Fourier Transform of Gaussian *

We wish to Fourier transform the Gaussian wave packet in (momentum) k-space $A(k)=\left({2\alpha\over\pi}\right)^{1/4}e^{-\alpha (k-k_0)^2}$ to get $f(x)$ in position space. The Fourier Transform formula is

$$\bgroup\color{black}f(x)={1\over \sqrt{2\pi}}\left({2\alpha\o... ...mits_{-\infty}^\infty\; e^{-\alpha(k-k_0)^2} e^{ikx} dk.\egroup$$

Now we will transform the integral a few times to get to the standard definite integral of a Gaussian for which we know the answer. First,

$$\bgroup\color{black}k'=k-k_0\egroup$$

which does nothing really since $dk'=dk$.

$$\begin{eqnarray*} f(x)&=&\left({\alpha\over 2\pi^3}\right)^{1/4}e^{ik_0x}\int\li... ...}\int\limits_{-\infty}^\infty\; e^{-\alpha k'^2} e^{ik'x} dk'\\ \end{eqnarray*}$$

Now we want to complete the square in the exponent inside the integral. We plan a term like $e^{-\alpha k''^2}$ so we define

$$\bgroup\color{black}k''=k'-{ix\over 2\alpha}.\egroup$$

Again $dk''=dk'=dk$. Lets write out the planned exponent to see what we are missing.

$$\bgroup\color{black}-\alpha\left(k'-{ix\over 2\alpha}\right)^2=-\alpha k'^2+ik'x+{x^2\over4\alpha}\egroup$$

We need to multiply by $e^{-{x^2\over4\alpha}}$ to cancel the extra term in the completed square.

$$\bgroup\color{black}f(x)=\left({\alpha\over 2\pi^3}\right)^{1... ...alpha(k'-{ix\over2\alpha})^2} e^{-{x^2\over4\alpha}} dk'\egroup$$

That term can be pulled outside the integral since it doesn't depend on $k$.

$$\bgroup\color{black}f(x)=\left({\alpha\over 2\pi^3}\right)^{1... ...a}}\int\limits_{-\infty}^\infty\; e^{-\alpha k''^2} dk''\egroup$$

So now we have the standard Gaussian integral which just gives us $\sqrt{\pi\over\alpha}$.

$$\begin{eqnarray*} f(x)&=&\left({\alpha\over 2\pi^3}\right)^{1/4}\sqrt{\pi\over\a... ...er 2\pi\alpha}\right)^{1/4} e^{ik_0x} e^{-{x^2\over 4\alpha}}\\ \end{eqnarray*}$$

Lets check the normalization.

$$\bgroup\color{black}\int\limits_{-\infty}^\infty\; \vert f(x)... ...2\alpha}}dx =\sqrt{1\over 2\pi\alpha}\sqrt{2\alpha\pi}=1\egroup$$

Given a normalized $A(k)$, we get a normalized $f(x)$.

The RMS deviation, or standard deviation of a Gaussian can be read from the distribution.

$$\bgroup\color{black}P(x)={1\over\sqrt{2\pi\sigma^2}}e^{-{(x-X)^2\over 2\sigma^2}}\egroup$$

Squaring $f(x)$, we get

$$\bgroup\color{black}P(x)=\sqrt{1\over 2\pi\alpha} e^{-{x^2\over 2\alpha}}. \egroup$$

Reading from either the coefficient or the exponential we see that

$$\bgroup\color{black}\sigma_x=\sqrt{\alpha}\egroup$$

For the width in k-space,

$$\bgroup\color{black}P(k)= \sqrt{2\alpha\over \pi}e^{-2\alpha(k-k_0)^2}. \egroup$$

Reading from the coefficient of the exponential, we get

$$\bgroup\color{black}\sigma_k={1\over\sqrt{4\alpha}}.\egroup$$

We can see that as we vary the width in k-space, the width in x-space varies to keep the product constant.

$$\bgroup\color{black}\sigma_x\sigma_k={1\over 2}\egroup$$

Translating this into momentum, we get the limit of the Heisenberg Uncertainty Principle.

$$\bgroup\color{black}\sigma_x\sigma_p={\hbar\over 2}\egroup$$

In fact the Uncertainty Principle states that

$$\bgroup\color{black}\sigma_x\sigma_p\geq{\hbar\over 2}\egroup$$

so the Gaussian wave packets seem to saturate the bound!