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If we know a set of $K$ states $\{\lvert\psi_j\rangle:j\in[K]\}$ such that they are pairwise orthogonal, and we are given an unknown state $\lvert\psi_i\rangle$, what sort of projective measurement would determine $i$ with probability $1$?

This question is from Ronald de Wolf's lecture notes (chapter 6).

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    $\begingroup$ Hint: For any state $|\psi\rangle$, you can show that $|\psi\rangle \langle \psi |$ is a projector. $\endgroup$
    – Rammus
    Jul 25 at 10:36
  • $\begingroup$ So just $\lvert\psi_i\rangle\langle\psi_i\lvert$? @Rammus $\endgroup$ Jul 25 at 11:27
  • $\begingroup$ I don't think I understand the question completely. What do they mean by "given an unknown state"? $\endgroup$ Jul 25 at 11:28
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    $\begingroup$ You are given one of the states from the set, you have to determine which one you got. $\endgroup$
    – Rammus
    Jul 25 at 12:54

2 Answers 2

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If the states $|\psi_j\rangle$ are orthogonal, then the set of projections $\{|\psi_j\rangle\!\langle \psi_j|\}_j$ form a projective measurement. Performing this measurement, if the input state is $|\psi_j\rangle$, you'll always find the $j$-th outcome. Therefore, you can deterministically discriminate between the elements of your ensemble using this measurement.

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Just measure in [K] basis. Since only one $|\psi_i\rangle$ among $i=1\ldots K$ can be measurement outcome, its probability $P_i=1$.

I wonder what answer is expected. Maybe you should characterize measurement in [K] basis by K projectors $|\psi_i\rangle\langle \psi_i|, i=1\ldots K$

PS: I assumed that K vectors $|\psi_i\rangle$ form a basis; this is not needed, the K vectors $|\psi_i\rangle$ may be a part of orthonormal basis.

PPS: It is not easy to explain what is asked in the textbook question because quantum measurement is very different from classical one. In classical physics, when we measure some physical value, the value always exists before measurement, and the measurement just reveals this value. Quantum measurement is very different. State before measurement may not exist at all, and we usually say about a state after measurement. But here we are said that the state before measurement exists, is one of $|\psi_i\rangle$, $i=1\ldots K$, just unknown, and with probability 1 this is the same state as we get after measurement, as measurent outcome. Hope it helps.

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  • $\begingroup$ So by measuring $\lvert\psi_i\rangle$ with the $K$ projectors $\lvert\psi_i\rangle\langle\psi_i\rvert,i=1,\ldots,K$, we get $\lvert\psi_i\rangle$ with probability $1$? $\endgroup$ Jul 26 at 18:16
  • $\begingroup$ We get as measurement outcome $|\psi_j\rangle$ with probability 1, where j is unknown index from $i=1,\ldots,K$. Think what is really asked here, what is the unknown state $|\psi_j\rangle$. Using $i$ everywhere is ambiguous, it may mean the particular unknown index and the set of indexes $1\ldots, K$ $\endgroup$
    – kludg
    Jul 27 at 0:01

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