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I have the understanding that the outcome of measuring the same observable twice in a row on the same state is getting the square of the eigenvalue if and only if the state is an eigenstate of the observable. Is it true?

Otherwise $\langle O \rangle \psi_n = n \psi_n$ would not hold, right?

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  • $\begingroup$ True. The onservable should have an eigenvalue. $\endgroup$ – Aman pawar Oct 9 '18 at 18:18
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    $\begingroup$ You do not get the square, but the same eigenvalue. $\endgroup$ – DanielC Oct 9 '18 at 18:22
  • $\begingroup$ If you get a different value in the second measurement, you weren't paying attention to the experimental setup and something else happened between measurements, or your measurement tool is broken. $\endgroup$ – ohwilleke Oct 9 '18 at 18:27
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    $\begingroup$ I suggest you take a look at the Stern-Gerlach experiment. en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment That page has a simple diagram showing the outcome of spin measurements. $\endgroup$ – R. Rankin Oct 10 '18 at 0:37
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I have the understanding that the outcome of measuring the same observable twice in a row on the same state is getting the square of the eigenvalue

If you measure the observable $O$ and get a value $o_1$, then you immediately measure $O$ again, you'll get the value $o_1$ and not $o^2_1$. This is because the first measurement leaves the system in the state $|o_1\rangle$.

Note: measurement of an observable $O$ is not equivalent to acting on the state with $O$

This is most easily seen by acting on a superposition of eigenstates of $O$

$$O(c_1|o_1\rangle + c_2|o_2\rangle) = c_1o_1|o_1\rangle + c_2o_2|o_2\rangle$$

which is not an eigenstate of $O$. But, according to the measurement postulate, immediately after measuring $O$, the state will be an eigenstate of $O$.

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  • $\begingroup$ What do you mean by 'acting on the state with O'? $\endgroup$ – JD_PM Oct 9 '18 at 18:28
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    $\begingroup$ @JD_PM, 'acting on the state with O' means the operator $O$ takes as input the state vector and produces as output another state vector, e.g., $|\psi_b\rangle = O|\psi_a\rangle$ $\endgroup$ – Alfred Centauri Oct 9 '18 at 18:36

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