In this video, we are going to focus on the propagation of a short pulse. This means that we are going to assume that the light beam has a transverse dimension large enough for assuming plane-wave like behavior. We thus assume a quasi-plane wave with negligible dependence on x or y. And so the only relevant variables are the position z and of course the time t or its Fourier conjugate the frequency omega. And what we are going to see is that, because of the response from the material medium, which depends on the frequency, the different spectral modes will not propagate at the same speed. This is the so-called chromatic dispersion. In order to describe the propagation of our short or ultrashort pulse, we will of course use the solution that we have found from the propagation equation omitting transverse variables x and y. We had already seen that in this case, E is a function of z and t which is given by this integral, with E(omega) exp(i( k(omega)z - omega*t)), where k(omega) is given by the refractive index of the material which itself is determined by the linear susceptibility Now, what we are going to do is to rewrite this directly as the Fourier transform of a function E(z, omega) It is the value of the field at z = 0 times a phase factor, exp(ik(omega)z). What we will do with the complex field E(z, omega) is to separate its modulus and its phase phi(z, omega). And now the phase will evolve with z. We will get on one hand an initial phase and on the other hand the effect of the wavevector. Let us remark that in the case of a transparent material, we know the susceptibility to be a real number so that k(omega) is also real. Therefore only the phase evolves and the spectrum, i.e. the square of the modulus of E(z, omega) will be invariant. Thus all physical quantities which depend on frenquency will be conserved such as the center frequency and the spectral width. The only evolving quantity is the phase we have just introduced, phi(z, omega), which is equal to the initial phase + k(omega)z. It is this phase that will evolve with z and will modify the pulse shape. Before moving on, I want to remind what we have seen last week about the role of the spectral phase on the Fourier transform of a function. Remember that the complex field E(t) can be written as A times epsilon(t) where A is chosen so that epsilon(t) is normalized which means that the integral of the square of the modulus of epsilon(t) is equal to one. Similarly, the integral of the squared modulus of epsilon(omega)will also be one. This is nothing but the Parseval-Plancherel theorem. We have introduced quite a few quantities. The group delay, tau g(omega), which is equal to the frequency derivative of the phase. Obviously, the phase will depend on z, so that the group delay will also depend on z. We have also introduced the center of the pulse t0, or the mean time value defined as the integral of t weighted by the squared modulus of epsilon(t). What you are going to prove in the exercise session is that, this expression can be written as the integral of d phi/d omega, weighted by |epsilon(omega)|Â². So let us leave this part for the exercices session. We also introduced the pulse duration. We have chosen to use the RMS (Root Mean Squared) i.e. the square root of the variance. And, what you are also going to prove in the exercice session is, by using this definition, that the duration of the pulse can be written as the square root of the sum of the square of two terms. So the variance will be the sum of two variances: the variance without the spectral phase, and a term resulting from the fact that different spectral components don't have the same group delay, or, in other words, that different spectral components do not arrive at the same time. Finally, this result is quite intuitive. We know there are two contributions to the pulse duration The first one you've seen last week by proving that Delta t times Delta omega is greater than one half. This means that getting a short pulse requires a broad spectrum. So that the spectrum will limit the pulse duration. va limiter la durÃ©e de l'impulsion. It is this effect that corresponds to the first contribution to the variance. And of course, there is a second contribution. When different components do not arrive at the same time, the pulse gets longer. We've seen some examples last week with chirped pulses. To sum up the results from last week and this week about the spectral phase, plus what you'll see in the exercice session, remember that for a given spectrum, which means for a fixed modulus of epsilon(omega), the shortest pulse will be obtained when this phase-dependent quantity vanishes. This happens when the phase is zero or when the phase is a linear function of frequency so that the variance of the group delay is zero. By definition such a pulse will be called "Fourier-transform limited" or simply "transform-limited" By that we mean that the duration is only determined by the first term which is a fonction of the pulse spectrum. Conversely, if the group delay is not constant avec la frÃ©quence, on dira que l'impulsion prÃ©sente une dÃ©rive de frÃ©quence. with respect to freqency, the pulse is said to be chirped. As in a bird song the frequency then varies with frequency. If you remember the signals that I produced last week using a slide whistle, then indeed when the group delay depends on frequency, we can hear something similar to a bird chirp, i.e. a signal whose instantaneous frequency varies with time. This is a so-called chirped pulse. Now, let's see how the material's response like the one shown here, similar to what we discussed during first week, affects the spectral phase and thus the pulse shape. So what we are going to do is to take a light beam and let it propagate in the transparent spectral region of the medium. corresponding to the region shaded in cyan here. So we assume that the spectrum is entirely encompassed in this transparent area. Here is the central frequency omega zero. Here is the spectrum |E(omega)|Â². z does not appear since, as you might remember, the spectrum is z-independent. We also assume that the spectral width is small enough to make possible a Taylor expansion of the material response. And indeed, if you consider here the variation of the refractive index, with respect with frequency, it can be said to be fairly linear. Thus in all region where the spectrum takes non-negligible values we will assume a linear variation of the refractive index with frequency So what does this imply ? If the refractive index is a linear function of frequency, and since the wavevector is the refractive index times omega over c, the wave number will be a quadratic function. So we can write the wavevector as a second-order polynomial of omega. More generally, if we don't use this approximation, we could still write the Taylor expansion shown here, where higher-order terms could be taken into account as well. However, this week, we will stop at second-order terms. So let us ignore higher-order terms. So, as suggested by the linear variation of the refractive index, we assume that the wavevector is a quadratic function of frequency expanded around the central frequency omega zero. Here we have the zero-order term, here the linear term proportional to omega minus omega zero, and the second-order term, proportional to omega minus omega zero squared. And each time, you have the value at omega zero, of the wavevector, of its first derivative at omega zero, k prime zero, and its second derivative at omega zero. What is relevant is the phase phi(z, omega). As already discussed, this phase is the sum of two terms the initial phase phi(0, omega), plus k(omega) times z. So, by replacing k(omega) by its expression we can eventually make a Taylor expansion of the spectral phase. OK, we still have the initial phase of the pulse. So we will have a constant term, k zero times z, plus a linear term with respect to frequency, (omega - omega zero) times k prime zero z, plus a quadratic term, one half times omega minus omega zero squared times k zero double prime z. And we stop here because we assumed a Taylor series up to second order only. So, the linear term here can be written as k zero minus omega zero k prime zero times z plus a term truely linear with omega, omega k prime zero z. By doing this we are going to see shortly why it is interesting to write the phase in this form.