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I am wondering if very short optical light pulses can have a Gaussian envelope?

When I describe the pulse shape with a Gaussian than the frequency distribution has also a Gaussian shape. But if the envelope of the short pulse will have a pulse width of $1\,\text{as}=10^{-18}\,\text s$, then the frequency bandwidth will roughly be the inverse of the pulse width (since its a Gaussian), which leads to a frequency bandwidth of $10^{18}\,\text{Hz}$. But when the mean frequency of the pulse is in the optical regime, say $10^{15}\,\text{Hz}$, the frequency distribution will have enormous negative frequency contributions. Almost half of the frequency distribution will be in the negative range.

How can one avoid this problem?

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The problem lies in the premise that a light pulse that is $1\,\text{as}$ long can have mean frequency around $10^{15}\,\text{Hz}$. Electric field has to rise and fall within $1\,\text{as}$, which can be achieved with oscillations that are at least of the same time scale - that is, at least $10^{18}\,\text{Hz}$.

This means that $1\,\text{as}$ pulses are not possible with visible light, one would have to use at least $c \times1\,\text{as}=0.3\,\text{nm}$ radiation which is in the range of soft X-rays. The shortest pulses reported in the visible range are still longer than $1\,\text{fs}$, corresponding to about one period of electromagnetic wave oscillation.

Update: the relation of Gaussian pulse width to Gaussian frequency spectrum is valid for $\Delta\nu \ll \nu_0$, where $\nu_0$ is the central frequency and $\Delta\nu$ is the bandwidth (as Floris assumes in his answer). For $\Delta\nu \approx \nu_0$ the frequency spectrum will be very non-Gaussian. Why it cannot simply shift to negative frequencies? Because there is a relation for a Fourier transform of a real signal: $$\hat{f}(-\omega)=\hat{f}^*(\omega)$$ where * refers to a complex conjugate. This is why negative frequencies are not considered - they mirror the positive side of the frequency spectrum.

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  • $\begingroup$ Hm, can you explain why a attosecond pulse can only be produced by appropriately high frequencies? For example I found this jur.byu.edu/?p=1123 and that iopscience.iop.org/0034-4885/67/6/R01 . So I think it will be possible (no matter when). $\endgroup$ – thyme Nov 4 '14 at 10:23
  • $\begingroup$ First link contains no information about attosecond pulses (besides the title), second link explicitly states that extreme UV (not visible) was used to produce single 650 as pulse (nowhere close to 1 as). $\endgroup$ – gigacyan Nov 4 '14 at 16:03
  • $\begingroup$ Ok, but in the first link they write "Producing a laser with pulse durations on this time scale composed of wavelengths centered in the visible regime is a long sought goal of laser physicists." I have two more refs: Here (arxiv.org/pdf/1206.0965v2.pdf) they write on pdf page 7: "..and recent rapid progress in the preparation of attosecond optical pulses may provid...". Or here (www.ros.hw.ac.uk/bitstream/10399/2702/1/McCrackenRA_1013_eps.pdf) on pdf page 11: "Generation of optical attosecond pulses is an appealing prospect". I just want to understand why you think that its impossible. $\endgroup$ – thyme Nov 4 '14 at 16:50
  • $\begingroup$ I have updated my answer, hopefully it is more clear now. The paper from arxiv cites two sources regarding "preparation of attosecond optical pulses". Looking through ref. 23 I see XUV (extreme ultraviolet, < 100 nm) mentioned everywhere. For example: "pulses as short as 80 as have been produced in neon (at a central photon energy of 80 eV)". 80 eV corresponds to 16 nm wavelength. $\endgroup$ – gigacyan Nov 5 '14 at 8:01
  • $\begingroup$ The second PDF also starts with a statement that all attosecond pulses up to date were produced in XUV/X-Ray region. The author then proposes a method for producing short optical pulses and concludes saying that the shortest pulse they could theoretically produce would be 1.2 fs, very close to the limit I gave in my answer. $\endgroup$ – gigacyan Nov 5 '14 at 8:11
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If you think about it, a pulse that is much narrower than one pulse of the "center frequency" is really no longer related to that center frequency and just becomes a delta function.

