1. Why pulses shorter than single cycle cannot propagate in free space ? Is there fundamental reason or it is engineering ?

  2. Is it possible for a laser pulse to have bandwidth that can support half cycle pulse and how it will propagate in free space ?

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    $\begingroup$ why do you think there is a limit single cycle pulses can not propagate. where this info come from? $\endgroup$ – Andreas H. Oct 27 '13 at 22:44

This does not have anything to do with the bandwidth of your gain medium. Considering the one-dimensional wave equation any traveling wave is a solution, including half-cycle pulses. You can virtually plug any solution $f(t-\frac z c)$ into the one-dimensional wave equation and it will be an exact solution, it does not matter if it's a sine-wave or a wave shaped like your butt. It is also certainly true that the plane wave assumption is an approximation of underlying experimental conditions. However, the single-cycle limit has nothing to do with the generation process per se, but the geometry that is being considered.

For a finite-size source confined to a two-dimensional plane (a circular aperture, for example) the source field can be considered as Fourier series in the transverse k-space. Physically, this means that the source is represented as a sum of infinitely many plane waves which make individual angles with respect to the propagation axis. There is a vacuum cut-off frequency for such sources, typically defined by a trivial dispersion relation that you can find by using a traveling wave Ansatz in the three-dimensional wave equation. Thus, not all frequencies can propagate as traveling waves, some frequencies are damped and exist only in the near-field of the source. This is particularly true for the DC field, which is not only a damped mode, but an evanescent mode, the propagation constant is purely imaginary. In the far-field all the DC of the initial signal is gone (it stays close to the source), and the only way to satisfy this requirement is to have at least one oscillation of the field. Hence, the single-cycle "limit".

A simple analogy occurs in electrostatics. If you consider an infinitely large charged plane there is a uniform electric field covering all space. In this way DC is "everywhere". For a finite-size plane the field typically drops off with a certain algebraic dependence. Thus, we begin to understand that the limit you are referring to is simply a matter of how fast the other modes are damped. That being said, for an infinitely large source (which is implicit in the plane wave assumption), you have zero width in transverse k-space and DC will propagate, the propagation constant suddenly becomes real (equal to zero, since DC has infinite wavelength). If you find such a source, you can do near-field probing in the "far-field", and you will be stinking, filthy rich. Obviously, you will strongly irradiate anything standing between you and your target, and an account of manslaughter might accompany your achievement. Geeky humor aside, if you suddenly see propagation of subcycle pulses you should start questioning your theoretical model.

Going back to your original question though, less than single-cycle pulses can, strictly speaking, still propagate. However, they will never propagate with a stable waveform since the low-frequency components travel with large angles with respect to the propagation axis. In this way the pulse is under continuous change and always becomes at least single-cycle in the far-field.

You might argue that, in principle, you could make a subcycle pulse propagate over long distances given a source of sufficient size (assuming you can excite a subcycle to begin with). In this case "sufficient size" would imply that the spatial dimensions of the source are much larger than the propagation length. For optical wavelengths such a scenario might be possible for fabricated nanostructures where propagation lengths are only hundreds of nanometers long. Obviously, on the other side of the room any traces of a subcycle pulse will be gone.

All of this being said, it's quite possible that this situation might change in a medium. For example can a pulse be reshaped by nonlinear processes so that it contains DC? In that case DC might be measurable at the output facet of your gain medium. This is an intriguing question for which I do not have an answer. I have never observed such a process (all transients eventually die out and things go back to thermal equilibrium), but you're right, it might be possible. Congratulations on your new thesis topic.


I suspect, that the statement that is implied in your question is not true.

Maxwells equations in vacuum allow for plane waves of arbitrary (positive) frequency as solution.

From Fourier transform argument hence all pulse forms can propagate, completely regardless of their shape, provided they have a zero mean (because by weighting plane waves arbitrary zero mean pulse shapes can be created).

Since there is no dispersion in free space (all plane waves propagate at the same velocity), pulses will not change their shape while propagating.

So from this reasoning half waves or quarter waves can propagate when they is no offset i.e no mean electric field.

In fact, electric wave propagation is not bound to any carrier waves, even though this is most often happening in the optical domain (but not in the electrical for instance).

I suspect, the limit you are referring to is not due to propagation but somehow lies in the specific method for pulse creation.


Please have a look at the so-called "Fourier limit":


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    $\begingroup$ Link only answer are not good answers. $\endgroup$ – dmckee --- ex-moderator kitten Oct 27 '13 at 17:51
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    $\begingroup$ Thanks. But why is the one cycle the limit. What would be wrong with half cycle ? Is it bandwidth problem or propagation that sets the limit ? $\endgroup$ – Anonymous Oct 27 '13 at 18:19
  • $\begingroup$ Actually, I think that the Mode-Locking article explains it more closely: It takes at least one full cycle for the phase-locked resonance modes to appear in the resonant cavity. $\endgroup$ – RBarryYoung Oct 27 '13 at 19:10
  • $\begingroup$ Sorry I was out for a conference and couldn't be more detailed. I will edit my answear $\endgroup$ – Mattia Oct 31 '13 at 8:35

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