Is the intensity given by $I = E_{tot}(\omega)E_{tot}(\omega)^\ast$? Or should the intensity $I$ be each $I = \sum(|E(\omega)|^2)$, squared separately?

The detector will integrate over time, but the pulses arrive with delay.

  • 1
    $\begingroup$ There are a few questions here, and it would help if your broke them up a bit more and elaborated. I can only really provide comments. "The max of I in this case will scale as number of pulses squared" only if you integrate the intensity over time, which isn't usually the type of intensity people talk about in nonlinear optical processes like high-harmonic generation. People usually mean peak intensity, which would be proportional to $E(t)E^{*}(t)$, rather than average intensity. Also, high-harmonic generation is a highly coherent process, so your comment about coherence is correct. $\endgroup$
    – user113857
    Commented Apr 12, 2016 at 18:55

1 Answer 1


You're conflating two different views on the description of the attosecond pulse train; in particular, you're flitting back and forth between the time-domain and frequency-domain descriptions, and it's not doing you many favours.

Let's look at a few things first:

The total field is $$ E(t) = \cdots + E(t-2\pi/\omega)-E(t-\pi/\omega)+E(t)-E(t+\pi/\omega)+E(t+2\pi/\omega)- \cdots, $$

This sort of description is likely to lead you astray, fast, because you have a total electric field $E(t)$ on the left-hand side and a per-event field $E(t)$ on the right, using the same symbol for the two different quantities. This makes it very hard to know exactly what your subsequent queries are referring to.

To make things a bit clearer, write $$ E_\mathrm{tot}(t) = \cdots + E(t-2\pi/\omega)-E(t-\pi/\omega)+E(t)-E(t+\pi/\omega)+E(t+2\pi/\omega)- \cdots, $$ with a per-event field $E(t)$, which is zero outside of a given half-cycle, on the right.

To begin with, there is in general a problem because the electric field will still contain the fundamental, as well as the below-threshold harmonics, and these are at least several orders of magnitude stronger than the attosecond pulse train you want to describe, so you need to do some spectral filtering to take these out. (Of course, this is mirrored in experiment, where you need to use different mirror radii for the IR and the XUV components if you're doing pump-probe between the two, or suitable metal-film filters on your HHG output if you want to do XUV-XUV pump-probe, or so on.) I will assume, however, that this has been taken care of, and that your electric field does not contain any unwanted spectral components.

After that, defining intensity ends up being more to do with what exactly you want to do with your experiment.

  • Very often, you want to use your pulse train to excite some specific transition in a target, so you look at the total energy in that harmonic. This means that you're looking at contributions from all the half-cycles in the train; alternatively, your target is narrow-band to below the driver frequency, which means that its (coherent) temporal response takes longer than one cycle and you therefore need to include all the pulses coherently.

    In this case the relevant measure is the spectral intensity $I(\Omega)=|E_\mathrm{tot}(\Omega)|^2$ at the particular frequency $\Omega=n\omega$ you're interested in. This can then be related to the Fourier transform of the per-event signal $E(t)$, since the Fourier transform of the displaced $E(t+k\pi/\omega)$ differs by a phase from $E(\Omega)$, as $e^{-ik\pi\Omega/\omega}E(\Omega)$, so \begin{align} |E_\mathrm{tot}(\Omega)|^2 &= \left|\sum_{k=K}^J (-1)^k E(\Omega) e^{-ik\pi\Omega/\omega} \right|^2 \\ & = \sum_{k=K}^J (-1)^k\sum_{k'=K}^J (-1)^{k'} E(\Omega)E(\Omega)^* e^{-i(k-k')\pi\Omega/\omega} \\ & = |E(\Omega)|^2\sum_{k=K}^J\sum_{k'=K}^J (-1)^{k-k'}e^{-i(k-k')\pi\Omega/\omega}. \end{align} Here the sum represents the combination of the contributions from the different half-cycles of emission, but the sum factorizes and in general it will return a factor of $N^2$, for $N=J-K+1$, the total number of cycles that contribute: $$|E_\mathrm{tot}(\Omega)|^2 =N^2|E(\Omega)|^2$$ when $\Omega=n\omega$. This is because the energy from the different half-cycles is adding coherently, producing an interference pattern in the frequency domain (i.e. the harmonic comb), by taking energy from the forbidden frequencies (first even harmonics, then everything that isn't a multiple of the driver frequency) and concentrating it on the harmonics that interfere constructively. From the target's point of view, these are the contributions of all the different half-cycles adding up coherently.

  • In most situations, however, what you really care about is the total pulse energy in each pulse of the train, as this will give you the most direct representation of how bright your pulse is. To get this, you need to integrate the spectral intensity over the entire spectrum, giving \begin{align} U=\int |E_\mathrm{tot}(\Omega)|^2 \mathrm d\Omega&= \int |E(\Omega)|^2\sum_{k=K}^J\sum_{k'=K}^J (-1)^{k-k'}e^{-i(k-k')\pi\Omega/\omega}\mathrm d \Omega \end{align} Here the interference no longer gives a quadratic effect, because you're counting regions with both constructive and destructive interference, so the energies from the different pulses add linearly. In general, it is easiest to go for the full energy in the total spectrum and then divide by $N$ to get the energy per pulse - but again, it depends on what you're actually going to do with your pulse.

  • $\begingroup$ Why would you use Etotal in the second case not E(w). The energy per pulse should be related to only E(w)? In the first case N^2 is not very intuitive. The pulses arrive with delay on the detector and does not interfere ? $\endgroup$
    – Anonymous
    Commented May 24, 2016 at 3:40

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