# How short can a pulse be?

1. Aug 24, 2008

### fisico30

Take a laser source with small bandwidth centered around the average frequency (or wavelength)
With an hypothetical external modulator that can switch as fast as we want, ca we make a pulse that is spatially shorter than 1 wavelength?
If not? Why? What would that entail?
Clearly the number of photons always need to be an integer in whatever pulse we have, right?

2. Aug 28, 2008

### n0_3sc

I think this is an awesome question. I spent half an hour debating this with other laser physicists (students off course :tongue2:)...
Here is my/our opinion:
With an external intensity modulator the absolute theoretical limit is a modulation frequency at the frequency of the laser (can be explained easily in the frequency domain). Thus the wavelength is the absolute minimum spatial width you can achieve.

Yes, this is exactly what we eventually got to, ie. you cannot "chop" a photon or a packet of photons into any fraction. We initially thought that modulating faster than the lasers carrier frequency would mean we have some output pulse of a shorter wavelength (generating a completely different pulse), but realising that we're dealing with photons too completely stumped us.

I'm keen to hear if anyone disagrees.

3. Aug 29, 2008

### Cthugha

The shortest possible length of a pulse - no matter how you modulate - is determined by the given bandwidth and the spectral shape of the pulse. This is a basic property of the Fourier transform.
The usual way to quantify how "good" a laser pulse is, for example when setting up a laser system, is to determine the time-bandwidth product. This is the product of the FWHM in the frequency and in the time domain. For a given pulse shape, there is a minimal value of this product (0.44 for Gaussian shape and around 0.315 for a sech^2 pulse), so the deviation from this product shows, how good the laser performs.

4. Aug 29, 2008

### n0_3sc

I have to disagree. That method is used to characterise pulses in things like mode-locking. If I have a pulse with a linewidth of 1nm it does not mean the time bandwidth product will give me the pulses width in the time domain. Optical pulses can look like anything you want in the time domain. It is dependent on the modulation technique. If you were doing some sort of pulse compression then you can use the time bandwidth product.

You also, cannot talk about frequency and time domain here because he refers to the pulse spatially.

5. Aug 29, 2008

### Cthugha

I did not intend to say it will give you the actual pulse width in the time domain, but the given pulse width gives you the Fourier transform limited minimal possible width in the time domain, which does not depend on the modulation technique. The tbp can then be used to say how close you are to that limit.

As it is not very clearly defined, what a pulse is (is a single emitted photon already a pulse?), I would say a pulse can only be as localized as a single photon can be localized, which is rather complicated as there is no generally agrred upon position operator for photons. Chapter 12.11 of Mandel and Wolf mentions some problems, which can be encountered when trying to localize photons. So from this point of view I would say the wavelength is the limit for a given mode.

6. Aug 29, 2008

### n0_3sc

good, so we all agree.

7. Aug 30, 2008

### fisico30

n0_3sc I can't wait to agree with all of you but first I need to get it.
The uncertainty relation between frequency band and time tells that a signal of a certain duration delta_t has a bandwidth delta_f. The shorter delta_t the bigger delta_f.
But the mentioned bandwidth has nothing to do with the linewidth of the source, right?

As far as the minimum spatial length of a signal, why is a wavelength the minimum length?
What if I had a chopper that could actually chop so fast, in theory, the resulting signal is less than lambda. Can you give me a basic, elementary explanation?
thanks

8. Aug 30, 2008

### n0_3sc

Bandwidth is related to linewidth by:
$$\Delta f = c\frac{\Delta\lambda}{\lambda_1\lambda_2}$$
Easy to prove using:
$$c = f\lambda$$
Although, I don't think that's the main issue you should think about...

The most "elementary" explanation i can think of is by the fundamental energy of a photon:
$$E = hf$$
which can be written as:
$$E = h\frac{c}{\lambda}$$
Therefore, each photon has an energy related to its wavelength, chopping the wavelength would mean you no longer can measure any wavelength (since you've chopped it). Hence no wavelength = no energy = no photon

9. Aug 30, 2008

### n0_3sc

I should also make it clear that what I've said is only my opinion based on what I know. Other people may have better explanations.

10. Sep 2, 2008

### fisico30

Hello Cthugha,
fisico 30 here. I have gone through your comments/teachings (Fock state) and answers on this post about minimum time and spatial pulse length. I wrote a little summary of what I know about "uncertainty" for you to validate if you have time, and a question about pulses at the end.

In classical deterministic physics, the probabilistic character of our forecasts is due to the precision in measuring initial conditions and in the complexity of the acting forces.
In classical physics probability is called subjective, but it is not objective or intrinsic in the system.
The properties of a system are independent from the measuring process:The system has properties, independently from the act of measuring.
The uncertainty relations:I think that in a book by Polkinghorne, it is said that to reach the most precise knowledge, we need to minimize the perturbation caused by the act of measuring. the effects of the perturbation could be quantified and analyzed, someone could observe, but this perturbation is "out of control". This is what brings to indetermination.
Maybe it is better to explain the indetermination by saying that there are incompatible observables such that it is impossible to reduce their uncertainty simultaneously to values that violate Heisenberg relations.these relations do not apply to compatible observables.
The indetermination(uncertainty) relations demonstrate that we need to give up he description of phenomena in terms of trajectories in space. we are so used to visualize the movement of macroscopic objects that we cannot do without it.

