# Repeated measurement question

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1. Sep 29, 2015

### mangokiller

A ket is expanded as $\vert \nu \rangle = c_1\vert \nu_1\rangle+c_2\vert \nu_{2}\rangle$.

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

$\vert \psi(t)\rangle = e^{-i \hat H t / \hbar}\vert \psi(0) \rangle =e^{-i \hat H t / \hbar}\vert \nu_1 \rangle= e^{-i \hat H t / \hbar} \big(c_1\vert \nu_1 \rangle+c_2\vert \nu_2\rangle \big) =e^{-i E_1 t / \hbar} c_1\vert \nu_1 \rangle+e^{-i E_2 t / \hbar}c_2\vert \nu_{2}\rangle.$

2. Sep 29, 2015

### Geofleur

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.

3. Sep 29, 2015

### Staff: Mentor

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 $\vert \nu \rangle = c_1\vert \nu_1\rangle+c_2\vert \nu_{2}\rangle$, 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