How can we classically describe the formation of absorption spectra?

When dealing with the problem that femtosecond laser pulses interact with atoms, the semi-classical method is usually used: we describe the atom as quantized and describe the pulse classically.

I find myself cannot classically describe the formation of absorption spectra! I'm reading a book about this. In the page 30 of this book, the author describes how the absorption spectra we detected can form: the incident pulse and the radiation emitted by the atom interfere in the detector, so we can detect the heterodyne signal, i.e. the absorption spectra. The formula is : $$\text{detected signal}=\int_0^{\infty}|E_\mathrm{incident}(t)+E_\mathrm{emitted}(t)|^2 \mathrm dt$$ I have two questions about this description:

1. Why does the electric field emitted by the atom transmit in the same direction as the incident laser pulse? (This seems to be implied by page 30 of these notes.)

2. Does the interference between the $E_\mathrm{incident}(t)$ and $E_\mathrm{emitted}(t)$ totally caused by the detector since they do not overlap in the space (the life time of the atom is tens of nanosecond. If the duration of the laser pulse is 10 femtosecond, then they hardly overlap in space in my opinion)?

I've also read that "the absorption spectra forms because the emitted radiation transmit through any direction so if we detect the incident light far away enough we get the absorption spectra". I understand this description, but which one is right?

• Who says that the emitted field is always collinear with the incident field, and who says that the two have no overlap in space? – Emilio Pisanty Oct 28 '17 at 6:37
• No one says the emitted field is always collinear with the incident field, but the picture in the page 30 of the book link, it imply this or I've misunderstood it. – Bettertomo Oct 28 '17 at 7:03
• Atoms do not emit fields, they emit photons. see hyperphysics.phy-astr.gsu.edu/hbase/mod5.html#c3 . classical em fields are built up by supersposition of innumerable photons – anna v Oct 29 '17 at 20:42
• @ anna v, Do you mean there is no actual field of the light, the field is a macroscopic approximation of the photon ensamble? – Bettertomo Oct 30 '17 at 2:55

The two components of your question are rather independent.

Propagation direction

The emitted field might be better called a scattered field, and it is important to note that (unless you are considering your driver beyond the dipole approximation) then the field emitted by each individual atom is pretty much isotropic, with no angular dependence other than the dipole angular distribution about the polarization direction.

However, your sample does not consist of an isolated atom; instead, it includes a macroscopically-sized gas jet and focus with a bunch of individual emitters distributed over scales larger than the wavelength, and all emitting at different phases. This means that as far as the emitted/scattered field is concerned, the sample acts as a phased array, and this is of course perfectly calibrated to collimate much of the scattered radiation onto the original direction of propagation of the incident laser pulse.

Spatial overlap

The spatial overlap is a much subtler question. I've done a more in-depth analysis of this in my answer to Why is interference there when pulses do not overlap in space and time? but the short of it is that if you actually want to observe the interference then you need to spatially stretch the pulses in a way that makes the incident and scattered pulses have a significant spatial overlap.

And, most importantly, the phrase "if you want to observe the interference" includes such basic things as observing the absorption spectrum and the presence of the "bite" taken out by the atomic resonance from the broader bandwidth of the femtosecond pulse:

In this situation, you will have some short pulse with femtosecond duration $\tau$ and a corresponding bandwidth $1/\tau$ (making the reasonable assumption of a Fourier-limited duration), and this is interacting with a transition with a much tighter bandwidth $\Gamma$ that might be in the 1/(tens of nanoseconds) regime. Borrowing a picture from the other answer,

if you want to observe the spectrum including the 'bite', then you need to have a spectrometer with a resolution comparable with $\Gamma$, which means that the spatial stretch of the pulses at the detector will be of the order of ($c$ times) tens of nanoseconds, so they will be perfectly posed to interfere at the pixels where they coincide.

Now, that sounds pedestrian and pragmatic but it is important to note that it is an essential part of the interaction, and it is constrained by the basic bandwidth theorem of Fourier transforms. One useful way of seeing this is thinking of your initial femtosecond pulse as a bunch of nanosecond pulses centered at frequencies throughout the femtosecond bandwidth: one of these is centered at the atomic transition frequency, and it is the part that actually gets absorbed, and the rest of them are at frequencies that don't interact with the atom, and they're just there to shape the time-domain electric field so that it vanishes outside of the femtosecond window of the pulse.

(That sounds weird, no? However, it's a valid way to decompose the pulse as a sum of other components, and because your interaction is linear, then it is an equally valid way to look at your pulse and how it interacts with your medium.)

• @ Emilio Pisanty, thank you very much! I've understood the "spatial overlap" part of your answer. The "propagation direction" part, the non-elastic of the photon, I need a little more time to digest. I've found your newest paper about the lorentz force experienced by the re-combined electron in HHG, it's really interesting ! – Bettertomo Oct 29 '17 at 15:16
• @Bettertomo That's not my newest paper ;-), but I'm glad you found it interesting! – Emilio Pisanty Oct 29 '17 at 15:17
• @ Emilio Pisanty, wow, I wish you be more successfull in your work! – Bettertomo Oct 29 '17 at 15:20
• Regarding the propagation-direction bit, it's a crucial part of understanding the propagation direction of pulses in both HHG and ATAS, so make sure you understand it correctly (or ask for clarification if you don't). It comes down to the fact that the interaction region is spatially extended and that each emitter's phase is determined by the driver's, so at a given time $t=0$ it will be different at different parts of the focus; the macroscopic field is then determined by their coherent superposition including the phases. I couldn't find a clearer way to explain it in this answer, though. – Emilio Pisanty Oct 29 '17 at 15:21
• @ Emilio Pisanty, I will think about it seriously! – Bettertomo Oct 29 '17 at 15:25