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I can only find the proof for the reverse statement (i.e. here). However Nielsen Chuang Quantum Computation and Quantum Information p. 90 states the following:

Suppose we define $$E_m = M^\dagger_m M_m \tag{2.117} $$ Then from Postulate 3 and elementary linear algebra, $E_m$ is a positive operator such that $\sum_m E_m = I$ and $p(m) = \langle \psi| E_m| \psi \rangle$.

From the definition it is clear, that $E_m$ is self-adjoint but I am wondering what elementary linear algebra I am missing, since I cant understand why from there it follows, that $E_m$ is positive.

The postulate 3 (p.84-85 can be found in google books) reads in short:

Quantum measurements are described by a collection $\{M_m\}$ of measurement operators.
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If the state of the quantum system is $|\psi\rangle$ immideately before the measurement then the probability that result $m$ occurs is given by $$p(m) = \langle \psi| M_m^\dagger M_m| \psi \rangle, \tag{2.92} $$
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The measurement operators satisfy the completeness equation, $$\sum_m M_m^\dagger M_m = I. \tag{2.94} $$

I am guessing that I need to incorporate the completeness equation but I can't see how. Can someone please give me a hint to understand this elementary statement?

v.tralala
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1 Answers1

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Not all Hermitian operators are positive (just take the simple Hermitian $1\times 1$ matrix $H= (-1)$ as a counter example). The completeness has also nothing to do with the positivity.

However, an operator of the form $$E= M^\dagger M$$ is trivially positive. We can show that for any vector $\bf v$, we have that $$\langle \mathbf v , E \mathbf v\rangle = \langle \mathbf v , M^\dagger M \mathbf v\rangle = \langle M \mathbf v , M \mathbf v\rangle = \langle \mathbf w, \mathbf w\rangle \geq 0 $$ where $\mathbf w = M \mathbf v$.

Fabian
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