As we all know, there is no even-order peaks in HHG of noble gas. I understand this phenomenon by the knowledge here: there will be odd and even peaks in the frequency domain when we do fourier transform on a series of equally seperated pulses in time domain. Then the even order peaks dispear because that the even-order nonlinear susceptbility of noble gas is zero. However, the present explanation is really ugly. Is there more direct and resonable understanding of this phenomenon?
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Selection rules in the spectrum correspond directly with symmetries in the time domain.
For the case of even harmonics in HHG (or indeed any harmonic-generation configuration), the driving laser has a symmetry of the form $$ E(t+\pi/\omega) = - E(t)$$ (i.e. advancing time by half a period results in a spatially-inverted electric field) and for a noble gas the generating medium is symmetric under that inversion, which entails that the dipole moment of the harmonic emission shares that symmetry, $$ D(t+\pi/\omega) = - D(t).$$ The lack of even harmonic orders is simply the Fourier transform of this property.
Emilio Pisanty
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