TD 7：四能级系统，粒子数反转

1. 考虑在稳态条件下的粒子数和粒子数反转

• 根据平衡态有：$\frac{dN_i}{dt} = 0$

$$\begin{split}&\frac{dN_4}{dt} = R_p-S_{43}N_4 = 0\\&\frac{dN_3}{dt} = S_{43}N_4+W{23}N_2-A_{32}N_3-W_{32}N_3 = 0\\&\frac{dN_2}{dt} = A_{32}N_3+W_{32}N_3-W_{23}N_2-S_{21}N_2 = 0\\&\frac{dN_1}{dt} = S_{21}N_2-R_P\end{split}$$

• 得到：

$$\begin{split}&R_p = S_{43}N_4 = S_{21}N_2\\&N_3 = N_2\frac{S_{21}+W}{A_{32}+W}\\\end{split}$$

• 得到$\Delta N$:

$$\Delta N = N_3-N_2 = \frac{S_{21}-A_{32}}{A_{32}+W}\frac{R_p}{S_{21}} = R_p(\frac{1-\frac{A_{32}}{S_{21}}}{A_{32}+W}) \approx R_p(\frac{1}{A_{32}+W})>0$$

• 发现，对于四能级系统来说，只要有Le taux de pompage泵浦$R_p>0$，就有$N_3,N_2$粒子数反转
• 能量变化：

$$\frac{du}{dz} = \Delta NB_{32}gh\nu_{32}\frac ncu = \beta(\nu_{32})u$$

• $\beta$ ：放大、增益系数 le coefficient d’amplification du milieu

2. 谐振腔

两面反射率都是R，介质为四能级物质，入射光为$\nu_{14}$，放大系数为$\beta$，衰减系数$\alpha$，反射率导致的损耗$\gamma = ln(R)$

• 走完2L的能量：

$$u(\nu,2L) = u(\nu,0)e^{2\gamma}e^{-2\alpha L}e^{2\beta L} = u(\nu,0) e^{2(\gamma-\alpha L+\beta L)}$$

• 为了保证能量不损失，得到激光器增益阈值$\beta_s$：$\gamma-\alpha L+\beta L>0 \Rightarrow \beta_s = \alpha-\frac{\gamma}{L}$
• 同理得到粒子差阈值：$\Delta N sB{32}h\nu_{32}\frac ncg = \alpha-\gamma/L \Rightarrow \Delta N_s = 1.6\times10^{20}atoms/m^2$

3. 功率阈值，输入功率和输出功率之比

• 当功率小于阈值时，输出为0，大于阈值时，$\Delta N = \Delta N_s$为常数，输出功率与输入功率呈线性关系
• $Rp \le R_s, \quad W = 0, \quad R_s = \Delta N_sA{32} = 1.33\times10^{23}atoms/m^{-3}$
• 功率$Ps = R_s.h\nu{41}.V = Rs.h\nu{41}.S.L = 3.19mW$
• 只要种子光的功率超过$P_s$，则出光

输入功率为15W时，

• $P{in} = R_ph\nu{41}.V$
• $P{out} = (R_p-R_s)h\nu{32}.V$
• $\frac{P{out}}{P{in}} = \frac{v{32}}{v{41}}(1-\frac{Rs}{R_p})\approx\frac{v{32}}{v_{41}}=76\%$
• $P_out = 11.5W$