an affine subspace, because you can write any trace-preserving quantum map $\Phi$ as $\Phi(X)=\operatorname{tr}(X)I/\dim(W)+\tilde\Phi(X)$ with $\tilde \Phi$ a "trace-destroying" quantum map: one such that $\operatorname{tr}(\tilde\Phi(X))=0$ for all $X$. The point is that the first term ensures that the trace is preserved, and adding $\tilde\Phi$ never affects this constraint. The advantage of this writing is that now the set of maps $\tilde\Phi$ (that is, the set of Hermitian-preserving quantum maps whose range is the subspace of traceless Hermitian operators) is again a vector space. More precisely, it is the vector space of linear functions in $\operatorname{Lin}(\operatorname{Herm}(V),\operatorname{Herm}_0(W))$. But we already know this space has dimension $\dim(V)^2(\dim(W)^2-1)$.

In conclusion, the set of trace-preserving quantum maps between Hermitian operators in a $d$-dimensional space is an affine space of dimension $d^4-d^2=d^2(d^2-1)$. I should then also note that we didn't impose the further conditions of positivity or complete positivity on these maps. These constraints, while clearly having an important effect, do not affect the parameter counting, but rather set some more complex kinds of boundaries to the set. In other words, the set of quantum channels will be a nontrivial (compact, convex) subset of the $d^2(d^2-1)$-dimensional affine space discussed above. Kinda like the positivity of states does not affect the number of free parameters they have (but it affects how "free" the parameters are).

I also essentially ignored the fact that the question was specifically about the process matrix here, but rather discussed more generally about maps/channels. The reason is that the process matrix is just one way to represent maps/channels, cue e.g. this answer, and the parameter counting will be independent of the representation used. Nonetheless, we obtain the same result reasoning directly in terms of process matrices, because the process matrices of channels are $d^2\times d^2$ Hermitian (positive semidefinite) matrices $\chi$ such that $\sum_{\alpha\beta}\chi_{\alpha\beta}\sigma_\alpha\sigma_\beta=I$. The space of Hermitian matrices of this size has dimension $(d^2)^2$, and the constraint amounts to $d^2$ individual linear constraints on the elements of $\chi$, hence we again end up with the final tally of $d^4-d^2$.