# How can I compute the probability distribution by using channel through Choi matrix in quantum process tomography?

The probability of observing an outcome corresponding to $$M_j$$ (a positive measurement operator), when the quantum process has transformed some input state $$\rho_j$$ is

$$p_{ij}=tr[\mathcal{E}(\rho_i)M_j]$$.

The evolution of a generic quantum state ρ under the channel E is obtained through the Choi matrix as $$\mathcal{E}(\rho)=tr_{\sigma}[(\rho^T\otimes I_{\tau})\Lambda]$$, where $$\Lambda$$ is the Choi matrix.

How can I get $$p_{ij}$$ by substituting $$\mathcal{E}(\rho)$$, cause there is already a partical trace in $$\mathcal{E}(\rho)$$, which will be $$p_{ij}=tr\{tr_{\sigma}[(\rho^T\otimes I_{\tau})\Lambda]M_j\}$$, seems terrible.

## 1 Answer

The Choi is related to the channel via $$\Lambda = \sum_{ij} (E_{ij}\otimes \mathcal E(E_{ij})) = (I\otimes \mathcal E)\mathbb{P}_+,$$ using the shorthand notation $$E_{ij}\equiv |i\rangle\!\langle j|$$ and $$\mathbb{P}_+\equiv |+\rangle\!\langle +|$$, $$|+\rangle\equiv \sum_i|i,i\rangle$$.

As you already pointed out, you recover the channel $$\mathcal E$$ from $$\Lambda$$ via $$\mathcal E(\rho) = \operatorname{tr}_1[(\rho^T\otimes I)\Lambda].$$ Using these to compute probabilities you get $$\operatorname{tr}(M\mathcal E(\rho)) = \operatorname{tr}[M\operatorname{tr}_1[(\rho^T\otimes I)\Lambda]] = \operatorname{tr}[(\rho^T\otimes M) \Lambda].$$ Intuitively, you see this because the second space of $$\Lambda$$ is not affected by multiplication by $$\rho^T$$ nor by the partial trace, therefore the outer multiplication by $$M$$ directly connected to $$\Lambda$$.

More formally, you can show this expanding the partial trace explicitly: $$\operatorname{tr}_1[(\rho^T\otimes I)\Lambda] = \sum_i (\langle i|\otimes I) [(\rho^T\otimes I)\Lambda](|i\rangle\otimes I) = \sum_{ijk\ell} \rho_{ji} \Lambda_{jk,i\ell} E_{k\ell},$$ and thus $$\operatorname{tr}(M\mathcal E(\rho)) = \sum_{\alpha\beta} \bar M_{\alpha\beta} \mathcal E(\rho)_{\alpha\beta} = \sum_{ij\alpha\beta} \bar M_{\alpha\beta} \rho_{ji} \Lambda_{j\alpha,i\beta} = \operatorname{tr}[(\rho^T\otimes M)\Lambda].$$