Probability of obtaining general quantum measurement outcome The Fundamental Theorem of Quantum Measurement is stated as follows:
Every set of operators $\{ A_n \}$ $n =1,...,N$ that satisfies $\sum_n A_n^{\dagger}A_n = I$ describes a possible measurement on a quantum system, where the measurement has $n$ possible outcomes labeled by $n$. If $\rho$ is the state of the system before the measurement and $\tilde{\rho}_n$ is the state of the system upon obtaining measurement result $n$, and $p_n$ is the probability of obtaining result $n$, then $$\tilde{\rho}_n = \frac{A_n \rho A_n^{\dagger}}{p_n}~~\text{and}~~p_n = \text{Tr}[A_n^{\dagger}A_n \rho]$$
Question: Since $p_n = \text{Tr}[A_n^{\dagger}A_n \rho]$ represents the probability of obtaining measurement result $n$, I assume that this is a real number (in the interval $[0,1]$) rather than complex, but I fail to see how it is guaranteed that this will be a real number. Am I missing something?
 A: Since $\sum_n A_n^\dagger A_n =I$, we have. for $||\psi||=1$
$$\sum_n \langle \psi |A_n^\dagger A_n\psi \rangle = \langle \psi| \psi \rangle =1 \:,$$
that is 
$$\sum_n \langle A_n\psi | A_n\psi \rangle = 1$$
and thus, since  $\langle A_n\psi | A_n\psi \rangle =||A_n\psi||^2\geq 0$, we obtain
$$0\leq \langle A_n\psi | A_n\psi \rangle \leq 1\:. \tag{1}$$
Finally, by definition of density matrix $\rho$ we have $\rho= \rho^\dagger$, $\langle \phi|\rho \phi\rangle \geq 0$, and $$tr(\rho)=1\tag{2}\:.$$ Now suppose that  $\{\phi_m\}$ is a Hilbert  basis of eigenvectors of $\rho$, so that $$\rho \phi_m = q_m \phi_m\tag{3}\:.$$ Since $\rho = \rho^\dagger$ it holds $q_m \in \mathbb R$
and we have form (1), (2), and (3), 
$$tr(A_n^\dagger A_n\rho) = \sum_m \langle \phi_m | A_n^\dagger A_n \rho\phi_m\rangle = \sum_m q_n\langle \phi_m | A_n^\dagger A_n\phi_m\rangle 
= \sum_m q_n\langle A_n\phi_m | A_n\phi_m\rangle 
\leq \sum_m q_n = tr(\rho)=1\:. $$
Notice that, in particular, 
$$tr(A_n^\dagger A_n\rho) = \sum_m q_m\langle A_n\phi_m | A_n\phi_m\rangle \in \mathbb R$$
because the right-hand side is a sum of real numbers since $q_m\in \mathbb R$ and
$\langle A_n\phi_m | A_n\phi_m\rangle = ||A_n\phi_m||^2 \in \mathbb R$.
On the other hand  $p_m = \langle \phi_m| \rho \phi_m\rangle \geq 0$ so that
$$tr(A_n^\dagger A_n\rho) = \sum_m \langle \phi_m | A_n^\dagger A_n \rho\phi_m\rangle = \sum_m q_m\langle A_n\phi_m | A_n\phi_m\rangle = \sum_m q_m || A_n\phi_m||^2 \geq 0\:.$$
In summary $tr(A_n^\dagger A_n\rho) \in [0,1]$.
