Probability of obtaining general quantum measurement outcome

[Danny Boy]

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?

 

[kith]

You can use the basis of eigenstates of the density matrix $\{|\psi_i \rangle\}$ to evaluate the trace. So you only need to show that the $\langle \psi_i|A^{\dagger}_n A_n|\psi_i \rangle$ are real.

 

[Demystifier]

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?

Due to the cyclic property of the trace we have
$$\text{Tr}[A_n^{\dagger}A_n \rho]=\text{Tr}[\frac{A_n^{\dagger}A_n \rho+\rho A_n^{\dagger}A_n}{2}]$$
But the operator on the right-hand side is hermitian, so its trace is real.