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Let $\left\vert \phi\right\rangle=\frac 1{\sqrt2}\left\vert x_1,0^n\right\rangle+\frac1{\sqrt2}\left\vert 0^n,x_2\right\rangle$ be a $2n$-bit quantum state for some unknown $x_1,x_2\in\{0,1\}^n$. My question is: is it possible to extract $x_1$ and $x_2$ from $\left\vert \phi\right\rangle$ with non-negligible probability?

glS
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Henry
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  • From how you've written it, is it known that $x_1\neq x_2$? – DaftWullie Feb 22 '22 at 07:56
  • Yes, $x_1 \neq x_2$ holds. – Henry Feb 22 '22 at 08:23
  • From a single copy of the state? At a first glance, that seems information-theoretically impossible ... From a singlee measurement you can learn at most $n$ bits, so the best you can do is to guess the other $n$ bits. – Markus Heinrich Feb 22 '22 at 16:10
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    @MarkusHeinrich but $\phi$ is a $2n$-qubit state so you can measure $2n$ qubits and hence get $2n$ bits of information. – DaftWullie Feb 22 '22 at 16:14
  • Imagine you could learn $x_1$ and $x_2$ with finite probability. That would mean you could succeed with the task of cloning $|\phi\rangle$ with finite probability, which should not be possible (as the set of possible $|\phi\rangle$ are not orthogonal). But one has to be a bit more careful about the success probability. – DaftWullie Feb 22 '22 at 16:16
  • The case where $x_1=0$ reduces to "pick out a vector from this set where all vectors are mutually 60 degrees apart". It's possible that's just a known quantity written down somewhere. It's definitely going to drop exponentially with $n$. – Craig Gidney Feb 22 '22 at 21:46
  • @DaftWullie since you know that one half of the bitstring will always be 0, you don't learn $2n$ bits. – Markus Heinrich Feb 23 '22 at 14:39
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    @MarkusHeinrich but that's only if you measure in the standard basis. It's not obvious to me that such an argument still applies after I've done some arbitrary processing to the state. – DaftWullie Feb 23 '22 at 14:53
  • @DaftWullie true, but I suspect that's the best you can do. So the whole game is basically classical. But, to be honest, I don't have a good argument for this. – Markus Heinrich Feb 23 '22 at 14:54
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    @CraigGidney Isn't it the case that if $x_1=0$, you just measure the second register and always succeed in identifying $x_2$ with probability 1/2? – DaftWullie Feb 23 '22 at 14:57
  • @DaftWullie Good point. That works in the original case too, actually, although you only ever get one of them. I guess needing both is perhaps key to it being difficult. – Craig Gidney Feb 23 '22 at 18:25

1 Answers1

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I don't think that's possible to do it with a single copy of the state.

A general strategy for this problem can be described like this: apply an arbitray unitary $U$ to $|\varphi\rangle|0\rangle$, where $|0\rangle$ is an ancilla register and then measure the whole state to get $\left|x_1,x_2,c\right\rangle$ for some $c$. After having applied $U$, the resulting state is: $$\begin{align*}&U\left|x_1,0,0\right\rangle+U\left|0,x_2,0\right\rangle\\={}&\sum_{a,b,c}\alpha_{a,b,c,x_1}|a,b,c\rangle+\sum_{a,b,c}\beta_{a,b,c,x_2}|a,b,c\rangle\\={}&\sum_{a,b,c}\left(\alpha_{a,b,c,x_1}+\beta_{a,b,c,x_2}\right)|a,b,c\rangle.\end{align*}$$ Measuring this state yields the state $|x_1,x_2,c\rangle$ with probability $\left\|\alpha_{x_1,x_2,c,x_1}+\beta_{x_1,x_2,c,x_2}\right\|^2$. We now want to maximize this probability, that is, we want $\left\|\alpha_{x_1,x_2,c,x_1}+\beta_{x_1,x_2,c,x_2}\right\|^2=1$. This is, however, not possible. Indeed, recall that since $x_1$ and $x_2$ are uniformly random, $U$, and as such the $\alpha$ and $\beta$, is chosen independently of their values. In particular, it is not possible to choose a $U$ that maximizes this particular probability.

We can go even further to generalize this: choosing the coefficients of $U$ such that the probability of measuring $\left|x_1,x_2,x\right\rangle$ is maximal must be equal between all possible values of $\left(x_1,x_2\right)$. Thus, no strategy for getting both value can do better than guessing them.

Note that this reasoning also confirms Markus Heinrich's intuition in the comments. Indeed, even though we measure more than $n$ qubits, the set of states we are considering only has states carrying $n$ bits of information. This can be seen by applying the exact same reasoning to the following problem: from $\left|x_1\right\rangle+\left|x_2\right\rangle$, get the value of both $x_1$ and $x_2$.

Do not hesitate to point out any mistake I could have made!

Tristan Nemoz
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