Complete Positive Trace Preserving Map (CPTP) operator is the most general operation that can be performed on a quantum system. This post mentioned that a CPTP operator is nothing but a unitary operator on the system after adding few ancilla bits. I would like to know how to realize a given CPTP operator.

Choi's theorem states that any CPTP operator $\Phi(\cdot) : C^*_{n\times n} \rightarrow C^*_{m \times m}$ can be expressed as $\Phi(\rho) = \sum_{j=1}^r F_j^\dagger \rho F_j$, for some $n \times m$ matrices $F_j$ such that $\sum_j F_j F_j^\dagger = I_n$.

Using this fact, can we come up with unitary operation corresponding to the given CPTP operator $\Phi$?


1 Answer 1


An isometry $U:S\rightarrow S\otimes E$, where $S$ is your system and $E$ is an environment, such that

$$\mathrm{Tr}_E(U\rho U^\dagger)=\Phi(\rho)$$ is called a Stinespring dilation of $\Phi$. An easy way to construct a Stinespring dilation from the Kraus operators is to consider $\mathcal{H}_E=\mathrm{span}\{|j\rangle\}_{j=1}^r$ and

$$U=\sum_j F_j^S\otimes|j\rangle^E $$

it is easy to see that

$$\mathrm{Tr}_E(U\rho U^\dagger)= \mathrm{Tr}_E\left(\sum_jF_j\rho F_k^\dagger \otimes |j\rangle\langle k|\right)= \sum_jF_j\rho F_j^\dagger=\Phi(\rho)$$

but notice that since the set of Kraus operators for a channel is not unique, neither is the Stinespring dilation.

  • $\begingroup$ nice! could you please define F_j^s for the sake of completeness? $\endgroup$ Nov 12, 2021 at 16:57

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