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I have got a quantum conservative system whose Hamiltonian is $H$. I consider an selfandjoint operator $O$ whose eigenvalues and eigenvectors are: $$O|\psi _{n}\rangle = \lambda _{n}|\psi _{n}\rangle$$

At initial time, the system is at the state $|\psi_{i}\rangle$. At time $t= \delta t$, I measure with the operator $O$. The approximation of the measured state to second order in $\delta t$ is
$$|\psi(\delta t)\rangle=U(\delta t, 0)|\psi(0)\rangle \;,$$ with $$U(\delta t, 0) = e^{-\frac{i}{\hbar}\delta t H} \approx 1 - \frac{i}{\hbar}\delta t H - \frac{1}{2}\left ( \frac{1}{\hbar^2}(\delta t)^2 H^2 \right ) \;;$$ the probability of obtaining $ \lambda _{i}$ in the second order on $\delta t$ is

$$P(\lambda_{i}) \approx 1-\left ( \frac{\delta t}{\hbar}\Delta H_{\psi_{i}} \right )^2 \;,$$ where I denote $$\Bigl(\Delta H_{\psi_{i}}\Bigr)^2= \langle\psi_{i}|H^2|\psi_{i}\rangle-\langle\psi_{i}|H|\psi_{i}\rangle^2$$

Question. In what regime the quadratic approximation on delta $\delta t$ is good?

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Looks like a quantum Zeno effect scenario, so t must be short compared to the Zeno time. –  twistor59 Jan 3 '13 at 19:24
What is the Zeno time? –  user15940 Jan 3 '13 at 23:05
It's defined in Dima's answer to the question I linked - basically it's your $\Delta H_{\psi_i}$. You would probably need $\hbar$ to get the units right to compare it with $\delta t$ –  twistor59 Jan 4 '13 at 7:59

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