I have a function $f(r,\theta,\phi)$ which I am expressing in terms of spherical harmonics

$$ f(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} g_{l,m}(r) d_{l,m}(\theta,\phi) $$

where $d_{l,m}$ are real spherical harmonics (RSH - formula below taken from Wikipedia).

\begin{align} d_{\ell m} &= \begin{cases} \displaystyle {i \over \sqrt{2}} \left(Y_\ell^{m} - (-1)^m\, Y_\ell^{-m}\right) & \text{if}\ m<0\\ \displaystyle Y_\ell^0 & \text{if}\ m=0\\ \displaystyle {1 \over \sqrt{2}} \left(Y_\ell^{-m} + (-1)^m\, Y_\ell^{m}\right) & \text{if}\ m>0. \end{cases}\\ &= \begin{cases} \displaystyle {i \over \sqrt{2}} \left(Y_\ell^{-|m|} - (-1)^{m}\, Y_\ell^{|m|}\right) & \text{if}\ m<0\\ \displaystyle Y_\ell^0 & \text{if}\ m=0\\ \displaystyle {1 \over \sqrt{2}} \left(Y_\ell^{-|m|} + (-1)^{m}\, Y_\ell^{|m|}\right) & \text{if}\ m>0. \end{cases} \end{align}

If I would like to take the Fourier Transform (FT) of $f(r,\theta,\phi)$, is there a special relationship between $d_{l,m}$ and the FT? I feel like there is some nice property like

$$ FT(f(r,\theta,\phi)) = \sum \sum FT(g_{l,m}(r)) d_{l,m}(\theta, \phi) $$

but I can't remember it or find it. Is there a nice relationship between the FT and RSH which will make the FT of $f(r,\theta,\phi)$ easy to compute when expanded in RSH?

  • $\begingroup$ presumably, the coefficient functions $f(r)$ are intended to be different from $f(r,\theta,\phi)$? $\endgroup$
    – Bill Barth
    Apr 9, 2014 at 3:16
  • $\begingroup$ Yes. I have edited the one dimensional function to be $f_{lm}$ $\endgroup$
    – drjrm3
    Apr 9, 2014 at 3:19
  • $\begingroup$ That doesn't really help since it's the repeated use of $f$ on both sides of the equals sign that's confusing! $\endgroup$
    – Bill Barth
    Apr 9, 2014 at 3:21
  • $\begingroup$ Ok, I will change $f_{lm}$ to $g_{lm}$ to make that distinction more clear. They cannot be the same thing since $f(r,\theta,\phi)$ is defined on $R_3 \rightarrow R$ and $f_{lm}(r)$ is defined on $R \rightarrow R$ $\endgroup$
    – drjrm3
    Apr 9, 2014 at 3:49
  • 1
    $\begingroup$ No, I would like to the FT to be performed in all 3 dimensions. I am just wondering if there is a nice trick which the SHs can help with. $\endgroup$
    – drjrm3
    Apr 9, 2014 at 12:02

2 Answers 2


You may find the expansion of a plane wave in spherical waves to be helpful here:

$$ e^{i\mathbf{k}\cdot\mathbf{x}} = 4\pi\sum_{l=0}^\infty\sum_{m=-l}^l i^lj_l(kr)Y_{lm}(\theta,\phi)Y_{lm}^*(\vartheta,\varphi) $$

where $\theta$, $\phi$ are the angular variables for $\mathbf{x}$ and $\vartheta$, $\varphi$ for $\mathbf{k}$; the radial functions $j_l$ are the spherical Bessel functions; and $Y_{lm}$ are the spherical harmonics*. In that case,

$$ \mathscr{F}f(\mathbf{k}) = (2\pi)^{-3/2}\int d^3x\hspace{2pt} e^{i\mathbf{k}\cdot\mathbf{x}}f(\mathbf{x}) $$ $$ = (2\pi)^{-3/2}\int d^3x\left(4\pi\sum_{l,m}i^lj_l(kr)Y_{lm}(\theta,\phi)Y_{lm}^*(\vartheta,\varphi)\right)\left(\sum_{l',m'}g_{l'm'}(r)Y_{l'm'}(\theta,\phi)\right) $$

Then you can use the orthogonality of spherical harmonics to reduce this to

$$ \mathscr{F}f(k,\vartheta,\varphi) = \sqrt{\frac{2}{\pi}}\sum_{l,m}i^lY_{lm}^*(\vartheta,\varphi) \cdot \int j_l(kr)g_{lm}(r)r^2dr. $$

You now have an integral in only the radial dimension. If you have a book of special functions handy, some recurrence relations or asymptotics for $j_l$ will prove invaluable no matter how you choose to evaluate those integrals.


*Every time you use this formula, a kitten is born, a child smiles, and your lifespan increases by 8 minutes.

  • $\begingroup$ This seems to be exactly what I am searching for. I have edited my original question with a specific definition for the Real Spherical Harmonics I am using - does your formula work for RSH? In which case, will $Y_{lm}^*$ simply be $d_{lm}^*$? $\endgroup$
    – drjrm3
    Apr 17, 2014 at 16:45
  • $\begingroup$ You can write each $Y_{lm}$ in terms of $d_{lm}$ by inverting the formula you cited. However, note that the Fourier transform of a real function is not necessarily real. Instead, it has the symmetry property $\mathscr{F}f(\mathbf{k})^* = \mathscr{F}f(-\mathbf{k})$, which means some harmonics are zero in the expansion of the Fourier transform. $\endgroup$ Apr 17, 2014 at 17:33
  • 2
    $\begingroup$ if anyone else is interested, the Addition Theorem which is used states that it is true for both real and complex harmonics, so $d_lm$ can be interchanged with $Y_lm$ without a problem. $\endgroup$
    – drjrm3
    Apr 17, 2014 at 20:28

The FT of a three dimensional real function is not in general real, so there is no way you could express it in real spherical harmonics. I think however that if you only include harmonics with even $l$ the result should be real.

The FT of a spherically symmetric function $g(r)$ is $S(Q) = \int_{0}^{\infty} \frac{r}{Q} \sin(Qr) g(r)\,\mathrm{d}r$ (with some factors of $2\pi$ depending on how you define the FT).

A 3D function decomposed into spherical harmonics is a sum of products $g_{lm}(r) d_{lm}(\theta,\phi)$, so the FT will be a sum of convolutions $S(Q) \otimes \mathrm{FT} \left[ d_{lm}(\theta,\phi) \right]$. I can't think of an easy way to calculate these in general.


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