510.sagews

 

Enough for the excercise. Let's see how sage can help us understand it.

#defining the dalta-funcion(!) and "showing" that it converges to the distribution
delta(x2,ep) = 1/sqrt(pi*ep^2)*exp(-x2^2/ep^2)
plot(delta(x,1)); plot(delta(x,0.1)); plot(delta(x,0.01))

# Showing the normalization
assume(ep > 0)
integrate(delta(x2), x2, -infinity, +infinity)

1

%latex
% and printing it nicely
$\int _{-\infty}^{+\infty} \delta _{ep}(x_2) \mathrm{d}x_2 =  \sage{integrate(delta(x2), x2, -infinity, +infinity)}$

# Lets have a look at the function in the complex plane and the two integration paths.. The black lines add the difference between the green and the red path. The farther out they are the flatter the curve gets and the less it contributes.
y = var('y'); g = plot3d(e^(-x^2 - y^2),(-4,4),(-4,4)) + line3d([(-4,0,0.02),(4,0,0.02)],10,color='red') + line3d([(4,0,0.02),(4,2,0.02)],10,color='black') + line3d([(4,2,0.02),(-4,-2,0.02)],10,color='green')+ line3d([(-4,-2,0.02),(-4,0,0.02)],10,color='black') ; show(g)

Q.E.D.

---

And for completeness' sake the latex source code..

%latex
\textbf{Excercise}

Proove that for $f : \mathbb{R} \rightarrow \mathbb{C} \in \mathrm{L}^1(\mathbb{R})$
\begin{equation}
 f(x) = \int _{-\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \tilde f(k) \exp(+\mathrm{i}kx) \label{f}
\end{equation} with $\tilde f(k)$ as in eqn \ref{transform}. Thus one can revert the Fourier transformation.
\newline
\newline

\textbf{Hints}

 \begin{itemize}
  \item You may use $\delta_\epsilon (x) = \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi} \epsilon}
 \exp(-x^2 / \epsilon^2)$ as a valid representation of the $\delta$-distribution. 
  \item Calculate $\tilde \delta_\epsilon (x)$  and then by taking the integral over $k$-space and letting $\epsilon$ go
  $\to 0$ proove that $ \delta(x) = \int _{-\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \exp(+\mathrm{i}kx)$.
  \item Generalize to $\delta(x-x')$ and apply it to eqn \ref{f}.
 \end{itemize}
 
\textbf{Definition - Fourier Transform} 

For a function $f : \mathbb{R} \rightarrow \mathbb{C} \in \mathrm{L}^1(\mathbb{R})$ the Fourier transform is defined by
\begin{equation} \label{transform} \tilde f (k) = \int _{-\infty} ^{\infty} \mathrm{d}x f(x) \exp(-\mathrm{i}kx)
\end{equation}.

\textbf{Solution}

\begin{equation}
\tilde \delta_\epsilon (k) = \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi}\epsilon} \int _{-\infty} ^{\infty}
\mathrm{d}x \exp(-\frac{x^2}{\epsilon^2} - \mathrm{i}kx)
\end{equation}

By completing the square $\frac{1}{\epsilon ^2} \left(x^2 - \mathrm{i}kx\epsilon^2 \right) = \frac{1}{C}(x^2 - ax)$
with the usual adosubtraction of $Ca^2/4$ in the exponent one obtains

\begin{equation*}
 \tilde \delta_\epsilon (k) = \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi}\epsilon} \exp(-\frac{k^2 \epsilon^2}{4})
 \int _{-\infty} ^{\infty} \mathrm{d}x \exp\left(\frac{(x-\mathrm{i}k/2)^2}{\epsilon^2}\right).
\end{equation*}

Here one can substitute $z=\frac{x}{\epsilon} + \frac{ik\epsilon}{2}; \mathrm{dx} = | \mathrm{det} \frac{\partial
x_i}{\partial z_i} |\mathrm{d} z = \epsilon \mathrm{d} x $ and obtains 
\begin{equation*}
 \tilde \delta_\epsilon (k) = \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi}} \exp(-\frac{k^2 \epsilon^2}{4})
 \int _{\gamma_1} \mathrm{d}z \exp(-z^2).
\end{equation*}

Here $\gamma_1$ is a line in the complex plane connecting the two ``shifted'' infinities: $\gamma_1 : t/\epsilon+  
\mathrm{i}kt; t \in (-\infty,+\infty)$. Now we add another line going in the other direction on the real axis $\gamma_2
: -t/\epsilon$ and close them by perpendicular lines at infinity $\gamma_{3,4} : \pm t\epsilon \pm t_2\mathrm{i}kt ;
t_2 \in (0,1)$ to form a closed contour $\gamma = \sum_i \gamma_i$.

Using the analyticity of $\exp(-z^2)$ and the \textit{residue theorem} $\oint_\gamma f(z)\, dz =
2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}( f, a_k )$ we find that $\lim_{t \to\infty}
\int _{\gamma} \mathrm{d}z \exp(-z^2) = 0 $. Since with $\lim_{t \to\infty} \int _{\gamma_{3,4}} \exp(-z^2) = 0$
the two outer integrals vanish we get 
\begin{eqnarray*}
 \tilde \delta_\epsilon (k) & = & \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi}} \exp(-\frac{k^2 \epsilon^2}{4})
 \int _{\gamma_1} \mathrm{d}z \exp(-z^2) \\
 & = & - \lim_{\epsilon \to 0} \frac{1}{\sqrt{\pi}} \exp(-\frac{k^2
 \epsilon^2}{4}) \underbrace{\int _{\gamma_2} \mathrm{d}z \exp(-z^2)}_{-\sqrt{\pi}}  \\
 & = & \lim_{\epsilon \to 0} \exp(-\frac{k^2 \epsilon^2}{4}). 
\end{eqnarray*}

Now we take the inverse Fourier transform and find the weak limit
\begin{eqnarray*}
 \delta(x) & = & \lim_{\epsilon \to 0}  
 \int _{\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \exp(+\mathrm{i}kx) \tilde \delta_\epsilon (k) \\
 & = &   \lim_{\epsilon \to 0} \int _{\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \exp(+\mathrm{i}kx) \exp(-\frac{k^2
 \epsilon^2}{4}) \\
 & = & \int _{\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \exp(+\mathrm{i}kx) \underbrace{\lim
_{\epsilon \to 0} \exp(-\frac{k^2 \epsilon^2}{4})}_{\to 1} \\
 & = & \int_{-\infty} ^{+\infty} \frac{\mathrm{d}k}{2\pi}
 \exp(+\mathrm{i}kx).
\end{eqnarray*}

It is obvious that a linear transformation does not change the form 
\begin{equation}
\delta(x-x')= \int_{-\infty}
^{+\infty} \frac{\mathrm{d}k}{2\pi} \exp(+\mathrm{i}k(x-x')).                                                                
\end{equation}Now we can return to the initial question. Inserting eqn \ref{transform} in eqn \ref{f} we get
\begin{eqnarray*}
 f(x) & = & \int _{-\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} \int _{-\infty} ^{\infty} \mathrm{d}x' f(x')
 \exp(-\mathrm{i}kx') \exp(+\mathrm{i}kx) \\
 & = & \int _{-\infty} ^{\infty} \mathrm{d}x'  f(x') \underbrace{ \int _{-\infty} ^{\infty} \frac{\mathrm{d}k}{2\pi} 
 \exp(\mathrm{i}k(x-x'))}_{\delta(x-x')} = f(x)
\end{eqnarray*}