Heisenberg Bound Question
Verify $x(t) = e^{i\omega t}e^{-(t-\tau)^2}$ exactly satisfies the Heisenberg bound of $\sigma_t(x)\sigma_{\omega}(x)$.

Attempt:
I know $\sigma_t(x) = \int_{\mathcal{R}} (t-\mu_t(x))^2\rho_x(t)dt$ where $\rho_x(t) = \frac{|x(t)|^2}{||x||^2}$ and $\mu_t(x) = \int_{\mathcal{R}} t\rho_x(t)dt$
$||x||^2 = <x. x^*> = \int e^{-2(t-\tau)^2}dt$ (I didn't include the derivation for this)
I'm also guessing I have to take the Fourier transform at some point to find $\sigma_{\omega}(x)$
I realize that it sort of has the form of a Gaussian which is when the inequality is sharp.  I also know that the inequality is $\sigma_t(x)\sigma_{\omega}(x) \geq \frac{1}{4}$
Other thoughts:
I saw a similar question on the physics forum but not sure how to actually calculate/prove it.  I figure that if i find the variance with respect to time, take the Fourier transform o find $\hat{x}(\omega)$ and find the variance with respect to $\omega$, then their product should equal $\frac{1}{4}$.
 A: Some things here are merely decorations. Modulation (multiplication by $e^{i\omega t}$) does not change $\sigma_t(x)$ because it does not change $\rho_x(t)$. Translation (replacing $t$ by $t-\tau$) does not change $\sigma_t(x)$ either: $\rho_x(t)$ shifts by $\tau$, but so does $\mu_t$, so the net effect is zero. (Simply put, central moments of a distribution are not affected by shifts.) 
Under the Fourier transform, modulation in $t$ becomes translation in $\omega$, and translation in $t$ becomes modulation in $\omega$. Neither of these affect $\sigma_\omega(x)$.  
The above means that $x(t)=e^{-t^2}$ is all we actually have to compute with. Clearly, $\mu_t(x)=0$. Computing the variance takes a little integration by parts: 
$$ \int_{\mathbb R} t^2 e^{-2t^2}\,dt   = 
-\frac14 \int_{\mathbb R} t \,d(e^{-2t^2})  = \frac14 \int_{\mathbb R} e^{-2t^2}\,dt = \frac14 \|x\|^2
 $$
Hence $\sigma_t(x)=\frac14$. 
You did not specify your Fourier transform convention, but I'm going to guess that $\widehat{e^{-2t^2}} = e^{-\omega^2/2}$. For which you have
$$ \int_{\mathbb R} \omega^2 e^{-\omega^2/2}\,dt   = 
-  \int_{\mathbb R} \omega\, d(e^{-\omega^2/2})   
 =  \int_{\mathbb R} e^{-\omega^2/2}\,d\omega =  \|\hat x\|^2
 $$
Hence $\sigma_\omega(x)=1$. 
