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kmcguir
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Maria Goeppert-Mayer described the theoretical foundation for 2-photon excitation and it was later proven correct with the advent of lasers. Today, two-photon microscopy uses this quantum physics principle. For example, if you want to excite a fluorophore that has an excitation peak at 495, one could use the simultaneous collision of two 990 nm photons to get the same excitation energy as one 495 nm photon without destroying the tissue. I was reading, however, that two-photon excitation spectra are difficult to measure, and in practice one begins with a two-photon excitation wavelength of twice the one-photon peak absorption, and then tunes the laser around that region to maximize the two-photon excitation signal. I thought that photon absorption was an all or nothing process. Meaning, to excite the 495 nm fluorophore it is required to use a 495 nm photon. 490 nm, 494 nm, 496 nm, 500 nm, etc will not cause excitation, correct? Isn't this true as well for 2-photon absorption? A 400 nm fluorophore would require two 800 nm photons to be excited, not two 801 nm or two 799 nm photons, yes? We are confused because, although fluorophores have a spectra of excitation (example: FITC has an excitation spectra from 430 nm to 530 nm, peak excitation at 490 nm), doesn't it take two photons of specific wavelength to excite at 490, and those same two photons would not excite at say 430 nm? I hope this question was clear. There are no 2-photon microscopy forums, and this is a quantum physics question.
Source:
http://onlinelibrary.wiley.com/doi/10.1111/j.1365-2818.2011.03532.x/pdf
Source:
http://onlinelibrary.wiley.com/doi/10.1111/j.1365-2818.2011.03532.x/pdf