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So this question is sort of a fundamental one but I'm a bit confused for some reason. Basically, say I have a Hermitian operator [itex]\hat{A}[/itex]. If I have a system that is prepared in an eigenstate of [itex]\hat{A}[/itex], that basically means that [itex]\hat{A}\psi = \lambda \psi[/itex], where [itex]\lambda[/itex] is real, right? So can I say the following, because the system is prepared in an eigenstate of [itex]\hat{A}[/itex]

[itex]∫\psi^{*}\psi=1[/itex]?

The reason I'm asking is because [itex]\psi[/itex] is just a function of [itex]x[/itex] - in literature the normalization is always written in terms of the big psi ([itex]\Psi[/itex]), which is a function of [itex]x,t[/itex].

Also, while im at it - by saying that it is prepared in an eigenstate of [itex]\hat{A}[/itex] does that also mean that the probability of measuring this state is equal to 1? so that the wavefunction is collapsed to this eigenstate?

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# Expectation value of a hermitian operator prepared in an eigenstate

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