0
$\begingroup$

I have some doubts about the definiton of a Fourier transform limited pulse. For example if I consider a generic pulse: $$E(z,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}A(\omega)e^{i(-\beta(\omega)z+\omega t)}d\omega$$ Defining:

  • $S(\omega,z)=|\mathscr{F}[E(z,t)]_{\omega,z}|^{2}$

  • $\Delta \omega(z)$ as the range of frequencies which correspond at half height of $S(\omega,z)$ for a fixed $z$ (pulse bandwith).

  • $\tau_{p}$ (the duration of the pulse) as the range of time which correspond at half height of $|E(z,t)|^2$ at fixed $z$. I'm considering a pulse inside a dispersive medium so its duration depends on the $z$ you are at.

Done these definitions, is it right to say that a pulse is Fourier limited if $$\Delta \omega(z)\tau(z)=\alpha$$ for each $z$, where $\alpha$ is a number which changes according to the type of pulse (Gaussian, ecc...)?

$\endgroup$
  • $\begingroup$ I could guess what "Fourier transform limited pulse" means, but I've never heard that phrase before, so it would be nice to have a link to an example use case. $\endgroup$ – DanielSank Jan 17 at 17:39
  • $\begingroup$ @DanielSank This paper is one representative example. $\endgroup$ – Emilio Pisanty Jan 17 at 17:41
  • $\begingroup$ @EmilioPisanty Ugh, paywall. Does "transform limited" mean "short enough that the spectral width is larger than I want"? $\endgroup$ – DanielSank Jan 17 at 17:47
  • $\begingroup$ @DanielSank c'mon ;-). See my answer for the details. $\endgroup$ – Emilio Pisanty Jan 17 at 17:48
2
$\begingroup$

I'm considering a pulse inside a dispersive medium so its duration depends on the z you are.

... then the concept of transform-limited pulse does not hold globally for your setup. Transform-limited pulses are a 1D (generally time-domain) phenomenon, so in your configuration the question "is the pulse transform-limited" would be asked and answered locally and independently at each different point. And, in the presence of dispersion, if the pulse is transform-limited at a given point $z_0$, then it will not be transform-limited at any other point in general.

Generically, given a locally-defined electric field $$ E(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}A(\omega)e^{+i\omega t}d\omega, $$ with spectral amplitude $A(\omega) = |A(\omega)| e^{i\phi(\omega)}$, the pulse is said to be transform-limited if its duration is minimal over the set of pulses that have an identical power spectrum $|A(\omega)|^2$. The reason we use this definition is that the power spectrum $|A(\omega)|^2$ is both

  • fixed by the gain profile of the laser gain medium and the details of the cavity, and
  • easily measurable by using a conventional spectrometer,

whereas the time duration is

  • extremely hard to measure,
  • not determined by the laser source, since the introduction of any dispersive optics will affect the pulse duration without affecting the power spectrum, and
  • accessible (given enough money, time, and dedication) to experimental modification via a number of pulse-shaping schemes.

For a given laser source, the power spectrum is basically fixed, and therefore so is the bandwidth $\Delta\omega$, and this puts a limit, via the Fourier bandwidth theorem, on the minimal pulse duration that's achievable with your laser source. However, unless you've done a lot of work, the pulse that comes out of your source will not be that short - instead, it will contain chirp and other types of dispersive features which make it longer than that minimal pulse duration. That problem can be fixed by using pulse shapers to introduce additional spectral phases (i.e. additional terms $e^{i\phi_\mathrm{shaper}(\omega)}$ multiplying the spectral amplitude) which cancel out the chirp and other dispersive behaviours to minimize the pulse duration.

The transform-limited pulse duration is the minimal pulse duration that's achievable using this procedure.

If you want to get truly technical, then this also depends on the choice of measure for the duration of the pulse (i.e. choosing the FWHM, as you've done with your $\tau$, or some other measure which e.g. takes into account some pre-defined sensitivity to pre- or post-pulses), but if you're arguing about that then you're well and truly into the weeds by that point.

The concept of a transform-limited pulse is of extreme relevance in on-the-ground experimental situations, where the spectrum of your pulse is some jagged beast instead of some nice smooth spectrum (say, take fig. 2(b) of this paper). To evaluate the transform-limited duration, you basically take a set of reasonably-smooth spectral phases $\phi(\omega)$ that's as expansive as you can, and you select the one that gives you the smallest pulse duration. (And yes, by duration you use the FWHM by default, but really you should use whatever is the best descriptor of the temporal resolution limits in your experiment, which will depend on the process you're using.)

