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I'm reading a quantum mechanics book which states that:

"For ω close to Ω, where ω is the frequency of the external field and Ω is the transition frequency, all states with other frequencies are negligible because they have negligible probabilities. Hence we can treat the system as a two-state system."

I do not understand why for ω far away from Ω, the probabilities are negligible. This is clear to me if we use time-dependent perturbation theory, but in this case we are assuming that the external field is very small - which seems to contradict the use of Rabi oscillations (for strong external field).

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  • $\begingroup$ Your statement: "I do not understand why for ω far away from Ω, the probabilities are negligible." is in contradiction with the quote from Sakurai: "For ω close to Ω, where ω is the frequency of the external field and Ω is the transition frequency, all states with other frequencies are negligible because they have negligible probabilities." When $\Omega$ is close to $\omega$ (resonance) then all other transitions associated with the other states are negligible. $\endgroup$
    – Sha
    Commented Jan 1, 2023 at 15:02

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This is clear to me if we use time-dependent perturbation theory, but in this case we are assuming that the external field is very small - which seems to contradict the use of Rabi oscillations (for strong external field).

Nothing can be strong and small by itself, but only in comparison to something else. In this case the matrix elements are small in respect to the detuning from the other levels, $$ \left|\frac{V_{0n}}{\omega - \omega_n}\right|\ll 1 $$ but the transition rate (or Rabi frequency) is not small compared to the time of observation $$ \Gamma t, \Omega_R t \sim 1 $$

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  • $\begingroup$ Okay I understand this, but still how can we "prove" that for frequencies far from Ω, the transition probabilities are negligible ? $\endgroup$
    – MTYS
    Commented Dec 31, 2022 at 12:45
  • $\begingroup$ @RosTT by checking the inequality - if the detuning (the difference between the transition and the driving frequencies) is much smaller than the matrix element for this transition (to required precision), we neglect it. (Perhaps, you are confused by my notation in the answer.) $\endgroup$
    – Roger V.
    Commented Dec 31, 2022 at 13:12

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