# How are probabilities different from expectation value for projective measurements

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)

$$$$p(m) = \langle\psi|P_m|\psi\rangle$$$$ But this is same as expectation value of operator $$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 $$$$p(m) = \langle\psi|M_m^\dagger M_m|\psi\rangle$$$$ Define $$E_m = M_m^\dagger M_m$$, then $$$$p(m) = \langle\psi|E_m|\psi\rangle = \langle E_m\rangle$$$$ So does this mean probability of getting a result $$m$$ and expectation value of that operator is same?

• 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. Commented Feb 17, 2023 at 14:58

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.
• Is there a more rigorous proof of question 2? Commented Feb 17, 2023 at 18:11
• 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 $\hat{A}$. Commented Feb 17, 2023 at 18:48