# Experimental Study on Partially Coherent Optical Coherent Detection
## Theoretical Analysis
### Polarization States and Mixing Efficiency
The signal light's phase modulation is represented by \(\phi(t)\) and the local oscillator light's phase is represented by \(\varphi(t)\). Following the transmission of the local oscillator light and the linearly polarized signal light to the optical mixer, the output of the orthogonal branch photocurrent, for loop control, and the DC branch photocurrent, for restoring modulation information, can be expressed as follows:  
\[I_{I}=2k_{2}k_{3}|E_{s}E_{LO}|\cos\phi(t) \tag{25}\]  
\[I_{Q}=2k_{1}k_{4}|E_{s}E_{LO}|\cos\left(\phi(t)+2(\tau_{\parallel}-\rho_{ \perp})+\frac{\pi}{2}\right) \tag{26}\]  
It is evident from the photocurrents of the \(I\) and \(Q\) channels that the polarization components of the signal light and local oscillator light, denoted as \(k_{1}\), \(k_{2}\), \(k_{3}\), and \(k_{4}\), will influence the output photocurrent's intensity, thereby impacting the output signal's amplitude and the power output of the mixer, altering the mixing efficiency and the signal light's energy distribution in the process. Equations (25) and (26), which simplify to yield two input optical signals, can be simplified by substituting the polarization components \(k_{1}\), \(k_{2}\), \(k_{3}\), and \(k_{4}\) with \(\frac{\sqrt{2}}{2}\), in an ideal scenario (where the angle \(\theta\) between the fast axis of the wave plate is 45\({}^{\circ}\)). \[I_{I}=\sqrt{2}\sin\theta|E_{s}E_{LO}|\cos\phi(t) \tag{27}\] \[I_{Q}=\sqrt{2}\cos\theta|E_{s}E_{LO}|\cos\Bigl{(}\phi(t)+\frac{ \pi}{2}\Bigr{)} \tag{28}\]  
where \(\theta\) is the angle formed between the horizontal direction and the light vector. The angle \(q\) only modifies the light's energy distribution, according to the analysis in the preceding text. Thus, efficient mixing is possible when there is an ideal phase difference, i.e., when the \(I\) and \(Q\) phases stay synchronized. Output power is taken into consideration when examining the impact of the angle \(q\) between the horizontal direction and the optical vector on mixing efficiency. It is possible to express the mixing efficiency as the ratio between the output signal's orthogonal and in-phase component amplitudes. By assuming that the input signal's in-phase and orthogonal components are represented as \(I_{lin}\) and \(I_{Qin}\), respectively, and the output signal's in-phase and orthogonal components are represented as \(I_{lout}\) and \(I_{Qout}\), respectively, the mixing efficiency can be calculated as follows:  
\[\eta=\frac{I_{lout}^{2}+I_{Qout}^{2}}{I_{lin}^{2}+I_{Qin}^{2}} \tag{29}\]  
Formula (29) can be simplified by substituting Formulas (27) and (28) into it. \[\eta(\theta)=\frac{\left(\sqrt{2}\sin\theta+\sqrt{2}\cos\theta\right)^{2}}{4} =\frac{\left|1+\sin 2\theta\right|}{2} \tag{30}\]  
The spatial optical mixer's mixing efficiency can be calculated using the mathematical model formula [19]. \[\eta_{\mathrm{c}}=\frac{\left[\int_{0}^{t_{0}}E_{s}(\mathrm{r})E_{LO}(\mathrm{ r})\exp(i\Delta\varphi)\mathrm{rdr}\right]^{2}\xrightarrow[\text{and phase matching.}]{\text{Energy change.}}}{\int_{0}^{t_{0}}\left[E_{LO}(\mathrm{r})\right]^{2} \mathrm{rdr}\int_{0}^{t_{0}}\left[E_{s}(\mathrm{r})\right]^{2}\mathrm{rdr}} \tag{31}\]  
The radius of the detector is denoted by \(r_{0}\). The Formula (31) indicates that the mixing efficiency can be expressed by Formula (30), provided that the phase difference deviation is disregarded and both light fields are perfectly matched. Figure 3 displays the simulation calculation of Formula (30). Figure 3 shows that the maximum optical power of the output signal occurs when the angles of the signal light vector and the local vibrating light vector are 45 degrees with respect to the horizontal direction. As a result, the mixing efficiency increases by the greatest amount. Accordingly, the maximum mixing efficiency occurs when the electromagnetic Gaussian Schell beam is linearly polarized and the angle between the horizontal direction and the light vector is 45\({}^{\circ}\).