3D Fourier Transform for 2p state Hydrogen Atom

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align}&\color{#66f}{\large% \int_{{\mathbb R}^{3}} z\exp\pars{\ic\vec{k}\cdot\vec{r}}\exp\pars{-\,{r \over 2a}}\,\dd^{3}\vec{r}} =-\ic\,\partiald{}{k_{z}}\int_{{\mathbb R}^{3}} \exp\pars{\ic\vec{k}\cdot\vec{r}}\exp\pars{-\,{r \over 2a}}\,\dd^{3}\vec{r} \\[5mm]&=-\ic\,\partiald{}{k_{z}}\int_{0}^{\infty} \exp\pars{-\,{r \over 2a}}\ \overbrace{% \bracks{\int\exp\pars{\ic\vec{k}\cdot\vec{r}}\,{\dd\Omega_{\vec{r}} \over 4\pi}}} ^{\dsc{\sin\pars{kr} \over kr}} 4\pi r^{2}\,\dd r \\[5mm]&=-4\pi\ic\,\partiald{}{k_{z}}\int_{0}^{\infty} \exp\pars{-\,{r \over 2a}}\,{\sin\pars{kr} \over kr}\,r^{2}\,\dd r =-4\pi\ic\,{k_{z} \over k}\,\partiald{}{k}\ \overbrace{\int_{0}^{\infty} \exp\pars{-\,{r \over 2a}}\,{\sin\pars{kr} \over kr}\,r^{2}\,\dd r} ^{\ds{\dsc{kr}=\dsc{\xi}\ \imp\ \dsc{r}=\dsc{\xi \over k}}} \\[5mm]&=-4\pi\ic\,{k_{z} \over k}\,\partiald{}{k}\braces{{1 \over k^{3}}\ \overbrace{\int_{0}^{\infty} \exp\pars{-\,{\xi \over 2ka}}\,\sin\pars{\xi}\xi\,\dd\xi} ^{\dsc{16\pars{ka}^{3} \over \bracks{1 + 4\pars{ka}^{2}}^{2}}}} \\[5mm]&=-4\pi\ic\,{k_{z} \over k}\,a^{4}\partiald{}{\mu}\bracks{{1 \over \mu^{3}}\ {16\mu^{3} \over \pars{1 + 4\mu^{2}}^{2}}}_{\mu\ =\ ka} \\[5mm]&=\color{#66f}{\large-4\pi\ic\,{k_{z} \over k}\braces{% {48\pars{ka}^{2} \over \bracks{4\pars{ka}^{2} + 1}^{2}} -{256\pars{ka}^{4} \over \bracks{4\pars{ka}^{2} + 1}^{3}}}a^{4}} \end{align}