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I was reading a book and then I found this statement. I will put the text as well as a screenshot of the text.

The expectation value of an operator is the mean or average value of that operator with respect to a given quantum state. In other words, we are asking the following question: If a quantum state $|\psi\rangle$ is prepared many times, and we measure a given operator $A$ each time, what is the average of the measurement results? This is the expectation value and we write this as $\langle A\rangle =\langle ψ|A|ψ \rangle$ (3.39)

Here is the screenshot,

enter image description here

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Any operator Hermitian $A$ can be described using its eigenvalue decomposition $$ A=\sum_\lambda\lambda P_\lambda, $$ where $\{\lambda\}$ are the distinct eigenvalues and $P_{\lambda}$ are the projectors onto the corresponding eigenspaces. So, to talk about measuring an operator means to make a measurement using the basis $\{P_{\lambda}\}$. It also assigns a label of $\lambda$ to an outcome from the projector $P_{\lambda}$ and, since this is a numerical value, you can ask for the expected value of that number. For an initial state $|\psi\rangle$ you get answer $\lambda$ with probability $\langle\psi|P_\lambda|\psi\rangle$, so the expected value is $$ \sum_{\lambda}\lambda\langle\psi|P_\lambda|\psi\rangle=\langle\psi|A|\psi\rangle. $$

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  • $\begingroup$ Ok. So nice to know this. I am happy now. Thanks and take care. $\endgroup$
    – user27286
    Dec 9 '20 at 11:51
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In this case, "measuring an operator" is meant to describe measuring the observable associated with the operator. More commonly one would put a "using" or "with" or "with respect to" between "measure" and "a given operator". As usual, the operator's eigenvalues correspond to measurement results for the observable, and measuring the observable collapses the quantum state to a corresponding eigenstate of the operator, with a probability for each orthogonal eigenstate equal to the probability amplitude of the eigenstate multiplied by its complex conjugate. The Pauli Z operator is the operator for measuring on the 1-qubit computational basis, with a measurement of $\left|0\right>$ being associated with an eigenvalue of 1 and $\left|1\right>$ being associated with an eigenvalue of -1, and the Pauli X operator is similarly associated with the Hadamard basis, to name two.

The reason the expectation value for this is given by $\left<\psi\right|A\left|\psi\right>$ can be seen through noticing what happens when $A$ acts on each of its associated eigenstates, which $\left|\psi\right>$ is in general in a superposition of. For the finite case, with the set of eigenvalues of the operator being denoted $S$, you can represent the quantum state as $\left|\psi\right> = \sum_{\lambda \in S}a_\lambda\left|s{_\lambda}\right>$, where $A\left|s_{\lambda}\right> = \lambda\left|s_{\lambda}\right>$ and $a_\lambda$ are the associated probability amplitudes. Then $A\left|\psi\right> = \sum_{\lambda \in S} \lambda a_\lambda \left|s_\lambda\right>$. With $\left<\psi\right| = \sum_{\lambda \in S}\bar a_\lambda\ \left<s_\lambda\right|$ and all the different $\left|s_\lambda\right>$ being orthogonal, we get $\left<\psi\right|A\left|\psi\right> = \sum_{\lambda \in S} \lambda a_\lambda \bar a_\lambda$. With the probability amplitudes multiplied by their complex conjugates being the probability of measurement, this is the traditional expected value formula.

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  • $\begingroup$ Thanks for answering in detail.Feeling good. Take care. $\endgroup$
    – user27286
    Dec 9 '20 at 11:52

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