Why does the process matrix $\chi$ have dimensions $d^2\times d^2$?

Question

A quantum map on a $d$-dimensional space has the general representation: $$ \mathcal{S}(\rho)=\sum_{\alpha,\beta}^{d^2}\chi_{\alpha\beta}\Gamma_{\alpha}\rho \Gamma_{\beta}^{\dagger}, $$ where $\chi$ is the $d^2\times d^2$ process matrix, which is positive semidefinite and trace preserving. The $\{\Gamma_{\alpha}\}$ is an orthonormal matrix basis set, and $tr(\Gamma_{\beta}^{\dagger}\Gamma_{\alpha})=\delta_{\alpha\beta}$.

My question: Why there are $d^2$ orthonormal basis? Are there any examples of two-qubit systems?

Answer 1

States lie in Hilbert space $\mathcal{H_S}$.

• $|\psi\rangle \in \mathcal{H_S}\,.$
• If you have $n$ qubits, then $\text{dim}(\mathcal{H_S}) = 2^n = d\,.$

Operators, density operators lie in the bounded operator space of $\mathcal{H}_S$.

• $\rho \in \mathcal{B}(\mathcal{H}_S)\,. $
• $\text{dim}(\mathcal{B}(\mathcal{H}_S)) = d \times d = d^2\,.$

Kraus operators are operators acting on $\mathcal{H_S}$. You have $d^2$ degrees of freedom, which requires $d^2$ basis to describe it.

If your map $\mathcal{E}$ is $$ \mathcal{E}(\rho) = \sum_{n} K_{n} \rho K_{n}^{\dagger}\,, $$

Then you can write your Kraus operators in $\{\Gamma_{\alpha}\}$ basis as $$ K_{n} = \sum_{\alpha=0}^{d^2-1} b_{n\alpha} \Gamma_{\alpha}\,, $$

where $b_{n\alpha}$ are the coefficients for the linear combination using the new basis; therefore, you will have total $d^2$ terms in the summation. $K_n$ is a $d \times d $ matrix.

Now, rewriting our map in terms of Kraus operators in the $\{\Gamma_{\alpha}\}$ basis gives us

$$ \begin{align} \mathcal{E}(\rho) &= \sum_{n} K_{n} \rho K_{n}^{\dagger}\\ &= \sum_n \bigg(\sum_{\alpha=0}^{d^2-1} b_{n\alpha} \Gamma_{\alpha} \bigg) \rho \bigg(\sum_{\beta=0}^{d^2-1} b_{n\beta}^* \Gamma_{\beta}^{\dagger}\bigg)\\ &= \sum_{\alpha=0,\beta=0}^{d^2-1, d^2-1} \big(\sum_n b_{n\alpha} \cdot b_{n\beta}^* \big) \Gamma_{\alpha} \rho \Gamma_{\beta}^{\dagger}\\ &= \sum_{\alpha=0,\beta=0}^{d^2-1, d^2-1} \chi_{\alpha \beta}\Gamma_{\alpha} \rho \Gamma_{\beta}^{\dagger} \\ \end{align} $$

So, as you can see above, Kraus operators being $d \times d$ results in $\chi$ having $d^2 \times d^2$ degrees of freedom. The process matrix is just

$$ \chi = bb^{\dagger} $$

where $b$ is the matrix with linear combination coefficients for your Kraus operators in $\{\Gamma_{\alpha}\}$ basis.

Answer 2

A slight rewording of what's already been said in the other answer: if the underlying space $\mathcal H$ has dimension $d$, then the (real) vector space of Hermitian operators acting on $\mathcal H$ has dimension $d^2$. The elements of the $\chi$ matrix are the coefficients attached to the terms $\Gamma_\alpha \rho \Gamma_\beta^\dagger$ for all $\alpha,\beta$, and $\Gamma_\alpha$ are a basis of Hermitian operators for the space. Thus these indices go from $1$ to $d^2$, and thus there's $d^2\times d^2$ pairs of such indices.