So I have the 2p state Hydrogen wavefunction, $$\psi(r)=Nz\exp(-\frac r{2a})$$, where $N$ is the normalization constant, and $a$, is the Bohr radius. Now I want to calculate the Fourier Transform of the above wavefunction.
I have done it to the $r$ integral. It would be nice if someone, would could post a more elegant solution. Also please check through my calculation.
My work:
Basically I have to compute, $$\frac N{(2\pi)^\frac32}\int z\exp(-\frac r{2a})\exp(i\vec k \cdot \vec r)d^3r$$
I have plugged in $z=r\cos\theta$. (Is it right? I think so, because in spherical polar coordinates, $z=r\cos\theta$). Also, I have aligned $\vec k$ along the $z$ axis. And also for the moment, I am forgetting the normalizations, I will tack them in, in the end. So,
$$\iiint r\cos\theta\exp(\frac r{2a})\exp(ikr\cos\theta)r^2\sin\theta d\theta d\phi dr$$
$$=2\pi\int_{r=0}^\infty r^3\exp(-\frac r{2a}) \left(\int_{\alpha=-1}^1 \alpha\exp(ikr\alpha) d\alpha\right) dr$$, where $\alpha=\cos\theta$, and can manipulate the integral, by plugging in this. Now taking by parts of the second integral we have,
$$=2\pi\int_{r=0}^\infty r^3\exp(-\frac r{2a}) \left( \frac{\exp(ikr)+\exp(-ikr)}{ikr} – \frac{\exp(ikr)-\exp(-ikr)}{(ikr)^2} \right)dr$$
Now from here I can individually take the terms and integrate it, and get the answer! (Too much work!). Isn't there a simpler way to achieve this?
Please go through my work, before making any suggestions.
Best Answer
The transform is
$$\begin{align}\Psi(k) &=\frac{N}{\sqrt{2 \pi}}\int_0^{\infty} dr \, r^3 \, e^{-r/(2 a)} \int_0^{\pi} d\theta \, \sin{\theta} \cos{\theta} \, e^{i k r \cos{\theta}} \\ &= \frac{N}{\sqrt{2 \pi}}\int_0^{\infty} dr \, r^3 \, e^{-r/(2 a)} \int_{\pi}^0 d(\cos{\theta}) \cos{\theta} e^{i k r \cos{\theta}} \\ &= \frac{N}{\sqrt{2 \pi}}\int_0^{\infty} dr \, r^3 \, e^{-r/(2 a)} \left [-i \frac{\partial}{\partial (k r)} \int_{-1}^{1} du \, e^{i k r u} \right ] \\&= N \sqrt{\frac{2}{\pi}} \int_0^{\infty} dr \, r^3 \, e^{-r/(2 a)} \frac{\partial}{\partial (k r)} \frac{\sin{k r}}{k r} \\ &= N \sqrt{\frac{2}{\pi}} \int_0^{\infty} dr \, r^3 \, e^{-r/(2 a)} \left [\frac{\cos{k r}}{k r} - \frac{\sin{k r}}{(k r)^2} \right ]\end{align}$$
So we are at the same place. Now we can move forward - it is not that bad. Consider, for $\operatorname{Re}{z} \gt 0$,
$$\int_0^{\infty} dr \, r e^{-x r} = \frac1{z^2}$$
$$\int_0^{\infty} dr \, r^2 e^{-x r} = \frac{2}{z^3}$$
Here, $z=1/(2 a) - i k$. Thus, using $\cos{k r} = \operatorname{Re}{e^{i k r}}$, etc.,
$$\begin{align}\Psi(k) &= N \sqrt{\frac{2}{\pi}} \frac{2}{k} \operatorname{Re}{\left [\frac1{\left (\frac1{2 a} - i k\right )^3} \right ]} - N \sqrt{\frac{2}{\pi}} \frac{1}{k^2} \operatorname{Im}{\left [\frac1{\left (\frac1{2 a} - i k\right )^2} \right ]} \\ &= N \sqrt{\frac{2}{\pi}} \frac{2}{k} \frac{\frac1{8 a^3} - \frac{3 k^2}{2 a}}{\left (\frac1{4 a^2} + k^2 \right )^3} - N \sqrt{\frac{2}{\pi}} \frac{1}{k^2} \frac{\frac{k}{a}}{\left (\frac1{4 a^2} + k^2 \right )^2} \\ &= -N \sqrt{\frac{2}{\pi}} 256 a^4 \frac{k a}{\left (1 + 4 k^2 a^2 \right )^3} \end{align}$$
I have the feeling you would normalize this using the above wave function, i.e., the FT of the square of the wave function at $k=0$ or something similar.