If you just take the Fourier Transform of the pulse shape, you end up with the actual frequency components - an essentially flat response. The "negative frequencies problem" you created for yourself comes from the fact that you are trying to treat the signal as "mostly $10^{15}$Hz with an envelope - but the envelope is so narrow that the "thing inside" doesn't matter.

There is a very relevant discussion of the question of negative frequencies in the answers to this earlier question

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  • $\begingroup$ Hm , sorry I don't understand that. What do you mean with an "an essentially flat response"? How can I see that the negative frequency contributions do not come in? When I fourier transform a delta I get negative frequency distributions, too. And what do you mean with "just becomes a delta function"? If I want to distinguish attosecond pulses from femtosecond pulses I can not just treat both as a delta function, since they have different pulse widths. I think they always have a shape, even if they are very short... $\endgroup$ – thyme Oct 29 '14 at 16:13
  • $\begingroup$ Maybe I just don't understand what your problem is with negative frequencies. My point was that when the pulse width is much shorter than the period of the wave, then the latter doesn't matter any more and you just have the transformation of a pure Gaussian (which looks "more and more" like a delta function) and whether that is centered at f=0 or f=$10^{15}$ doesn't matter when its width is so much greater than the offset. $\endgroup$ – Floris Oct 29 '14 at 17:12
  • $\begingroup$ Ok, maybe my problem with negative frequencies is solved, when you can explain their meaning. I thought any negative frequency contribution is unphysical... or am I wrong? $\endgroup$ – thyme Oct 29 '14 at 18:14
  • $\begingroup$ There is nothing unphysical about negative frequency - you can write $\sin(\omega t)$ or $\sin(-\omega t)$ and will see immediately that one is just the inverse of the other - if you are looking at just one point, it's got 180° phase shift, or if you look at a wave, it's traveling in the opposite direction. Does that clear it up for you? $\endgroup$ – Floris Oct 29 '14 at 18:40
  • $\begingroup$ Oh, I see. The fourier transform of my wave just shows the frequency contributions of a decomposition of my wave into plane waves. And negative frequencies then just mean plane waves propagating in the opposite direction... right? $\endgroup$ – thyme Oct 30 '14 at 10:22
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We sometimes forget about negative frequencies but they are always there. All real signals are composed equally of negative frequency and positive frequency components. If you take an oscillating pulse with Gaussian envelope, $$E(t) = A \exp(-\tfrac 12 t^2/\tau^2)\cos(\omega_0 t)$$ and convert it into its spectral representation, then you get two Gaussians: $$E(\omega) = \tfrac{A\tau}{\sqrt{8\pi}} \exp(-\tfrac 12(\omega - \omega_0)^2\tau^2) + \tfrac{A\tau}{\sqrt{8\pi}} \exp(-\tfrac 12(\omega + \omega_0)^2\tau^2) $$ Note that I have defined that $E(t) = \int E(\omega) e^{-i\omega t} d\omega$. [Though I am possibly off by a prefactor here]

This is exact, but let's not get confused about the meaning of $\tau$ and $\omega_0$. $\tau$ and $\omega_0$ only roughly correspond to pulse duration and peak frequency. You can see this if you closely examine plots of $|E(t)|^2$ and $|E(\omega)|^2$ for various cases.

In general the two Gaussians mix together in frequency space---the tail of the negative frequency one extends to positive frequencies, and vice versa. As a result, $\omega_0$ is only the peak energy frequency in the case of long pulses ($\omega_0\tau \gg 1$) where the mixing is negligible. In this case can we approximate $|E(\omega)|^2$ as the sum of two Gaussians centered at $\pm \omega_0$. If I choose short pulses ($\omega_0 \tau \lesssim 1$) then the peak energy frequency is less than $\omega_0$ for this example. Though that frequency shift does on the phase that I have chosen -- if I replaced $\cos$ by $\sin$ in the first formula then the peak energy would be at higher frequencies than $\omega_0$ and the pulse would last slightly longer and contain less overall energy. So do not expect $\omega_0$ and $\tau$ to have precise intuitive meanings in these sorts of cases --- they are just pulse shaping parameters.

One final exception to keep in mind --- if $\omega_0 = 0$ then $\tau$ again exactly corresponds to pulse duration and the peak energy is again at $\omega_0$.

Short answer: Gaussian pulses do not necessarily have single-Gaussian spectra, and negative frequencies are real.

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