I did a Fourier transform of a gaussian function $\scriptsize \mathcal{G}(k) = A \exp\left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2}\right]$
$$
\scriptsize
\begin{split}
\mathcal{F}(x) &= \int\limits_{-\infty}^{\infty} \mathcal{G}(k) e^{ikx} \, \textrm{d} k = \int\limits_{-\infty}^{\infty} A \exp \left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2}\right] e^{ikx}\, \textrm{d} k = A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2} \right] e^{ikx}\, \textrm{d} k =\\
&= A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{i(m+k_0)x}\, \textrm{d} m = A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{imx} e^{ik_0x}\, \textrm{d} m =\\
&= A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{imx}\, \textrm{d} m = A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 \right] e^{iu \sqrt{2} {\sigma_k} x} \sqrt{2} {\sigma_k} \textrm{d} u = \\
&=\sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 \right] e^{iu \sqrt{2} {\sigma_k} x}\, \mathrm{d} u = \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 + i u \sqrt{2} {\sigma_k} x \right]\, \mathrm{d} u =\\
&= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\left(u + \frac{i {\sigma_k} x}{\sqrt{2}} \right)^2 – \frac{i^2 {\sigma_k}^2 x^2 }{2}\right]\, \mathrm{d} u =\\
&= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\left(u + \frac{i {\sigma_k} x}{\sqrt{2}} \right)^2 + \frac{{\sigma_k}^2 x^2 }{2}\right]\, \mathrm{d} u = \\
&= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} e^{-z^2} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right]\, \mathrm{d} z = \sqrt{2} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right] \underbrace{\int\limits_{-\infty}^{\infty} e^{-z^2} \, \mathrm{d} z}_{\text{Gauss integral}}=\\
&= \sqrt{2} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right] \sqrt{\pi}\\
\mathcal{F} (x)&= \sqrt{2\pi} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right]\\
\end{split}
$$
It can be seen that Fourier transform equals $\scriptsize \mathcal{F} (x)= \sqrt{2\pi} {\sigma_k} A e^{ik_0x} \exp \left[ ({{\sigma_k}}^2 x^2) / 2\right]$. It is said on Wikipedia that the Gauss will be normalized only if $\scriptsize A=1 /(\sqrt{2 \pi} \sigma_k)$. I used this and got a result which corresponds with a result on Wikipedia – Fourier transform and characteristic function:
$$
\mathcal{F} (x)= e^{ik_0x} e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\
$$
If i use a centralized Gauss whose mean value is $k_0=0$ i get:
$$
\mathcal{F} (x)= e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\
$$
Which can be written as a:
$$
\mathcal{F} (x)= e^{\frac{x^2 }{2 \left(1/\sigma_k \right)^2}}\\
$$
And i can see that $1/\sigma_k = \sigma_x$ BUT from this it follows that i get the Heisenberg uncertainty principle like this:
$$
\begin{split}
\sigma_k \sigma_x &= 1\\
\Delta k \Delta x &= 1\\
\Delta p / \hbar \, \Delta x &= 1\\
\Delta p \Delta x &= \hbar\\
\end{split}
$$
And this is a wrong result because i should get $\hbar/2$ in place of $\hbar$.
Question: On our university professor derived this in a simmilar way but in the beginning in Gaussian he used $4{\sigma_k}^2$ instead of $2 {\sigma_k}^2$. This contributed to the right result $\hbar/2$ in the end. But i want to know why do we use factor $4$ instead of $2$?
Best Answer
You're treating this like a probability distribution instead of a wavefunction. Instead of assuming $2 \sigma_k^2$ vs. $4 \sigma_k^2$, I suggest setting the denominator equal to some constant and then finding the true variances in both position and wavenumber space directly--i.e. through the relation
$$\sigma_a^2 = \langle \psi | (\hat a - \bar a)^2 | \psi \rangle$$
for any observable $\hat a$ with expectation $\langle \hat a \rangle = \bar a$. Break this into two integrals and see what you get for $\hat a = \hat k$ and $\hat a = \hat x$.