I met a question asking 3-dimensional Fourier transformation of some wave functions in my QM text.

When I saw the question first, I felt it's simple question that I could solve plugging the wavefunction in the formula. But, it made me hard that 3 Cartesian coordinates are coupled in integral.

The target wavefunction is

$$ \psi(\vec{r}) = \frac{e^{- \mu r}}{r} $$

where $ \mu $ is a positive constant, and $ r=\sqrt{x^{2}+y^{2}+z^{2}} $.

I tried to get the Fourier-transformed function $\phi(\vec{k})$ in $ \vec{k} $-space by the formula as follows:

$$ \phi(\vec{k}) = \frac{1}{(2 \pi)^{3/2}}\int d^{3}r e^{-i \vec{k} \cdot \vec{r} } \frac{e^{- \mu r}}{r} $$

But, the integration is not simple for me because of the inner product. It seems that it's hard to calculate this integration in the spherical coordinate.

How can I solve it?

  • 1
    $\begingroup$ You can choose a reference frame with the z axis along the $\vec{k}$ direction, and write $\vec{k}\cdot\vec{r}=kr\cos \theta$. Now the integral is easy in spherical coordinates. $\endgroup$
    – GCLL
    Jan 5, 2017 at 0:07

1 Answer 1


Problem solved in the way suggested by @GCLL.

The given wavefunction is spatially isotropic, so is the transformed wavefucntion in $ \vec{k}$-space. WLOG, let z axis is along the $\vec{k}$, that is, $\vec{k} \cdot \vec{r} = kr \cos \theta$.

Therefore, the integration described in the spherical coordinate can be expressed as follows:

\begin{align*} \phi(\vec{k}) = \phi(k) &= \frac{1}{(2 \pi)^{3/2}} \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty}r e^{-ikr \cos \theta - \mu r}\sin \theta dr d \theta d \varphi\\ \\ &= \frac{1}{(2 \pi)^{3/2}}\int_{0}^{2 \pi} d \varphi \int_{0}^{\infty} r e^{- \mu r}\int_{0}^{\pi} e^{-ikr \cos \theta} \sin \theta d \theta dr \\\\ &= \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} r e^{-\mu r} \left[ \frac{e^{-ikr \cos \theta}}{ikr} \right]_{0}^{\pi} d r \\\\ &= \frac{1}{ik \sqrt{2 \pi}} \int_{0}^{\infty} (e^{(ik-\mu)r}-e^{-(ik+\mu)r})dr \\\\ &= \frac{1}{ik \sqrt{2 \pi}} \left[\frac{e^{(ik-\mu)r}}{ik-\mu}+\frac{e^{-(ik+\mu)r}}{ik+\mu} \right]_{0}^{\infty}\\\\ &= \frac{1}{ik\sqrt{2 \pi}}\left(\frac{1}{\mu - ik} - \frac{1}{\mu + ik}\right) \\\\ &= \sqrt{\frac{2}{\pi}} \frac{1}{\mu^{2}+k^{2}} \end{align*}



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