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In the book Quantum Computation and Quantum Information, there is a discussion about how if states are not orthonormal then there is no quantum measurement capable of distinguishing the states.

I am interested in the consequences of this. What does this mean physically? How is this applied when manipulating quantum states?

I am sure there are other questions that stem from this that I am not even thinking of, but that would be interesting to explore. So any help in understanding the consequences of this would be greatly appreciated.

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  • $\begingroup$ Are you sure you're quoting this correctly? $|p\rangle$ and $|x\rangle$ are certainly not orthogonal but they are definitely physically distinguishable states. Do you perhaps means that states related by a complex phase are physically indistinguishable? i.e. $|\phi\rangle\sim \alpha|\phi\rangle$ for $\alpha\in \Bbb C$? $\endgroup$
    – Charlie
    Jun 12, 2021 at 21:12
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    $\begingroup$ @Charlie I think the statement is that no one quantum measurement can reliably distinguish the states. Suppose the states are $|a\rangle$ and $|b\rangle$ with $\langle a|b\rangle\neq 0$. If we measure the observable $P_a\equiv|a\rangle\langle a|$ and get the result $0$, then we know that the initial state was definitely not $|a\rangle$. So in this case, we distinguished between the two states with one measurement. But we can't do this reliably, because if we measure the same observable $P_a$ and get the result $1$, then the initial state could have been either $|a\rangle$ or $|b\rangle$. $\endgroup$ Jun 12, 2021 at 21:34
  • $\begingroup$ See for example this QC SE thread and the respective exercise in the book mentioned in the question. $\endgroup$ Jun 12, 2021 at 21:37
  • $\begingroup$ @ChiralAnomaly Ah ok I haven't seen this before, fair enough thanks $\endgroup$
    – Charlie
    Jun 12, 2021 at 21:48
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    $\begingroup$ I think this is a bit too broad. What does this mean physically? well, exactly what you said? That there are states which cannot be deterministically distinguished. You could try looking up quantum state discrimination, it's kind of a whole field in itself. $\endgroup$
    – glS
    Jun 13, 2021 at 14:29

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Let assume two states $|x\rangle$ and $|y\rangle$ and their inner product $\langle x|y\rangle$. If $\langle x|y\rangle = 0$, then both state can be perfecly distinguished and there is no uncertainty which is which. Examples of such states are $|0\rangle$ and $|1\rangle$ or $|+\rangle$ or $|-\rangle$. As $\langle x|y\rangle$ tends to $1$, the states are more and more similar and less distinguishable. In extereme case when $\langle x|y\rangle$ is 1, it holds that $|y\rangle = \mathrm{e}^{i\theta}|x\rangle$. Such states differ in global phase only and they are absolutely indistinguishable. This all means that not only two orthogonal states are distinguishable but others as well. However, as $\langle x|y\rangle$ goes to one, the states are more and more similar.

Note that you can employ so-called swap test to calculate value $|\langle x|y\rangle|^2$ and decide how two state are similar to each other.

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    $\begingroup$ "Such states differ in global phase only and they are absolutely indistinguishable." -- Those states don't differ. Quantum states are vectors in a projective space. $\endgroup$ Jun 14, 2021 at 21:19
  • $\begingroup$ @NorbertSchuch: Thanks for the comment. I know that such states are not physically distinguishable but they still have different global phase...so they are not exactly the same from mathematical point of view, they are rather equivalent...or do I miss something? $\endgroup$ Jun 15, 2021 at 6:25
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    $\begingroup$ The only reason why they are mathematically different objects is because physicists tend to use the wrong mathematical objects to describe quantum states. Quantum states should be described by vectors in a projective vector space, where a global phase (and scaling = normalization) is divided out. Alternatively, the phase disappears when you write pure states as density operators. $\endgroup$ Jun 15, 2021 at 11:10
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    $\begingroup$ @NorbertSchuch you make a good point, but to nitpick, the elements of the projective Hilbert space are not vectors because it is not a vector space. $\endgroup$
    – J. Murray
    Jun 15, 2021 at 13:22
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I'm not sure if this will help but a very sharp former colleague once posed this question.

You have two boxes. The first contains 1000 horizontally polarized photons and 1000 vertically polarized photons while the second contains 1000 left circularly polarized photons and 1000 right circularly polarized photons. How do you tell which is which?

Since this is sort of a riddle, I'll make the answer below not visible by default.

Use a vertical polarization filter. If it lets exactly 1000 photons through then you opened the first box. If there's a slight imbalance then you opened the second box. Note than even though 1000 is the most likely value for the second box, binomial distributions are such that it is more likely for you to not see the most likely value :).

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  • $\begingroup$ You can also include spoilers. $\endgroup$ Jun 12, 2021 at 21:46
  • $\begingroup$ Thanks! Edited. $\endgroup$ Jun 12, 2021 at 21:52

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