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Consider a quantum operation described by Kraus operators $K_1, ..., K_n$. As I understand the effect of this operation on a density matrix $\rho$ can be described as $ \mathcal{E}(\rho)= \sum_{i}p(i)\rho_i$, where $\rho_i$ is a possible state of the system after the operation and $p(i)$ is the probability of that state.

If I only have Krauss operators, can I still infer the possible states and their probabilities? Each term $K_i\rho K^{\dagger}_i$ in operator-sum representation of the quantum operation seems to incorporate both the potential outcome and its probability. Is there a way to extract each of them?

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We can indeed rewrite $\mathcal{E}(\rho)=\sum_iK_i\rho K_i^\dagger$ as $\mathcal{E}(\rho)=\sum_ip(i)\rho_i$ by setting $p(i):=\mathrm{tr}(K_i\rho K_i^\dagger)$ and $\rho_i:=\frac{K_i\rho K_i^\dagger}{p(i)}$. Note that $\mathrm{tr}(\rho_i)=1$ and $\sum_ip(i)=\mathrm{tr}(\rho \sum_iK_i^\dagger K_i)=1$, so we can interpret $\rho_i$ as states and $p(i)$ as probabilities.

That said, the probabilities $p(i)$ generally depend on the input state $\rho$. However, if $\mathcal{E}$ is a unitary mixture, i.e. if every $K_i$ is a scalar multiple of a unitary operator $K_i=\alpha U_i$ then $p(i)=\mathrm{tr}(K_i^\dagger\rho K_i)=|\alpha|^2$ is independent of $\rho$.

Finally, note that $\rho_i$ and $p(i)$ are not unique since Kraus representation is not unique.

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