1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ $\rho$ 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. $\endgroup$ – secavara Feb 5 '18 at 13:49
  • $\begingroup$ @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? $\endgroup$ – Moses Feb 5 '18 at 13:54
  • 2
    $\begingroup$ 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. $\endgroup$ – secavara Feb 5 '18 at 13:56
  • 1
    $\begingroup$ 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. $\endgroup$ – secavara Feb 5 '18 at 14:05
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.