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I'm struggling with showing both results in the question and help will be very much appreciated.

Thanks.

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  • $\begingroup$ What are your assumptions? You need to give us the setting of your course. Some approaches accept that an observable will not change while being measured. This is one of John von Neumann's axioms. $\endgroup$ – superAnnoyingUser Nov 24 '13 at 22:50
  • $\begingroup$ @Ivan: Clearly it's just the Copenhagen interpretation, where each measurement causes the wavefunction to collapse to a particular, randomly chosen eigenstate. $\endgroup$ – mjqxxxx Nov 24 '13 at 22:56
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It looks like there are two eigenstates $\chi_{\pm}$ of $Q$, and a measurement yielding $Q=\pm1$ leaves the system in the eigenstate $\chi_+$ for $Q=+1$. The time evolution operator being $e^{-iHt/\hbar}$, after time $T/n$ where $n$ is large this state becomes $e^{-iHT/(n \hbar)} \chi_+ = \left(I - \frac{i H T}{n\hbar}\right) \chi_+ + {\cal O}(1/n^2)$. So apparently (and this will be the result of whatever "the previous example" was), we must have $$ \langle \chi_+, H \chi_+ \rangle= \dfrac{E_1 + E_2}{2}$$

If $A_n$ is as given, show that $|A_n|^2 = 1 + {\cal O}(1/n^2)$. What does this say about $\log(|A_n|^2)$ and then about $\log(|A_n|^{2n}) = n \log(|A_n|^2)$ and $|A_n|^{2n}$?

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  • $\begingroup$ i think I can go from here to get the answer, thanks $\endgroup$ – Raul Nov 24 '13 at 23:31

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