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If we consider a two level system, generally we see exponential decay from the excited state to the ground state. Why is this? Some assumption about the noise must yield this result, but what assumption?

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    About what noise? Please describe the system you are talking about in more detail. – ACuriousMind Jun 22 '16 at 14:40
  • I mean a two level system with no interactions with it's environment won't decay, but in normal physical two level systems at finite temperature we see exponential decay to the ground state. When I say noise I just mean interactions with the environment. –  Jun 22 '16 at 14:58
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    It's exponential for a two-level system with white noise, if the noise is not white, it's not exponential but shows correlations. – CuriousOne Jun 22 '16 at 19:23
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    Possibly relevant question here. It's about when the exponential decay law breaks down, but also includes your question. – knzhou Jun 22 '16 at 21:13

1 Answers1

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Indeed, the quantum mechanical evolution of a strict 2-level system won't be exponential: it will be sinusoidal i.e. periodic. The probability of being in one of the states is $a+b\cos\omega t$ for some constants $a,b$. The coefficient $b$ is nonzero whenever the two states deviate from the energy eigenstates.

The exponential decay only appears when the system is composed of the 2-level system and the environment degrees of freedom where the energy may be deposited. The interactions between the 2-level system and the environment have to be nonzero. Another condition is that there has to be a continuum of the states of the environment.

With this setup, the rule that implies the exponential decay is known as Fermi's golden rule

https://en.wikipedia.org/wiki/Fermi%27s_golden_rule

and the decay rate may be derived from the laws of quantum mechanics, perturbative methods, the density of states in the continuum, and the strength of the interactions.

The first surprising fact is that the exponential $\exp(-\Gamma t)$ may be linearized as $1-\Gamma t$ for a small $t$. So a linear drop exists right away. This seemingly differs from the omnipresent quantum mechanical behavior in which the amplitudes evolve linearly and the probability of dropping to the lower state should therefore scale like $t^2$ rather than $t$ because of the Born's rule. But the extra factor of $1/t$ basically arises from the total number of accessible states of the continuum, $\Delta E \sim \hbar/t$, by the "energy-time" uncertainty principle. At very short time, many final states are available because the error margin of the energy may be higher.

When some inequalities are satisfied, the rate $\Gamma$ may be calculated from the squared matrix element of the interaction Hamiltonian. The behavior of the probability of "no decay" is linear for small but not too small $t$. The exponential behavior may be obtained by composing several transformations like that.

Luboš Motl
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