A textbook approach, attributed to Charlie Bennett, for creating reversible circuit which outputs the input qubits and the initialized ancilla qubits involves copying the function output between the two blocks - the forward and the reverse function. As can be seen in this figure

enter image description here

My question is, aren't the copied function qubits entangled to the internal qubits which are being zeroed by the reverse circuit?

Does this trick only work when the function output qubits are in the eigenstates (either |0⟩ or |1⟩)?


1 Answer 1


Yes, it will work fine. If a classical reversible circuit works on every classical basis state, it will also work on superpositions of those states. Try it in a simulator.

Of course, the output qubit at the bottom will end up entangled with the inputs. But that is intended.

  • $\begingroup$ Thanks for your answer. When you say "classical reversible circuit" do you mean that the circuit can only use Boolean gates, i.e., no Hadamard gate? As for the entanglement of the output qubit to the input qubits, it sounds like a form of entanglement swapping (starting with input-output entanglement, and ending with input-copy(output) entanglement) $\endgroup$
    – inq
    Apr 24, 2022 at 0:06
  • $\begingroup$ @hanan I mean that the circuit is equivalent to a permutation matrix. The implementation can use quantum gates like Hadamards, but the overall function is classical. I just know that particular subset is safe. For more general tasks it's not always true (e.g. a circuit that turns classical inputs into an entangled state may not turn superposed inputs into entangled states). $\endgroup$ Apr 24, 2022 at 1:29
  • $\begingroup$ Another example of a circuit that will work on essentially all classical inputs, but can fail if you feed it a superposition, is the initial state of the ancillary register of Shor's algorithm. $\endgroup$ Apr 24, 2022 at 1:35
  • $\begingroup$ Following a comment on a similar question, link I see that: "Uncomputation trick works only if the circuit C is a classical reversible circuit. In other words, we cannot have Hadamard gate in the circuit." $\endgroup$
    – inq
    Apr 24, 2022 at 14:49

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