Does anybody know about an accurate quadrature rule over three Euler angles $\theta, \phi, \chi$? I am trying to calculate the average value of an arbitrary function $f(\theta, \phi, \chi)$ for a given probability distribution $\rho(\theta, \phi, \chi)$:

$$ \langle f\rangle_{\rho} = \int_0^{2\pi} d\chi \int_0^{2\pi} d\phi\int_0^{\pi} \sin\theta d\theta f(\theta, \phi, \chi)\rho(\theta, \phi, \chi) \approx \sum_{l} w_l g(\theta_l, \phi_l, \chi_l) $$ where $f(\theta, \phi, \chi)$ is not sum-of-products of single angle-coordinate functions. $w_l$ are quadrature weights and $(\theta_l, \phi_l, \chi_l)$ represents a single grid point. $g(\theta, \phi, \chi) = \sin\theta f(\theta, \phi, \chi)\rho(\theta, \phi, \chi)$ and an appropriate quadrature weight-function is implicitly included in $f$.

Such an integration is sometimes needed in orientation averaged calculations of material/molecule's properties.

What we can assume about $\rho(\theta, \phi, \chi)$ is that it is a linear combination of the products of the Wigner D-matrices: $\rho(\theta, \phi, \chi) = \sum_{K,K'} c_{KK'} D^{(J)}_{KM}(\theta, \phi, \chi)D^{(J')}_{K'M'}(\theta, \phi, \chi)$.

So to sum up. It seems there is a need for a simple quadrature scheme for integration over three Euler angles: $\theta, \phi, \chi$.

  • Update:

Ways around the need for explicit three-Euler angle $\theta, \phi, \chi$ quadrature:

  1. Expanding $f(\theta, \phi, \chi)$ in the Wigner matrices basis is one way of proceeding here. The appropriate integrals can be calculated analytically. But for some functions $f(\theta, \phi, \chi)$ this method is very inefficient, as the Wigner expansion has many terms.

  2. Another possibility, if we are dealing with a quantum system, is to solve the Schroedinger equation for wavefunctions in three separate system-fixed embeddings of the coordinate frame. Then one can choose the system-fixed z-axis to be the axis in the system that we want to quantify (average over). In such a case only $\theta, \phi$ are needed. The down-side to this approach is that one needs to repeat calculations for three independent embeddings, which for some embeddings and some external potentials can be very unnatural. In the case of rotational dynamics problems the complete symmetric-top basis sets guarantee accuracy of the solutions regardless of the embedding chosen. Rotational wavefunction representations among different embeddings differ only by the coefficients. In good embeddings the rotational wavefunction can be reprsented compactly in the basis, but if the embedding is poor, often many basis functions are needed. In the case of problems which add coupling of rotational degrees of freedom to some other internal or external degrees of freedom, the choice of embedding is often critical for quick convergence of the variational procedure.

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    $\begingroup$ +1. Nice first question Emil! I have also advertised it here: chat.stackexchange.com/rooms/1878/computational-science. You can draw more attention to it by hovering your mouse on the question there, and then pressing the star. $\endgroup$ Jul 14, 2020 at 18:45
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    $\begingroup$ Do you really need three angles? If you have a system, then one axis is already fixed, and you're left with quadrature on a sphere which is a well-known problem that has several accurate rules. $\endgroup$ Jul 15, 2020 at 9:47
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    $\begingroup$ Thanks, however I'm not sure what you mean by one axis being fixed. To determine the orientation of a body which is not linear three angles are required. The $f$ function is often represented through rational functions of the elements of the rotation matrix (en.wikipedia.org/wiki/Euler_angles#Rotation_matrix) - it is a function of three Euler angles. For a linear rotating body Lebedev quadratures are good. I'm searching for a generalization to non-linear rotating bodies. $\endgroup$
    – Emil Zak
    Jul 15, 2020 at 20:15
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    $\begingroup$ @EmilZak I mean that if you have an input orientation, then one axis is already fixed implicitly. You only need two angles to compute the rotational average..? $\endgroup$ Jul 17, 2020 at 10:27
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    $\begingroup$ In case integration of the three Euler angles is really what you need, I haven't done it myself but if I am stuck at this for too long, I will try the inner 2D integral over $(\theta, \phi)$ first with the available quadrature-on-sphere methods like Lebedev quadrature as you mentioned, and then do the integral over $\chi$ with the usual one-variable quadrature. The potential downside of this technique is that we probably can't guarantee that the function of $\chi$ resulting from the $(\theta,\phi)$ integration is well-behaving enough for the subsequent quadrature to work with few points. $\endgroup$
    – nougako
    Aug 6, 2020 at 15:40

1 Answer 1


I don't know of any exact methods, but quadpy is a really good place to start, and possibly ask the related question there.

Not really an answer, but more of a hint.


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