A Spontaneous emission and coherence

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When preparing a linear superposition of states between a ground and excited level, the decay dynamics of the excited state play a crucial role. If the excited state has a long enough lifetime and decays to multiple levels, the outcome after a significant time will be that half of the atoms remain in the ground state while the others decay to various other levels. The negligible probability of returning to the ground state ensures that the population is effectively redistributed among the other levels. This scenario confirms that not all atoms will decay back to the ground state. The discussion highlights the importance of understanding spontaneous emission and coherence in quantum systems.
kelly0303
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Assume I prepare a linear superposition ##\frac{1}{\sqrt{2}}(|g>+|e>)## between a ground and excited level for a large number of "atoms" (it can by any multilevel system, not necessarily an atom). We can assume that the lifetime of the excited level is long enough to allow us to create this superposition, but it is not infinite. Assume also that the excited state can decay to many other levels, beside ##|g>##, such that the probability of ##|e>## decaying back to ##|g>## is negligible for the purpose of this question. If I wait for a time much longer than the lifetime of the excited state (the ground state is stable), will I find half of my initially prepared "atoms" in ##|g>## and the other spread among the other levels, or will all the "atoms" decay to the other levels? Thank you!
 
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kelly0303 said:
will I find half of my initially prepared "atoms" in |g>
Yes. What made you think that it might not be the case?
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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