Question:
In quantum theory, the wave function can be convertible either in momentum space or in configuration space. Suppose 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.
\[ \Phi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dp \frac{N}{p^2+\alpha^2}e^{ipx/\hbar} \]
\[ = \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{0}_{-\infty} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \]
\[ = \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{\infty}_{0} \frac{e^{-ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \]
\[ = \frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{e^{ipx/\hbar}+e^{-ipx/\hbar}}{p^2+\alpha^2}dp \]
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, \(p=k\hbar\).
\[ \Phi(x) = \frac{2}{\hbar}\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \]
\[ = \sqrt{\frac{2N^2}{\pi \hbar^3}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \]
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, \(I\).
\[ A \int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk = \frac{A}{2}\int^{\infty}_{-\infty} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \]
\[ = \frac{A}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
Therefore, we finally have the wave function in configuration space:
\[ \Phi(x) = \sqrt{\frac{2N^2}{\pi \hbar^3}}\frac{1}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
\[ = \frac{N}{\alpha}\sqrt{\frac{\pi}{2\hbar}}e^{-\frac{\alpha}{\hbar}x} \]