Why are there only odd harmonics at High Harmonic Generation?

Question

I'm trying to understand why in High Harmonic Generation (HHG) the spectrum only consist out of odd multiples of the driving frequency. Many sources state "because of symmetry" but they do not clearly explain why this symmetry is needed or symmetry in what? Wikipedia seems to claim it has to do with a symmetry of a monatomic gas where the HHG is generated:

In monatomic gases it is only possible to produce odd numbered harmonics for reasons of symmetry - High Harmonic Generation Wikipedia

Another source states that the spectrum of a HHG pulse train can be described accordingly:

$g(\omega) = \Big[2\omega_d\sum_{n=0}^{\infty}f(n2\omega_d)\delta(\omega-n2\omega_d)\Big]*f_{env}(\omega)$

With $\omega_d$ being the driving frequency of the pump laser, $f(\omega)$ a single HHG pulse in frequency space, $f_{env}(\omega)$ the envelope of the driving pulse in frequency space. In the document it claims that because of this equation it is "clearly visible" we only have odd harmonics. But I would say: we only have even harmonics due to the $2n$ part.

Also, if only odd harmonics are possible then what is Second Harmonic Generation? Is this a different mechanic compared to HHG?

EDIT: The source I mentioned before, does state that there should be symmetry in the polarisation of a monatomic atom.

Meaning that the electron cloud displacement is independent of the direction of the electric field (up to a sign change).

This means that the even-orders of $\chi$ should be zero and thus no even harmonics. This part I can understand why atom symmetry leads to the odd harmonics and added it as an answer.

Now the only thing which remains is the equation mentioned above which "clearly" shows odd harmonics.

Answer 1

Actually, the source mentioned in my question does answer partially the question. Consider a monatomic gas with an electron cloud. Its response to an electric field is typically given in terms of an expansion.

$P(t) = \varepsilon_0\Big[\chi^{(1)}E(t)+\chi^{(2)}E^2(t)+\chi^{(3)}E^3(t)+..\Big]$

When there is spherical symmetry, for a monatomic case, we can expect the reaction of the media (polarisation) for opposite electric fields to be symmetric, i.e. $P(E) = -P(-E)$. This can only be true for odd orders and thus the even $\chi$'s need to be zero. When you plug in a plane wave into the equation, with zero for the even $\chi$, it does create only odd harmonics.

Answer 2

One way to explain the existence of only odd harmonics for HHG in a gas is to think about what is happening in the time domain, then Fourier transform back to the frequency domain. Temporally, the high-harmonics are a pulse train of "attosecond bursts" with durations of, say, 200 attoseconds. The Fourier transform of a pulse train in a frequency comb. So, we know that, in the frequency domain, we are going to have a comb of harmonic peaks.

In the Three-step Model of HHG, the strong electric field removes and electron from the atom, accelerates it, then slams it back into the ion, generating a high-harmonic photon. This process occurs twice every optical cycle of the driving field. In the frequency domain, the fact that the attosecond bursts are occuring with half the period of the driving field means that the harmonics are spaced by twice the original photon energy. So, this means that we are going to have either only even or only odd harmonics, but not both!

Now, we just need to decide if we are going to have even or odd harmonics. To figure this out, we need to look the "carrier-envelope offset" in the time domain, which determines how much the carrier is offset from the envelope. The key is that, in the three-step model, the electrons are returning from different directions for each subsequent attosecond burst. So, the carrier-envelope phase flips from each attosecond burst to the next. This means that the carrier-envelope offset is equal to the frequency of the driving field. In the frequency domain, this shifts our peaks away from zero frequency by one photon energy of the driving field, giving us only odd harmonics.

Answer 3

Higher harmonic generation always takes place within a nonlinear transparent medium and the summetry constraints are always due to the symmetry properties of the medium. There are vastly different symmetries between homogeneous gases and nonlinear crystal structures. This is what your sources are referring to. To go into the symmetry constraints of a particular medium you have to go into the mathematical weeds of nonlinear optics.