1
$\begingroup$

For projective operators, Suppose we have set of operators $\{P_m\}$ then according to measurement postulate, probability of getting result $m$ is (I believe $m's$ here are eigenvalue. Please correct me if I'm wrong)

\begin{equation} p(m) = \langle\psi|P_m|\psi\rangle \end{equation} But this is same as expectation value of operator $P_m$. Does this mean that probability and expectation value are same for projective operators?

Edit: The same is in the case of POVM. If there is a set of measurement operators $\{M_m\}$, then probability for getting result $m$ is \begin{equation} p(m) = \langle\psi|M_m^\dagger M_m|\psi\rangle \end{equation} Define $E_m = M_m^\dagger M_m$, then \begin{equation} p(m) = \langle\psi|E_m|\psi\rangle = \langle E_m\rangle \end{equation} So does this mean probability of getting a result $m$ and expectation value of that operator is same?

Improve this question
$\endgroup$
1
  • $\begingroup$ A projection $P$ satisfies $P^2=P$ and can therefore only have the eigenvalues $0$ or $1\,.$ The expectation of something that can have only those values can be considered a probability. $\endgroup$
    – Kurt G.
    Commented Feb 17, 2023 at 14:58

1 Answer 1

Reset to default
2
$\begingroup$

Suppose you have an operator $\hat{A}$ with eigenvalues $a$. Then, in the non-degenerate case (this is easily generalized to the degenerate case), there is a set of projection operators $\hat{P}_a$ s.t. that the probability of getting the outcome $a$ upon measurement of the physical variable associated with $\hat{A}$ is $$ p(a|\psi) = \langle\psi |\hat{P}_a | \psi\rangle\,. $$

We can ask one of two questions:

  1. "If I measure the variable associated with $\hat{A}$, what are the possible outcomes of the measurement and with what probabilities?"

  2. "If I measure the variable associated with $\hat{A}$, will I get outcome $a$ or not?

The answer to the question 1 is

  • $\{a\}$ and $p(a|\psi)$.

The answer to question 2 is

  • We get $a$ with probability $p(a|\psi)$ and not $a$ with probability $1-p(a|\psi)$. Put a different way, I can get the value 1 meaning "I got $a$ as the result" or I can get the value 0 meaning "I didn't get $a$ as the result." Then, 1 and 0 are the eigenvalues of $\hat{P}_a$ corresponding to the eigenvectors $|a\rangle$ and everything else, respectively.
Improve this answer
$\endgroup$
2
  • $\begingroup$ Is there a more rigorous proof of question 2? $\endgroup$
    – QuantumOscillator
    Commented Feb 17, 2023 at 18:11
  • $\begingroup$ Rigorous proof? What do you mean? $\hat{P}_a$ is a Hermitian operator, and as such, it follows the standard measurement prescription, which I've outlined, and it all works because the eigenvectors of $\hat{P}_a$ are also eigenvectors of $\hat{A}$. $\endgroup$
    – march
    Commented Feb 17, 2023 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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