The following back-of-the-envelope calculations do not lead to any useful result, but you might be interesting in reading through them if you want to get a better understanding of what happens during two-photon excitation microscopy.
The basic idea of two-photon microscopy is to direct so many photons onto a single confined location in the sample that two photons interact with a fluorophore roughly at the same time, leading to fluorescence. The confinement in time seems to be given by the duration of the laser pulse (ca. 50-500 fs). The confinement in space is in the best case given by the resolution limit (let’s say ca. 0.3 μm in xy and 1 μm in z).
However, since the laser beam is moving around, I wondered whether this may influence the excitation efficiency (spoiler: not really). I thought that his would be the case if the scanning speed in the sample is so high that the fs-pulse is stretched out so much that it spreads over a distance that is greater than the lateral beam size (0.3 μm FWHM).
For normal 8 kHz resonant scanning, the maximum speed (at the center of the FOV) times the temporal pulse width is, assuming a large FOV (1 mm) and a laser pulse that is strongly dispersed through optics and tissue (FWHM = 500 fs):
Δx1 = vmax × Δt = 1 mm × π × 8 kHz × 500 fs = 0.01 nm
This is clearly below the critical limits. Is there anything faster? AOD scanning can run at 100 kHz (reference), although it can not really scan a 1 mm FOV. TAG lenses are used as scanning devices for two-photon point scanning (reference) and for two-photon light sheet microscopes (reference). They run at up to 1000 kHz sinusoidal. This is performed in the low-resolution direction (z) and usually covers only few hundred microns, but even if it were to cover 1 mm, the spatial spread of the laser pulse would be
Δx1 = 1 mm × π × 1000 kHz × 500 fs = 1 nm
This is already in the range of the size of a typical genetically expressed fluorophor (ca. 2 nm or a bit more for GFP), but clearly less than the resolution limit.
However, even if the infrared pulse was smeared over a couple of micrometers, excitation efficiency would still not be decreased in reality. Why is this so? It can be explained by the requirement that the two photons arriving at the fluorophor have to be absorbed almost ‘simultaneously’. I was unable to find a lot of data on ‘how simultaneous’ this must be, but this interaction window in time seems to be something like Δt < 1 fs (reference). What does this mean? It reduces the true Δx to a fraction of the above results:
Δx2 = 1 mm × π × 1000 kHz × 1 fs = 0.003 nm
Therefore, smearing the physical laser pulses (Δx1) does not really matter. What matters, is the smearing of the temporal interaction window Δt over a spatial distance larger than the resolution limit (Δx2). This, however, would require a line scanning frequency in the GHz range – which will never, ever happen. The scan rate must always be significantly higher than the repetition rate of pulsed excitation. The repetition rate, however, is limited to <500 MHz due to fluorescence lifetimes of >1-3 ns. Case closed.