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Suppose the electron of a single electron atom is in the state $ a|{E_{1}}\rangle + b|{E_{2}}\rangle $. If it decays to the ground state $ |E_{0}\rangle $ which spectral lines do I expect to see? The ones that correspond to the transition from $ |E_{1}\rangle $ to $ |E_{0}\rangle $, the ones that correspond to the transition from $ |E_{2}\rangle $ to $ |E_{0}\rangle $ or neither? My guess is that "right before the decay" the electron's wave function collapses to $ |{E_{1}}\rangle $ with a probability $ \lvert a \rvert ^{2} $ or $ |{E_{2}}\rangle $ with a probability of $ \lvert b \rvert ^{2} $ and then it decays to $ |E_{0}\rangle $, therefore I expect to see the spectral line of the transition $ |E_{1}\rangle $ to $ |E_{0}\rangle $ with a probability of $ \lvert a \rvert ^{2} $ and the other one with a probability of $ \lvert b \rvert ^{2} $. Is my reasoning correct?

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You will observe transition $E_1\rightarrow E_0$ with probability $|a|^2$, and transition $E_2\rightarrow E_0$ with probability $|b|^2$. Of course, if you literally made one measurement on a single electron, you would observe only one transition. However, in practice you either observe an ensemble of identical systems or the same system multiple times.

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  • $\begingroup$ Exactly. The collapse (decoherence) happens due to the measurement, not as long before it as before emission. $\endgroup$
    – Ruslan
    Commented Apr 12, 2020 at 21:45

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