$\endgroup$
  • $\begingroup$ Ok, so this is what I've understood: given a pulse $E=E(z,t)$ first of all I fix $z=0$ and I compute $\Delta \nu \Delta \tau_{p}$ (using FWHM). This value will depend on the type of pulse (=0.44 for Gaussian) and I define Fourier limited each pulse of the same shape for which it holds. Then if I evaluate $\Delta \nu \Delta \tau_{p}$ at a generic $z$ inside a dispersive medium, we can say that the value will be different since $\Delta \tau_{p}$ increases, while I have no idea of how $\Delta \nu $ will change. So the pulse won't be T.F. anymore. Do you confirm what I have written? $\endgroup$ – Landau Jan 20 at 13:17
  • $\begingroup$ Not particularly. You're fixing the concept to a fixed time-domain shape, and that's wrong. The concept says that the duration is minimal for a fixed spectral shape. $\endgroup$ – Emilio Pisanty Jan 20 at 14:15
  • $\begingroup$ See edited answer; hopefully that'll make the concept clearer. $\endgroup$ – Emilio Pisanty Jan 20 at 14:27
  • $\begingroup$ Ok, I've read it carefully. However there is still one thing that I don't understand. Fixed $A(w)$ (and so $\Delta w$) then it should be fixed also $E[0,t]$ (and so $\Delta \tau_{0}$) simply by putting $A(w)$ inside the definition of the pulse. And if compute $\Delta \tau _{0} \Delta w$ I get a number let's say $\alpha$. I don't understan why, how you say, it should hold for a laser that $\Delta \tau_{0} \Delta w>\alpha$. I mean it can't be possible if we are basing on the previous definition of pulse... $\endgroup$ – Landau Jan 20 at 18:32
  • 1
    $\begingroup$ Well, you can define it, but depending on the set of spectral phase functions you (implicitly or explicitly) set, the infimum may or may not be achieved (and it may or may not be achieved with a physical pulse). In practice, yes, you want phi to be restricted to some set of reasonable, well-behaved functions. But seriously, you're over-stressing. It's a fuzzy term. The details depend on the set of spectral phases and the measure of pulse length that you choose, but if your choices are reasonable then the result is robust, and the shape and duration of the minimal pulse will be roughly similar. $\endgroup$ – Emilio Pisanty Jun 2 at 18:15
0
$\begingroup$

Given a generic pulse: $$E[z,t]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} A(w)e^{i[\beta (w)z -wt]}dw$$ You have: $$A(w)=\mathscr{F}(E(0,t))_{w}$$ $$S(w)=|A(w)|^2$$ $$S(w)_{FWHM}=\Delta w$$ $$I(z,t)=|E(z,t)|^2$$ $$I(z,t)_{FWHM}=\tau_{p}(z)$$ Fixed $S(w)$ ( and consequently $\Delta w$), which is automatically fixed once you choose a particular laser, you have: $$\tau_{p}(0)=I(0,t)_{FWHM}=|E(0,t)|^2_{FWHM}$$ $$=\vert \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \sqrt{S(w)} e^{-iwt+i\phi}dw \vert ^2_{FWHM}$$ So $\tau_{p}(0)$ depends on $\phi$. If our laser produces a pulse with that partcular $\phi$ for which $\tau_{p}(0)$ is minimum then we call it transform limited pulse. For example if $S(w)$ is fixed as a Gaussian function then this pulse will be T.L. only if $\tau_{p}(0)\Delta w =2.44$.

Finally, for $z>0$ since we are considering a dispersive media then: $$\tau_{p}(z)>\tau_{p}(0)$$ So we are sure that the pulse won't be T.L. anymore.

$\endgroup$
  • $\begingroup$ If nothing else: $w$ ($w$) is not the same thing as $\omega$ ($\omega$). You keep using the former when you should be using the latter. That's kind of equivalent to someone writing Panda instead of Landau and claiming that they're the same thing. Language matters. Use it correctly. $\endgroup$ – Emilio Pisanty Jan 20 at 20:12
  • $\begingroup$ Other than that, yeah, this is mostly correct (though pretty muddled in its presentation - I imagine it makes sense to you, but for a general reader it's a pretty confused exposition). Note that in your $z>0$ final paragraph, that's only true if the pulse is transform-limited at $z=0$ and the medium is dispersive. It's obviously perfectly possible for the pulse to be transform-limited at $z=z_0 \neq 0$, in which case it won't be transform-limited at $z=0$. $\endgroup$ – Emilio Pisanty Jan 20 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.