GVD, Chirp, and Pulse Spreading

The Fourier transform of a gaussian pulse shape is also gaussian.  If $f(x) =  \exp(- (\frac{x}{a})^2)$, then we can develop the transform as

$$\begin{align}F(k) &= \frac{1}{2 \pi} \int f(x) e^{-i k x} {dx} \\ &= \frac{1}{2 \pi} \int e^{- ( \frac{x}{a})^2}   e^{-i k x} {dx} \\ &= \frac{1}{2 \pi} \int \exp \left[ - \left( \frac{x}{a} \right)^2 -i k x \right] {dx} \\ &=  \frac{a \sqrt{\pi}}{2 \pi} \exp \left( - \frac{a^2 k^2}{4} \right)\\ &=  \frac{a}{2 \sqrt{\pi}} \exp \left[ - \left(\frac{a k}{2} \right)^2 \right] \end{align}$$

Consider now the time-evolution of this pulse, assuming this shape (in $x$), at $t=0$ and allowing each $k$ component to evolve according to $e^{-i w(k) t}$.  That is, evolve through linear media with $\omega \equiv \omega(k)$

$$f(x,t) = \int  \frac{a}{2 \sqrt \pi} \exp \left[ - \left(\frac{a k}{2} \right)^2 \right] \exp (i k x - i \omega(k) t){dk}$$

Let $\omega(k)  =  \omega_0 +  \omega_1 k  + \omega_2 k^2 +$ H.O.T.

$$\begin{align} f(x,t) &= \frac{a}{2 \sqrt \pi}  \int  \exp \left( - \frac{a^2}{4} k^2  + i k x  - i \omega_0 t -i \omega_1 k t - i \omega_2 k^2 t \right)  {dk}\\ &= \frac{a e^{-i \omega_0 t}}{2 \sqrt \pi}  \int  \exp \left[ -  \left(\frac{a^2}{4} + i \omega_2 t \right)  k^2  + i (x   -  \omega_1  t) k  \right]  {dk} \\ &= \frac{a e^{-i \omega_0 t}}{2 \sqrt \pi} \sqrt \frac{ \pi}{ \left(\frac{a^2}{4} + i \omega_2 t \right)}   \exp \left( - \frac{( x   -  \omega_1  t) ^2}{4 \left(\frac{a^2}{4} + i \omega_2 t \right) }   \right) \\ &= \frac {a e^{-i \omega_0 t}}{\sqrt {{a^2} + 4i \omega_2 t }}   \exp \left( - \frac{( x   -  \omega_1 t)^2}{(a^2 + 4 i \omega_2 t ) } \right) \\ &= \frac { e^{-i \omega_0 t}}{\sqrt {1 + \frac{4i \omega_2 t} {a^2} }}   \exp \left( - \frac{( x   -  \omega_1  t)^2}{a^2 \left(1+ \frac{4 i \omega_2 t}{a^2} \right) } \right) \end{align}$$

We can see the pulse moves with velocity $\omega_1$, the group velocity, and spreads with time according to a factor controlled by $\omega_2$, the group velocity dispersion.  To see this more clearly, let's look at the spectrum of a gaussian with a complex width parameter.

Since
$$\int \exp(- ax^2) \exp (-bx) dx = \sqrt \frac{\pi}{a} \exp \left( \frac {b^2}{4 a }\right)$$

We can write
$$\int \exp(- \Gamma k^2) \exp (ikx) dk = \sqrt \frac{\pi}{\Gamma} \exp \left( -\frac {x^2}{4 \Gamma } \right)$$

Or
$$\exp(- \Gamma k^2)   \iff   \sqrt \frac{\pi}{\Gamma} \exp \left( -\frac {x^2}{4 \Gamma } \right)$$

If $\Gamma = a + ib$ then

$$\begin{align} \exp(- (a+ib) k^2) &= \exp(- a k^2) \bbox[ border:2px solid yellow ] {\exp(-ib k^2)} \\ & \iff   \sqrt \frac{\pi}{(a+ib)} \exp \left( -\frac {x^2}{4 (a+ib) } \right) \\ &= \sqrt \frac{\pi}{(a+ib)} \bbox[ border:2px solid cyan ] {\exp \left( -\frac {a x^2}{4 (a^2+b^2) } \right)} \exp \left( i \frac {b x^2}{4 (a^2+b^2) } \right) \end{align}$$
and we can see that the imaginary part of $\Gamma$ both controls the $\bbox[ border:2px solid yellow ] {\text{chirp}}$ in the $k$ domain, and influences $\bbox[ border:2px solid cyan ] {\text{width}}$ in the $x$ domain.