Understanding Expectation Values in Quantum Mechanics

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In quantum mechanics, the expectation value of an observable A for a state |psi> is not simply calculated as A acting on |psi>. Instead, A|psi> results in another state, unless |psi> is an eigenstate of A, in which case it equals the eigenvalue multiplied by |psi>. The proposed formula for the expectation value, <A> = Integral[P(x) A psi(x) dx], is incorrect because it does not align with the standard definition, which is <psi|A|psi>. The correct expectation value requires using the inner product of the state with the operator acting on it. Understanding these distinctions is crucial for accurate calculations in quantum mechanics.
dudy
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Let A be an observable (opeator), and we're assuming that for a given state psi(x), the value of A is given by A acting on psi(x), namely - A|psi>.
Also we assume that - P(x) = |psi(x)|^2
So, I'de expect the Expectation value of A to be defined like so:
<A> = Integral[-Inf:+Inf]{ P(x) A psi(x) dx} = Integral[-Inf:+Inf]{ |psi(x)|^2 A psi(x) dx} , which is not <psi|A|psi>, and that's clearly not right. where did i go wrong here?
 
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dudy said:
Let A be an observable (opeator), and we're assuming that for a given state psi(x), the value of A is given by A acting on psi(x), namely - A|psi>.

No. A|\psi\rangle is not a value. It's another state, in general.

If |\psi\rangle happens to be an eigenstate of the operator A, then A|\psi\rangle = a|\psi\rangle, where a is an eigenvalue of the operator A.
 
got it, thank you!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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