What is the simplest quantum circuit which takes states $|a\rangle$ and $|𝑏\rangle$ as its input, and outputs their (normalized) sum $c (|𝑎\rangle+|𝑏\rangle$)? Here $c$ is a normalization constant to make the output a valid quantum state. The use of Ancilla qubits is allowed. It might also be necessary to use some kind of "encoding" for states $|a\rangle$ and $|𝑏\rangle$.
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$\begingroup$ What does "encoding" mean? And can |a> and |b> be arbitrary states, or orthogonal basis states? In either case, what would a valid "encoding" be? $\endgroup$– Norbert SchuchApr 14, 2020 at 16:10
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$\begingroup$ I’m voting to close this question because it is an exact cross-post of a question on QuantumComputing.SE. If you do not get an acceptable answer on QC.SE after a reasonable amount of time (>= 1 week), feel free to ping one of the mods and we can consider re-opening it here. $\endgroup$– tpg2114Apr 15, 2020 at 12:28
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3$\begingroup$ did you have a look at Oszmaniec et al. (2015)? $\endgroup$– glSApr 15, 2020 at 12:28
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2$\begingroup$ @glS Nice catch! Theorem 1! (Also, that's how one states the question precisely!) $\endgroup$– Norbert SchuchApr 15, 2020 at 13:26
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$\begingroup$ @tpg2114 This cross-posting between pse and qc.se has become kind of a habit. Do we have a general policy on that? $\endgroup$– Norbert SchuchApr 15, 2020 at 13:27
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1 Answer
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Such a circuit cannot exist. To see why, assume $|b\rangle = e^{i\phi}|a\rangle$. Then, the input $|a\rangle|b\rangle$ is mapped to $|a\rangle + e^{i\phi}|b\rangle=(1+e^{i\phi})|a\rangle$, which (i) is not normalizable independent of $\phi$, (ii) is identically zero for $\phi=\pi$, and (iii) whose output normalization, and whose output state modulo a phase depends on the unphysical state of the input.
In brief, it is incompatible with linearity.
After all, if the map - let's call it $U$ - were linear, on input $|v\rangle$, it should produce the output $U|v\rangle$, and therefore due to linearity, on input $e^{i\phi}|v\rangle$ the output $e^{i\phi}U|v\rangle$. (As you can see: You don't even need linearity in this step, you just need homogeneity over the complex numbers!) Clearly, the argument above shows that for the map you quote, this is not the case.
answered Apr 14, 2020 at 16:14
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$\begingroup$ Hi Norbert, thanks for attempting to answer, but obviously we assume that the state is normalized, so your counterexample isn't valid. Nevertheless, a different linearity argument is likely the explanation why the problem cannot be solved :) $\endgroup$– SamuelApr 14, 2020 at 16:36
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$\begingroup$ I added a normalization constant c to the output, which is what I initially meant with "normalized" $\endgroup$– SamuelApr 14, 2020 at 16:36
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$\begingroup$ @SamuelBosch My counterexample implies that linearity is violated: If the map would be linear, applying it to $|v\rangle$ and $e^{i\phi}|v\rangle$ would give the same output up to a global phase. $\endgroup$ Apr 14, 2020 at 17:34
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$\begingroup$ I am not sure I understand what you mean with this argument. Could you please explain it? $\endgroup$– SamuelApr 14, 2020 at 18:16
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2$\begingroup$ Dear @SamuelBosch, how about this: You write down clearly what you mean. And by "clearly" I mean formulas, not just words. Make dependencies clear, i.e. if $c$ depends on $a$ and $b$, write $c(a,b)$. Make clear what $|a\rangle$ is, i.e.: $|a\rangle\in\mathrm{set}$. Then I will be happy to explain why such a map cannot exist. Now you want me to state assumptions and conclusions, just to risk that you later complain that those weren't your assumptions. $\endgroup$ Apr 14, 2020 at 19:14