Since the continuous spectrum lies above the discrete spectrum, E f − − E i > 0 and it is clear from the previous section, major role is played by the energies E f lying near the resonance energy E i + ℏ ℏ ω ω , i.e., when ω ω f i ≈ ≈ ω ω . In this case, it is sufficient to keep only the first term for a f ( t ) . Assuming that perturbations are turned on from time t = 0 , we have then a f ( t ) = − − i ℏ ℏ ∫ ∫ 0 t ⟨ ⟨ f | H ′ ( t ′ ) | i ⟩ ⟩ e i ω ω f i t ′ d t ′ = − − ⟨ ⟨ f | F | i ⟩ ⟩ e i ( ω ω f i − − ω ω ) t − − 1 ℏ ℏ ( ω ω f i − − ω ω ) The squared modulus of a f is | a f | 2 = 4 | ⟨ ⟨ f | F | i ⟩ ⟩ | 2 sin 2 ⁡ ⁡ ( ( ω ω f i − − ω ω ) t / 2 ) ℏ ℏ 2 ( ω ω f i − − ω ω ) 2