# How to extract probabilities from Kraus representation?

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?

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.