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Suppose we have a quantum circuit that contains an arbitrary number of quantum gates and takes as an input more than a single qubit, say three. What are the restrictions on the quantum gates and the way that they are composed (in parallel or in sequence) so that when we partially trace out for all the wires except one, we get a pure state? This is an example of the set-up I have in mind.

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  1. Input State: Pure
  2. Apply Quantum Circuit Operation
  3. Output State: Pure
  4. Partially trace out qubits 2 and 3.
  5. Induced State 1: Pure

Cross-posted on physics.SE

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    $\begingroup$ You want to avoid generating entanglement between the system you are keeping and the system you are tracing out. Any entangled state will have a mixed marginal -- consider the Schmidt decomposition. $\endgroup$
    – Rammus
    Jul 7 at 16:32
  • $\begingroup$ Have you any idea about how is the avoidance of entanglement between the two subsystems reflected on the choice of gates and the way that they are composed? $\endgroup$ Jul 7 at 17:43
  • $\begingroup$ crossposted to physics. $\endgroup$ Jul 7 at 23:12
  • $\begingroup$ Is it a problem? The audience might not be the same. $\endgroup$ Jul 8 at 0:08
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    $\begingroup$ @GeorgeSmyridis not necessarily a problem, but please link the two posts together, to avoid people wasting time writing up something that might have already been explained somewhere else $\endgroup$
    – glS
    Jul 8 at 9:16
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The goal is equivalent to asking for there to be no entanglement between system $1$ and the other states at the end of the circuit. Luckily, nomenclature helps us: if there are no "entangling gates" between system $1$ and the other systems, this will be the case. However, that answer is trivial, as it means there were never any interactions between system $1$ and the other systems, so it is still a good question to ask what nontrivial interactions can lead to zero entanglement at the end of the circuit.

I don't think there's a more general rule, but I can give a list of examples of nontrivial circuits that generate no entanglement between system $1$ and the others:

  1. Allow for controlled operations but enforce that the control mode is not in a superposition state. For example, the interaction $CNOT|0\rangle_2\otimes|\psi\rangle_1=|0\rangle_2|\psi\rangle_1$ generates no entanglement. This is the trivial example in disguise, because it is using an entangling gate in a way that does not generate entanglement.
  2. Use a set of entangling operations that end up performing a swap. For example, denoting by $cX_{ij}$ the act of a CNOT with control $i$ and target $j$, we have that $cX_{12}cX_{21}cX_{12}|\psi\rangle_1|\phi\rangle_2=|\phi\rangle_1|\psi\rangle_2$. This only works if the state being swapped into system 1 is not entangled with any other state, so again is a trivial example in disguise because it pushes the requirement of not generating entanglement to some other state.
  3. Use a set of entangling operations that cancel each other. The trivial example is to use $CNOT^2=\mathbb{I}$. A more complicated example can be found by realizing that $cX_{13}=cX_{23}cX_{12}cX_{23}cX_{12}$ (eg https://arxiv.org/abs/1110.2998v1) such that $cX_{13}cX_{23}cX_{12}cX_{23}cX_{12}=\mathbb{I}$.

And many many more. But in general, if there are entangling gates between system 1 and the others, the final result in system 1 will not be pure.

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    $\begingroup$ Are you sure (2) is correct in this situation? If I start with $|0\rangle_1 |\Phi\rangle_{23}$ for some entangled $|\Phi\rangle$ and then apply $SWAP_{12}$, then it is now system $2$ that is unentangled and the local state over system $1$ is no longer pure. $\endgroup$
    – forky40
    Jul 7 at 23:56
  • $\begingroup$ Agreed with forky40, this seems wrong. $\endgroup$ Jul 8 at 9:37
  • $\begingroup$ (2) is correct as stated but you are right that it requires the state being swapped to not be entangled with anything else; will clarify $\endgroup$ Jul 8 at 14:50
  • $\begingroup$ Well, then this does not seem to answer the question as intended by OP (i.e. valid for all inputs - see comments on physics.se) $\endgroup$ Jul 10 at 19:52
  • $\begingroup$ @NorbertSchuch before the posts were linked I only answered this question, which did not mention validity for all inputs. (3) still works for any separable input (obviously if you don't know whether the input is separable or not you can't do anything in general) $\endgroup$ Jul 10 at 20:34

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