In the book Quantum Computation and Quantum Information (Nielsen & Chuang) the section 2.2.4 discusses quantum distinguishability and arrives at the conclusion that if someone were to show me a set of states and give me one of them, I can only tell which one it is if the set is orthogonal. So I came up with a counterexample to this. Let's say we have these two states: \begin{align} |\psi_1\rangle & = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\\ |\psi_2\rangle & = |0\rangle \, . \end{align}

If I make a measurement on the unknown state and the result is the eigenvalue associated with $|1\rangle$, I know with certainty that the given state was $|\psi_1\rangle$, so I was able to distinguish them even though they are not orthogonal. Clearly there's something wrong with my thinking, but I can't tell what it is.

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    $\begingroup$ what if you got $0$? $\endgroup$ – spaceisdarkgreen Sep 14 '17 at 0:57
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    $\begingroup$ You get one shot at telling what the state is and you have to be guaranteed correct. $\endgroup$ – spaceisdarkgreen Sep 14 '17 at 1:08
  • $\begingroup$ I'm not sure if this matters; but in this example you also destroy the state you started with; so you no longer have it $\endgroup$ – Joshua Lin Sep 14 '17 at 1:15
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    $\begingroup$ I don't have the book in front of me, so can't say for sure what the authors were thinking, but if I had to bet, I'd bet that @spaceisdarkgreen is exactly right. $\endgroup$ – WillO Sep 14 '17 at 1:26
  • $\begingroup$ @spaceisdarkgreen Why don't you turn this into an answer? $\endgroup$ – Norbert Schuch Sep 15 '17 at 6:04

As mentioned in the comments, this protocol requires you to be able to distinguish them with a single measurement, on a single copy of the system; moreover, you're not allowed to fail, and you're not allowed to skip or give any non-committal answer.

This is why your protocol fails: if you're given $|\psi_1\rangle$, you have a 50% chance of getting the measurement outcome $|0\rangle$, in which case you don't know which one to choose and you're forced to provide an answer that might be wrong.

In general, the answer to quantum distinguishability problems can depend quite sensitively on the details of the formulation of the requirements on the protocol. Two other examples:

  • You might be allowed an unlimited number of copies of the states to do your measurements (though you might want to keep the number of required copies to a minimum), in which case you can do quantum state tomography and provide very detailed answers about the given states.

  • You might be given a single copy of the system, and you're not allowed to make mistaken identifications, but you allow the protocol to fail, i.e., to refuse to give an answer (often called unambiguous discrimination). In this case your measurement is OK but it can be optimized, and in the optimal case you can provide definite answers that it must be $|\psi_1\rangle$ or that it must be $|\psi_2\rangle$, no matter how similar the two are ─ but the closer they get, the more often the protocol fails to produce an answer.


I'm guessing that the book just has a higher standard for "distinguishable" than you do. Notice that in your example, you could not tell what state it was if you had measured a zero rather than a one. So you cannot guarantee to know what state it was in any outcome, only in the case that you "get lucky" and measure a one.


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