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Repeated measurement question

  1. Sep 29, 2015 #1
    A ket is expanded as [itex] \vert \nu \rangle = c_1\vert \nu_1\rangle+c_2\vert \nu_{2}\rangle [/itex].

    A measurement results in the eigenvalue a1. Is it possible to measure the other eigenvalue a2 at a time t after the first measurement?

    Could I write something like

    [itex] \vert \psi(t)\rangle = e^{-i \hat H t / \hbar}\vert \psi(0) \rangle =e^{-i \hat H t / \hbar}\vert \nu_1 \rangle= e^{-i \hat H t / \hbar} \big(c_1\vert \nu_1 \rangle+c_2\vert \nu_2\rangle \big) =e^{-i E_1 t / \hbar} c_1\vert \nu_1 \rangle+e^{-i E_2 t / \hbar}c_2\vert \nu_{2}\rangle. [/itex]
     
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  3. Sep 29, 2015 #2

    Geofleur

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    You could measure a2 after the first measurement, but the equations in that last line are not quite right. You cannot expand ## | \nu_1 \rangle ## in terms of itself and ## | \nu_2 \rangle ##, because these vectors are linearly independent of each other. Instead, the time evolution operator could transform ## | \nu_1 \rangle ## into a different vector, which is itself expandable in terms of ## | \nu_1 \rangle ## and ## | \nu_2 \rangle ##. That is, you could have ## e^{-i\hat{H}t/ \hbar} | \nu_1 \rangle = c_1 | \nu_1 \rangle + c_2 | \nu_2 \rangle ##. Of course, this assumes that the two eigenstates form a basis, so that any state can be expanded in terms of them.
     
  4. Sep 29, 2015 #3

    Nugatory

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    It depends on what operator ##\vert \nu_1 \rangle## and ##\vert \nu_2 \rangle## are eigenstates of. If that operator commutes with the Hamiltonian, then the answer is no; if it does not, the answer is yes.

    Even though the initial state would have been [itex] \vert \nu \rangle = c_1\vert \nu_1\rangle+c_2\vert \nu_{2}\rangle [/itex], after a measurement produces the result ##a_1## (and assuming that that's the eigenvalue associated with ##\vert \nu_1 \rangle##) the state is going to be ##\psi = \vert \nu_1 \rangle##, so that's the state that you want to be evolving forward in time.
     
    Last edited: Sep 29, 2015
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