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The transition rate corresponding to the first-order probability of absorption is given in the dipole approximation as $$W_{ba}=\frac{dP_{ba}^{(1)}}{dt}=\frac{\pi I(\omega_{ba})}{\hbar^2 c \varepsilon_0}\left|\hat{\epsilon}\cdot \bf{D}_{ba}\right|^2=\frac{\pi I(\omega_{ba})}{\hbar^2 c \varepsilon_0}\cos^2\theta\left|\bf{D}_{ba}\right|^2$$

where $\hat{\epsilon}$ is the direction of polarisation of the incident radiation. (I'm only mentioning those quantities which I think are relevant to this question.)

If the incident radiation is unpolarised and isotropic, the orientation of the polarisation vector $\hat{\epsilon}$ is at random, in which case $\cos^2\theta$ can be replaced by its average value.

Now my question is why $\langle\cos^2\theta\rangle=\frac{1}{3}$ here and not $\frac{1}{2}?$
Does polarisation has anything to do with it?

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For uniformly random orientations of the polarisation vector over a whole sphere enclosing the dipole, the average of $\cos^2\theta$ is given, in the usual spherical polar coordinates, by $$<\cos^2\theta> =\frac{\int^{2\pi}_0 \int ^{\pi}_0 \cos^2\theta\ \sin\theta\ d\theta\ d\phi}{\int^{2\pi}_0 \int ^{\pi}_0 \sin\theta\ d\theta\ d\phi} = \frac{1}{3}\ ,$$ where the solid angle element $d\Omega = \sin \theta\ d\theta\ d\phi.$

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Averaging is done over all possible spatial orientations of the dipole moment matrix element vector with equal probability. So the function $\cos^2\theta$ is integrated over whole unit sphere, and then the result is divided by surface area of that sphere ($4\pi$), and the resulting average is $\frac{1}{3}$.

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  • $\begingroup$ It would be helpful if you could show some steps. The "over all possible spatial orientations of the dipole moment matrix element vector" is really puzzling me. $\endgroup$ Jan 31, 2023 at 16:01
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    $\begingroup$ $\mathbf D_{ab}$ is a vector, it has three components. Depending on the state of the atom/molecule, it can have any orientation in space. The task is to average the factor $\cos^2 \theta$ over all directions of this vector in space. $\endgroup$ Jan 31, 2023 at 16:06

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