I don't know how to evaluate an integral of the form

$$\int d^3 r \exp(-i \vec r\cdot\vec q)\exp(-a^2 r^2)$$ where $a\in \mathbb R$.

Could anyone please teach me how to do this integral?

Many thanks.


Note that this integral is separable. Assuming $\vec{r}=(r_1,r_2,r_3)$ and by $r^2$ you mean $\vec{r} \cdot \vec{r}=r_1^2+r_2^2+r_3^2$ and $\vec{q}=(q_1,q_2,q_3)$ the integral can be reformulated as:

\begin{equation} \int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3 \end{equation}

This equals to the product of evaluating each integral separately: \begin{equation} \int{e^{-ir_1q_1-a^2r_1^2}}dr_1\int{e^{-ir_2q_2-a^2r_2^2}}dr_2\int{e^{-ir_3q_3-a^2r_3^2}}dr_3 \end{equation}

Hence, you got three Fourier transform of a Gaussian.

A Fourier transform of each Gaussian is given by: \begin{equation} \int_{-\infty}^\infty{e^{-irq-a^2r^2}}dr=\sqrt{\frac{\pi}{a}}e^{-\frac{q^2}{4a}} \end{equation}

So your integral all in all is equal to a product of those three Gaussians in $(q_1,q_2,q_3)$. Assuming I didn't make any mistake along the way:

\begin{equation} \int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3=(\frac{\pi}{a})^{1.5}e^{-\frac{q_1^2+q_2^2+q_3^2}{4a}} \end{equation}

  • $\begingroup$ Many thanks, Gil $\endgroup$
    – Glen
    Feb 18 '13 at 14:53
  • $\begingroup$ There is a typo in the formula for the Fourier transform of the one-dimensional Gaussian. All $a$ in the right hand side should be squared. $\endgroup$ Apr 8 '15 at 11:01

Use spherical coordinates: $x = r \sin{\theta} \cos{\phi}$, $y = r \sin{\theta} \sin{\phi}$, $z = r \cos{\theta}$, and

$$\mathrm{d^3}\vec{r} = r^2 \sin{\theta} \, dr \, d \theta \, d \phi $$

Assume a plane wave expansion of the form

$$e^{-i \vec r\cdot\vec q} = e^{-i q r \cos{\theta}} = \sum_{l=0}^\infty\mathrm i^l(2l+1)j_l(q r)P_l(\cos\theta)$$

where $j_0$ is a spherical Bessel function and $P_l$ is a Legendre polynomial.

Note that, as the integral is over a spherically symmetric function, the only term in the sum that does not vanish is the $l=0$ term. Integrating over $\theta$ and $\phi$ produces a factor of $4 \pi$. We then have

$$\begin{align}\int_{\mathbb{R}^3} d^3 r \exp(-i \vec r\cdot\vec q)\exp(-a^2 r^2) &= 4 \pi \int_0^{\infty} dr \: r^2 e^{-a^2 r^2} j_0(q r) \\ &= \frac{4 \pi}{q} \int_0^{\infty} dr \: r e^{-a^2 r^2} \sin{q r} \\ &= \frac{\sqrt{\pi^3}}{a^3} e^{-\frac{q^2}{4 a^2}} \end{align}$$

  • $\begingroup$ Many thanks, rigordonma $\endgroup$
    – Glen
    Feb 18 '13 at 14:56
  • $\begingroup$ There is a typo in the last formula. All occurrences of $a$ in the right hand side should be squared. Indeed, both $q$ and $a$ have the dimension of the inverse of a length. $\endgroup$ Apr 8 '15 at 11:04
  • $\begingroup$ @GiuseppeNegro: you are correct. Thanks. $\endgroup$
    – Ron Gordon
    Apr 8 '15 at 11:17
  • $\begingroup$ You are welcome! I have a couple of questions on your interesting approach, if you don't mind. The first (and most important) is: how do you compute that expansion of the plane wave into spherical harmonics? $\endgroup$ Apr 8 '15 at 13:04
  • 1
    $\begingroup$ @GiuseppeNegro: this is a well-known expansion, believe it or not. You may find a quick derivation here: arxiv.org/pdf/0909.0494v4.pdf . Really, the idea is to expand an arbitrary plane wave in terms of spherical harmonics, and then use symmetry to sum over the azimuthal coordinate of the spherical harmonics to produce the Legendre. (Remember that the spherical harmonics are associated Legendres, more or less.) $\endgroup$
    – Ron Gordon
    Apr 8 '15 at 15:21

The Fourier Transform of each Gaussian should be(referring to http://mathworld.wolfram.com/FourierTransformGaussian.html) $$ \int e^{−irq}e^{−a^2r^2} dr = \sqrt{\frac{\pi}{a^2}}e^{-\frac{q^2}{4a^2}} $$

The 3d Fourier Transform is

$$ (\frac{\pi}{a^2})^{3/2}e^{-\frac{q_1^2+q_2^2+q_3^2}{4a^2}} $$

  • 1
    $\begingroup$ Sorry, but how is your answer correct when it is dimensionally wrong? The answer should have units of length^3, but yours has units of length. $\endgroup$
    – Ron Gordon
    Apr 7 '15 at 6:08

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