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I need to calculate the spectral full width at half max (FWHM) of a Gaussian light pulse.

The frequency spectrum of a Gaussian light pulse is

$$ \tilde{E}(\omega)\propto \exp{-\frac{(\omega-\omega_0)^2}{4\Gamma}}$$

with the complex Gauss parameter $\Gamma=\Gamma_1-i\Gamma_2$ from the Gaussian pulse

$$E(t)=\exp{(i\omega_0 t)\cdot \exp{(-\Gamma t^2)}} \, .$$

The FWHM of the spectrum $\omega_{F}$is defined as $|\tilde{E}(\omega_{F})|^2=\frac{1}{2}|\tilde{E}(0)|^2$. The solution should be

$$\omega_F=2\sqrt{2 \ln 2}\cdot \sqrt{\Gamma_1+\frac{\Gamma_2^2}{\Gamma_1}} \, .$$

How do I get there?

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closed as off-topic by DanielSank, user191954, John Rennie, Aaron Stevens, Jon Custer Oct 22 '18 at 23:04

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First, we write the frequency spectrum as

$$\tilde{E}(\omega)=\exp{\bigg[\frac{-(\omega-\omega_0)^2}{4\cdot (\Gamma_1-i\cdot \Gamma_2)}\bigg]}$$

thus, the intensity that we need via the condition $|\tilde{E}(\omega_{F})|^2=\frac{1}{2}|\tilde{E}(0)|^2=\frac{1}{2}$ can be written as

$$|\tilde{E}(\omega_{F})|^2=\exp{\bigg[\frac{-(\omega-\omega_0)^2}{4\cdot(\Gamma_1-i\cdot \Gamma_2)}\bigg]}\cdot \exp{\bigg[\frac{-(\omega-\omega_0)^2}{4\cdot(\Gamma_1+i\cdot \Gamma_2)}\bigg]} $$

or

$$|\tilde{E}(\omega_{F})|^2=\exp{\bigg[\frac{-(\omega-\omega_0)^2(\Gamma_1+i\cdot \Gamma_2)}{4\Gamma}\bigg]}\cdot \exp{\bigg[\frac{-(\omega-\omega_0)^2(\Gamma_1-i\cdot \Gamma_2)}{4\Gamma}\bigg]} $$

which gives

$$|\tilde{E}(\omega_{F})|^2=\exp{\bigg[\frac{\Gamma_1}{2}\frac{-(\omega-\omega_0)^2}{4\Gamma}\bigg]}$$

Now, with the condition for the FWHM, and the arbitrary choice $\omega_0=0$

$$\frac{1}{2}=\exp{\bigg[\Gamma_1\frac{\big(\frac{\omega_F}{2}\big)^2}{2\Gamma}\bigg]} $$

solving for $\omega_F$, we get the result

$$\underline{\omega_F=2\sqrt{2 \ln{2}}\cdot \sqrt{\Gamma_1\bigg(1+\bigg(\frac{\Gamma_2}{\Gamma_1}\bigg)^2\bigg)}}$$

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