I have a question about a passage in the Book Quantum Computation and Quantum Information Textbook by I. Chuang and M, Nielsen. My query is about a passage on page 86 of the 10th edition:

Distinguishability, like many ideas in quantum computation and quantum information, is most easily understood using the metaphor of a game involving two parties, Alice and Bob. Alice chooses a state $|\psi_i\rangle ~(1 \le i \le n)$ from some fixed set of states known to both parties. She gives the state $|\psi_i\rangle$ to Bob, whose task it is to identify the index $i$ of the state Alice has given him.

Suppose the states $|\psi_i\rangle$ are orthonormal. Then Bob can do a quantum measurement to distinguish these states, using the following procedure. Define measurement operators $M_i\equiv |\psi_i\rangle\langle\psi_i|$, one for each possible index $i$, and an additional measurement operator $M_0$ defined as the positive square root of the positive operator $I - \sum_{i\ne 0}|\psi_i\rangle\langle\psi_i|$. These operators satisfy the completeness relation, and if the state $|\psi_i$ is prepared then $p(i) = \langle \psi_i|M_i|\psi_i\rangle = 1$, so the result $i$ occurs with certainty. Thus, it is possible to reliably distinguish the orthonormal states $|\psi_i\rangle.$

Why does $M_0$ need to be the positive square root of the positive operator $\mathbb{1}-\sum_{i \neq 0} |\psi_i\rangle\langle\psi_i|$?

If $M_i:= |\psi_i\rangle\langle\psi_i|$ wouldn't $\mathbb{1}-\sum_{i \neq 0} |\psi_i\rangle\langle\psi_i|+ \sum_{i=1}^n |\psi_i\rangle\langle\psi_i|=\mathbb{1}-\sum_{i \neq 0} |\psi_i\rangle\langle\psi_i|+\sum_{i=1}^n M_i=\mathbb{1}$ already?

I feel like I am missing something super obvious here, but I just cannot figure it out.

  • $\begingroup$ Are there no states past $n$, or are Alice and Bob just choosing to use states up to $n$? $\endgroup$ – BioPhysicist Sep 6 '19 at 15:52
  • $\begingroup$ There is no further information about this in the section, so I would hazard a guess that there are no more states. $\endgroup$ – F6uWJTHhK8 Sep 6 '19 at 15:54
  • $\begingroup$ It sounds like two different notions of "measurement operator" are being mixed here - one satisfying $\sum M_i = I$, and the other $\sum N_i^\dagger N_i=I$, with $M_i=N_i^\dagger N_i$. --- I also have the vague feeling that I have seen precisely the same question here previously. $\endgroup$ – Norbert Schuch Sep 6 '19 at 21:01
  • $\begingroup$ If $M_i \equiv |\psi_i\rangle\langle\psi_i |$, that is, if $M_i$ is a Hermitian projector, doesn't $M_i=M_i^\dagger M_i$? $\endgroup$ – F6uWJTHhK8 Sep 7 '19 at 22:16

So there are certainly some really strange aspects of that passage, for example that $i \ne 0$ is redundant.

The basic reason to define $M_0$ as $$M_0 = I - \sum_{i=1}^n M_i $$ is that the $n$ different states $|\psi_i\rangle$ probably do not form a complete basis for your Hilbert space.

The square root described is actually redundant; that is, if these operators are orthonormal then there should be some basis $|n\rangle$ where $|\psi_i\rangle = |i\rangle$ but $i$ continues on past this over the whole Hilbert space, and then $I - \sum_i |i\rangle\langle i| = \operatorname{diag}(0,0,\dots 0, 1, 1, \dots)$ in this basis, and that operator is always going to be equal to its square root.

For example if we have two qubits we might set up the situation with $$|\psi_1\rangle = \sqrt{\frac12} |00\rangle + \sqrt{\frac12} |11\rangle\\ |\psi_2\rangle = \sqrt{\frac12} |01\rangle - \sqrt{\frac12} |10\rangle$$ and then we would have in matrix notation$$ |\psi_1\rangle\langle\psi_1| = \frac12\begin{bmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{bmatrix}\\ |\psi_2\rangle\langle\psi_2| = \frac12\begin{bmatrix}0&0&0&0\\0&1&-1&0\\0&-1&1&0\\0&0&0&0\end{bmatrix}\\ $$ But if you want to build a measurement apparatus out of detecting these two states as "1" and "2" you need to be able to define what that measurement apparatus will do if I hand you instead the state $$\sqrt{\frac12} |00\rangle - \sqrt{\frac12} |11\rangle,$$ even though that state will not occur when Alice and Bob are playing this “game.” Like, I steal your apparatus after the game and just feed this state into it, what does it read out?

The authors choose to just map all of those “other” states to the measurement outcome 0. So they form the matrix $$M_0 = \frac12\begin{bmatrix}1&0&0&-1\\0&1&1&0\\0&1&1&0\\-1&0&0&1\end{bmatrix}$$based on a general idea that when we add together the squares of all of the measurement operators we should get $I$, and then they take a trivial square root of it (in this case it can be seen quite directly that $(M_0)^2 = M_0$ and therefore defining $M_0^{1/2} = M_0$ is perfectly adequate as a square root of itself).

  • $\begingroup$ Thank you for your great reply! Why do we add together the squares of measurement operators to get $I$, though? The completeness relation depends on the sum of the operators, not the sum of the squares of the operators, no? $\endgroup$ – F6uWJTHhK8 Sep 6 '19 at 16:06
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    $\begingroup$ @F6uWJTHhK8 sharp thinking! Yeah I realized that after I had worked out the example, the fact that I found that $M_0$ was its own square root seemed highly suspicious so I went to see if it was always true and it always is, so I think the prescription to take the square root is another error. I edited my answer to include all of that now. $\endgroup$ – CR Drost Sep 6 '19 at 16:11
  • $\begingroup$ Thank you so much! $\endgroup$ – F6uWJTHhK8 Sep 6 '19 at 16:13

Completeness means that all operators associated with measurements outcomes sum to the identity $I$. For an arbitrary set of $\left|\psi_{i=1,...,n}\right\rangle $ the operators $$M_i=\left|\psi_i\right\rangle\left\langle \psi_i \right|$$ need not sum to $I$. For that purpose we add one more operator, namely $$M_0=I-\sum_{i=1,...,n}M_i$$ and then $$I=\sum_{i=0,...,n}M_i$$


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