# 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?

• $\rho$ is hermitian, semipositive and it's trace is 1. We can check directly that $A^{\dagger}_n \rho A_{n}$ is hermitian. As a hermitian operator, its eigenvalues will be real. The trace is the sum of the eigenvalues and so $\mathrm{Tr}[A^{\dagger}_n \rho A_{n}]$ is real. The trace of $\frac{A^{\dagger}_n \rho A_{n}}{p_n}$ is then 1 as well. – secavara Feb 5 '18 at 13:49
• @secavara Thanks for your response. Are you saying that we can show that $A_n^{\dagger}A_n \rho$ is Hermitian and this implies that $\text{Tr}[A_n^{\dagger}A_n \rho]$ is real? – Moses Feb 5 '18 at 13:54
• Yes, $\left(A^{\dagger}_n \rho A_{n}\right)^{\dagger} =A_{n}^{\dagger} \rho^{\dagger} (A^{\dagger}_n)^{\dagger} = A_{n}^{\dagger} \rho A_n$. So $A^{\dagger}_n \rho A_{n}$ is hermitian. – secavara Feb 5 '18 at 13:56
• You can also show in addition that $A^{\dagger}_n \rho A_{n}$ is semipositive. $\rho$ is semipositive so by definition, $\langle \psi |\rho|\psi\rangle \geq 0$ for all $|\psi\rangle$. Now if we compute $\langle \psi |A^{\dagger}_n \rho A_{n}|\psi\rangle$ we get $\langle \psi'|\rho |\psi'\rangle$, where $|\psi'\rangle=A_{n}|\psi\rangle$. But $\rho$ is semipositive so this is a non-negative quantity and hence $\langle \psi |A^{\dagger}_n \rho A_{n}|\psi\rangle$ is semipositive as well. – secavara Feb 5 '18 at 14:05

## 1 Answer

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 $tr(A_n^\dagger A_n\rho) \in [0,1]$.