Spectral full width at half max of a Gaussian light pulse [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – DanielSank, Community, John Rennie, Aaron Stevens, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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