# What determines the form of the intensity curves in Laser-Induced Fluorescence (LIF) measurements?

What determines the form of the intensity spectra of different particle species in Laser-Induced Fluorescence (LIF) measurements? See e.g.

I figure that bigger particles have more ways to get excited and so the intensities accumulate and make the curve wider? But how exactly do I derive an expected curve for a given molecule type? Why the steep rise and the slower fall?

• This is a great question and I'm pretty sure for most people the practical answer is one simply looks up experimentally measured curves. So I'd really like to see an answer: I envisage something like a Fermi golden rule calc or semiclassical antenna from orbital calc like one sees in Feynman's QED (the lecture series, not the popular book) but I have never seen anyone draw Feynman diagrams (real ones, with fundamental rather than pseudo particles) for these kinds of calcs. I'm sure that someone would have had a go at numerically calculating the fluorophores in your graph ... – WetSavannaAnimal Oct 16 '13 at 12:37
• ...because they are fairly simple. As for the breadth, I'm altogether stumped and the reason must be something like you say as fluorescence lifetimes are of the order of nanoseconds, so the Lorentzian linewidth for a Wigner-Weisskopf spontaneous radiator is of the order of $10^{-3}\mathrm{nm}$. – WetSavannaAnimal Oct 16 '13 at 12:40

The intensity of a transition is determined by its probability amplitude $P$ $$P = \langle \psi ' | \hat{\mu} | \psi \rangle$$ where $\hat{\mu}$ is the molecular dipole moment operator and $\psi$ and $\psi'$ are the wavefunctions of the initial and final states, respectively. The Franck-Condon principle tells you that electronic transitions to and from the lowest vibrational states (0-0 transitions) are most probable. These transitions are responsible for the steep rise in the signals that you mention. The "slower fall" corresponds to multiple vibronic transitions from your excited electronic state in the lowest vibrational state to excited vibrational states in a lower electronic energy level. These transitions are noticeable in fluorescence spectroscopy because, after the molecule is excited by the laser, there is enough time for some of the the energy to be dissipated as heat (i.e. rotations, translations and vibrations) before it emits a photon.
All of these transitions are of course strictly quantized in energy and, in principle, you should be able to observe individual, discretized, lines instead of a single broad signal. However, because of the uncertainty principle, spectral lines always show line broadening. This uncertainty (in energy units) is given approximately by $$\Delta E = \hbar \tau^{-1}$$ where $\tau$ is the lifetime of the chemical species. This lifetime can be increased by reducing the temperature and, if you take the spectra that you show above at very low temperatures, you should be able to resolve the lines of the different vibronic transitions that conform your broad signals.
• What is the interpretation of the object $\hat \mu|\psi \rangle$? Why don't you just write $\langle\psi'|\psi\rangle$ as the transition amplitude? – Nikolaj-K Oct 23 '13 at 7:01
• The $\tau$ in your equation is awfully long - 4.5ns for fluorescein - so the individual transitions are going to be very, very narrow (less than a GHz) or less than 0.01nm wavelength. So is it true to say that the main broadening is because the transition is between two "uncertain" levels - because there are many possible vibronic levels that the transition could jump between? Actually if you want a fuller picture of the broadening of individual lines, see my answer here in the section "The shape of the spectrum without the cavity. – WetSavannaAnimal Oct 23 '13 at 8:54
• @NickKidman The term $\langle \psi ' | \hat{\mu} | \psi \rangle$ is the transition dipole moment; it represents the overlap between the states $| \psi \rangle$ and $| \psi ' \rangle$ when $| \psi \rangle$ is polarized by the incident light. In contrast, $\langle \psi '| \psi \rangle$ is just the overlap between two eigenstates of the Hamiltonian, which are orthogonal an thus $\langle \psi '| \psi \rangle = 0$, reflecting the fact that transitions don't occur spontaneously. – Goku Oct 24 '13 at 3:41
• @WetSavannaAnimalakaRodVance Thanks for the link. Indeed, there are many possible vibronic states that the transition could jump between. Each of these transitions have its own broadening (depending on the lifetime of the vibronic, not just electronic, state) and are close to the 0-0 transitions, leading to an overlap of signals and a single broad peak. The spectra of Li$_2$O above in particular has a very characteristic vibronic-like structure. I should also note that $\hbar / \tau$ approximates the best theoretical resolution possible, but in practice other factors may affect resolution. – Goku Oct 24 '13 at 4:14
• Why is $\hat\mu|\psi\rangle$ the state affected by light? Does $\hat\mu$ couple to the photon in the Hamiltonian? Where is the Hamiltonian, which one is it? - Also, in this LIF measurements, you go in with specific wavelengths for each molecule right? Are there more wavelengths per molecule, do you have to choose these extremely sharp? And if yes to both, are there many values of wavelengths where you get nothing from the molecules back to the detector? – Nikolaj-K Oct 24 '13 at 6:39