Momentum space to configuration space

Question:
In quantum theory, the wave function can be convertible either in momentum space or configuration space. We have a wave function in momentum space: $\phi(p) = N/(p+\alpha^2)$. Find the equivalent wave function in configuration space.

Answer:
We can use the Fourier transformation to obtain the wave function in configuration space.
\begin{eqnarray*}
\Phi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dp \frac{N}{p^2+\alpha^2}e^{ipx/\hbar}  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{0}_{-\infty} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right)  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{\infty}_{0} \frac{e^{-ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right)  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{e^{ipx/\hbar}+e^{-ipx/\hbar}}{p^2+\alpha^2}dp
\end{eqnarray*}
Use the following formula.
\[ \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \]
We can rewrite it as
\[ \Phi(x) = 2\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos(px/\hbar)}{p^2+\alpha^2}dp \]
Replace the momentum with the wavenumber; namely, $p=k\hbar$.
\begin{eqnarray*}
\Phi(x) &=& \frac{2}{\hbar}\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk    \\
   &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk
\end{eqnarray*}
Let us put $\sqrt{\frac{2N^2}{\pi \hbar^3}}$ as $A$. Now, integrate the above in the complex plane. Consider the path of half circle in the upper half plane. The integral becomes
\[ I = A\oint \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk
     = A\left( \int^r_{-r} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk + \int_{C_r}\frac{e^{izx}}{z^2+\alpha^2/\hbar^2}dz \right) \]
where $z$ is the complex variable of $k$.
The second term will become zero because $1/(z^2+\alpha^2/\hbar^2) = 0$ as $|z|\rightarrow \infty$. Due to Jordan's lemma, when $r \rightarrow \infty$, the integral becomes zero. Namely, only the first term survives.
\[ I = A \int^{\infty}_{-\infty} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk \]
The pole inside the contour is $i\frac{\alpha}{\hbar}$. Using the residue theorem, we obtain
\[ \mathrm{Res} [f, i\alpha/\hbar] = \lim_{z\rightarrow i\frac{\alpha}{\hbar}}\frac{e^{izx}}{z+i\frac{\alpha}{\hbar}} =  \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\]
Thus
\[ I = A \left(2\pi i \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\right) = A \frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
Take the real part of the integral, $I$.
\begin{eqnarray*}
A \int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk &=& \frac{A}{2}\int^{\infty}_{-\infty} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk     \\
   &=& \frac{A}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar}
\end{eqnarray*}
Therefore, we finally have the wave function in configuration space:
\begin{eqnarray*}
\Phi(x) &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\frac{1}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \\
       &=& \frac{N}{\alpha}\sqrt{\frac{\pi}{2\hbar}}e^{-\frac{\alpha}{\hbar}x}
\end{eqnarray*}

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