\\
 +{\rm e}^{+\tfrac{l}{3}\zeta^2\Omega^2u}\,8l\Bigl[2u[1+\cos(2\varphi)\cos(2\beta)]-(1+u^2) \sin(2\varphi)\sin(2\beta)\Bigr]\biggr\}, \label{eq:rhop1}\end{gathered}$$ and $$\begin{gathered}
 h_{(1)}^{3\Omega}(u,\varphi)=\frac{\alpha}{874800\pi}\,3(3\Omega\zeta)^2\,{\rm e}^{-\frac{\zeta^2\Omega^2+(\frac{w}{2})^2(3\Omega)^2}{6}}\,{\rm e}^{-\frac{[\zeta^2-(\frac{w}{2})^2](3\Omega)^2}{6}u^2}  \\
 \times
\biggl\{(1-u^2)\bigl[\cos(2\beta) + \cos(2\varphi)\bigr]\\
+\frac{1}{2}\sum_{l=\pm1}{\rm e}^{+l\zeta^2\Omega^2u}(1-ul)^2[1+\cos(2\varphi-2l\beta)] \biggr\}. \label{eq:rhop2}\end{gathered}$$ In [Eq. ]{} we have pulled out an overall factor, such that, apart from $u$ and $\varphi$, the functions $h_{(1)}^{\omega}(u,\varphi)$ only depend on the dimensionless combinations $\xi^2\Omega^2$ and $(\frac{w}{2})^2\Omega^2$. The results for $p=2$ again follow by shifting the angle $\beta\to\beta+\frac{\pi}{2}$, i.e., $\rho_{(2)}^{\omega}=\rho_{(1)}^{\omega}\big|_{\beta\to\beta+\frac{\pi}{2}}$.

As $\cos(\chi\pm\pi)=-\cos\chi$ and $\sin(\chi\pm\pi)=-\sin\chi$, the [*total number densities*]{} $\rho^{\,\omega}(u,\varphi)=\sum_{p=1}^2\rho^{\,\omega}_{(p)}(u,\varphi)$ of photons of frequency $\omega$ obtained in a polarization insensitive measurement obviously become independent of $\beta$, i.e., independent of the specific polarization basis used, as they should: The resulting expressions are effectively obtained by multiplying Eqs.  and with a factor of two and setting all trigonometric functions involving $\beta$ in their arguments to zero. They read $$\rho^{\omega}(u,\varphi)=\left(\frac{e{\cal E}}{m^2}\right)^6 \left(\frac{w}{2}m\right)^4\frac{\Omega \tau}{2\pi}\, h^{\omega}(u,\varphi)\,, \label{eq:rhosum}$$ with $$\begin{gathered}
 h^{\Omega}(u,\varphi)=\frac{\alpha}{874800\pi}(\Omega\zeta)^2\,{\rm e}^{-\frac{[\zeta^2+(\frac{w}{2})^2]\Omega^2}{6}}\,{\rm e}^{-\frac{[\zeta^2-(\frac{w}{2})^2]\Omega^2}{6}u^2}  \\
 \times\sum_{l=\pm1}\biggl\{
 \sum_{j=0,2} c_j\,{\rm e}^{+\tfrac{lj}{3}\zeta^2\Omega^2u}(1-u^2)\cos(2\varphi)
 +\sum_{j=1,3}c_j\,{\rm e}^{+\tfrac{lj}{3}\zeta^2\Omega^2u}(1-ul)^2  \\
 +16\,{\rm e}^{+\tfrac{l}{3}\zeta^2\Omega^2u}\,l u\biggr\}, \label{eq:rhosum1}\end{gathered}$$ and $$\begin{gathered}
 h^{3\Omega}(u,\varphi)=\frac{\alpha}{874800\pi}\,3(3\Omega\zeta)^2\,{\rm e}^{-\frac{\zeta^2\Omega^2+(\frac{w}{2})^2(3\Omega)^2}{6}}\,{\rm e}^{-\frac{[\zeta^2-(\frac{w}{2})^2](3\Omega)^2}{6}u^2}  \\
 \times
\biggl\{2(1-u^2)\cos(2\varphi)+\sum_{l=\pm1}{\rm e}^{+l\zeta^2\Omega^2u}(1-ul)^2 \biggr\}. \label{eq:rhosum2}\end{gathered}$$

The number of photons emitted in a given solid angle interval characterized by $u_2\leq u\leq u_1$ and $\varphi_1\leq\varphi\leq\varphi_2$ is obtained by integration of [Eq. ]{} or , respectively. Note that $\int_{\vartheta_1}^{\vartheta_2}{\rm d}\vartheta\,\sin\vartheta=\int_{u_2}^{u_1}{\rm d}u$, with $u_i=\cos\vartheta_i$ and $i\in\{1,2\}$.

Hence, the total 