## 3 Analysis and Discussions
Here, \(\Gamma_{d}\) contains charged particles recombination and decay, and the rate \(\Gamma_{d}=\Gamma_{t}+2N_{g}\sqrt{8K_{B}T/(\pi m_{e})}\sigma_{r}\rho_{c}\) represents the losses in the charged particle. \(\Gamma_{t}\) is the charged particle diffusion rate. The most obvious contribution is that \(\Gamma_{t}\) describes the motional decay parameters of charged particles leaving the interaction volume due to thermal motion, which vary slightly with temperature. The value of \(\Gamma_{t}\) is \(\sim\)0.25 MHz. The second term in \(\Gamma_{d}\) is represented by \(2\Gamma_{r}\), which describes the recombination of ions with electrons into neutral particles. Here, \(K_{B}\) is the Boltzmann constant, \(m_{e}\) is the electron mass, \(\rho_{c}\) is the charged particles population, and we take \(\rho_{c}\approx 1\) in our model. The expected value for \(\sigma_{r}\) can be estimated [21] in a semi-classical treatment with [PERSON]'s formula. Compared with the ion diffusion rate of 0.25 MHz, \(2\Gamma_{r}\) is very small and can be ignored. We obtain the densities of charged particles and Rydberg atoms according to the rate equations, as shown in Figure 4a. It can be seen that the Rydberg atomic density increases first nearly linear with increasing temperature, but at relatively high temperature, this changes to a slowly increase. This may be caused by the different two-step Rydberg excitation (i.e., laser Rabi frequency remains constant, but the single-photon detuning is quite different). The single-photon detuning is relatively small under low temperature conditions, so that the two-step Rydberg excitation is relatively strong, which is just the opposite situation under high-temperature conditions. According to Rydberg atomic density, we can calculate Rydberg level shift for the \(50S_{1/2}\) state [22], within a short-distance range, it is a dipole-dipole interaction, which is proportional to \(1/R^{3}\). The theoretical result is shown in Figure 4b. In the long-distance range, it is Van der Waals interaction, which is proportional to \(1/R^{6}\). It can be seen that the Rydberg level shift due to dipole-dipole interaction is much smaller than our measured values. Therefore, it is impossible to cause a Rydberg level shift of several hundred MHz. Therefore, the Rydberg level shift caused by the interaction between Rydberg atoms is not the main factor than our measured values. Here, considering the quantum defect of the Rydberg atoms, we know that there are two types of interaction between the outermost electron of the alkali-metal Rydberg atom and its nucleus: penetration and polarization. As a result of these two effects, the atomic binding energy increases and total energy reduces (quantum defect) [23,24]. However, in general, the energy loss caused by the quantum defect is very small, so the effect of quantum defect can also be ignored. Based on the calculated density of charged particles, the inhomogeneous electric field is \(\varepsilon=2.603\cdot|e|/(4\pi\varepsilon_{0})\cdot N_{C}^{2/3}\)[25,26], where \(\varepsilon_{0}\) is the vacuum dielectric constant, \(e\) is the charge of electron. In low electric fields, Stark shift of the Rydberg state \(\Delta_{S}\) is given by second order perturbation theory [27] to be \(\Delta_{S}=-(1/2)\alpha\varepsilon^{2}\), where \(\alpha\) is the atomic polarizability of Rydberg state. Referring to the related literature ref. [28], the value for \(\alpha\) can be estimated; the polarizability of \(50S_{1/2}\) Rydberg state is \(\sim\)54.33 MHz\(\cdot\)cm\({}^{2}\)/V\({}^{2}\). The dependence of Rydberg level shift on the density of charged particles can be given as follows [25,26,27]:  
\[\Delta_{S}=-\frac{1}{2}\alpha(\frac{2.603\cdot|e|}{4\pi\varepsilon_{0}}\cdot N _{C}^{2/3})^{2} \tag{3}\]  
We calculate Rydberg level shift \(\Delta_{S}\) according to Equation (3), as shown in Figure 4c. with \(\Omega_{p}/2\pi=5.0\) MHz, \(\Omega_{c}/2\pi=36.0\) MHz.