In Nielsen and Chuang's text, p.86, they discuss distinguishing states:

Suppose states $|\psi_i \rangle$ are orthonormal. Define measurement operators $M_i = |\psi_i \rangle \langle \psi_i|$ for each possible index $i$, and another operator $M_0$ as the positive square root of $I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|$. These operators satisfy the completeness relation [...] if the state $\psi_i$ is prepared then $p(i) = \langle \psi_i| M_i | \psi_i \rangle = 1$.

I don't see why they use the positive square for defining $M_0$. Isn't it enough to define $M_0 = I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|$?

If we do define $M_0 = I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|$ then we have

\begin{align*}\sum_{i=0,1,\ldots} M_i^+ M_i &= \left(I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\right)^+ \left(I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\right) + \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\\ &= \left(I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\right) \left(I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\right) + \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i| \\ &= \left(I - \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i|\right) + \sum_{i \ne 0} |\psi_i \rangle \langle \psi_i| \\ &= I \end{align*}

So what's wrong with this?

  • $\begingroup$ The eigenvalues of $M_0$ are $0$ and $1$, so indeed it does not matter whether you take the square root or not. However, in a more general context (e.g. non-orthogonal $\psi_i$) you want to use the square root. $\endgroup$ Oct 27, 2016 at 21:33


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