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I have seen sometimes that in some protocols, one can uncompute a circuit using only Clifford + measurements. See for example

Erase the two qubits containing $p_{w,r}$ and $p_{w,s}$, then the register containing component w of p. These erasures can be done with measurements and Clifford gates.

before eq 73 in Su et al. For context, these are qubits that were "copied" via C-Nots from another register, and have been used as a control in a C-Z gate.

When can one erase registers using measurements and Clifford gates in general?

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I'm not sure what the exact general requirement is, but a sufficient requirement is:

  1. The qubit's computational basis must be a function of other qubits' computational basis value. For example, the state should have the form:

    $$\sum_k \alpha_k |k\rangle |\dots\rangle |f(k)\rangle$$

    where $f$ is some classical function that outputs a single bit.

  2. The fixup operation $U$ which performs $U(\sum_k \alpha_k |k\rangle) = \sum_k (-1)^{f(k)} \alpha_k |k\rangle$ must be possible to do using only stabilizer operations. Because you're going to measure the qubit storing $f(k)$ in the X basis, and if the result is $|-\rangle$ then you need to apply $U$ to finish the uncomputation.

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  • $\begingroup$ Thanks! Would $U$ not amount to an unimportant global phase now that we have measured $f(k)$ (the recovery operation does no longer depend on $k$)? $\endgroup$
    – Pablo
    Dec 22, 2021 at 22:27
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    $\begingroup$ @Pablo No, because it has negated states matching $f$ relatives to states that don't. $\endgroup$ Dec 22, 2021 at 22:43
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    $\begingroup$ For example, if $f$ was "an even number of bits are ON", then $U$ would negate the state $|101\rangle$ but not the state $|111\rangle$. $\endgroup$ Dec 22, 2021 at 22:45
  • $\begingroup$ I think I get it: starting from $|\psi\rangle|k\rangle|f(k)\rangle$, - If $f(k) = 0$, then one applies X and gets $|\psi\rangle|k\rangle(|+\rangle+|-\rangle)$. Since controlling on $k$ the operation $U$ does nothing (because $f(k) = 0$), it stays the same in either case. - If $f(k)=1$ one would have $|\psi\rangle|k\rangle(|+\rangle-|-\rangle)$. I guess the idea is that if one measures $|-\rangle$ then applying $f(k)$ controlled on register $|k\rangle$ recovers the original state. I guess the main limitation is the need to control on $|k\rangle$. Is this right? $\endgroup$
    – Pablo
    Dec 22, 2021 at 22:51
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    $\begingroup$ @Pablo Yes, that's the main limitation. It's usually easier to do $|k\rangle \rightarrow (-1)^f(k) |k\rangle$ than it is to do $|k\rangle \rightarrow |k\rangle|f(k)\rangle$, e.g. you'll be able to apply a CZ to the qubits of $k$ instead of a Toffoli controlled by the qubits of $k$. $\endgroup$ Dec 23, 2021 at 0:20

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