Expectation value of a hermitian operator prepared in an eigenstate

[Dixanadu]

Hey guys,

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 $\hat{A}$. If I have a system that is prepared in an eigenstate of $\hat{A}$, that basically means that $\hat{A}\psi = \lambda \psi$, where $\lambda$ is real, right? So can I say the following, because the system is prepared in an eigenstate of $\hat{A}$
$∫\psi^{*}\psi=1$?

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

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

 

[kith]

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 $\hat{A}$. If I have a system that is prepared in an eigenstate of $\hat{A}$, that basically means that $\hat{A}\psi = \lambda \psi$, where $\lambda$ is real, right?

Yes. However, you should write something like ψ_λ to make clear that ψ is the eigenstate for the eigenvalue λ. Usually, there are many different eigenvalues and corresponding eigenstates.

So can I say the following, because the system is prepared in an eigenstate of $\hat{A}$
$∫\psi^{*}\psi=1$?

No. Why do you think this normalization is related to the property of ψ being an eigenstate of A?

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

If your wavefunction is normalized at some time, it will remain so at subsequent times. This is guaranteed by the Schrödinger equation (we say the time evolution is"unitary").

Also, while I am at it - by saying that it is prepared in an eigenstate of $\hat{A}$ does that also mean that the probability of measuring this state is equal to 1?

More precisely, the probability of getting the corresponding eigenvalue as measurement outcome is 1.