Show that the Fourier transform of \(f(t)=e^{-(t / \tau)^{2}}\) is \(e^{-(\omega \tau / 2 \pi)^{2}}\). Show that the duration of \(f(t)\), defined by \((12.24)\), is \(\tau / 2\) and that the width \(\Delta \omega\) of its energy spectrum, defined by \((12.2 .5)\), is \(1 / \tau\) and hence that the product \(\Delta l \Delta \omega\) has precisely the minimum value \(\frac{1}{2}\) permitted by the uncertainty relation (12.2.7).

Recall the definition of the Fourier transform, which can be mapped from a function of time \(f(t)\) to the function of angular frequency \(F(\omega)\): $$F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \, dt$$ Now, let's substitute the given function \(f(t)\): $$F(\omega) = \int_{-\infty}^{\infty} e^{-(t/\tau)^2} e^{-i\omega t} \, dt$$ To solve this integral, we make the substitution \(u = t/\tau\), resulting in \(dt=\tau du\), and the limits of the integral remain the same: $$F(\omega) = \tau \int_{-\infty}^{\infty} e^{-u^2} e^{-i\omega\tau u} \, du$$ This integral is a Gaussian-type integral with an imaginary component in the exponent. We can solve it by using the following property of Gaussian functions: $$\int_{-\infty}^{\infty} e^{-ax^2} e^{-ibx} \, dx = \sqrt{\frac{\pi}{a}} e^{-\frac{b^2}{4a}}$$ In our case, \(a=1\) and \(b=i\omega\tau\), so we find: $$F(\omega) = \tau \sqrt{\pi} e^{-\frac{(i\omega\tau)^2}{4}} = \tau \sqrt{\pi} e^{-(\omega\tau / 2\pi)^2}$$ The function \(F(\omega)\) is the Fourier transform of \(f(t)\).

The duration of \(f(t)\), denoted as \(\Delta l\), is defined by formula (12.24) as: $$\Delta l = \sqrt{2\pi\,\frac{\int_{-\infty}^{\infty} (t - t_{0})^2 \lvert f(t) \rvert^2 \, dt}{\int_{-\infty}^{\infty} \lvert f(t) \rvert^2 \, dt}}$$ In our case, \(t_{0} = 0\) since the function \(f(t)\) is centered at time \(t = 0\). Therefore, we can write the formula as: $$\Delta l = \sqrt{2\pi\,\frac{\int_{-\infty}^{\infty} t^2 \lvert f(t) \rvert^2 \, dt}{\int_{-\infty}^{\infty} \lvert f(t) \rvert^2 \, dt}}$$ Now substitute \(f(t) = e^{-(t/\tau)^2}\) into this formula: $$\Delta l = \sqrt{2\pi\,\frac{\int_{-\infty}^{\infty} t^2 e^{-2(t/\tau)^2} \, dt}{\int_{-\infty}^{\infty} e^{-2(t/\tau)^2} \, dt}}$$ We can solve these integrals with the substitution \(u=t/\tau\), and obtain: $$\Delta l = \sqrt{2\pi\,\frac{\tau^3 \int_{-\infty}^{\infty} u^2 e^{-2u^2} \, du}{\tau \int_{-\infty}^{\infty} e^{-2u^2} \, du}}$$ Cancel the \(\tau\)'s and notice that the integrals are simply constants. Using the Gaussian integrals properties, we obtain: $$\Delta l = \sqrt{2\pi\,\frac{\tau^2\sqrt{\pi}/4}{\tau \sqrt{\pi}/2}} = \frac{\tau}{2}$$

The energy spectrum width, \(\Delta \omega\), is defined by formula (12.2.5) as: $$\Delta \omega = \sqrt{2\pi\,\frac{\int_{-\infty}^{\infty} (\omega - \omega_{0})^2 \lvert F(\omega) \rvert^2 \, d\omega}{\int_{-\infty}^{\infty} \lvert F(\omega) \rvert^2 \, d\omega}}$$ Similar to Step 2, we have \(\omega_{0} = 0\) since the function \(F(\omega)\) is centered at \(\omega = 0\). Substitute \(F(\omega) = \tau \sqrt{\pi} e^{-(\omega\tau /2\pi)^2}\) into the formula: $$\Delta \omega = \sqrt{2\pi\,\frac{\int_{-\infty}^{\infty} \omega^2 \lvert \tau \sqrt{\pi} e^{-(\omega\tau /2\pi)^2} \rvert^2 \, d\omega}{\int_{-\infty}^{\infty} \lvert \tau \sqrt{\pi} e^{-(\omega\tau /2\pi)^2} \rvert^2 \, d\omega}}$$ Solving these integrals with the substitution \(u=\omega\tau/2\pi\), we obtain: $$\Delta \omega = \sqrt{2\pi\,\frac{\tau\vphantom{\pi} (\frac{2\pi}{\tau})^3 \int_{-\infty}^{\infty} u^2 e^{-2u^2} \, du}{\tau \int_{-\infty}^{\infty} e^{-2u^2} \, du}}$$ Notice that the constants cancel out, and using Gaussian integrals properties, we get: $$\Delta \omega = \frac{1}{\tau}$$ Finally, the product of \(\Delta l\) and \(\Delta \omega\) is: $$\Delta l \Delta \omega = \frac{\tau}{2} \cdot \frac{1}{\tau} = \frac{1}{2}$$ This shows that the duration and energy spectrum width of \(f(t)\) satisfy the uncertainty relation, as required by the problem statement.