What is the unitary operator realizing a given CPTP operator - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2019-11-15T08:13:25Z https://quantumcomputing.stackexchange.com/feeds/question/6911 https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/6911 4 What is the unitary operator realizing a given CPTP operator satya https://quantumcomputing.stackexchange.com/users/8014 2019-07-31T22:59:33Z 2019-07-31T23:38:15Z <p>Complete Positive Trace Preserving Map (CPTP) operator is the most general operation that can be performed on a quantum system. This <a href="https://quantumcomputing.stackexchange.com/questions/6906/are-cptp-operators-and-unitary-operators-the-same-thing">post</a> 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.</p> <p>Choi's theorem states that any CPTP operator <span class="math-container">$\Phi(\cdot) : C^*_{n\times n} \rightarrow C^*_{m \times m}$</span> can be expressed as <span class="math-container">$\Phi(\rho) = \sum_{j=1}^r F_j^\dagger \rho F_j$</span>, for some <span class="math-container">$n \times m$</span> matrices <span class="math-container">$F_j$</span> such that <span class="math-container">$\sum_j F_j F_j^\dagger = I_n$</span>. </p> <p>Using this fact, can we come up with unitary operation corresponding to the given CPTP operator <span class="math-container">$\Phi$</span>?</p> https://quantumcomputing.stackexchange.com/questions/6911/-/6913#6913 2 Answer by user2723984 for What is the unitary operator realizing a given CPTP operator user2723984 https://quantumcomputing.stackexchange.com/users/5125 2019-07-31T23:38:15Z 2019-07-31T23:38:15Z <p>An isometry <span class="math-container">$U:S\rightarrow S\otimes E$</span>, where <span class="math-container">$S$</span> is your system and <span class="math-container">$E$</span> is an environment, such that</p> <p><span class="math-container">$$\mathrm{Tr}_E(U\rho U^\dagger)=\Phi(\rho)$$</span> is called a Stinespring dilation of <span class="math-container">$\Phi$</span>. An easy way to construct a Stinespring dilation from the Kraus operators is to consider <span class="math-container">$\mathcal{H}_E=\mathrm{span}\{|j\rangle\}_{j=1}^r$</span> and</p> <p><span class="math-container">$$U=\sum_j F_j^S\otimes|j\rangle^E$$</span></p> <p>it is easy to see that</p> <p><span class="math-container">$$\mathrm{Tr}_E(U\rho U^\dagger)= \mathrm{Tr}_E\left(\sum_jF_j\rho F_k^\dagger \otimes |j\rangle\langle k|\right)= \sum_jF_j\rho F_k^\dagger=\Phi(\rho)$$</span></p> <p>but notice that since the set of Kraus operators for a channel is not unique, neither is the Stinespring dilation.</p>