TD 7：四能级系统，粒子数反转
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
- \(R_p \le R_s, \quad W = 0, \quad R_s = \Delta N_sA_{32} = 1.33\times10^{23}atoms/m^{-3}\)
- 功率\(P_s = R_s.h\nu_{41}.V = R_s.h\nu_{41}.S.L = 3.19mW\)
- 只要种子光的功率超过\(P_s\)，则出光
输入功率为15W时，
- \(P_{in} = R_ph\nu_{41}.V\)
- \(P_{out} = (R_p-R_s)h\nu_{32}.V\)
- \(\frac{P_{out}}{P_{in}} = \frac{v_{32}}{v_{41}}(1-\frac{R_s}{R_p})\approx\frac{v_{32}}{v_{41}}=76\%\)
- \(P_out = 11.5W\)
