Is the absorption of photons instantaneous? I'd like to know if the excitation of chlorophyll by photons $$Chl+ h\nu \rightarrow Chl^*$$
is instantaneous. I imagine a photon arriving $0.5$ Angstroms away from the molecule, and then disappearing as its energy is absorbed and converted to an electronic excitation. If you could slow things down (very small) frame by frame, how would you describe the process?  
I'll leave the question open-ended like this for now.
 A: 
If you could slow things down (very small) frame by frame, how would you describe the process?

The process is both instantaneous and not instantaneous. 
Heuristically, one can construct quantum states by taking the set of possible naive classical states, and then allowing superpositions of those states. Naively, we can have an integer number of photons. You might think that during an absorption event, the number of photons goes smoothly down from one to zero, passing through fractional values alone the way. However, what actually happens is that we evolve through a superposition of having either zero or one photons,
$$|\psi(t) \rangle = c_0(t) |0\rangle + c_1(t) |1 \rangle$$
where the initial state has $(c_0, c_1) = (0, 1)$, the final state has $(c_0, c_1) = (1, 0)$, and the intermediate states have fractional values of $c_0$ and $c_1$.
During the process, the probability of finding one photon (if you interrupt the process by making a photon number measurement) gradually decreases. However, when you perform the measurement, you will always see an integer number of photons.
A: No the absorption of photons is not instantaneous.
This answer includes an intuitive "cartoon" for how to think of a transition. Then it provides a rough sketch for how one might calculate how long that transition might take involving a discussion of Rabi frequencies. This answer does not go into detail about where the Rabi frequency comes from or how one should think about it but that can be found in many texts on quantum and atomic physics.
In simple electronic transitions such as $s\rightarrow p$ atomic transitions (ignoring transitions that adjust electron or nuclear intrinsic spin states) the picture one should have in their head is that the transition happens because the electric field of the light wave (photon) puts a force on the electron. This force is able to "push" the electron from one state into another state. After all, there must be some force which pushes the electron around because the spatial distribution of the electron cloud is different before and after the transition.
Imagine an electron ground state in an $s$-orbital (spherically symmetric) and an excited state in a $p$-orbital (dumbell shape, say along the $z$-axis, $p_z$.). This system can be thought of sort of like a harmonic oscillator with frequency $\omega$ corresponding to the splitting between this $s$ and $p$ state. Imagine a photon comes along with a linear polarized electric field along the $z$-axis. This will put a force on the atom up and then down as the electric field moves by. However, if the frequency is too low or too high then the spherically symmetric $s$-orbital electron could will wiggle up and down in response to the field but not very much. It is like a harmonic oscillator being driven off resonance. BUT, there is a special frequency $\omega$ where if the electric field is changing at that frequency it can have a great effect on the electron orbital. It will now wiggle very hard and at if the field is on for the right length of time (Rabi oscillation time, ignoring spontaneous decay effects) it can wiggle itself into the shape of a $p$-orbital.
Now, if the electron is driven with a large electric field (many photons) then the energy needed to excite the electron will not really have any effect on the electric field. However, when the electric field is reduced to the single photon level, all of the energy of the field must be converted into the energy of the new electron configuration. This means that during this process of electron excitation the electric field is weakening in strength.
Thus the whole process of the electron being excited and the field diminishing in amplitude takes place over some length of time which is related to the inverse of the Rabi frequency (this is related to the strength of the electric field and the geometric properties of the two electron orbitals in question). 
Reasonable Rabi frequencies can range anywhere from sub-kHz up to GHz meaning the transition time can take from ns up to ms depending on experimental conditions. Here is a rough formula for how one might calculate the Rabi frequency for an atom transition.
$$
\Omega = \frac{d E}{\hbar}
$$
Here $d$ is the transition dipole moment for the $s\rightarrow p$ transition in question. $d$ is approximately equal to $e a_0$ for many atomic transitions. $e$ is the electron charge and $a_0$ is the Bohr radius. $E$ is the electric field amplitude and $\hbar$ is Planck's reduced constant.
Suppose the electric field is generated by a focused Gaussian beam. The intensity at the center of a Gaussian beam is given by
$$
I_0 = \frac{2P}{\pi w^2}
$$
Here $I_0$ is the peak intensity, $P$ is the total power in the beam and $w$ is the mode waist of the Gaussian beam.
Electric field is related to intensity by
$$
I = \frac{1}{2} c \epsilon_0 E^2
$$
$$
E = \sqrt{\frac{2 I}{c\epsilon_0}}
$$
Here $c$ is the speed of light and $\epsilon_0$ is the permittivity of free space.
Putting this all together we get
\begin{align}
\Omega \approx \frac{e a_0}{\hbar}\sqrt{\frac{4}{\pi \epsilon_0 c}} \sqrt{\frac{P}{w^2}}
\end{align}
The only experimentally controllable parameters here are $P$ and $w$. for $P=1W$ and $w=1mm$ we get $\Omega = 1.7 GHz$. $P$ could range from $\mu W$ (or basically zero) up to 10s of $W$ for certain experiments and $w$ can range from a micron up to a few centimeters (or basically infinity) for certain experiments.
The transition time is
$$
\tau \approx \frac{1}{\Omega}
$$
A: 
If you could slow things down (very small) frame by frame, how would you describe the process? 

With a Feynman diagram, and in this mathematical sense, the diagram shows an instantaneous  transfer of energy:


Feynman diagrams demonstrating how electrons (denoted by e–) can accelerate (change direction of motion) by  (a) absorbing or (b) emitting a photon (denoted by the Greek letter gamma: γ).

Replace the electron with a Chl atom.
In the diagrams, one uses a given instantaneous t, but it is a mathematical artifact. In order to get a value to be compared with experiment, one has to integrate over the variables of the integral which a given diagram represents, and that introduces widths in energy lines and therefore , by the Heisenberg Uncertainty Principle (HUP) also in time.
So a single photon hitting an atom will be an instance in the probability distribution that one has to accumulate in order to fit the measurement of photon Chl scattering.


The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time.

The HUP is a useful envelope for the probabilistic nature of the solution of quantum mechanical problems. The solutions of the differential equations, called wave functions because the equations are wave equations, $Ψ$ are complex numbers. The wave function postulate 

Turns the complex value of the wavefunction to a probability value, which can be checked by measurement.
This means that the energy level you are asking about, Chl+ hv --> Chl* has an uncertainty in the values of time and energy within the bounds of HUP. This means that nothing is in reality instantaneous.
