What is the law of spatial dispersion of a photon in an optical fiber?

Say I have a femto second photon ($\lambda$ around $700\,\text{nm}$) with gaussian shape ($\sigma = c\cdot 1\,\text{fs}\approx 3\cdot 10^{-7}\,\text m$). How will it look at the end when I send it through a 50 km long optical fiber?

I guess it will still have a gaussian shape but with a much bigger $\sigma$.

  • 2
    $\begingroup$ The spectral bandwidth of a $1\text{fs}$ pulse is gigantic, so it will be heavily dispersed due to group velocity dispersion and third order dispersion, irregardless of whether the center wavelength is at the zero-GVD wavelength of the fiber material. So yes, in general it will spread out and become much longer, probably several thousand times longer. $\endgroup$ May 15, 2014 at 12:46

2 Answers 2


A $1\text{fs}$ pulse is extremely short, and unlikely to be sent down an optical fiber for any distance, let alone 50 kilometers, and often femotosecond pulses are produced at high enough powers that it can result in nonlinear effects when in contact with matter.

Ignoring these concerns for the moment and assuming linear behavior, the dispersion of the pulse is easiest to determine in the frequency domain. Working in one dimension, let $$\widehat{E}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty E(t)e^{-i\omega t}\,dt$$ be the Fourier transform of the pulse before it enters the fiber. Then after passing through the fiber, the pulse becomes $$\widehat{E_L}(\omega)=\widehat{E}(\omega)\Phi_L(\omega)$$ where $$\Phi_L(\omega)=\exp\left[-iL\sum_{n=0}^\infty\frac{k^{(n)}(\omega_0)}{n!}\left(\omega-\omega_0\right)\right]$$ is the frequency domain transfer function of the material, where $L$ is the length of the fiber, $\omega_0$ is the center frequency of the pulse, and $k^{(n)}(\omega_0)$ is the $n^\text{th}$ derivative of the spatial wavenumber $k$ evaluated at the center frequency. At that point, after exiting the fiber, you can obtain the pulse in the time domain by inverse Fourier transforming, $$E_L(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\widehat{E_L}(\omega)e^{i\omega t}\,d\omega.$$

$k^{(0)}$ and $k^{(1)}$ do not affect the width of the pulse, and can be ignored. $k^{(2)}$ creates group velocity dispersion (GVD), $k^{(3)}$ creates third-order dispersion (TOD), and so on, all of which will widen the pulse. If you have the derivatives (which can be obtained by differentiating the Sellmeier curves of the material), then you can just use the above to numerically compute the output pulse.

The simplest (and most useful) case is when you ignore everything except $k^{(2)}$ and use a Gaussian pulse, in which case RPPhotonics cites the formula $$\tau\approx4\ln(2)\frac{D_2}{\tau_0}$$ where $\tau_0$ is the initial pulse length, $\tau$ is the final pulse length and $D_2=Lk^{(2)}(\omega_0)$ is the group delay dispersion for the fiber. For Corning HPFS fused silica, $k^{(2)}=450\text{fs}^2/\text{cm}$ at $700\text{nm}$, and using $L=50\text{km}$ gives $$\tau=6.2\times 10^9\text{fs}.$$

So in short, the pulse will be over a billion times longer!


Since I don't have time to expound on the theory from DumsterFDoofus's excellent answer, I thought I'd add to it by showing typical results for this kind of pulse.

First of all, note that the output of any length of fibre to such an extremely short Gaussian Pulse is highly non-Gaussian. There are a huge number of factors to your answer: whether the fibre is one-moded, what its refractive index profile is (zero dispersion point shifting through refractive index profile design is a key concept here).

Your pulse has a bandwidth roughly 1000THz. This is huge. To get an exact answer to your question you will need to simulate your fibre system with a commercial optical link design software. But, as DumpsterDoofus says, 50km of transmission will utterly scramble and attenuate the signal severely. To give you a taste of what you will be up against, consider the phase versus frequency response I plot below. It is taken from a patent of mine "Optical Scanner and Scanned Lens Optical Probe" and arises from 20 metres, i.e. three orders of magnitude shorter than your fibre, of a certain one-moded fibre together with a lens system mounted on the fibre.

Phase Response

The input pulsewidth for even this length of fibre is pretty irrelevant if it is less than about 10fs. Just as the destructive interference far from a stationary phase point (SPP) means that one needs only consider the response near the SPP to calculate the interference of a system of intereferers through a method-of-stationary-phase-evaluated integral, or, in the same line of thought, one needs only to consider paths near those minimising the Lagrangian in a Feynman path integral, here we find that the output is pretty much independent of bandwidth as long as we simulate a band of about 200THz in width. When we do so, here is what the pulse looks like at the output (the spread out, uncompensated one):

Pulse Spread

This squarish looking pulse with the fine ripples on top is HIGHLY typical of the kind of pulseshape one gets on simulating the linear pulse propagation through a fibre link and one needs only generally to consider up to quartic phase versus wavelength variation for accurate results. Manufacturers quote dispersion in figures of ps per nanometre wavelength per kilometre of fibre: this corresponds to a cubic phase against frequency dependence and often this is all one needs for accurate results. You pulse after 50km will be of the order of tens of nanoseconds in duration.

Two ways of overcoming this are:

  1. Dispersion compensators that with a fibre Bragg grating add the phase response to cancel the dispersion, either at the beginning or end of the link;

  2. Use of nonlinear fibres to set up soliton pulses: the nonlinear tendency for pulses to gather themselves up into shorter and shorter wavepackets actively compensates for the linear dispersion, and the feedback loop thus established allows stable, propagation invariant shape solitons.


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