Back to the pulse:
We have a source with linewidth delta_f. This source emits a pulse of a certain time duration delta_t. The pulse can be short or long (depending on the modulator speed).
Once created ,this pulse can be spectrally analyzed and the uncertainty relation just tells us that the longer the time duration, the bigger the frequency band. This frequency band is not the same as the linewidth of the source.
Why can the pulse not be spatially longer than the center wavelength of the source?( you say that the wavelength is he minimum)
In telecom, they would like "short pulse", meaning short in time, to get high data rates right? The energy in each of those pulses can be more than one photon.
In vacuum, can we have two pulses with the same time duration but different spatial extent? Would those two pulses have the same usefulness in telecom applications?
In general, distance=velocity*time, so if time is small, distance is small....

thanks again Cthugha!

11. Sep 2, 2008

### Cthugha

Well, let me just say, that there are several points of view on the how and why of uncertainties. A Bohmian will disagree on discarding trajectories while others will point out, that complementarity is the key. I think the actual formulation of uncertainty, which will cause the least disagreement at the moment is the impossibility of preparing an ensemble more well defined than uncertainty allows. As soon as you want to get to single particles the opinions start to differ.

Do you maybe mean longer duration and shorter frequency band or the other way round? And the pulse can of course be longer than the wavelength is, but getting shorter is a problem.

Ok, so first of all do you mean spatially short in terms of short in 1D (like a pulse, which propagates in a direction and is simply cut, which is more or less rather equivalent to being short in time) or in terms of 3D (like a superposition of several pulses, which form a huge and short superposition at one spot or like in focusing a beam)? I must admit, that this is not really my area of expertise. I know that there are experiments with light and subwavelength slits, if this is also your area of interest. That is near field optics and shows interesting behaviour like extremely high transmission efficiencies and subwavelength spatial resolution like in near field scanning optical microscopy, but iirc in this area you have to take things like surface plasmon interactions or the small slits showing microcavity-like behaviour into account, but as I said - this is not the area I know much about.

My assumption that the minimal spatial length is on the order of the wavelength comes more or less from the wave-like character of light. Mandel and Wolf once showed, that if you define the probability to have a photon inside a certain volume (already larger than the cubic wavelength) and examine a rather localized probability distribution, the energy density distribution and the detection probability distribution will differ significantly and be far more widespread. As I said before, there is no generally agreed upon position operator for photons, so even what it means for a photon, and therefore also for a pulse, to be localized that good is not too well defined.

12. Sep 2, 2008

### n0_3sc

Yes you are right in saying the full freq band is not the same as the linewidth. (I should've been clear in saying the FWHM in the freq domain can be used to find the linewidth).

Did you mean "Why can the pulse not be spatially shorter than the center wavelength of the source?"
You were only specifically referring to modulation (ie. chopping) techniques. But you can focus any beam to a spatial width less than the wavelength - this is the diffraction limit.

In telecom, they prefer "high rep rate" pulses not "short pulses". (it's actually debatable because having shorter pulses means you can decrease the time between consecutive pulses). Yes, you can have two pulses (same time width) with different spatial extent. You must remember that you're talking about independent dimensions here. You cannot neccessarily switch between domains/dimensions so easily.

The other thing is that telecom have a limit to the repetition rate because pulses in optical fiber's will disperse and broaden (temporally and spatially) - so enough room must be given between pulses to allow for these effects.

13. Sep 3, 2008

### Andy Resnick

I'm late to this thread, but I don't think the OP's question ever got answered.

Sentence two potentially mixes up two concepts- the temporal extent of a pulse and the spatial confinement of a pulse. It's possible to define a pulse duration as the product of group velocity and center frequency (http://www.rp-photonics.com/pulse_duration.html), but that's usually not what is meant by spatial confinement- focusing is. And photon number doesn't really enter into any of this- these are bright pulses, not dim pulses.

Assuming the OP was not asking about spatial widths, the spatial extent of a wavefront is also related to certain source characteristics via a Fourier transform- it's spatial rather than temporal, but the concept is the same. Essentially, the fourier transform of the source gives the so-called 'point spread function'. Technically, it's the spatial fourier transform of the exit pupil of an optical system, but basically, larger sources give smaller focal volumes. That's why microscope lenses have high numerical apertures- a high NA gives a small focal volume. The Rayleigh limit- something usually associated with resolution limits- is actually a measure of the Fourier transform of a disk function- lenses are typically disk-shaped. The minimum spot size is about wavelength/NA, but there's tricks to get very small focal volumes in, for example, nonlinear optics.

So, the limit of spatial confinement of a pulse depends on spectral badwidth indirectly. First, the wavelength comes into the fourier transform as a scale factor, but more critically, the construction of lenses to minimize the chromatic aberrations over a large bandwidth is very difficult, and why microscopy using pulsed sources is so tricky.

Now spectral bandwidth- as others have posted, the spectral bandwidth and pulse duration are related by a Fourier transform, and there are some simple transform pairs (Gassian/Gaussian and sech^2/?) that have a minimum product, and are useful metrics to compare real pulses against. The fastest pulses I know of are tens of attoseconds long with a center frequency in the x-ray and created by harmonic generation of UV pulses. A 10 as pulse has a large spectral bandwidth and lasts about 1/100 of an optical cycle.

Narrowband sources make poor pulsed sources- pulsed sources usually have a dispersive element in order to broaden the initial pulse and allow for compression.