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Expectation value of a hermitian operator prepared in an eigenstate

  1. Nov 5, 2013 #1
    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 [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?
     
  2. jcsd
  3. Nov 5, 2013 #2

    kith

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    Science Advisor

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

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

    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").

    More precisely, the probability of getting the corresponding eigenvalue as measurement outcome is 1.
     
    Last edited: Nov 5, 2013
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