Graduate Two-Photon Excitation: Theory and Applications

[kmcguir]

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

 

[Spinnor]

excitation peak at 495

An excitation peak means that photon energies just above and just below the peak will still cause excitation but they are not as likely, right?

 

[blue_leaf77]

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?

In reality, due to finite lifetime of excited states an excitation peak is never a perfectly single line in shape, it always has certain non-zero width. Any wavelength which falls within this width can still have non-zero probability to excite the system.

 

[HAYAO]

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?

This is not true. The first two people have already mentioned that the absorption of a fluorophore is not a delta function, and it has some finite width in terms of wavelength in which the fluorophore can be excited. Let me elaborate on this.

In spectroscopy, we usually deal with two different widths (i.e. spectroscopic width of the excitation source and the sample) that cause the spectra to be broader than just a single spike. The wavelength width of the excitation source can be improved depending on the machine configuration. The wavelength width of the fluorophore is usually quite broad. The latter is the main reason why the excitation spectra looks broad.

In a typical spectroscopy, we usually use some very wide range emitting light source (e.g. xenon lamp, Halogen lamp) and then use the monochrometer to take out very little portion of the wavelength of that light source. The ability to minimize the width depends entirely on the machine. Those with physically large monochromater, double monochrometer, and/or extremely small bandpass will enhance the chromaticity of the output light. However, no matter how good the machine is, it always have some width in wavelength. In this case, we are talking about 2-photon microscopy in particular, which is why we are using lasers. Now the configuration of the machine so that the excitation spectra is measurable entirely depends on the type of laser itself: dye lasers, excimer laser, semi-conductor laser, solid-state lasers, gas laser, etc. These lasers usually have high chromaticity, but the same discussion still applies. They still have some width in wavelength, but in the order of few nanometers to sub-nanometers.

As for the fluorophore, due to the uncertainty principle, it will have a finite width spectra. Additionally, most of these fluorophore will have multiple excited states, both electronic (zero phonon line) and vibrational (phonon-sideband). Have you ever looked at the UV-Vis absorption spectrum of benzene (google it)? It is very broad in the order of hundred nanometer. At any wavelength of the spectrum of benzene, as long as there is an absorption there, excitation will lead to benzene to emit.