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

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?

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.

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.