Negative frequency contributions for very short pulses?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.